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Differential Geometry- curves and surfaces -
I.
Introduction
it.
What is
DifferentialGeometry
How can you tell if you"
live"
on the surface of a ball
--
- --
.
.
§called a sphere
-or 2- sphere
or the plane
?one way is to look at"
straight lines' '
in the plane of it two people walk in the
✓"
same direction"
from different
points they stay a fixed
distance app art
( parallel lines don't intersect )
but on the sphere
µ the distances get
..closer together
that is,
the"
geometry" of
"
lines"
on the sphere is different
from the geometry of the plane .
SO we see the"
curvature" of the sphere by looking at
straight lines in the space
True scion : What about the 3 - dimensional space in which we live ?
isit' ' flat Euclidean space" ?
is it a
"
3 - dimensional sphere" ?
something else ?
General Relativity postulates that gravity can be understood as
a
"
curvature "in space lain e)
.
The language to study all these ideas is RiemannianGeometry
or more generally DifferentGeometry
and it all starts with studying curvesandstraightlines
This course is an introduction to Riemann coin Geometry
through curves and surfaces in Euclidean spaceC see list of topics on the web page )
B.
The geometry of Euclidean Space
IR "= { I p, , . . . , p n ) I pi a real number
,i.e
. p ,E IR }
we can think of D= Ip , , . . .
, Pn ) as a pointin IR"
or atech ai IR"
• F
when thinking of B as a vector we will frequently write it as a
column vector
op
given pig EIR"
then their dot product is
F.9- = pig ,t pzgze . . . t paan = ⇐
,
Pi 9 i
The
we sometimes write 48,97 for F.5 and thisgives an inner
product on IR"
,that is l .
,. ) satisfies
D 4 pig ) = II. p )symmetric
2) tap , g) = a sp. 97=48 , af )
I Ft9- if ftp.ry-sg.ry ) linear
'
3) I F, p ) Z O
I F , p > = o ⇒ p } positive definite
geometry is about lengthsandangles ,with a dot product we
can define the length of F to be
lip H - LEFTand the angle between B and of to be
cos a = CBI it
1181111911 €p
note for this to be well-defined,
we need
lemma I ( Cauchy - Schwartz inequality)
for all F. g- E IR"
KpiIHe 1101111511
with equality it and only if I and g- are linearly dependent
Proof nice trick : compute the length of a linear combination
Of Hap tbg IE tap tbg , apt bop )
= a 'll pl ft b 'll g H 't Zab 9.5 )
so it a = Hghand b = I HFH,
then we have
0 I 2118112119712IZHPTIH-gtkp.gl= 21181111911 ( lip 1111911 I 6,5I ) } ④
so if Hp Ht 0411911,
then
± tph > E 11,51111511 ( if either HFH -
- o or 11511=0
then E is obvious )and ltpllllglkmax ftp.T )
, -45,57 )= 145.9731
thus the E is the lemma is true
note : assuming HFIH of 11911 then
trig 7 = Itp1111911⇐
we have equality in ④
⇐
Http49 I
HgYp-H" on -
i
aeaeracy⇒
of I er product
krillg- I Hgtlp = O
-tie
. p and g- are
linear- ly dependent #
The standard distance between points in' IR
"
is .
d I pig ) = Itp - g- It
a metric on a set X is a function
( metrics describegli stanced : X x X → IR
between pointssuch that
i ) dlp ,g) Z O with equality ⇒ p= q
a ) dip . 9) =D (9. P )
triangle inequality3) dcp , g) E d Cp .
rt t dlr, 9 )
forexercise . Show that d (pig ) above is ap #
qmetric on IR "
given two metric spaces I M,
,d , ) and ( Mr ,da ) an isometry
is a surjective function
to : M,
→ Me
such thatdip ha
,fly ) ) = d. I x
, y) for all X. y EM,
Isometries identify points of M, with points of Mz so that
distances are preserved . They are" symmetries " of
spaces with metric 's
We are interested in isometries from DR"
,d ) to itself
Notice any"
geometric quantity" should not change
under isometries leg length of a curve . . . )
An orthogonal transform is a linear map
A : IR"
→ pinsuch that
ftp.Agl-hp.gl for all pig
Theorem 2 :
If f : IR"
→ IR"
is an isometry ,then there is
some I E IR"
and orthogonal transform A
such that tip ) = a- t Ap
Proof let I (B) = fcp ) - f- CE )
it we show Iis ① linear andn ~
② satisfies Lf IF),
f ta ) ) =L pig 7?
