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Department of Mathematics, IIT Madraspresents...
March 15th and 16th, 2014
Souvenir
TABLE OF CONTENTS
S. No. Title Author Page No.
1. Message from Head, Department of
Mathematics
Prof. S. H. Kulkarni 1
2. Report of Forays 2014 Dr. T. E. Venkata Balaji 2
3. Sponsors of Forays 2013 12
4. Completeness and Invertibility S. H. Kulkarni 14
5. On Continuity... P. Veeramani 19
6. Is there a Focus for a Concave Spherical
Mirror?
M. T. Nair 25
7. Population Dynamics and Games A. J. Shaiju 27
8. Rotations and Quaternions Vijay Kodiyalam 33
9. Hamming’s Original Paper written in
Symbolic Form: A Preamble to Coding
Theory
H. Gopalakrishna Gadiyar
and R. Padma
39
10. Every non-constant polynomial with
real coefficients has a root in C
K. N. Ranganathan 43
11. Topological Complexity and Topologi-
cal Robotics
Tulsi Srinivasan 50
12. Representation Theory and Particle
Physics
Girish M. Kulkarni 55
13. An Introduction to Fractional Calculus Madhukant Sharma 57
14. Rubik’s Cube and Group Theory Prajakta Sahasrabuddhe 60
15. Forays 2014 Prize Winners 65
16. Forays 2014 Team 67
MESSAGE FROM THE HEAD, DEPARTMENT OF MATHEMATICS
PROF. S. H. KULKARNI
I am very happy to note that the students of the Department of Math-
ematics, IIT Madras are organizing FORAYS 2014 on March 15 and 16,
2014 under the able guidance of Dr. T. E. Venkata Balaji, the Faculty
Coordinator. This interesting annual event has been going on for last
several years, since 1992. Apart from talks by experts, this festival has
many competitive events like Olympiads, quizzes, puzzle solving and
many others. A new feature since last year has been the inclusion of
school students. This year another attraction is added, namely a video
lecture by the Fields Medal winner Prof. David Mumford.
I am sure the participants will find FORAYS 2014 interesting, full of
fun and also highly educative. I appreciate the efforts by students and
faculty members of the Department in organizing this festival and wish
FORAYS 2014 a grand success.
1
REPORT OF FORAYS 2014
DR T. E. VENKATA BALAJI
Forays 2014, the current edition of the annual festival of the Dept of Maths, IIT-
Madras, was celebrated successfully on 15th and 16th March, 2014. The festival was
inaugurated by IIT-Madras Director Prof. Bhaskar Ramamurthi on 15th March, a day
which was filled with events for college-level students.
College students registering
Prof. Bhaskar Ramamurthi
2
The competitions involving cash prizes (First, Second and Third prizes of Rs. 5000/=,
3000/= and 2,000/= respectively) consisted of an Olympiad and two quizzes. The
Olympiad was conducted by faculty members Arindama Singh, Sounaka Mishra and
Thamban Nair.
College Olympiad
A masters-cum-doctoral level quiz, newly introduced this year, was conducted by
faculty members A. V. Jayanthan, Kalpana Mahalingam, S. H. Kulkarni, Neelesh S.
Upadhye, K. C. Sivakumar and R. Usha. It turned out to be an interesting event and
was well received.
Clash of the Masters
3
The bachelor’s level quiz was conducted as usual with much enthusiasm by Dr. K. N.
Ranganathan and K. N. Visveswaran who have been conducting the event for a decade.
K N Visveswaran Dr. K N Ranganathan
Battle of the Brains (College Level)
An invited lecture on Cryptography was given by Dr. Pramod Kumar Saxena, Di-
rector SAG DRDO New Delhi and a presentation on the opportunities for research,
with an emphasis on applications of Maths, in DRDO and Armed Forces was given by
Mr. Bhupendra Singh, Scientist, CAIR DRDO Bangalore.
Dr. Pramod Kumar Saxena
4
A star event of the day was an interesting video lecture by Emeritus Professor David
Mumford, Fields Medallist, Brown University Dept of Maths, USA, on Maths for Com-
puter Vision. This drew a large audience that jam-packed Studio -1 of the ICSR.
Prof. David Mumford’s Video Lecture
The prizes for college level events were given away by Dr. V. Ramanujachari, Direc-
tor RIC, IIT-M Research Park.
College Prize Distribution
5
The festival continued into its second day on 16th March, a day of events exclusively
designed for school students.
School students registering
Apart from an interesting quiz and a stimulating olympiad which were conducted by
the masters and doctoral students of IIT-Madras Dept of Maths, there was a Mathemat-
ical Working Model Exhibition Competition juried by faculty members P. Veeramani
and Thamban Nair, a screening of the Math movie ”Flatland” and a Workshop on
Cryptography for school students by Mr. Bhupendra Singh.
Battle of the Brains (School Level)
6
School Olympiad
Math Works
7
An interesting talk on the concept of infinity was given by Prof. Thamban Nair.
Prof. Thamban Nair
The prizes were given away by Prof. R. Nagarajan, IIT-M Dean Alumni Affairs.
School Prize Distribution
The major sponsors for Forays 2014 were IIT-M Office of International Relations
and Alumni Affairs, DRDO and Dell Networking R&D India, Chennai.
8
Prof. Bhaskar Ramamurthi
Prof. S.H. Kulkarni, Prof. Bhaskar Ramamurthi, Dr. T. E. Venkata Balaji
9
Invocation
Workshop on Cryptology by Bhupindra Singh
10
Refreshments
Prof. R. Nagarajan
11
SPONSORS OF FORAYS 2014
Chief Sponsors
• Dell Networking R&D India, Chennai
• Defence Research and Development Organisation, India
• Office of International and Alumni Affairs
• Department of Physics, IIT MADRAS
12
Alumni Contributors
• Mr. S. V. Bharanedhar, Visiting Scientist, ISI Chennai
• Tulsi Srinivasan, Graduate School Fellow at the Department of Mathematics,
University of Florida
• Manasi Kulkarni, Ph.D. student at the Department of Computer Science, Uni-
versity of Western Ontario
• Banhita Maitra, EUROSPIN student at NCBS
• Dr. Sachindranath Jayaraman
• Dr. Natarajan, IIST Trivandrum
• Sriram Balasubramanian, IISER-Kolkata
• Kumar Balasubramanian, IISER Bhopal
• Kaashyap Rajeevsarathy, IISER Bhopal
• Girish Kulkarni, IISER Pune
• Ramesh Manna, HRI Allahabad
• Nijjwal Karak Graduate Student, Department of Mathematics, University of
Jayvaskyla
• Chandini, Department of Mathematics, NIT Suratkal
13
COMPLETENESS AND INVERTIBILITY
S. H. KULKARNI
Abstract. We show that two very important concepts in Functional Analysis, namely
the completeness of a normed linear space and invertibility of a bounded linear map
are related to each other. This gives a possibly new characterizations of completeness.
1. Introduction
Answers to many important questions in Functional Analysis depend upon knowing
whether a certain normed linear space is complete or/and whether a certain bounded
linear map has a bounded linear inverse. Usually students do not think that these two
important ideas in Functional Analysis, namely completeness and invertibility, have
anything to do with each other. In this note, we try to draw the attention of students
to connections between these ideas.
The following well known theorem is given in many textbooks of Functional Analysis.
(See for example, [1].)
Theorem 1.1. Let T be a bounded(continuous) linear map from a Banach space X to
a normed linear space Y . Then the following are equivalent:
1. T has a bounded inverse.
2. T is bounded below and the range of T is dense in Y .
It is natural to ask what happens if the hypothesis of completeness of X is dropped.
Somehow, this question is not discussed in the textbooks. It is obvious that (1) would
still imply (2) even without completeness. But the converse is false and it is easy to
construct a counterexample. We give such an example. Further, it is interesting to
note that (2) is equivalent to the following even without the completeness of X.
3. The transpose T ′ of T has a bounded inverse.
Even more interesting is the fact that the completeness of X is equivalent to the
invertibility of every bounded linear map satisfying (2).14
2. preliminaries
We recall a few standard notations, definitions and results that are used in the next
section. For normed linear spaces X, Y , we denote by BL(X,Y ) the set of all bounded
linear operators from X to Y . For an operator T ∈ BL(X,Y ), N(T ) denotes the null
space of T and R(T ) denotes the range of T . Thus
N(T ) = {x ∈ X : T (x) = 0} and R(T ) = {T (x) : x ∈ X}.
Further, T is said to be bounded below if there exists α > 0 such that ‖T (x)‖ ≥ α‖x‖for all x ∈ X and T is said to be invertible if there exists S ∈ BL(Y,X) such that
ST = IX , the identity map on X, and TS = IY , the identity map on Y . The dual space
X ′ of X, is the set of all bounded linear functionals on X, that is, X ′ = BL(X,K),
where K is the underlying field of real or complex numbers. For a subset A ⊆ X, the
annihilator A0 is the set of all continuous linear functionals that vanish on A, that is,
A0 := {φ ∈ X ′, φ(a) = 0 for all a ∈ A}. If A is a subspace of X, then it follows by the
Hahn-Banach Theorem, that A is dense in X, if and only if A0 = {0}. The transpose
T ′ of T ∈ BL(X, Y ) is the operator in BL(Y ′, X ′) defined by
(T ′ψ)(x) := ψ(T (x)) for all x ∈ X and ψ ∈ Y ′. All the other notations (including the
notations for sequence spaces c00, `1 etc.) are as in [1] and [2]. We shall make use of
the following well known results:
1. (R(T ))0 = N(T ′).
2. Every normed linear space X can be viewed as a dense subspace of a Banach space
which we shall denote by Xc. (More precisely, there is a linear isometry of X onto a
dense subspace of Xc.) The Banach space Xc is called the completion of X. These
results can be found in any book on Functional Analysis, for example [1] and [2].
3. Notes
We begin with the example.
Example 3.1. Let X := (c00, ‖.‖1), Y := `1 and T : X → Y be given by T (x) = x for
x ∈ X. Clearly, T is bounded below, range of T is dense in Y , but T is not onto and
hence not invertible. More generally, we can consider the inclusion map from a proper
dense subspace of a normed linear space.
Remark 3.2. Note that in the above example, though T is not invertible, its transpose
T ′ is invertible. In fact, both the dual spaces X ′ of X and Y ′ of Y can be identified
with `∞ in the usual way (See [2] for details.) and with respect to this identification
T ′ becomes the identity operator on `∞.15
This leads to some natural observations. First we consider some elementary results.
The following elementary result is given as an Exercise in some books.
Lemma 3.3. Let T be a bounded(continuous) linear map from a normed linear space
X to a normed linear space Y . Then R(T ) is dense in Y if and only if T ′ is injective.
Proof. Recall that R(T ) is dense in Y if and only if {0} = (R(T ))0 = N(T ′) if and only
if T ′ is injective. �
Lemma 3.4. Let T be a bounded(continuous) linear map from a normed linear space
X to a normed linear space Y . Then the following statements are equvivalent:
1. T has a bounded inverse from R(T ) to X.
2. T is bounded below.
3. T ′ is onto.
Proof. (1) implies (2): This is easy. Suppose S : R(T ) → X is a bounded inverse of T .
Then for each x ∈ X,
‖x‖ = ‖ST (x)‖ ≤ ‖S‖‖T (x)‖, that is, ‖T (x)‖ ≥ 1‖S‖‖x‖.
(2) implies (1) and (3): Since T is bounded below, there exists α > 0 such that
‖T (x)‖ ≥ α‖x‖ for all x ∈ X. In particular, T is injective. Hence we can define a
map S : R(T ) → X by S(y) = x for y = T (x) ∈ R(T ). This is well defined since T is
injective. It is easy to see that S is linear. Also
‖S(y)‖ = ‖x‖ ≤ 1α‖T (x)‖ = 1
α‖y‖.
Hence S is bounded. This proves (1).
Next let φ ∈ X ′ and y ∈ R(T ). There exists unique x ∈ X such that y = T (x).
Define ψ by ψ(y) := ψ(T (x)) = φ(x). This defines ψ as a linear functional on R(T ).
Further,
|ψ(y)| = |ψ(T (x))| = |φ(x)| ≤ ‖φ‖‖x‖ ≤ ‖φ‖ 1α‖T (x)‖ = ‖φ‖ 1
α‖y‖.
This shows that ψ is bounded on R(T ) and hence has a bounded (norm preserving)
extension to Y by the Hahn-Banach Theorem. We denote this extension also by the
same symbol ψ. Thus ψ ∈ Y ′ and φ = T ′(ψ). This shows that T ′ is onto.
(3) implies (2): Let x ∈ X. By the Hahn-Banach Theorem, there exists φ ∈ X ′ such
that φ(x) = ‖x‖ and ‖φ‖ = 1. Further, since T ′ is onto, there exists ψ ∈ Y ′ such that
φ = T ′(ψ). Now
‖x‖ = φ(x) = T ′(ψ)(x) = ψ(T (x)) ≤ ‖ψ‖‖T (x)‖, that is, ‖T (x)‖ ≥ 1‖ψ‖‖x‖
This shows that T is bounded below.
�16
We now give the main theorem.