then we are done since we can set A = t and I = flotto get
tipi' Atta
note LI - I , I -77=115114117112-245,77
So245
,7--115112+117112 - HI-FIT
thusz C
ftp.FGD-xfiptlftHFGTH-HFirst-
Fight= Http ) - f- CEN HI f Cgt - f IoTH
'
- It f Cpt - f GT ITisometry? ftp.olf-illq-olf-llp -9112
-
= 110112+119112 - lip - g- If = up . I >
so f satisfies ②
now let E,
. . . ,En be an orthonormal basis for R
"
leg. It:o) , E = LI:o) ,
. -. )
- -
exercise : f let), . . . ,
f ten ) isalso an orthonormal basis
for IR" because of ②
so for an
c peg ,,Ice, > = Lptg, = IF .-9745.57- ~ ~ ~
= http ),
f ( Edt Hlf ),
f let ) )~ a
~
= ( f (f) tf Cg- ),
f (Ed ) for all i
~ ~~
and thus f ( ptg ) = ftF)t f CE)
exercise . Prove this it it is not clear to you
Hint : To,
. . . In an orthonormal basis ,then
E = I ⇐ 48,5;> stew ,Ii > for all i←
~
scimitar ly I flop ),
flea ) ) = top ,Ez > = Csp , -97
- - - ~
= ch f ( p ),
f (Ee ) ) = L of IF ),
f IED- -
so t ( c p ) = c f CF )~
and thus f is linear#
So any isometry of IR"
( also called a rigidmotion ) is
a composition of
① an orthogonal transformation
f- C p ) = Af and
② a translationf- I ft F ta
we understand ②.
let 's explore ①
Recall : given a linearmap A : IR "
→ IR"
we can express it as an nxn matrix
e. g .let I ,
. . . En bethe standardbasis for IR"
AE,= a
, f-it . . .
t an ien
I etma .
. laida
:
any vector can be written
I = Viet . . . then =
then AT corresponds to the vector MAL ?;)from now on we will think of A as the matrix above
that rep resets it in this basis
now with Tr and I =
" te
L f,-w , = f. in = Ftw where IT means the transposeof v
re.
switch rows and columns
for any matrix A
( Atv,
I > = LATE) = TEA'T tu = E' AT = to,
AE >
if A is an orthogonal transform,
then
IT,
AT ) = LATE,
I ) - LA Atv,
AT )
SO
( T . A ATF,
AT) = O = L 8
,AT ) for a1
wi. if we let I run through an orthonormal basis I
,. . .
Ten
we seeJ - AAT F =D
fun identitymatrix
soA ATF = I = I dnt
and
A AT = Idn
this implies 1- - detlldn ) = det ( AA' ) - (detA)
'
SOdef A =
I 1
it def A = I,
we call A a special orthogonal transform
A side . Old = I orthogonal transforms of IR "
}
SO In = I special " " }
are examples of Liegroups_ ,the study of these is
a beautiful and deep area of math
Isometries of IR ?
If A = ( Edb) is a special orthogonal transformation
then I = det A -
- ad - be
and too .tl :ballscat %a
so we have a 2+5=1at d 2=1
.( d. c)
act bd -
- Ono
ad - be = I
← unit circleF unique angle 0 set
.
D= cos o
( = sin'
0
now [ ba) . Idc) = O so [ ba ) is a unit vector
orthogonal to [ I ]
f- sure , lose )• .
I cost,
Sino )b a µ
dc
•
(
cost,
- Sino )
b a
finally ad - be = 1 ⇒ a = co so
b =- Sino
so A = ( cost - sin a
Sino cost)
and A corresponds to arotationabout the origin by angle 0
exercise it A not special, but just orthogonal, then
A = ('
o? , ) . ( cos o - since
~sin cos o
) some a
reflect about x - axis
so rigid motions of 1122 are compositions of :
rotationstranslations
,
and
reflectionsaboutx-axis
exercise . Isometries of IR'
are compositions of
rotations about some line,
translations,
reflections about xy - planereflections through the origin
exercise . let E . . . En be anyorthonormal basisfor IR
"based
at a point p E IR"
and I . . . In be another orthonormal basis for IR"
based
at a point g- E IR"
Then there is an isometry to : IR"
→ IR"
suchthat
e-a lol p ) =
gif -HE
,I Dotplez ) -
-f
,%I we total derivative of 4 at I£
Recall . given a function
F : IR"
→ Mm
we can write it
FIX, , . . . Xn ) = ( f,
I x, . . .Xu )
,. . .
, fmlx, , . . . xn ))then
D Ep : IR"
→ IRm
r avectors based actors basedat p at g-
Hint . consider the case
where E, . . .En is the is a linear map
that can be expressedstandard basis and f -
- O as the mxn matrix
then consider 2 f - ,1- I p- ( p
¢ to) = g- t AT
me . nai:÷s÷i÷÷.
. . . . .
÷÷÷÷÷÷¥⇐÷m
I Curves
A. Curves in IR"
A curve C is the coinage
ofa coat in in a ou s function
I : I → IR"
where
I = I a. b) or Ca. b ]
example . I It ) = C t,
t ) t E lo,
I ]
§ I t ) = ( It,
I t ) t E Eo ,
' Iz ]
@
J It ) = I t'
,t
' ) TECO , D c
@
note .
D all 3 functions have the same linage ,so a given
curve can be described by many different functions
2) we say that I I or I or J . . . ) parameterize the
curve C
3) We can think of C as the pathofaparticle or
a piece of wire in IR"
4) we frequently confuse C and I but remember we
are really interested in C,
I is just a convenient
way to describe C mathematically5) Curves do not need to be given by a parameterization
e. g . g I e y'
= I
of course we can also parameterize it : I HI =L cost ,sin t )
t E [ o,
21T ]
Remark There is a subtlety in the definition of curve
really it is not just the linage of I but the"
trajectory "
of the particle traveling along C
<
example .
the coinage of I and f are the same
but the order in which parts of the pathare traversed is different so we will
say cheese are differentcurve
So really we should think of a curve C as
the image of some I :C a. 53 → IR "
together with the order in which the
points on C are encounter e
examples :
it
Straight-line
given points p and g- in IR"
I think of them as vectors )then
I th = (I - t ) f t tf t Eco,
I ]
parameterize the line from p togy z
e. g .
C o
@
x I ,@
ICH -
- l-f) lift 13 ) x
-
- ft
I;] Titled -A t tf?)
"
circles
=L' ¥ I
given r EIR, r > o
F C- IR2
then I Ltt = Ft ( r cost,
r s int ) t t lo .2T ]
parameterize s the
firdeofradiusraboutpe
's .
a-us .
- I ? ) - if =L ? I iii.ITy
:x
3) Helix
given r,
h E IR,
r > 0,
h Z O
set I I t ) = ( r cost,
r s int, ht ) t e IR
2-
↳9 y
x
4) I It ) = ( t ? t' ) cusp
t e IR ✓T
given a curve C, parameterized by a function
I.I → HM
IIt ) = It,
Itt, Htt
,. . .
,data )
recall from calculus that at the point I Lto ) c- C a tangentvector to C is given by
I ' Ito) =L dittos,- . .
,a
'
nIto D
Tito,
•→ It
Remarks : it Actually I'
Hot is a vector based at to,
o,
. . . ,o )
When we say a vector T is based at f,
then we
shift I so Its' ' tail
"
is at p
? off to based atp✓
so we should say I '
Ito ) based at Itto) is tangent to C
2) If T to,
then the line spanned by I is
↳ = { rt I r E IR }
it E is based at F then the line through F in
the direction of T is
{ r Ttp I re IR }
fir• pI
so the tangent line to the curve C at D= It to )
is given byTp C = { It to ) tr I
'
Ito ) I r E IR }
Tpc"
Tpc
Recall : the tangent line to C at p is the"
line that best
approximates C at B"
a parameterized curve I HI is called regular if I' let
existsand isnon-Zero for all t
we need this to talk about tangent lines and
manyother thingsso we usually assume curves are regular
( except maybe at afinite number of points )
note . let I : I → IR"
parameterize a curve
if I ' '
It ) = O for all t,
then I parameter I Ees ( part
of ) a line
to see this note
I' '
Its = I Ii' Itt
,. . . , In
"
I t ) ) = I O,
. . .