Theorem 3.5. Let T be a bounded(continuous) linear map from a normed linear space
X to a normed linear space Y . Consider the following statements:
1. T has a bounded inverse.
2. T is bounded below and the range of T is dense in Y .
3. T ′ is invertible.
Then (2) and (3) are equivalent and each is implied by (1). If, in addition, X is a
Banach space, then all the three statements are equivalent.
Proof. (1) implies (2): Obvious. Since T has a bounded inverse, R(T ) = Y . Also T is
bounded below by Lemma 3.4.
(2) if and only if (3): By Lemma 3.3, R(T ) is dense in Y , if and only if T ′ is injective.
Further, by Lemma 3.4, T is bounded below, if and only if, T ′ is onto. Thus (2) is
equivalent to the following:
T ′ : Y ′ → X ′ is a bijection.
Hence T ′ is invertible by the Closed Graph Theorem as X ′, Y ′ are Banach spaces.
Finally, if X is a Banach space, then (2) implies (1) by Theorem 1.1 and hence all
the three statements are equivalent. �
Remark 3.6. We may further note that completeness of X is , in fact, equivalent to
the invertiblity of every bounded linear map satisfying (2). In other words, a normed
linear space X is a Banach space if and only if every bounded linear map T from X to
any normed linear space Y such that T is bounded below and the range of T is dense
in Y , is invertible. The only if part is already proved above. To prove the if part,
consider Y = Xc, the completion of X. Then there is a linear isometry T of X onto a
dense subspace Y0 of Y . (See [2] for details.) Obviously, this T is bounded below and
R(T ) = Y0 is dense in Y . Hence by the hypothesis, T is invertible and, in particular,
onto. Thus X is linearly isometric to Y and hence complete.
Remark 3.7. It is known that the invertibility of an operator is closely related to
its spectrum. Let X be a complex normed linear space and T ∈ BL(X,X). Recall
that the spectrum σ(T ) of T is the set of all complex numbers λ such that λI − T
is not invertible. Applying Theorem 3.5 to λI − T , we obtain the known result that
σ(T ′) ⊆ σ(T ) and the equality holds if X is a Banach space. (See [2]) (The inclusion
can be strict if X is not a Banach space. See the next example.) A natural question is
whether the converse holds. In other words, can the completeness be also characterized
in terms of spectra as follows: A complex normed linear space X is complete if and only17
if σ(T ′) = σ(T ) for all T ∈ BL(X,X) ? Another formulation of the same question is as
follows: Given an incomplete normed linear space X, does there exist T ∈ BL(X,X)
such that T is bounded below, its range is dense in X and T is not invertible (that is,
not onto)? Note that the above examples and remarks do not answer this question as
the spaces X and Y considered there are different.
Example 3.8. This example shows that the inclusion σ(T ′) ⊆ σ(T ) can be strict if X
is not complete.
LetX := (c00, ‖.‖2), and T : X → X be the right shift operator given by T (x1, x2, . . .) =
(0, x1, x2, . . .) for x := (x1, x2, . . .) ∈ X. Then the dual space X ′ can be identified with
`2 and the transpose T ′ of T can be identified with the left shift operator. (See [2] for
details.) Then it can be shown that
σ(T ′) = {z ∈ C : |z| ≤ 1}, the closed unit disc. On the other hand, it is easy to see
that the equation (λI−T )x = e1 = (1, 0, 0, . . .) has no solution x ∈ X for any complex
number λ. In other words, λI − T is not onto. Thus σ(T ) = C.
References
[1] B. Bollobas, Linear analysis, Cambridge Univ. Press, Cambridge, 1990. MR1087297 (92a:46001)
.
[2] B. V. Limaye, Functional analysis, Second edition, New Age, New Delhi, 1996. MR1427262
(97k:46001)
———————————————————-
Department of Mathematics, Indian Institute of Technology - Madras, Chennai
600036
E-mail address: [email protected]
18
ON CONTINUITY...
P. VEERAMANI
1. Continuous functions
Let X, Y be nonempty sets and f : X → Y . That is f is a function with domain
X and codomain Y . Now fix an element x0 in X. Let us try to understand intutively
what do we mean by saying the given function f is continuous at the point x0?
It is quite natural to expect that if an element x in X is closer to x0 then after the
transformation the element f(x) is closer to f(x0).
Think of X as a set of people living in a particular place and suppose all these people
are asked to relocate to another place. Let us say that the same set of people in the
new place is denoted by Y . Due to this process x in X is identified as f(x) in Y (that
is we have a function f : X → Y ). We say that the function f (or the process of
relocation) is continuous at x0 in X if it satisfies:
We are given a set (say V ) in Y of people closer to f(x0) (in such a case we say that
V is a neighbourhood of f(x0) in Y ).
Having given this neighbourhood V of f(x0) in Y our task is to find at least one
neighbourhood, say U , of x0 in X satisfying:
x is in the neighbourhood U of x0 =⇒ f(x) is in the neighbourhood V of f(x0).
So we define:
A function f : X → Y is said to be continuous at a point x0 in X if for each given
neighbourhood V of f(x0) in Y , there exists a neighbourhood U of x0 in X satisfying:
x ∈ U =⇒ f(x) ∈ V.
Of course, if X and Y are nonempty sets then we can define a function f : X → Y .
Now we should also make it clear the term ”neighbourhood of a point x0 in X”.
Take X = R2 = {(x, y) : x, y ∈ R}, where R is the set of all real numbers.19
Here (x, y) ∈ X = R2 is an element in R2. It is absolutely correct if we say x0 ∈ R2.
That is x0 ∈ R2 implies x0 = (a, b) for some a, b ∈ R.
Now what do we mean by a neighbourhood of (a, b) in R2. The concept neigh-
bourhood is a relative term. So we fix a radius r > 0. Then we say that all those
(x, y) ∈ R2 which are within r distance from (a, b), are in our neighbourhood say U of
(a, b). That is U is an open ball center at (a, b) and radius r > 0, which is defined as
B((a, b), r) := {(x, y) ∈ R2 : d((a, b), (x, y)) < r}.
Here d((a, b), (x, y)) is the distance between (a, b) and (x, y) (or think d as the Eu-
clidean distance).
So, let us take X = Y = R2, x0 = (a, b) ∈ X and f : X → Y . Then f is said to be
continuous at x0 if for each ε > 0, there is a δ > 0 satisfying:
x ∈ B(x0, δ) =⇒ f(x) ∈ B(f(x0), ε).
That is for a given ε > 0, we should able to find a δ > 0 satisfying:
x ∈ R2, d(x, x0) < δ =⇒ d(f(x), f(x0)) < ε.
Now suppose A, B are nonempty subsets of R2 and f : A → B. Then f is said to be
continuous at a point x0 in A if for each ε > 0, there is a δ > 0 satisfying:
x ∈ A, x ∈ B(x0, δ) =⇒ f(x) ∈ B(f(x0), ε).
In this case, domain of our function f is A, a subset of R2. That is our world is
restricted to A. Yes it is possible to have x ∈ R2 \A and x ∈ B(x0, δ). But in this case
we can not talk about f(x) and hence our neighbourhood U of x0 is restricted to our
world A.
That is U = {x ∈ A : d(x, x0) < δ}. Once we take x ∈ U , then x ∈ A and
d(x, x0) < δ and in this case f(x) ∈ B (note that f : A→ B).
Example 1. Let A = N × N = {(m,n) : m,n ∈ N} (where N = {1, 2, 3, ....}) and
f : A→ R2 be a function. Fix (2, 3) ∈ A. Is f continuous at (2, 3)?
Start with an ε > 0 and consider the neighbourhood V = B(f(2, 3), ε) of f(2, 3) in
R2. What is our aim ? We will have to find a neighbourhood U of (2, 3) satisfying:
x ∈ U =⇒ f(x) ∈ V .
Note that it is enough to find at least one neighbourhood U satisfying the above condi-
tion.20
Are you able to visualize the Euclidean plane R2 and the subset A = N × N of R2?
Take δ = 14. Of course B((2, 3), 1
4) contains infinitely many points of R2. But what is
our domain ? Is A = R2? No.
So, our neighbourhood U of (2, 3) is B((2, 3), δ) restricted to A. That is,
U = {(x, y) ∈ A : d((x, y), (2, 3)) < δ}.
Now (x, y) ∈ A = N × N, d((x, y), (2, 3)) < 14
iff x = 2, y = 3. So our U contains
only the element (2, 3).
Therefore for ε > 0, if we take δ = 14
then the condition (x, y) ∈ U =⇒ f(x, y) ∈ Vis satisfied. Hence every function f : A→ R2 is continuous. �
Remark 1. If x0 ∈ R and r > 0 then B(x0, r) = {y ∈ R :| y−x0 |< r} = (x0−r, x0+r)
That is for r > 0, (x0 − r, x0 + r) is a neighbourhood of x0 ∈ R.
But if A is a nonempty subset of R, x0 ∈ A and r > 0, then the neighbourhood of x0in A is given by U = {x ∈ A :| x− x0 |< r} = (x0 − r, x0 + r) ∩ A.
For example if A = [0,∞) and x0 = 0 then, for r > 0, U = (−r, r) ∩ [0,∞) = [0, r)
is a neighbourhood of 0 in [0,∞) (there is no negative numbers in [0,∞)). �
2. Uniformly continuous functions
When do we say that a function f : X → Y is uniformly continuous ?
The given function f : X → Y is said to be uniformly continuous if for each ε > 0
there exists a δ > 0 satisfying: x, y ∈ X, d(x, y) < δ =⇒ d(f(x), f(y)) < ε.
Remark 2. Here we assume that d is a given distance function on X. We also use
the same symbol d to denote the given distance function on Y . �
Note that, we study the behaviour of functions under the concepts namely continuous
functions and uniformly continuous functions. One can ask the following questions.
(1) If f is uniformly continuous, does it imply f is continuous at each point in X?
(2) If f is a continuous function, is f uniformly continuous?
Suppose f : X → Y is a uniformly continuous function. Hence for a given ε > 0
there exists a δ > 0 satisfying: x, y ∈ X, d(x, y) < δ =⇒ d(f(x), f(y)) < ε.
Now fix x0 ∈ X. Note that for y ∈ X, d(x0, y) < δ =⇒ d(f(x0), f(y)) < ε.
That is y ∈ B(x0, δ) =⇒ f(y) ∈ B(f(x0), ε). This proves that f is continuous at
x0 ∈ X.21
Hence uniform continuity is stronger than continuity. That is, if f : X → Y is a
uniformly continuous function, then f is continuous at each point in X.
Now the following example shows that if a function f is continuous, then in general
it need not be uniformly continuous.
Example 2. Suppose f : R→ R is given by f(x) = x2 for all x ∈ R.
Start with an ε > 0. Start with what you want to prove(or disprove!). Can we find
a δ > 0 satisfying: x, y ∈ R, d(x, y) < δ =⇒ d(f(x), f(y)) < ε (when we say f is
continuous at x0, we need to consider only those x within δ−distance from x0).
Here x, y can be anywhere in the domain. No such restriction that x is closer to a
particular x0.
If at all, there is a δ > 0 satisfying: x, y ∈ R, d(x, y) < δ =⇒ d(f(x), f(y)) < ε.
Then we have the freedom to take x = n and y = n+ δ2.
Note that d(x, y) =| x− y |= δ2< δ, but d(f(x), f(y)) = (n+ δ
2)2 − n2 = nδ + δ2
4.
Is nδ + δ2
4< ε? If so, nδ < ε, for every n ∈ N. That is, n < ε
δfor every n ∈ N.
Is this relation true for every n ∈ N ? Recall the Archimedean principle namely for
each α ∈ R, there exists n ∈ N such that n > α.
Note that, in particular, if we take ε = 1 then we will not be in a position to find a
δ > 0 satisfying n < 1δ
for all n ∈ N. Therefore f : R→ R defined as f(x) = x2 is not
uniformly continuous.
(Here note that d(f(x), f(y)) ≥ nδ →∞ as n→∞, though d(x, y) =| x− y |= δ2< δ.)
It remains to show that f is continuous at each x ∈ R.
Note that, if we fix x0 ∈ R, then our x ∈ (x0 − δ, x0 + δ) and in this case x cannot
move right side of x0 + δ or left side of x0 − δ.
What is important to note here is, for the time being, we have restricted our domain
to (x0 − δ, x0 + δ) which is a bounded domain.
Also we have to find at least one δ > 0 and hence we have the freedom to seek a δ
such that 0 < δ < 1. Now | f(x)− f(x0) |=| x2 − x20 |=| x+ x0 || x− x0 |. Hence
| f(x)− f(x0) | < | x+ x0 | δ(1)
Here due to | x−x0 |< δ, we have | x+x0 |≤| x−x0 | +2 | x0 |< δ+2 | x0 |< 1+2 | x0 |.
Hence from eqn.(1), | f(x) − f(x0) |< (1 + 2 | x0 |)δ ≤ ε. That is, we can further
restrict our δ such that δ ≤ ε1+2|x0| .
22
So, for given ε > 0, if we choose δ = min{1, ε1+2|x0|} then for x ∈ R,
| x− x0 |< δ =⇒| f(x)− f(x0) |< ε is satisfied.
Hence f(x) = x2 is continuous at each x0 ∈ R. �
Though f : R → R, for checking the continuity of f at a point x0 ∈ R we do not
require to analyze the behaviour of the function on the entire domain of f . What is
important to note here is, it is enough to restrict our domain to a subdomain of the
type (x0 − α, x0 + α) for some real number α > 0. Of course such a restricted domain
depends on x0.
Are you able to realize the relation between continuity and uniform continuity?
Well, now take a nonempty bounded subset A of R (A is a bounded subset of Riff there exists M > 0 such that | x |≤ M for all x ∈ A) and define f : A → R as
f(x) = x2 for all x ∈ A. Let us leave it as an exercise to prove that f is a uniformly
continuous function.
3. Equicontinuous family of functions
Let X and Y be nonempty sets. Let us assume that for x ∈ X neighbourhoods of x
in X and for y ∈ Y neighbourhoods of y in Y are known. We may as well assume that
(X, d) and (Y, ρ) are metric spaces.
Now take a nonempty set J , known as index set. For each α ∈ J , we have a function
fα : X → Y . Also fix an element x0 ∈ X.
Now suppose for a given ε > 0, there exists a δ > 0 satisfying:
x ∈ X, d(x, x0) < δ =⇒ ρ(fα(x), fα(x0)) < ε for all α ∈ J .
That is, we have a collection of neighbourhoods of the type {Bρ(fα(x0), ε) : α ∈ J}where Bρ(fα(x0), ε) = {y ∈ Y : ρ(fα(x0), y) < ε}. These are the neighbourhoods with
possibly different centers fα(x0) but with same radius ε.
For these collection of neighbourhoods in Y we should be in a position to find a
neighbourhood U = Bd(x0, δ) satisfying:
x ∈ X, x ∈ U =⇒ fα(x) ∈ Bρ(fα(x0), ε), for each α ∈ J .
In such a case we say that the collection of functions {fα : α ∈ J} is equicontinuous
at x0 ∈ X.23
Now note that if our family {fα : α ∈ J} of functions is equicontinuous at x0 ∈ X,
then, in particular, for each ε > 0 and for each fixed α0 ∈ J there exists a δ > 0
satisfying:
x ∈ X, d(x, x0) < δ =⇒ ρ(fα0(x), fα0(x0)) < ε.
This proves that for each α0 ∈ J the function fα0 : X → Y is continuous at x0.
Example 3. Consider the family of functions fn : R → R, for n ∈ N, defined by
fn(x) = xn
for x ∈ R. Fix x0 ∈ R. Now for ε > 0,
d(fn(x), fn(x0)) =| xn− x0
n|= 1
n| x− x0 |≤| x− x0 |= d(x, x0)
That is, we have d(fn(x), fn(x0)) ≤ d(x, x0) < δ. Therefore, if can take δ ∈ (0, ε] then
for x ∈ R satisfying d(x, x0) < δ, we have d(fn(x), fn(x0)) < ε. Hence the family {fn}is equicontinuous at every point x0 in R. �
Example 4. Define, for n ∈ N, fn : R → R by fn(x) = nx for x ∈ R. Is this family
of functions equicontinuous at every point of R?
Fix x0 ∈ R and take ε > 0. Is it possible to prove that for every n ∈ N, d(fn(x), fn(x0)) ≤αd(x, x0) for some α > 0?
For ε = 1, can we find a δ > 0 satisfying:
| x− x0 |< δ =⇒ d(fn(x), fn(x0)) = n | x− x0 |< 1 for all n ∈ N?
What will happen if such a δ exist? Take x = x0 + δ2
and see what happens. �
Acknowledgement: The author would like to thank S. Rajesh for careful reading
and helping to prepare this manuscript.
Department of Mathematics, Indian Institute of Technology - Madras, Chennai
600036
E-mail address: [email protected]
24
IS THERE A FOCUS FOR A CONCAVE SPHERICAL MIRROR?
M.T. NAIR
Abstract. This article is essentially meant for under graduate students dealing with
a property of concave spherical mirrors which either they may not have noticed or
they may have ignored. The purpose is to answer the question raised in the title
negatively using elementary geometry
First let us recall the definition of a spherical mirror that appear in school text
books.
A spherical mirror is a mirror which is a part of a sphere and with
circular boundary. It is concave if the reflecting surface is in the inside
part, and convex if the reflecting surface is outside.
We shall be concerned with a spherical concave mirror and the question raised in the
title.
Is there a focus, and hence, a focal length for a concave spherical mirror?
In the school text books on physics, you will find a definite answer: There is a focus
and, and the focal length is half the radius of the sphere of which the mirror is a part.
I intend to refute the above answer by using ideas from elementary geometry.
First of all, what we mean by the focus or the focal point of a concave spherical
mirror, if at all such a point exists?
Suppose the spherical mirror is part of a sphere of radius R and centre at a point
C. Let C0 be the centre of the mirror. Suppose a ray of light is travelling parallel to
the axis of the mirror (i.e., the line joining C and C0 in the direction of the mirror.
If reflection of all such rays pass through a single point, say F , on the axis, then F is
called the focus of the mirror, and the distance from C0 to F is called the focal length.
We show:
A concave spherical mirror does not have a focus, and hence there is no
focal length.25
Let us denote the distance between points P and Q in the space by |PQ|, and let
the line segment joining P to Q be denoted by PQ. Then we have
R = |CC0| = |CP |,
for any point P on the mirror.
Suppose that a ray parallel to CC0 falls on the mirror at a point P , and let D be
the point of intersection of the ray with the perpendicular to CC0 passing through C.
Thus, the ray can be thought of as the line segment DP .
Suppose the reflected ray cuts the line segment C0C at F , and let the perpendicular
to C0C through P cuts C0C at X. Since PD is parallel to C0C, and since the angles
that the incident ray and the reflected ray make with the radius CP of the sphere are
equal,
∠DPC = ∠CPF = ∠FCP.Therefore, FC = FP . Now, by Pythagoras theorem,
R2 = |XC|2 + |XP |2,
|XP |2 + |XF |2 = |FP |2 = |FC|2.Hence,
R2 = |XC|2 + |XP |2 = |XC|2 + |FC|2 − |XF |2.But, |XF | = |XC| − |FC|. Thus,
R2 = |XC|2 + |FC|2 − (|XC| − |FC|)2
= 2|XC| |FC| = 2(R− x)(R− f),
where x = |C0X| and f = |C0F |. Therefore,
f = R− R2
2(R− x)=R
2
[2− R
R− x
]=R
2
[1− x
R− x
].
The above expression for f shows that f <R
2for every incident ray, i.e., with x > 0,
and hence there does not exist a (fixed) focus.
Further, we observe that following:
• If x ≈ 0, then f ≈ R2.
• If R ≈ ∞, then 1− x
R− x≈ 1 so that f =
R
2
[1− x
R− x
]≈ R
2.
Department of Mathematics, I.I.T. Madras, Chennai-600 036, INDIA
E-mail address: [email protected]
26
POPULATION DYNAMICS AND GAMES
DR A. J. SHAIJU
1. Introduction
Consider a large population of individuals each of which chooses to be either ‘good’
or ‘bad’, at each time t. Let n1(t) and n2(t) be respectively the number of good and bad
individuals in the population, at time t. Then p1(t) =n1(t)
n1(t)+n2(t)and p2(t) =
n2(t)n1(t)+n2(t)
are the proportions of good and bad individuals respectively. Note that both p1(t) and
p2(t) lie in [0, 1] and their sum is 1. In other words, for each t, p(t) = (p1(t), p2(t)) is a
probability vector. So it is natural to interpret p1(t) as the probability that a random
individual. at time t, is good. Similarly p2(t) may be interpreted as the probability
that a random individual is bad at time t.
We refer to p(t) as the population state at time t. One natural question arises here;
what is the large-time behavior of the population state p(t)?.
To answer this question, we need to make some assumptions on how the population
grows/declines in time. We begin with the assumption that the relative growth raten1(t)n1(t)
of good individuals is a function r1(n1(t), n2(t)) of n1(t) and n2(t). Similarly, we
assume that the relative growth rate n2(t)n2(t)
of bad individuals is r2(n1(t), n2(t)). Using
these assumptions, we can try to derive differential equations for p1(t) and p2(t), as
follows.
p1(t) =n1(t)
n1(t) + n2(t)− n1(t)
(n1(t) + n2(t))2(n1(t) + n2(t))
=n1(t)r1(n1(t), n2(t))
n1(t) + n2(t)− n1(t)
(n1(t) + n2(t))2(n1(t)r1(n1(t), n2(t)) + n2(t)r2(n1(t), n2(t)))
= p1(t)r1(n1(t), n2(t))− p1(t)(p1(t)r1(n1(t), n2(t)) + p2(t)r2(n1(t), n2(t))).(1)
In addition to our previous assumptions on the relative growth rates, if we also assume
that these relative growth rate functions depend only on frequencies/proportions of
good and bad individuals, then r1(n1, n2) = f1(p1, p2), r2(n1, n2) = f2(p1, p2). This27
helps us to obtain, from (1), the following differential equation for p1(t):
p1(t) = p1(t)[f1(p1(t), p2(t))− p1(t)f1(p1(t), p2(t))− p2(t)f2(p1(t), p2(t))].
As usual, we suppress the t variable in this differential equation, and rewrite as
(2) p1 = p1[f1(p1, p2)− p1f1(p1, p2)− p2f2(p1, p2)].
In a similar fashion, we can derive the differential equation
(3) p2 = p2[f2(p1, p2)− p1f1(p1, p2)− p2f2(p1, p2)].
Thus we have a system of differential equations, given by (2) and (3), which describes
the evolution of the population state p(t) = (p1(t), p2(t)).
Remark 1. Since p1 + p2 ≡ 1, it follows that p1 + p2 = 0. This may also be derived
from (2) and (3).
Example 1. Consider the case where ri(n1, n2) = cin1n2
)n1+n2)2; i = 1, 2. (Here c1 and
c2 are known constants.) In this case, it is obvious that fi(p1, p2) = cip1p@; i = 1, 2.
Therefore, the population state is governed by the systems of equations:
p1 = p21p2[c1 − c1p1 − c2p2],
p2 = p1p22[c2 − c1p1 − c2p2].
Before trying to answer the question concerning the large-time behavior of the pop-
ulation state, we give a slightly different view regarding the population dynamics given
by (2) and (3) based on game theory.
2. Game theory interpretation of population dynamics
Imagine pairs of individuals (from the same population) participating in a game or
competition described below.
At any given time, if the population is in the state p = (p1, p2), then a good individ-
ual receives a reward/fitness quantified by the number f1(p1, p2) and a bad individual
receives f2(p1, p2).
Clearly, if the population is in the state p = (p1, p2), then the total population aver-
age fitness is p1f1(p1, p2) + p2f2(p1, p2). In view of this fact, the differential equations
(2) and (3) may be interpreted as saying that the proportion p1 of good individuals
grows (resp. declines) if their fitness f1 is more (resp. less) than the total population28
average fitness.
In the sequel, we discuss only the linear case. That is, the case when
(4) f1(p1, p2) = a11p1 + a12p2, f2(p1, p2) = a21p1 + a22p2.
2.1. Games with linear fitness functions. In this subsection, we are concerned
with the population dynamics for games with linear fitness functions given by (4). We
start by observing that the population dynamics may now be written as
(5)p1 = p1[(e
1)′Ap− p′Ap]
p2 = p2[(e2)′Ap− p′Ap].
}Here ′ indicates transposition, e1 (resp. e2) is the vector with first coordinate 1 (resp.
0) and second coordinate 0 (resp. 1), and A =
[a11 a12a21 a22
]. Note also that, for popu-
lation states q and p, q′Ap =∑2
i,j=1 aijqipj.
As explained earlier, the population dynamics (5) has the underlying 2 × 2 matrix
game in which an individual has only two options-good (e1) or bad (e2)-when compet-
ing/playing against another individual. When good competes with good, each receives
fitness a11. when good competes with bad, good receives a12 and bad receives a21.
When bad meets bad, each gets a22.
In the next section, we discuss briefly the general case with each individual, at any
time, having k options.
3. Population dynamics associated with k × k matrix games
We now consider the general case where each individual can choose from k op-
tions e1, e2, · · · , ek. For 1 ≤ i ≤ k, pi(t) denotes the proportion of individuals choos-
ing the ith option ei, at time t. The population state at time t is the k-vector
p(t) = (p1(t), p2(t), · · · , pk(t)). The underlying game is having the fitness matrix
A = [aij]1≤i,j≤k. Here again the interpretation is that, when ei competes against ej,
the former gets fitness aij.
As in the previous section, we can show that the evolution of the population state
is governed by the following system of differential equations:
(6) pi = pi[(ei)′Ap− p′Ap], i = 1, 2, · · · , k.
29
Remark 2. Since∑k
i=1 pi = 0, it follows that the set
∆ = {p = (p1, · · · , pk) :k∑i=1
pi = 1 and p1, · · · , pk ≥ 0}
of all possible population states, is invariant under the dynamics (6). As a result, to
study the large-time behavior of the population state p(t), we may restrict our attention
to the behavior of the system (6) on ∆.
3.1. Stationary states of the population dynamics. A population state p ∈ ∆ is
stationary if at this point the R.H.S. of the system (6) vanishes. This means that p is
a stationary state if and only if
(7) (ei)′Ap = p′Ap whenever pi 6= 0.