,O )
SO
x'j It ) = O di . I → IR
integrate to get
ditty = f ai'
It ) d t t vz = Vi for some constant Vi
SO
wtf Sai Hdt t pi = vet t pi
that is
* , ftp.fef?rnI--p- + it
so the second derivative tells us how far I is
from being a line
Problem : I' '
is not a geometric quantity !
i.e .it depends on the parameterization not just C
e. g .
I It ) = I t,
t ) te Eo ,it
f- Lt ) = It ? E ) te Eo.
, ]
I "
I t ) = I o.o ) t I"
I t ) = ( 2,
2)
also HI " 11=0 117" 11=252
but both give .
⑧
so I" doesn't necessarily give us information
about the curve C it parametrizes
( e.g .can't tell example is a line from I " )
to fix this we need to consider arelengthRecall from calculus that if C is parameterized by
I :[a. b ] → IR "
then the length of C is given by
length (c) = Sba HI ' # Hdt
or more generally the length along C from
the end point I Ca ) to Ics ) is
lemmaI .
lls ) = Sas HIGH It
It I is a regular parameterization of a curve C,
then we can re parameterize C by another function
To:[ o
, e) → IR"
such thatup
'
I s ) It = 1 for all s
Remarks :
, ) if I :[ a. b) → IR " parameterize s a curve C and
B- :c c. d) → IR" is another parameterization of C
then we say I is a reparamecerizats.co f C
note : if I is one - to - one ( re.
I CE,
) = I Cta )
then E,
= ta )
then for each s E Ead ],
there is a unique
ts Ela . b) such that Icts ) = J Cs )
F fist Icts )
[ c ,d ] - ~• C
t f[ a. b ]
so set f : I c. d) → Ca . b ] : s t t,
and we see
f- C s ) = I ( fist ) = I of C s )
exercise : Convince yourself that you can
find f- as long as I is regular( i.e
.don't need one - to - one )
Conversely , given anyfunction h :[ he ] → La. b ]
Joh : ( h.e) → IR
"
is a re parameterization of I
2) We say that I :{ a. b) → IR"
parametrizes C by arc length
it It Its)H=1 for all s e [ o.
l ]
note given such an I we have
list = S !HI'
lxllldx = s
i.e. length of C from Ilo ) to Its ) is s
so lemma says regular curves can alwaysbe parameterized by arc length
Proof given I :[ a , 63 → IR"
parameter it ing C
let f CE ) = Sat II I'M Hdx
the fundamental theorem ofcalculus says
date = HI' kill -5%35%7so f is increasing on [ a. b ]
:. f is one - to - one
it f- fl b) = length of C then f also onto [ o,
l ]
e -
J t 't '
I I t
a b
exercise fat has an inverse
I recall this is a function
g : [ o, e) → La ,
b ]
Sf. fog C s ) = s and got Lt ) = t )
chain rule givesdads Htt ) ) date ) -
- dat ( go f) HI -
- Fett = I
SO dates ) = daft, where s - Ht ) and t =
g Cs)
now set B Csl = Ilg HD
so B : foie ] → IR"
parameterize s C
and
Itp'
It -
- HI '
lgcsb g' Coll = HI '
Ist dat, all = " I ' kN ¥*y= HI ' IHH ¥ HIT
= 1EH
example . Helix
Itt ) = ( r cost,
r suit,
be ) te Co ,x )
I 'It ) = ( - r suit
, r cost ,b)
It I '
HN -
- r'sin2ter7os7t5# = FELT
so t
f- It ) = So ¥5 dx - FEE t
the inverse of f is
f- II s ) = bzS
SOBcs ) = I tf - '
I s )) = ( r costs s,
r sin Is , bf¥ )
is a parameterization by arc length
Notation .