Clearly the pure states e1, · · · , ek are stationary states. There may be other interesting
stationary states as well.
Example 2. Consider the case where A is a 2×2 diagonal matrix with diagonal entries
a, b having same sign. In this case, it can be verified that p = ( ba+b
, aa+b
) is a stationary
state.
Example 3. Consider the population dynamics associated with A =
0 1 −1
−1 0 1
1 −1 0
.It can be verified that p = (1/3, 1/3, 1/3) is a stationary state.
The stationary states are closely related to the associated matrix game and various
concepts in game theory such as Nash Equilibrium.
3.2. Stationary states and Nash equilibria. In a matrix game defined by the k×kmatrix A, p = (p1, · · · , pk) is said to be a symmetric Nash equilibrium if
(8) (ei)′Ap ≤ p′Ap for 1 ≤ i ≤ k.
The proof of the next lemma is left as an exercise to the reader.
Lemma 1. If p is a symmetric Nash equilibrium, then it is a stationary state of the
population dynamics (6). Conversely, if p (with pi > 0, ∀i) is a stationary state, then
it is a symmetric Nash equilibrium.
This lemma says that a symmetric Nash equilibrium is necessarily a stationary state
of the associated population dynamics. It also provides a (partial) sufficient condition
for a stationary state to be a symmetric Nash equilibrium.
30
We end this short note with the next theorem which gives a necessary condition for
a population state to be a limiting state of the population dynamics trajectory with
initial population state having all k types of individuals.
Theorem 1. Let p(t) be the solution of (6) where all components of p(0) = (p1(0), p2(0), · · · , pk(0))are strictly positive. Let p∗ = (p∗1, · · · , p∗k) = limt→∞ p(t). Then p∗ is a symmetric Nash
equilibrium.
Proof. If possible, let p∗ be not a symmetric Nash equilibrium. This would imply that
(ei)′Ap∗ > (p∗)′Ap∗ =∑j:p∗j 6=0
p∗j(ej)′Ap∗ for some i.
This implies that, for some ` with p∗` 6= 0,
(ei)′Ap∗ > (e`)′Ap∗.
Let ε be the difference between L.H.S. and R.H.S. of this inequality. Since limt→∞ p(t) =
p∗, it immediately follows that, for t large enough (say t ≥ t0)
(9) (ei)′Ap(t)− (e`)′Ap(t) ≥ ε
2.
Since p(t) is a solution of (6), it follows that
(10)d
dt
pi(t)
p`(t)=pi(t)
p`(t)[(ei)′Ap(t)− (e`)′Ap(t)].
From (9) and (10), we get
d
dt
pi(t)
p`(t)≥ ε
2
pi(t)
p`(t), ∀t ≥ t0.
This implies that
pi(t)
p`(t)≥ pi(t0)
p`(t0)e
ε2(t−t0), ∀t ≥ t0.
Letting t→ ∞, we have a contradiction because p∗` 6= 0. This contradiction shows that
our initial assumption, that p∗ is not a symmetric Nash equilibrium, is wrong. Hence
the proof. �
Remark 3. The above thoerem tells us that, if initially all options/behavioral types
are present, the limit of the associated population trajectory is necessarily a symmetric
Nash equilibrium. This partially answers the question we considered in the beginning
of the discussion about population dynamics.31
4. Concluding remarks
The theorem proved in the previous section is just a tip of an iceberg. There are
several interesting results and unsolved problems in the broad area of population dy-
namics and games. Interested readers may refer to the following articles/books for a
deeper insight.
References
[1] J. Hofbauer and K. Sigmund, Evolutionary Games and Population Dynamics, Cambridge Uni-
versity Press, 1998.
[2] J. Hofbauer and K. Sigmund, Evolutionary game dynamics, Bulletin of the American Mathemat-
ical Society, 40(4):479-519, 2003.
[3] R. Cressman, Evolutionary Dynamics and Extensive Form Games, The MIT Press, 2003.
[4] W. H. Sandholm, Population Games and Evolutionary Dynamics, The MIT Press, 2010.
Department of Mathematics, I.I.T. Madras, Chennai-600 036, INDIA
E-mail address: [email protected]
32
ROTATIONS AND QUATERNIONS
VIJAY KODIYALAM
This article is an excursion into the geometry of rotations, the algebra of quaternions
and the intimate relationship between these two.
1. The geometry of rotations
Let us begin by doing some experiments with a cube with each face painted a different
colour. Orient this cube so that there is a face that is facing you which we call the
“Front” face and a face behind that which we’ll call the “Back” face. There is a face
on your right which will be called the “Right” face and a face on your left which will
be called the “Left” face. The remaining two faces will be the “Up” and “Down” faces.
In order to describe what we will do clearly, consider a three dimensional coordinate
system whose origin is at the centre of the cube and whose positive x-, y- and z-axes
go through the centres of the Front, Right and Up faces of the cube respectively - see
Figure 1.
L
R
B
D
U
x
y
z
F
Figure 1. Cube and associated notations
Let us now make two 90◦ turns one about the positive x-axis and one about the
positive z-axis. By a turn on some angle about some axis we will always mean the turn
given by the right hand rule, i.e., curl your fingers around the axis with thumb pointing
in the direction of the axis and make a turn in the direction given by the other fingers.33
What happens after the first turn? What colour is the Up face? Where is the blue
face? What happens to the white face? Note that the right face is now coloured green
- the face opposite the blue one.
Now make the second turn and consider the position of the cube after this. A little
thought or experiment should convince you that the new position of the cube is as
below:
Figure 2. Final position
What is really remarkable is the following. This position is obtainable from the
original position by a single rotation. And what is this rotation? Consider an axis
through the pair of opposite vertices of the cube that are at the meeting of the FRU
and BLD faces. Regard this axis as oriented from the BLD corner to the FRU corner.
Observe that a 120◦ rotation through this axis does the same thing as the two rotations
combined.
Let us now change slightly what we do. The first is still a 90◦ turn about the positive
x-axis but let the second be a 180◦ turn about the positive z-axis. Now we see that
the final position of the cube is (where the yellow face is opposite the white one):
Figure 3. Position after reversing order of the moves34
Again we see that this position is obtainable from the initial position by a single
rotation - this time of 180◦ about an axis going from the centre of the Down-Left to
the centre of the Up-Right edge.
Now these examples raise at least two natural questions. First, is it possible that
following one rotation (about some axis) by another (through a possible different axis)
is equivalent to some other rotation (through a possibly third axis). It turns out that
this is indeed the case though it is far from obvious. Now we can ask the second
question. How may we find the amount and axis of the rotation equivalent to two
others ? Note that the axis of rotation of the final rotation was not dependent on
just the axes of rotation of the two initial rotations. This leads to a very interesting
number system called the quaternions. And the connection between the rotations and
quaternions is a pleasant mix of analytic geometry, trigonometry and algebra.
2. The algebra of quaternions
It turns out that there are “numbers” called (unit) quaternions such that each de-
scribes a rotation and such that multiplying two corresponds to multiplying the two.
Before we get there, notice one very important and interesting feature of rotations -
they do not commute.
If we started with the cube in its standard position and first did a 90◦ turn about the
z-axis followed by a 90◦ turn about the x-axis, the final position is (where the orangish
face is opposite the red one):
Figure 4. Position after reversing order of the moves
This is clearly different from the position we obtained earlier.
What we can deduce from all this is that we should not expect for quaternions
that zw and wz are the same - since these are supposed to be the results of rotations
performed in different orders.35
So what really are quaternions ? These are certain mathematical objects which
are rare in the sense that their exact birth date is known. Quaternions were born on
16th October 1843 in Dublin, Ireland. It seems that the Irish mathematician William
Hamilton was walking along a canal with his wife when these occurred to him and he
was so thrilled with his discovery that he promptly took a penknife and carved the
following on one of the stones of Brougham bridge.
i2 = j2 = k2 = ijk = −1.
What exactly does this rather cryptic formula mean ? In the plane, which is 2-
dimensional space, every point is represented by a pair of real numbers - normally
(x, y). We can also think of the point (x, y) as the complex number x + iy. Here i is
the so-called imaginary unit, a “number” whose square is−1. We add complex numbers
by adding their real and imaginary parts separately and multiply them according to
the rule:
(a+ bi)(c+ di) = (ac− bd) + (ad+ bc)i.
Now, what Hamilton realised was that a similar thing could be done in 4-dimensional
space, every point of which represents a “quaternion” w+xi+ yj+ zk. Again addition
is easy and as expected. It is multiplication that is the surprise and the rules for which
can be derived from his carving on Brougham bridge. In addition to what he carved,
the following can be derived:
ij = k = −ji
jk = i = −kj
ki = j = −ik
Using these and the “usual rules for arithmetic” any two quaternions can be multiplied
to give a third, and of course, this multiplication is not commutative. The point is that
it can be shown that except for this lack of commutativity, all other things are as usual
- in particular, that every non-zero quaternion has a reciprocal. Also, interestingly
there is a square root of −1 for every direction in 3-space.
3. The link between quaternions and rotations
Now where is the connection with rotations ? Take a quaternion w + xi + yj + zk.
We say that it is a unit quaternion if w2 + x2 + y2 + z2 = 1. It is an exercise to show
that the reciprocal of this unit quaternion is w−xi−yj− zk. Another very interesting
exercise is to show that the product of two unit quaternions is a unit quaternion. This
has applications to expressing a positive integer as a sum of 4 squares.36
We may associate a unit quaternion with a rotation as follows. A rotation has an
angle θ and an axis of rotation. This axis makes angles say α, β, γ with the positive
x-, y- and x-axes, say. To this rotation, we associate the unit quaternion:
cos(θ
2) + sin(
θ
2)(cos(α)i+ cos(β)j+ cos(γ)k)
Recall that cos(α), cos(β), cos(γ) are called the direction cosines of the axis of rotation
and satisfy the relation cos2(α) + cos2(β) + cos2(γ) = 1. For instance, for a rotation
of 90◦ about the positive x-axis, θ = 90◦, α = 0◦, β = γ = 90◦. The associated unit
quaternion is therefore1√2+
1√2i.
Similarly for a rotation of 90◦ about the positive z-axis, the associated unit quaternion
is1√2+
1√2k.
The really amazing fact is the following. Let p be a point in 3-dimensional space
represented as a quaternion xi + yj + zk - where its co-ordinates are (x, y, z). Let q
be the unit quaternion associated to a rotation as above. Then qpq−1 is still in 3-
dimensional space and represents the result of the rotation on the point p. Further,
the only other unit quaternion that has the same property is −q.
It follows that the result of rotation by 90◦ about the x-axis followed by that of 90◦
about the z-axis is represented by the quaternion product
(1√2+
1√2k)(
1√2+
1√2i) =
1
2(1+ i+ j+k) = cos(60◦)+ sin(60◦)(
1√3i+
1√3j+
1√3k)
This says that the angle of final rotation is 120◦ and that the axis of rotation makes
equal angles with all the positive axes - as we already knew, from our experiment.
There is very little that we have proved in this article. We’ll take the viewpoint
that mathematics is not about proofs but about ideas. With what we have done so far
it should be easy to calculate using quaternions where a point goes under rotation of
some given angle about a given axis. Also if we are willing to do some 3-dimensional
co-ordinate geometry and a bit of trigonometry, it should also be possible to prove the
‘amazing fact’ that is stated above.
Let us finish describing an experiment of the physicist Paul Dirac related to the
same circle of ideas. This is related to the following question. Why is there a θ2in the
rotation quaternion?
Imagine that a cube is connected to the corners of a room which we also imagine is
a cube. The eight vertices of this little cube are connected to the eight corners of the37
room by, let’s say, elastic strings. Now give the little cube a complete 360◦ rotation
about the z-axis. It returns to its original position, of course, but the strings are in
a bit of a mess. We can unravel the mess if we rotate it back 360◦ in the opposite
direction. Suppose however that we make another 360◦ rotation. You’d imagine the
strings would be tangled up even worse. However here is an amazing thing - if you try
imagining it, it really is amazing - it is possible to unravel all the strings holding the
little cube fixed. Remember that the strings are elastic. But - and this is not very easy
to see - if you stopped with a 360◦ rotation it would not be possible to so unravel the
strings.
For a little more explanation I’ll refer you to a short video illustrating this experiment
except that it is done with a belt and not with a pair of cubes. This beautiful video is
available at: http://www.youtube.com/v/CYBqIRM8GiY.
The Institute of Mathematical Sciences, Chennai, India
E-mail address: [email protected]
38
HAMMING’S ORIGINAL PAPER REWRITTEN IN SYMBOLICFORM:
A PREAMBLE TO CODING THEORY
H. GOPALAKRISHNA GADIYAR AND R. PADMA
Abstract. In this note we try to bring out the ideas of Hamming’s classic paper on
coding theory in a form understandable by undergraduate students of mathematics.
1. Introduction
Long ago Brinn [5] made an appeal for introducing algebraic coding theory in the
undergraduate curriculum. This goal is more urgent now than ever with the ubiquity
of computers and communication devices. This article is still worth reading though it
was written thirty years ago.
The aim of this note is to write Hamming’s paper in symbolic form. Hamming used
the deceptively simple idea of interleaving parity check which helps to locate and hence
correct errors.
This note will enable readers to move on to the now classic books by V. Pless [9]and
Berlekamp [1]. Hamming’s classic paper [6] is difficult to read because the mathematics
is written in words and tables. This was because the audience of mathematicians
and engineers in those days who worked in applied fields preferred to avoid symbolic
notation and algebra as far as possible. This situation has completely changed due to
various reforms in the curriculum.
In the books by Birkhoff [3] and Pless [9], a more sophisticated approach is used
where the group property of the codes is emphasized as this leads to the more re-
cent developments. The parity check bits are not interleaving but an identity matrix
appended to the end.
Our simple minded approach of translating Hamming’s paper into symbolic form is to
enable undergraduate students to understand the elegance and simplicity of Hamming’s
construction which is a mix of engineering thinking and mathematical thinking. The
concepts of interleaving and parity check have their origin in engineering. The idea
of coding belongs more to pure mathematics. Understanding Hamming’s paper is
essential for reading the book “From error correcting codes through sphere packings to39
simple groups” by Thomas M. Thompson [10] which is a delightful mix of history and
pedagogy. This would enable undergraduate students to appreciate the unity of pure
and applied mathematics, interdisciplinary and multidisciplinary research through this
concrete example given in historical form. It would also enable them to take the more
standard route for pursuing further developments in algebraic coding theory.
2. Hamming’s construction
Parity bits are an engineering trick to detect errors in a string of 0’s and 1’s. In its
simplest form the number of 1’s is counted and then computed modulo 2. The answer
would be 0 if the number of 1’s is even and 1 if the number of 1’s is odd. This is
appended to one end of the binary string. In Hamming’s case he interleaves the parity
check bits in a clever way for error correction. This is a conceptual leap beyond error
detection which was well known then.
Let m be the number of information bits, k the number of error correction bits and
n = m+k. Since any k bits represent numbers from 0 to 2k− 1, we need the condition
that 2k − 1 ≥ n = m + k. Hence if a single error has occurred, one can determine its
position from the k-bit binary representation of its position number (Hamming calls it
Checking number.) Hamming interleaves the k check bits in positions x20 , x21 , · · ·x2k−1 .
He places them = n−k information bits at the remaining positions. Hamming analyzed
the case of (7, 4) code with the rate 47∼ 0.571 in the modern notation. In the notation
given above, k = 3, m = 4 and n = 7.
The k check bits are calculated as follows. At the encoding end, x1 = x20 is deter-
mined by the partial parity check equation
(1) x1 + x3 + x5 + x7 + · · · = 0
Notice that all these bits have their position numbers 1, 3, 5, 7 · · · which when they are
written in their binary representation have the least significant bit equal to 1. Hence
if the single error has occurred in any one of the odd positions, then at the decoding
end, the partial parity check equation will give
(2) x1 + x3 + x5 + x7 + · · · = 1
Next, x2 is determined (at the encoding end) by the equation
(3) x2 + x3 + x6 + x7 + · · · = 0
Notice that the binary representations (10, 11, 110, 111, · · · ) of the position numbers of
2, 3, 6, 7, · · · have 1 as their second bit from the right and 2 is the smallest of these40
numbers. Hence if the single error has occurred in a position whose second bit is 1,
then at the decoding end, we would get
(4) x2 + x3 + x6 + x7 + · · · = 1
Similarly x22 , . . . x2k−1 are determined by the corresponding partial parity check equa-
tions. Notice that 1, 2, 4, 8, · · · , 2k−1 (1, 10, 100, 1000, · · · 100 · · · 0)are the smallest num-
bers having 1 in the first, second, third, fourth, · · · kth positions in their binary rep-
resentations. Thus the position number of the error bit is determined bit by bit from
right to left. The least significant bit is zero if (1)is true and 1 if (2) is true. Similarly
the previous bit is zero if (3) is true and 1 if (4) is true and so on. Once the position
of the error bit is found, the bit can be corrected as a bit can take only two values: 0
or 1.
At this point we would encourage the reader to look at the classic paper of Hamming
paper [6] which is freely down loadable from the Internet and then read the standard
books listed below.
3. Pedagogical and historical comments
[11] uses symbolic notation for bits with parity check matrix but the idea of inter-
leaving is missed. [8] also does not talk about interleaving parity check bits. [3] and
[11] discuss Hamming codes as a special case of group codes.
References
[1] E. R. Berlekamp, Algebraic coding theory, McGraw - Hill, 1968.
[2] E. R. Berlekamp, Key papers in the the development of Coding theory, Ed: E. R. Berlekamp,
IEEE Press, 1974.
[3] G. Birkhoff and T. C. Bartee, Modern applied algebra, McGraw - Hill, 1970.
[4] Ian F. Blake, Algebraic Coding theory: History and Development, Dowden, Hutchinson & Ross,
1973
[5] L. W. Brinn, Algebraic coding theory in the undergraduate curriculum, American Math. Monthly,
91, 8 October, 1984, 509-513.
[6] R. W. Hamming, Error detecting and error correcting codes, The Bell System Technical Jour-
nal, 29 April 1950, 147-160. http://www3.alcatel-lucent.com/bstj/vol29-1950/articles/
bstj29-2-147.pdf
[7] R. Hill, A first course in coding theory, Oxford University Press, 1986.
[8] W. C. Huffman and V. Pless, Fundamentals of error correcting codes, Cambridge University
Press, 2003.
[9] V. Pless, Introduction to the theory of error correcting codes, Third Edition, Wiley - Interscience
Series in Discrete Mathematics and Optimization, 1998.
[10] T. M. Thompson, From error correcting codes through sphere packings to simple groups, Carus
Monograph No. 21, Math. Assoc. Amer., Washington, D.C., 1983.41
[11] J. P. Tremblay and R. Manohar, Discrete Mathematical Structures with applications to computer
science, McGraw-Hill Interamericana, 1975.
School of Advanced Sciences, V. I. T. University, Vellore 632 014, India
E-mail address: [email protected], [email protected]
42
EVERY NON-CONSTANT POLYNOMIAL WITH REALCOEFFICIENTS HAS A ROOT IN C
K. N. RANGANATHAN
Abstract. In this article we give an almost algebraic proof of ”The Fundamental
Theorem of Algebra” [1]. We observe that there are polynomials with real coefficients
that do not have a root in R. In fact, we do not even have a square root for any t < 0
in R. In other words
x2 + a = 0
has no roots in R when a > 0. Can we make R the set of real numbers into a bigger
set K where we can extend the usual addition and multiplication in R to the bigger
set K, so that all quadratic polynomials with coefficients in K have roots in K ? In
other words can we have a field extension K of the field R of real numbers such that
all quadratic polynomials can be factored into two linear polynomials? We observe
that we do have a square root in C the set of complex numbers, for every complex
number ω ∈ C.
Lemma 1. Every complex number ω ∈ C has a square root in C.
Proof. If ω = a+ ib = z2 with a, b ∈ R, then we have
(1) ω = a+ ib = z2 = (x+ iy)2 = (x2 − y2) + 2ixy
where z = x+ iy with x, y ∈ R. This gives
(2) x2 − y2 = a and 2xy = b.
Solving we get
(3) x2 =
√a2 + b2 + a
2, y2 =
√a2 + b2 − a
2, 2xy = b
and these equations definitely have a solution for x, y ∈ R. Thus every complex number
has a square root in C. �
This enables us to solve any quadratic equation with complex coefficients, completely
in C. In fact, we see that the roots of
(4) az2 + bz + c = 043
has the roots
(5) z1 =−b+
√b2 − 4ac
2a, z2 =
−b−√b2 − 4ac
2a.
Consider a polynomial P (x) of degree n > 0 with real coefficients. First, we ask the
question
” Do we have a field extension K of R such that P (x) has a root in K?”
We answer this question in the following lemma.
Lemma 2. Let P (x) be a non-constant polynomial in R[x], the ring of polynomials in
x with coefficients in R. Then there exists a field extension K of R such that P (x) has
a root in K.
Proof. Let degree of P (x) be n and since P (x) is a non constant polynomial we see
that n > 0. If n = 1, then P (x) = ax + b for some a, b ∈ R with a 6= 0. In this case,
we have
(6) P (x) = a
(x+
b
a
)and x = − b
a∈ R is a root of P (x). We prove our lemma by induction on n. For n = 1
we have seen that the result is true. Let now n > 1. Assume that the lemma is true for
all non constant polynomials of degree less than n and let P (x) ∈ R[x] be a polynomial
of degree n > 1. If P (x) admits a non trivial factorization in R[x]
(7) P (x) = Q(x) R(x)
with deg Q(x) > 0 and deg R(x) > 0, Q(x), R(x) being polynomials in R[x], then we
note that deg Q(x) < n and hence by induction hypothesis there exists an extension
field K of R and α ∈ K such that Q(α) = 0. This means that P (α) = 0. In other
words, by induction, the lemma is true for all reducible polynomials. Assume now that
P (x) is an irreducible non constant polynomial in R[x] of degree n > 1. Define K as
the quotient ringR[x]
(P (x))where (P (x)) is the ideal consisting of all multiples of P (x)
in the ring R[x]. The non-zero elements of K are of the form (P (x))+ q(x) where q(x)
is a non-zero polynomial in R[x] of degree k < n. As P (x) is irreducible it is clear that
g.c.d(P (x), q(x)) = 1. Hence we can find polynomials a(x), b(x) and q1(x) with real
coefficients and degree q1(x) < n such that
(8) a(x)P (x) + (b(x)P (x) + q1(x))q(x) = 1.44
This means that in the quotient ringR[x]
(P (x))we have
(9) ((P (x)) + q1(x)) ((P (x)) + q(x)) = (P (x)) + 1.
In other words, every non-zero element in the quotient ring K =R[x]
(P (x))has a multi-
plicative inverse or K is a field. Clearly, t ∈ R can be identified with (P (x)) + t ∈ K.
This identification establishes K as an extension of R. Now, consider the the quotient
map
(10) Π : R[x] −→ R[x](P (x))
= K
defined by
(11) Π(p(x)) = (P (x)) + p(x)
for any polynomial p(x) ∈ R[x]. We note that
(12) Π(P (x)) = (P (x)) = 0 ∈ K
and the kernel of Π is the ideal (P (x)). Suppose we denote Π(x) = (P (x)) + x by
x0 ∈ K. Then
(13) Π(a0xn + a1x
n−1 + · · ·+ an−1x+ an) = a0xn0 + a1x
n−10 + · · ·+ an−1x0 + an.
In other words for any polynomial f(x) ∈ R[x] we have
(14) Π(f(x)) = f(x0) = f(Π(x)).
In particular
(15) P (x0) = P (Π(x)) = Π(P (x)) = 0 ∈ K.
This means that we have constructed a field extension K of R in which P (x) has a
root. This proves the lemma. �
Lemma 3. If P (x) is a non constant polynomial in R[x] of degree n > 1, then there
exists a field extension K of R such that P (x) has all its n roots in K. In other words
P (x) can be completely factored into linear factors in K[x].
Proof. If degree P (x) is 1 then there is nothing to prove. We prove the lemma by
induction on degree of P (x). Assume that the lemma is true for polynomials of degree
k with 1 ≤ k < n and let P (x) be a polynomial in R[x] of degree n. If P (x) is reducible,as in Lemma 1, we are done by induction. So we may assume that P (x) is irreducible.
By lemma 1 we have an extension L of R such that P (x) has a root α1 ∈ L. We may
write
(16) P (x) = (x− α1)q(x)45
where q(x) ∈ L[x] is of degree n − 1. Now by induction hypothesis, we can find an
extension K of L such that q(x) has all its roots in K. (Note that, in Lemma 2 and
Lemma 3, we can replace R by any field F ) In other words q(x) can be completely
factored as
(17) q(x) = a(x− α2)(x− α3) · · · (x− αn)
in K[x]. This means that
(18) P (x) = a(x− α1)(x− α2)(x− α3) · · · (x− αn)
in K[x]. This proves Lemma 3. �
Definition 1. The polynomial function f(x1, x2, · · · , xn) is symmetric in n variables
if
(19) f(x1, x2, · · · , xn) = f(xσ(1), xσ(2), · · · , xσ(n)) ∀ σ ∈ Sn
where Sn = the set of all permutations of {1, 2, 3, ·, n}.
For example the following polynomials are symmetric polynomials:
s1 = s1(x1, x2, · · · , xn) =n∑i=1
xi
s2 = s2(x1, x2, · · · , xn) =∑i<j
xixj
s3 = s3(x1, x2, · · · , xn) =∑i<j<k
xixjxk
· · ·
sm = sm(x1, x2, · · · , xn) =∑
i1<i2<i3<···<im
xi1xi2xi3 · · ·xim
· · ·sn = sn(x1, x2, · · · , xn) = x1x2x3 · · ·xn(20)
These are known as elementary symmetric polynomials. If x1, x2, x3, · · · , xn are all the
roots of a polynomial P (x), then P (x) must be of the form
(21) P (x) = a{xn−s1xn−1+s2xn−2+· · ·+(−1)kskx
n−k+· · ·+(−1)n−1sn−1x+(−1)nsn}
for some a 6= 0, where s1, s2, s3, · · · , sn are the elementary symmetric polynomials in
x1, x2, x3, · · · , xn.