When we use the
variablesis a parameterization of
a curve we will always mean we have parameterized byarelength, where t is used for any parameterization
now given a parameterization To : lo, e ] → IN " of a curve C by
are length we say
IN = pics )
is the unit tangent vector
lemma 2 .
I '
Cs ) is perpendicularto Test
ProofTis ) . Fcs) = It Fcs ) It
'
= I
the product rule gives
O -
- Ist = Is IF = (IsF) It to lads'T)
= 2 F '
19 . Fcs)
so I '. F = O
¥7
We call NTS) = Its ) the unit normal vector to C at pics)
Fl s )
¥7 .
>
s)
F' 1st
Wecall Kis ) -
- It I '
Call the curvature of C at post
note : Kcs ) = HI"
Csl It is the ace Iteration of a particle movingon C with unit speedso you feel the curvature of a road while driving !
exercise :
1) Show the curvature of a curve C is independent of param .
this says 2) Show it I : IR"
→ M " is an isometry then the curvature
K isreally
ggE off' Ifis the same as the curvature of Icc ,
tellingyou
how 4
C sitsin IN
3) It I '.
Iab ] → IR"
is any regular parameterization of C
then show you can calculate the curvature of C at Ict )
BY* = attest
4) Show directly that the formula in. 3) is independent of
parameterizationTh " 3 .
aregular curve C is C part of ) a line
if
curvature of C is 0 at all points
Remark .
So curvature is preciselythe measure of how
far a curve is from being a line !
Proof :# If C is aline from p tog then
Test -
- tf p t IF
for s E lo ,e ] l -
- Hp - IH
is an are length parameterization of C
pts) =- tep t
tfS o
Kcs ) = It I " call -
- toll =
g
⇐ ) let Gls ) be an are length parameterization of C
assume KCSI = O so B "
Cs ) = 8
then we saw earlier that Bcs ) parameter ites part
of a lineat
example : recall an are length parameterization of the Helix is
I Csl = ( r cos ¥+5,
r sin SEE ,
"
is'
is ( -
asin ¥5
,
us Fees,
¥+E )
and
I" 1st =L - ¥5 cos ¥5 , ¥5 sin ¥I ,
o )
thus
KC s ) = Hf " call = rr¥b
Again let§ : lo . e) → IR
"
be an are length parameterizationof some curve C
recall Its ) = F' is ) and I Cs) =Fcs ) are unit
orthonormal vectors in IR"
so they span a plane in IR"
P = { a Flat b Fcs ) I a. b E IR)
translate P so thatit godsthroughBcs , =p
Ppc= { peso ) ta TT so ) t bit Csd I a. bEIR )
this is called the osculating plane to C at p = I Go)
.
i :c
note . Ppc contains the tangent line Tpc
exercise Convince yourself that Ppc is the plane that C
comes closest to lying in atp
I later we will see precisely when C lies in Ppc )
Recall given 3 points I ,I
, xj in IR"
that do not lie on a line
then they① determine a unique plane
PII ,I .
= I ,t span I Ii - I , ,
I ,- I ,
}
② deter min a unique circle Ctx , ,IT in PCI , E. Is )
" """" '
Facts : let F ! Lo, e) →IR" be a regular parameterization of C
suppose so C- [ oil ] Sf. K Cso ) ¥0
I ) for points s, , Sa
,s ,
C- Co,e ] sufficiently close to So
§ Is.)
, fish , f- Cs,) do not lie on a line
I ) the osculatingplane is
Pasok=
, .
!Ys,
→PCB 's
. ? Goal , ptsd )
HI ) The limit im CC fist , fish , ptsd )4.5.55 ' So
is a circle Cpcso ,in PBisnt
it is called the osculating circle and can be
parameterized byIN = pics , + ¥ ,
N' Iso) t IT, , Iint ) Icsd + lost ) Is;]
so the circle has radius ¥ ,
~ .
osculating circle
note i ) osculating circle is tangent to C at B
( has "
order 2"
contact with C )
2) if K is close to 0 then C is almost straightif K is large then C curves a lot
.
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