Lemma 4. A polynomial f(x1, x2, x3, · · · , xn) is symmetric if and only if it is a poly-
nomial in the elementary symmetric functions s1, s2, s3, · · · , sn.46
Proof. Consider the map
(22) π : K[x1, x2, x3, · · · , xn] −→ K[x1, x2, x3, · · · , xn−1]
defined by
(23) π(xj) =
{xj 1 ≤ j < n
0 j = n
This map π kills off xn. We note that
(24) π(f(x1, x2, x3, · · · , xn)) = f(x1, x2, · · · , xn−1, 0).
It is clear that when f(x1, x2, x3, · · · , xn) is a symmetric polynomial in n variables then
so is π(f) = f(x1, x2, · · · , xn−1, 0) symmetric in the n − 1 variables x1, x2, · · · , xn−1.
We also note that
(25) π(sj(x1, x2, · · · , xn)) ={sj(x1, x2, · · · , xn−1) j 6= n
0 j = n
where s1, s2, · · · , sn are the elementary symmetric polynomials. We use induction on
n the number of variables to get a polynomial P such that
(26) π(f(x1, · · · , xn)) = P (s1(x1, · · · , xn−1), · · · , sn−1(x1, · · · , xn−1)).
Using the same polynomial P , we now define a function g(x1, x2, · · · , xn) by
(27) g(x1, · · · , xn) = P (s1(x1, x2, · · · , xn), · · · , sn−1(x1, x2, · · · , xn)).
Then we get
(28) π(f(x1, · · · , xn)− g(x1, · · · , xn)) = f(x1, · · · , xn−1, 0)− g(x1, · · · , xn−1, 0).
This implies that xn divides f − g and as the polynomial f − g is symmetric in the n
variables x1, x2, · · · , xn we see that each xj divides f − g for j = 1, 2, 3, · · · , n. This
again implies that sn(x1, x2, · · · , xn) = x1x2x3 · · ·xn divides f − g. The degree of the
monomial axa11 xa22 · · ·xann is the sum a1+a2+ · · ·+an. The total degree of a polynomial
h ∈ K[x1, x2, · · · , xn] is the maximum of the degrees of its monomial summands. For
example, if
(29) h(x1, x2, · · · , xn) = x31x43 − x72x
43x
54 + x31x
42x
2n + x3x
5n
then
(30) degree of h(x1, x2, · · · , xn) = max{7, 16, 9, 6} = 16.
As we have seen, sn divides f − g and therefore the total degree of the polynomial
(31)f − g
sn=f(x1, x2, · · · , xn)− g(x1, x2, · · · , xn)
sn(x1, x2, · · · , xn)47
is less than the total degree of the symmetric polynomial f . Hence by induction on the
total degree of f we see thatf − g
sncan be expressed as a polynomial in the elementary
symmetric polynomials in x1, x2, · · · , xn. This means that f itself is expressed as a
polynomial in elementary symmetric polynomials. �
Lemma 5. Every polynomial P (x) of odd degree in R[x] has a root in R.
Proof. Each polynomial in R[x] is a continuous function on the whole of R and hence
by the intermediate value theorem has a real root. In fact degree of P (x) is odd implies
that
(32) limx→∞
P (x) = ∞ and limx→−∞
P (x) = −∞.
So, we find that there exist a > 0 and b < 0 such that f(a) > 0 and f(b) < 0. Hence by
the continuity of P (x) we must have a c lying in the interval [b, a] such that P (c) = 0.
This means that P has a real root. �
Armed with these five lemmas we can now prove the Fundamental Theorem of Al-
gebra.
Theorem 1. Every non-constant polynomial with real coefficients has a root in the set
of complex numbers C.
Proof. Let
(33) P (x) = a0xn + a1x
n−1 + a2xn−2 + · · ·+ akx
n−k + · · ·+ an−1x+ an
be a polynomial in R[x] with deg P (x) = n > 0. We want to prove that there exists
a complex number z0 such that P (z0) = 0. Without loss of generality we may assume
that a0 = 1 or in other words, P (x) is a monic polynomial. By Lemma 3, there exists
an extension field K of R such that P (x) has all its roots in K. This means that there
exist z1, z2, · · · , zn in K such that
(34) P (x) = (x− z1)(x− z2)(x− z3) · · · (x− zn).
This means that the elementary symmetric functions sj(z1, z2, · · · , zn) satisfy
(35) sj(z1, z2, · · · , zn) = (−1)jaj for j = 1, 2, 3, · · · , n.
Let deg P (x) = n = 2km where m is an odd integer. If k = 0, we note that P (x) has
odd degree and by Lemma 5 it has a real root. We prove by induction on k. Assume
that P (x) has a complex root when degree of P (x) is 2k−1m1 where m1 is odd. This is
our induction hypothesis. For any real number r ∈ R, we define
(36) Qr(x) =∏
1≤i<j≤n
{x− (zi + zj + rzizj)} .
48
The coefficients of the polynomial Qr(x) are all symmetric polynomials with real coef-
ficients in the n variables z1, z2, · · · , zn. Hence, by Lemma 4, the coefficients of Qr(x)
can be expressed as polynomials with real coefficients in elementary symmetric poly-
nomials, namely, −a1, a2,−a3, · · · , (−1)jaj, · · · , (−1)nan. This implies that Qr(x) is a
polynomial with real coefficients. Further
(37) degree of Qr(x) =n(n− 1)
2= 2k−1m(n− 1)
and m(n − 1) is odd. Therefore by induction hypothesis Qr(x) has a complex root.
This means that we have two indices ir < jr in {1, 2, 3, · · · , n} corresponding to r ∈ R,such that
(38) zir + zjr + rzirzjr ∈ C.
The number of pairs (i, j) with i < j is N =n(n− 1)
2. Hence we can find distinct real
numbers r, s such that for the same pair (i, j) we have
(39) zi + zj + rzizj ∈ C and zi + zj + szizj ∈ C.
This implies that zi+zj = ω1 and zizj = ω2 are in C. Now, by Lemma 1 every complex
number has a square root in C and hence the quadratic equation
(40) 0 = z2 − ω1z + ω2 = z2 − (zi + zj)z + zizj
has the roots zi and zj and it follows that zi, zj are complex numbers.
This proves our Fundamental Theorem of Algebra. �
Remark: In the above proof, the only place where we use Analysis is in the use
of Lemma 5 and intermediate value theorem. Our proof of Fundamental Theorem of
Algebra as stated above, can be extended to
Theorem 2. Every polynomial P (z) ∈ C[z] has all its roots in C or in other words
the field C of complex numbers is algebraically closed.
Proof. Let P (z) be a non constant polynomial in C[z]. Define F (z) = P (z)P (z). Then
F (z) is a polynomial with real coefficients and F (z) has a root in C if and only if P (z)
has a root in C. Thus in view of our Theorem 1, we deduce that C is algebraically
closed. �
References
[1] Pierre Samuel, Algebraic Theory of Numbers (1970)
49
TOPOLOGICAL COMPLEXITY AND TOPOLOGICAL ROBOTICS
TULSI SRINIVASAN
1. Introduction
Topological robotics is one of many emerging areas in the burgeoning field of com-
putational topology. Given a mechanical system, we consider the space X of all its
possible configurations. Typically, the configuration space is a subset of Euclidean
space, and so it naturally inherits a topology. We are concerned with the discontinu-
ities that arise when we attempt to define motion between the configuration states of
the system.
Let PX = {γ : [0, 1] → X : γ is continuous} be the path space of X, equipped
with the compact-open topology. There is a natural map π : PX → X × X that
sends a path γ to the pair (γ(0), γ(1)). We restrict our discussion to path connected
spaces, which means that π is surjective and, in fact, a fibration. In addition to the
paths between states being continuous, we would like the paths themselves to vary
continuously as we move from one initial-final configuration pair to another. Formally,
we would like for there to exist a section - a continuous map s : X × X → PX such
that π ◦ s = idX×X . Unfortunately, this turns out to be impossible for all but the most
prosaic of configuration spaces:
1.1. Theorem. The map π : PX → X × X defined above has a section iff X is
contractible. [2]
Proof. Suppose X is contractible, i.e., there is a continuous map H : X × [0, 1] → X
with H(x, 0) = x and H(x, 1) = x0 for all x ∈ X and some fixed x0 ∈ X. Given any
(a, b) ∈ X ×X, we let s(a, b) be the path given by H that first joins a to x0 and then
joins x0 to b. This is done by defining
s(a, b)(t) =
{H(a, 2t) if 0 ≤ t ≤ 1/2;
H(b, 2− 2t) if 1/2 ≤ t ≤ 1.
Conversely, suppose a section s : X × X → PX exists. Given any x0 ∈ X, define
G : X× I → X by G(x, t) = s(x, x0)(t). Then G is a continuous map with G(x, 0) = x
and G(x, 1) = x0, and so X is contractible.50
�
If we drop continuity as a requirement, then such an s always exists, and is called
a motion planning algorithm. If X is not contractible, then the best thing to hope for
is that X ×X can be broken down into a finite number of parts, upon each of which
s is continuous. In this case, the motion planning algorithm is said to be tame. The
minimum number of parts required for this can be thought of as the minimum number
of rules required to define motion in X, and is the intuition behind Michael Farber’s
definition of topological complexity.
This is not a new idea. The Lusternik-Schnirelmann category, introduced in 1934 and
still an object of study, is the smallest number of open subsets a space can be broken
down into so that each subset is contractible in the whole space. In fact, both this
and topological complexity are special cases of the Schwarz genus of a fibration, which
was used by Smale and Vassiliev to study the complexity of algorithms for solving
polynomial equations [3]. However, it was Michael Farber who first realised that a
similar homotopy invariant could be defined for configuration spaces, an insight that
has led to results and questions that are interesting from both a theoretical and an
applied point of view, a few of which we discuss in this article.
2. Topological complexity and some computations
2.1. Definition. A Euclidean neighbourhood retract (ENR) is a topological space X
that can be embedded in Rn for some n such that there is an open neighbourhood
U ⊃ X in Rn and a retraction r : U → X.
It is known that a subset of Rn is an ENR iff it is locally contractible and locally com-
pact. All locally well-behaved spaces like polyhedra, finite dimensional cell complexes
and manifolds are ENRs. The Hawaiian earring and fractal spaces like the Sierpinski
carpet are not.
2.2. Definition. Given a space X, its topological complexity, TC(X), is the smallest
integer k such that X × X =⋃k
i=1 Ui such that i) each Ui is open and ii) for each i,
there is a continuous section si : Ui → PX of the map π : PX → X×X defined above.
If no such k exists, we write TC(X) =∞.
For the remainder of this article we will assume that X is an ENR space, in which
case the Ui can be taken to be arbitrary ENR subsets rather than open ones.
2.3. Example. From Theorem 1.1, TC(X) = 1 iff X is contractible.51
We now compute the topological complexity of Sn, n ≥ 1. We will use the fact that
there is a non-vanishing vector field on Sn iff n is odd. Furthermore, if n is even, then
there is a vector field that vanishes at precisely one point.
2.4. Example. For odd n, TC(Sn) = 2, and for even n, TC(Sn) ≤ 3. [4]
Proof. Since Sn is never contractible, TC(Sn) > 1. Given two points on Sn that are
not antipodal, there is a unique geodesic joining them. Thus if we set F1 = {(a, b) ∈Sn × Sn : a 6= −b}, then there is a section s1 : F1 → PX sending each pair to the
geodesic between them.
Assume n is odd and let F2 = {(a,−a) ∈ Sn × Sn} be the set of antipodal pairs.
Choose a non-vanishing vector field V on Sn, and define a section s2 : F2 → PX that
sends each antipodal pair (a,−a) to the semi-circle tangent to V (a). Mathematically,
s2(a,−a)(t) = a cos(πt) +V (a)
|V (a)|sin(πt). Since the Fi are ENRs, we have TC(Sn) = 2
when n is odd.
If n is even, pick a vector field V that vanishes at exactly one point, say at a0 ∈ Sn.
Let F2 = {(a,−a) ∈ Sn × Sn : a 6= a0}, and let s2 be as above. Finally, let F3 =
{(a0,−a0)}, and let s3 map the pair to some fixed arc between the two points. This
yields TC(Sn) ≤ 3 when n is even. �
In fact, the topological complexity of an even dimensional sphere is precisely three,
a fact which follows from TC(X) being bounded below by cup-length in any cohomo-
logical ring. For the sake of accessibility, we do not discuss lower bounds at all, but
the interested reader is referred to [2].
3. Motion planning for a robot arm
We consider a robot arm that consists of straight bars L1, ..., Ln in R2, connected
via joints that can revolve freely in the plane (see the figure below). We assume that
one end of L1 is fixed. Let αi be the angle made by the bar Li with the x-axis.
L1
L2
L3
α1
α2
α3
52
Each state of this system is uniquely determined by the αi, so, if we allow the arm to
self-intersect, the configuration space is the n-torus T n = S1× ...×S1. Similarly, if the
Li lie in R3 and the joints can revolve in 3-dimensional space, then the configuration
space is S2 × ...× S2. In order to compute the complexity of these spaces, we use the
following:
3.1. Theorem. Let X, Y be ENRs. Then TC(X × Y ) ≤ TC(X) + TC(Y )− 1.
Sketch of proof. It can be shown that TC(X) ≤ k iff there is a motion planning algo-
rithm s : X ×X → PX and an increasing sequence ∅ = U0 ⊂ U1 ⊂ ... ⊂ Uk = X ×Xof open subsets of X ×X such that each restriction s|(Ui+1−Ui) is continuous [4].
Let TC(X) = n and TC(Y ) = m (the result follows trivially if either one is infinite),
and take motion planning algorithms s : X × X → PX and s′ : Y × Y → PY , and
towers of open sets U0 ⊂ ... ⊂ Un and V0 ⊂ ... ⊂ Vm of X ×X and Y × Y respectively,
satisfying the conditions above.
There is a natural product map S : (X × Y ) × (X × Y ) → P (X × Y ) given by
S((x, y), (x′, y′))(t) = (s(x, x′)(t), s′(y, y′)(t)). For k = 0, ..., n + m − 1, set Wk =⋃i+j=k+1 Ui × Vj. The tower W0 ⊂ ... ⊂ Wn+m−1 and the map S then satisfy the
conditions above, and so we get TC(X × Y ) ≤ n+m− 1. �
3.2. Corollary. Let X be the cartesian product of n copies of Sm. If n is odd, then
TC(X) ≤ n+ 1, and if n is even, then TC(X) ≤ 2n+ 1.
Proof. It follows from Theorem 3.1 that TC(X1× ...×Xn) ≤ TC(X1)+ ...+TC(Xn)+
1− n. This, together with Example 2.4, gives the required result.
�
It follows that the complexity of motion planning for a robot arm is at most n + 1
in the planar case and at most 2n + 1 in the spatial case. Again, an argument using
cup-lengths can be used to show that equality holds in both cases.
4. The Iwase-Sakai conjecture
This has only been a brief overview of topological complexity. There are numerous
other topologically interesting results, including upper and lower bounds in terms of
dimension, cup-lengths and the Lusternik-Schnirelmann category, and a relationship
between complexity and the immersion problem for projective spaces and lens spaces.
There are also, amongst others, applications to the complexity of simultaneously con-
trolling multiple systems and collision free motion-planning (for instance, the motion53
of automated guided vehicles). There is, however, one rather basic practical issue that
is yet to be resolved: if the final state is the same as the initial state, then is the path
provided by the motion planning algorithm the constant path? To this end, Iwase and
Sakai introduced the following:
4.1. Definition. The monoidal topological complexity, TCM(X), is the smallest k such
that X×X =⋃k
i=1 Ui, such that i) each Ui is open and ii) for every i, there is a section
si : Ui → PX such that si(a, a) is the constant path at a for all a ∈ X.
4.2. Conjecture (Iwase-Sakai). For any ENR space X, TC(X) = TCM(X).
A proof for this appeared in [5] and was later retracted, though the authors have
shown that the inequality TC(X) ≤ TCM(X) ≤ TC(X) + 1 always holds. There have
been several contributions towards proving the conjecture, including an equivalent
statement by Doeraene and El Haouari [6], a proof for Lie groups by Dranishnikov [1],
and a proof for weaker definitions of TC(X) and TCM(X) by Vandembroucq et al. [6],
but, for now, the problem remains open.
References
[1] A. N. Dranishnikov, On topological complexity and LS-category. arXiv:1207.7309, 2012.
[2] M. Farber, Topological complexity of motion planning. Discrete Computational Geometry,
29, no. 2, 2003.
[3] M. Farber, Topology of robot motion planning. Morse theoretic methods in nonlinear analysis
and in symplectic topology, NATO Science Series II: Math. Phys. Chem., 217, Springer,
Dordrecht, 2006.
[4] M. Farber, Invitation to topological robotics. Zurich Lectures in Advanced Mathematics,
European Mathematical Society (EMS), Zurich, 2008.
[5] N. Iwase and M. Sakai, Topological complexity is a fibrewise L-S category. Topology and its
Applications, 157, no. 1, 2010.
[6] L. Vandembroucq, Topological complexity and related invariants. Presentation at Applied
and Computational Algebraic Topology, Bremen, 2011.
Tulsi Srinivasan, Department of Mathematics, University of Florida, 358 Little
Hall, Gainesville, FL 32611-8105, USA
E-mail address: [email protected]
54
REPRESENTATION THEORY AND PARTICLE PHYSICS
GIRISH M. KULKARNI
Abstract. In this article, we take an overview of the connection between elementaryparticles in Physics and the irreducible representations of Poincare group. Here welook at the mathematical side of the connection.
Minkowski Space
In Physics, Minkowski space is the four dimensional space where three co-ordinatesgive position in the space and remaining one specifies position in time. Mathemat-ically, it’s a four dimensional real vector space with the inner product defined by〈v, w〉 = −v0w0+v1w1+v2w2+v3w3, where v = (v0, v1, v2, v3) and w = (w0, w1, w2, w3).The group of all isometries (distance preserving maps) of a space under which the spaceis invariant is called the symmetry group of the space. The usual n− dimensional Eu-clidean space has Euclidean group E(n), which is generated by orthogonal group O(n)and translational group T (n), as its symmetry group. While for Minkowski space sym-metry group is the Poincare group.
Poincare Group
As is the case of Euclidean space, the symmetry group of Minkowski space is gen-erated by translations and the isometries which leave the origin fixed, the latter arecalled Lorentz translations. Precisely the Poincare group is the semidirect productR1,3 o SO(1, 3) of the translational group and Lorentz group.
Representations of Poincare Group
The following is the method of little groups by Winger and Mackey to find irreduciblerepresentations of semidirect product of groups. In which one assumes that the normalsubgroup in the semidirect product is abelian.Let G = A o H, where A is normal subgroup of G and let A be abelian. Since A is
55
abelian its irreducible representations are of degree one and they form a group, say X.The action of G on X can be given by
(sχ)(a) = χ(s−1as) for s ∈ G,χ ∈ X, a ∈ A.
Then we consider (χi)i∈X/H the representatives of orbits of H in X and subgroupHi = {h ∈ H|hχi = χi} of H. Let Gi = AoHi. Extend χi to Gi by
χi(ah) = χi(a) for all a ∈ A, h ∈ Hi.
Since hχi = χi for all h ∈ Hi, χi is character of degree 1 of Gi. Now take ρ to bean irreducible representation of Hi, composing it with canonical projection Gi → Hi
we get irreducible representation ρ of Gi. Finally we take tensor product of these tworepresentations to get an irreducible representation of Gi and we get correspondinginduced representation θi,ρ of G. The following result states that all the irreduciblerepresentations of G can be obtained in this fashion.
Theorem 1. Irreducible representations of semidirect product [2, page 62]
(1) θi,ρ is irreducible.(2) θi,ρ and θi′,ρ′ are isomorphic then i = i′ and ρ is isomorphic to ρ′.(3) Every irreducible representation of G is isomorphic to one of the θi,ρ.
So we see that the irreducible representations of Poincare group are parametrized bytwo parameters.
The Connection
Eugene Winger in 1930 found a connection between Particle physics and represen-tation theory where he noted that the different quantum states of elementary particlegive rise to an irreducible representation of the Poincare group. These are parametrizedby mass and spin (which are the two parameters mentioned above). Hence each type ofparticle corresponds to an irreducible representation and we can classify the particles.
References
[1] Wigner, E. P. (1939), “On unitary representations of the inhomogeneous Lorentz group”, Annalsof Mathematics 40 (1): 149–204.
[2] Jean-Pierre Serre, Linear Representations of Finite Groups Springer 1977.
56
AN INTRODUCTION TO FRACTIONAL CALCULUS
MADHUKANT SHARMA
The concept of the differentiation operator D = d/dx is familiar to all who have
studied the elementary calculus. And for suitable functions f , the nth derivative of f ,
namely Dnf(x) = dnf(x)/dxn is well defined - provided that n is a positive integer. In
1695 Marquis de L’Hopital (1661 - 1704) inquired of Gottfried Wilhelm Leibniz (1646 -
1716) what meaning could be ascribed to Dnf if n were a fraction. On 30th September
1695, Leibniz wrote to L’Hopital ”...This is an apparent paradox from which, one day,
useful consequences will be drawn. ...”. The paradoxical aspects are due to the fact
that there are several different ways of generalizing the differentiation operator to non-
integer powers, leading to inequivalent results. Most authors on this topic will cite a
particular date 30 September 1695 as the birthday of so called ’Fractional Calculus’.
The fractional calculus has drawn the attention of many famous mathematicians,
such as Euler, Laplace, Fourier, Abel, Liouville, Riemann, and Laurent. But it was not
until 1884 that the theory of generalized operators achieved a level in its development
suitable as a point of departure for the modern mathematician. By then the theory had
been extended to include operators Dµ, where µ could be rational or irrational, positive
or negative, real or complex. Thus the name fractional calculus became somewhat of a
misnomer. A better description might be differentiation and integration to an arbitrary
order. Many mathematicians found, using their own notation and methodology, defini-
tions that fit the concept of a non-integer order integral or derivative. The most famous
of these definitions that have been popularized in the world of fractional calculus (not
yet the world as a whole) are the Riemann - Liouville and Caputo definitions.
Most of the mathematical theory applicable to the study of fractional calculus was
developed prior to the turn of the 20th century. However it is in the past 100 years
that the most intriguing leaps in engineering and scientific application (see [1, 2, 3])
have been found such as fluid flow, rheology, diffusive transport akin to diffusion,
electrical networks, electromagnetic theory, and probability. It seems that hardly a
field of science or engineering has remained untouched by this topic. Yet even though
the subject is old, it is rarely included in today’s curricula. Possibly this is because
many mathematicians are unfamiliar with its uses.
The definitions of Riemann - Liouville and Caputo’s fractional integral and differ-
ential operators are defined as follows: The Riemann - Liouville fractional integral of57
order n > 0 is defined by [1, 2, 3]:
Jna f(t) = D−na f(t) =
1
Γ(n)
∫ t
a
(t− s)n−1f(s)ds.
The Riemann - Liouville fractional derivative of the function f of order n as
aDnt f(t) =
1
Γ(m− n)
dm
dtm
∫ t
a
(t− s)m−n−1f(s)ds.
And the Caputo’s fractional derivative of the function f of order n is given by
caD
nt f(t) =
1
Γ(m− n)
∫ t
a
(t− s)m−n−1fm(s)ds.
where f is an abstract continuous function on the interval [a, b], m = dne and Γ(n) is
the Gamma function. For n = 0, Jna , aDnt , and
caD
nt are identity operators.
Properties
(i) Jna Jma φ = Jm+n
a φ, for m, n ≥ 0 and φ ∈ L1[a, b], similarly for operators aDnt ,
and caD
nt hold.
(ii) aDnt J
na f = f whereas Jna aD
nt f 6= f for n > 0.
(iii) Integral operator Jna and the differential operators aDnt ,
caD
nt are linear opera-
tors, i.e.,
Jna (c1f1+c2f2)(t) = c1Jna f1(t)+c2J
na f2(t), similarly for operators aD
nt , and
caD
nt
hold.
(iv) It can be seen that aDnt φ = DmJm−n
a φ and caD
nt φ = Jm−n
a Dmφ where m = dne.
Application
The Tautochrone Problem
A bead of mass m slips along a frictionless wire, from height
η = y, at time t = 0
to height
η = 0, at time t = T
We want to find the smooth curve (if there is such curve) along which the bead’s falling
time is the same for any y. To that end, we look for a differential equation that will
describe such motion.58
The Tautochrone Differential Equation :
In this case, one can see that the potential energy converts into the kinetic energy,
i.e.,
mg(y − η) =1
2m
(ds
dt
)2
where s = s(η) is the arc - length of the curve. Therefore,
ds
dt= ±
√2g(y − η)
Since the particle speed is in the direction of −y, we have
ds
dt= ±
√2g(y − η)
s′(η)dη√(y − η)
= −√2gdt.
The Tautochrone Integral Equation :
Integrating the right side over the time interval, and the left over the height inter-
val we obtain an integral equation for the curve of the bead’s motion:∫ η=y
η=0
s′(η)dη√(y − η)
=√
2gT = C.
The above integral equation can also be written in the following form:
D− 12 s′(y) =
C
Γ(12
) .Which is a fractional integral equation of order 1
2. On applying D
12 to both sides of
above equation, one can get
s′(y) =C
π√y
References
[1] Igor Podlubny, Fractional Differential Equations, Vol. 198, Mathematics in Science and Engineer-
ing, Academic Press, San Diego, California, USA, 1999.
[2] Kai Diethelm The analysis of Fractional Differential Equations, Springer-Verlag Berlin Heidelberg
2010.
[3] Saıd Abbas, Mouffak Benchohra, Gaston M. N’Guerekata, Topics in Fractional Differential Equa-
tions, Springer Science+Business Media New York 2012.
59
RUBIKS CUBE AND GROUP THEORY
PRAJAKTA SAHASRABUDDHE
”We turn the cube and it twists us” - Erno Rubik
Erno Rubik , a socialist burocrat who lived in Budapest, Hungury has puzzled millions
of people and even has inspired mathematicians with his invention. What inspired
Erno was a popular puzzle called a 15 puzzle ,invented in 1870’s. The ”15 puzzle”
consisted of 15 consecutively numbered flat squares that can be slid around inside a
square frame. This motivated Erno to create a 3-D version of the puzzle called magic
cube. It was later renamed in honor of its creator as a Rubiks cube.
A rubik cube consists of 6 sides with 9 individual pieces on each with the main ob-
jective to recreate its original position, a solid colour for each side, without removing
any piece from cube. Though it looks like a colorful children’s toy, solving it involves a
deep mathematical concept that have lead to several championships being conducted
to solve it.
Though there are many versions of the rubiks cube, this article will concentrate on
3 × 3 × 3 cube in particular. So a simple 3 × 3 × 3 cube consists of 27 small cubes,
typically called ”Cubies”. 26 of these cubies are visible, which consists of 8 corner
cubies, 12 edge or side cubies and 6 centre cubies. Remaining 1 is invisible middle
cubie.
Here we will develope a systematic way to figure out what the possible valid con-
figurations which can be achieved practically. We assume a basic knowledge of Group
theory. Now we are going to develope some notations which are helpful in referring to
individual cubies and their places.
Notations
Let’s name the faces as Front-’f’ , Back-’b’, Right-’r’, Left-’l’, Down-’d’. Similarly
clockwise rotations of these faces will be named same as above but by corresponding
capital letter. We will name corner cubie simply by listing its visible faces in clockwise
order. For example, the cubie in the upper, right, front corner is written as ’urf’. We60
can also call this cubie ’rfu’ or ’fur’. Sometimes we will care which face is listed first,
at these times we will talk about ”oriented cubies”. That is ’urf’, ’rfu’ and ’fur’ are
different. Otherwise we will talk about an oriented cubie. Similarly ”cubicle” is the
space in which the cubie lives. Thus if the cube is in the start configuration(i.e the
cube is solved) then each cubie lives in the cubicle of same name. So rotating a face of
cube will affect name of the cubie but not the cubicle, as the cubicle do not move but
the cubies do. Edge cubies and cubicals are recognized in same way.
Observations:
(1) Any move is a sequence of the six basic moves R,L,F,D,U,B and hence keep the
center cubie in its cubicle.
(2) Any move keeps corner cubies in corner cubicles and edge cubeis in edge cubi-
cles.
(3) If we calculate the number of possible configurations (”At least theoretically
possible”) of the cube then it turns out to be equal to 38 × 8! × 212 × (12)! =
5.19 × 1020 or 519 quintillion! (The number of possible positions for corner
cubies is 38 × 8!, and that for edge cubies is 212 × (12)!).
Valid Configurations of the Cube:
Although these configurations are theoretically possible, that doesn’t mean that these
could really occur. It turns out that some of the theoretically possible configurations
we have counted are actually not valid. (”Can you think of some?”) We need some
technique to figure out what are all the valid configurations. For this purpose we make
the set of moves of the cube into a group and denote it by (G,*). Where if M1,M2 G
are any two moves then M1*M2 will be a move in which you first do M1 and then do
M2 (verify that its a group!)
Observe that the set six basic moves R,F,L,D,U,B will be generating set of G. So this
restricts our problem of looking into all the moves to just looking for these basic moves.
(As they say-” to understand a group G, we should try to understand a small piece of
it”) So to describe each of the six basic moves mathematically, we take help of notion
of permutation. Any move will be described in a cycle notation because we want to
describe where each cubie moves and where each face of the cube moves.
For example if we unfold and draw down face it looks like,:
61
f f f
l d d d r
l d d d r
l d d d r
b b b
after rotating by 90
l l l
b d d d f
b d d d f
b d d d f
r r r
So D(dlf)=dfr, D(dfr)=drb, D(df)=dr and so on. So,
D= (dlf dfr drb dbl)(df dr db dl)
R=(rfu rub rbd rdf)(ru rb rd rf)
Note: here we are considering oriented cubies.
So a configuration of the cube is determined by 4 piece of data :
(1) Position of corner cubies.
(2) Position of edge cubies.
(3) Orientation of corner cubies.
(4) Orientation of edge cubies.
The first can be described as an element σ ∈ S8, second can be described by an element
τ ∈ S12.
Orientation of corner cubies and edge cubies :
We are going to number one face of each corner cubicle as follows:
1 on ’u’ face of ’ufl’ cbicle, 2 on ’u’ face of ’urf’ cbicle,
3 on ’u’ face of ’ubr’ cbicle, 4 on ’u’ face of ’ulb’ cbicle,
5 on ’d’ face of ’dbl’ cbicle, 6 on ’d’ face of ’dlf’ cbicle,
7 on ’d’ face of ’dfr’ cbicle, 8 on ’d’ face of ’drb’ cbicle.
62
Each corner cubie thus has one face lying in a numbered cubicle face. Lable this
cubie face ’0’, next face ’1’ and next ’2’ in clockwise manner.
So the down face will look like :
2 (1)
(1) (0)6 (0)7 2
2 (0)5 8(0) (1)
(1) 2
So the orientation of corner cubie is described like this:
For any i such that 1 ≤ i ≤ 8, xi is the number of the cubie face living in the cubicle
’i’. So xi ∈ {0, 1, 2}. Therefore we can think of xi in Z/3Z and hence x = (x1, ..., x8)
is in (Z/3Z)8.
Similarly we label edge cubies as :
1 on ’u’ face of ’ub’ cubicle, 2 on ’u’ face of ’ur’ cubicle,
3 on ’u’ face of ’uf’ cubicle, 4 on ’u’ face of ’ul’ cubicle,
5 on ’b’ face of ’lb’ cubicle, 6 on ’b’ face of ’rb’ cubicle,
7 on ’f’ face of ’rf’ cubicle, 8 on ’f’ face of ’lf’ cubicle,
9 on ’d’ face of ’db’ cubicle, 10 on ’d’ face of ’dr’ cubicle,
11 on ’d’ face of ’df’ cubicle, 12 on ’d’ face of ’dl’ cubicle.
We label 0 as the cubie
face lying in numbered cubicle face and 1 as the other. Let yi is the number of cubie
face in the cubicle ’i’. So, yi ∈ Z/2Z and hence y = (y1, ..., y12) ∈ (Z/2Z)12.
Thus any configuration of cube can be described by σ ∈ S8, τ ∈ S12, x ∈ (Z/3Z)8
and y ∈ (Z/2Z)12 and so it is represented by ordered 4-tuple (σ, τ, x, y). So the start
configuration is (1,1,0,0).
Again, D=(dlf dfr drb dbl)(df dr db dl), R=(rfu rub rbd rdf)(ru rb rd rf), D−1=(dbl
drb dfr dlf)(dl db dr df), R−1=(rdf rbd rub rfu)(rf rd rb ru) so [D,R]=DRD−1R−1=(dlf
dfr lfd frd fdl rdf)(drb bru bdr ubr rbd rub)(df dr br) (Be carefull about orientations!!)
Writing any configuration this way allows us to recognize patterns more easily (and
prove them). So as in the above example - [D,R] describes a permutation of oriented
cubies , if we consider unoriented cubies then ’dfr’ and ’frd’ are the same. So just to63
look into change of position of a cubie we define a map Φcube : G → S20, Which is
defined as
Φcube(M) = Φedge(M)× Φcorner(M)
where Φedge is a map from G to S12 and Φcorner is from G to S8. And Φcorner(M) is
an element of S8 which describes what M does to ”unoriented corner cubies”. And
Φedge(M) describes that about ”edge cubies.” Since any move results into rearrange-
ment of the unoriented cubies , so Φcube is a homomorphism. Moreover, we can observe
that Φcube(M) is an even permutation (Observe the fact by taking M in D,U,L,R,F,B).
And as Φcube(M) = Φcorner(M) × Φedge(M), either both Φcorner(M) and Φedge(M) are
odd or both are even. Hence in any valid configuration (σ, τ, x, y), σ and τ should have
same sign.
So from the above discussion we conclude that a configuration (σ, τ, x, y) is valid iff
sgnσ = sgnτ ,∑xi = 0(mod3), and
∑yi = 0(mod2).
Earlier we calculated that there were 212× 38× 8!× (12)! possible configurations of the
cube but now the above tells us that only 1/12th of those are valid. ”Of course they
are still more than ”4× 1019”; no small number!!!!!!”
64
FORAYS 2014 PRIZE WINNERS
Day 1
• Clash of the Masters:
– 1st place - Kousik Dhara, Samprita Das Roy and Anirban Kundu from IIT
Madras
– 2nd place - Amith Shastri K, P Thirumal and R Kamalakkannan from
RIASM Madras University
– 3rd place - Arpita Nayek, Digjoy Paul and Sayantan Saha from IIT Madras
• Olympiad (College Level):
– 1st place - Shubham Sinha from IISc. Bangalore
– 2nd place - Sahil Sharma from IIT Madras
– 3rd place - Shruti Sridhar from PS Senior Secondary School
• Battle of the Brains (College Level):
– 1st place - Goutham R from CMI, Mohamad Nehal Imam and Abdullah
from IIT Madras
– 2nd place - Sahil Sharma and Ajay Krishna from IIT Madras
– 3rd place - Saswata Chatterjee, Shubham Sinha and A Kushal from IISc.
Bangalore.
Day 2
• Math Works:
– 1st place - Maitreyi Vijay from Kendriya Vidyalaya CLRI, Rajni Soundarara-
jan from DAV Girls Senior Secondary School, M Arun Balaji from DAV
Boys Senior Secondary School65
– 2nd place - V Kiruthiskiran, Abdullah Sheriff, Gowthama Prasad from
Boston International School
– 3rd place - Sanyuktha B, Lavanya R Raja, Keerthika Ramesh from AMM
Metric Higher Secondary School
• Treasure Hunt:
– 1st place - Thanasekaran from KVASN, Shivani Chander from PSBB,
Pooja Hariharan from KVASN
– 2nd place - J Praveen, Shubhankar, Tejas Sivan from AMM Metric Higher
Secondary School
– 3rd place - Srivatsav K E from PSBB, Shruthi Sridhar and Thejaswini K
S from PS Senior Secondary School
• Battle of the Brains (School Level):
– 1st place - Srivatsav K E from PSBB, Siddharth Sridhar from Chettinad
Hari Shree Vidyashram
– 2nd place - Thejaswini K S and Shruthi Sridhar from PS Senior Secondary
School
– 3rd place - Pooja Kumar from Shri B S Mootha Girls Senior Secondary
School, A Nivasini from Chettinad Vidyashram
• Olympiad (School Level 1 (Class 8 to 10)):
– 1st place - Rajni Soundararajan from DAV Girls Senior Secondary School
– 2nd place - Siddharth Sridhar from Chettinad Hari Shree Vidyashram
– 3rd place - Shivani Chander from PSBB
• Olympiad (School Level 2 (Class 11 & 12)):
– 1st place - Shruthi Sridhar from PS Senior Secondary School
– 2nd place - Thejaswini K S from PS Senior Secondary School
– 3rd place - Srivatsav K E from PSBB
66
FORAYS 2014 TEAM
Faculty
• Chairman: Prof. S. H. Kulkarni, Head, Dept. of Mathematics, IIT Madras
• Faculty Coordinator: Dr. T. E. Venkata Balaji, IIT Madras
• Quiz Panel for ’Clash of the Masters’:
– Dr. R. Usha, IIT Madras
– Dr. K. C. Sivakumar, IIT Madras
– Dr. A. V. Jayanthan, IIT Madras
– Dr. Kalpana Mahalingam, IIT Madras
– Dr. S. H. Kulkarni, IIT Madras
– Dr. Neelesh S. Upadhye, IIT Madras
• Panel for ’Battle of the Brains’:
– Dr. K. N. Ranganathan, RKMVC Chennai
– Dr. K. N. Visveswaran, Hexaware Technologies
• Panel for College Olympiad:
– Dr. Thamban Nair, IIT Madras
– Dr. Arindama Singh, IIT Madras
– Dr. Sounaka Misra, IIT Madras
• NPTEL Web Studio Support: Mr. Kannan Krishnamurthy
Students
• Student Coordinators: Kousik Dhara, Namoijam Changningphaabi, Raisa Dsouza,
Roshima P B, Sushobhan Mazumdar
• Website Designer: Subramaniya Bharathi K
• Souvenir team: Arundhathi Krishanan, Raisa Dsouza
• Volunteers: Arpita Nayek, Sampa Dey, Bishwanath, Smruti Ranjan, Amar Ku-
mar, Sudipti, Anirban Tarafdar, Sarvesh, Prajakta Sahasrabuddhe, Sarika Chowd-
hury, Priya Jaiswal, Archita Mondal, Varsha, Dipa, Samprita Das, Ayushi Singh,
Roshima, Geetanjali, Viswanathan P, Subhajit Chanda, Jayanta Sarkar, Kapil Ku-
mar, Suhas, Poorna, Saswato, Kuntal, Sukanta, Anindya, Subhajit Saha, Sonika
Dhillon, Vinita, Ujjal Das, Paresh Baidya, Sumit KR Rano, Tanmay Sarkar, Ran-
jit, Sayantan, Ramesh, Amartya, Ajay, Aastha, Fathima Safikaa, Amartya, Anirban
Kundu, Digjoy, Deepak, Sagar, Abhay, Om Prakash, Gyan Ranjan, Amar Bahadur
Singh, Mahipal, Bidisha, Projesh Nath, Kousik Dhara.
67
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