Frobenius Algebrasgeillan/research/tqft_frob.pdfPart 3 Frobenius algebras are compatible over direct products. Part 4 By Wedderburn, every ﬁnite dimensional semi-simple algebra is

Background Frobenius Algebras Examples of Frobenius Algebras Results and Properties

Frobenius Algebras

Geillan Aly

May 13, 2009


Outline

Background

Frobenius Algebras

Examples of Frobenius Algebras

Results and Properties


Background

We recall the following definitions, constructions and theorems.


Theorem (1)Let K be a field, then the following are equivalent:

A is a ring and a K-vector space over K such that(ka)b = k(ab) = a(kb) for all k ∈ K, a, b ∈ A.

A is a K vector space with two K linear mapsµ : A⊗A → A (multiplication) η : K→ A (unit),such that the following diagrams commute:

AOO A⊗Aµoo

A⊗A

µ

OO

A⊗A ⊗Aµ⊗IAoo

IA⊗µ

OO

A A⊗Aµoo

A⊗A

µ

OO

K ⊗A � A � A⊗ Kη⊗IAoo

IA

jjUUUUUUUUUUUUUUUUUUUIA⊗η

OO


DefinitionA structure A that satisfies the conditions of theorem 1 is an algebra. Wewill assume that all algebras A have a unique identity element 1A.


Examples of commutative algebras

ExampleA field L such that L ⊂ K.

ExamplePolynomial algebra K[X1, . . . ,Xn].

Observe that a polynomial algebra satisfies the requirements of analgebra, but is an infinite dimensional vector space over K. Thus, there isno requirement that the algebra A be finite dimensional.

ExampleK valued functions on a nonempty set S.


Examples of non-commutative algebras

ExampleMatrix algebra Mn(K) with matrix multiplication being the productoperation.

ExampleHom K(V ,V) for any vector space V with composition being the product

operation.

Observe that this example and example 4 are equivalent with respect toa choice of basis.


DefinitionLet A be a K algebra, and M a vector space over K. Then M is aleft A-module if each a, a′ ∈ A, m, m′ ∈ M and k ∈ K a product am ∈ Mis defined such that

a(a + m′) = am + am′

(a + a′)m = am + a′m

(aa′)m = a(a′m)

1Am = m where 1A is the identity element in A

(ka)m = k(am)a(km)

A right A-module is defined analogously.


RemarkA is itself a left A-module, denoted AA, with canonical left multiplication.This is called the left regular A-module. Again, we can define theright regular A-module, AA as a right A-module with the canonical rightmultiplication.


DefinitionLet A be a K algebra and M a left A-module. Let M∗ be the dual spaceof M. Then M∗ becomes a right A-module if for ψ ∈ L∗, a ∈ A, l ∈ L ,

(ψa)(l) = ψ(al).

The right A module M∗ is the dual of M. Similarly, the dual of a rightA-module is a left A-module.


Theorem (2)Let A be a finite dimensional algebra over a field K, then the followingare equivalent:

1 For AA and (AA)∗, two left A-modules, there exists an A algebraisomorphism λ :A A → (AA)∗.

2 There exists a non-degenerate linear form η : A⊗A → K which is“associative” in the following manner

η(ab ⊗ c) = η(a ⊗ bc) for a, b , c ∈ A.

3 There exists a linear form f ∈ A ∗ whose kernel contains nonon-trivial left or right ideals.


DefinitionAn algebra A that satisfies any of the conditions of theorem 2 is aFrobenius algebra.

It is important to note that a Frobenius algebra is not a “type” of algebra,rather it is an algebra endowed with a given structure.


Proof.Part 1 (1)⇒ (2):For λ : AA → (AA)∗, λ(ab) = aλ(b)∀a, b ∈ A. Thus ∀x ∈ A,

λ(ab)(x) = (aλ(b))(x) = λ(b)(xa).

Define the linear form η : A⊗A → K

η(x ⊗ y) B λ(y)(x).

η is non-degenerate since λ is a K isomorphism. If η(· ⊗ y) = 0 thenλ(y) = 0 and thus y = 0. If η(x ⊗ ·) = 0⇒ x = 0.

Associativity η(xy ⊗ z) = η(x ⊗ yz) follows fromλ(ab)(x) = (aλ(b))(x) = λ(b)(xa):

η(xy ⊗ z) B λ(z)(xy)= yλ(z)(x)= λ(yz)(x)= η(x ⊗ yz)


Proof con’t.

Part 2 (2)⇒ (1):Let η be a non-degenerate linear form on A. Define λ : AA → AA as

λ(y)x B η(x ⊗ y).

λ is a K isomorphism since η is non-degenerate and is an A algebraisomorphism by the associativity of η.


Proof con’t.

Part 3 (2)⇒ (3):Given a linear form η(x ⊗ y), define a linear function f ∈ A∗ as

f(x) B η(x ⊗ 1A).

Then f(xA) = 0 implies η(xA⊗ 1A) = η(x ⊗A) = 0.

Thus x = 0 since η is non-degenerate.

Likewise, f(Ax) = 0 implies x = 0.

Thus, the kernel of f contains no non-trivial left or right ideals.


Proof con’t.

Part 4 (3)⇒ (2):Given a linear function f ∈ A∗ with no non-trivial left or right ideals, define

η(x ⊗ y) B f(xy).

It is clear that η satisfies the conditions of (2).

f does not have a non-trivial left or right ideal. If y is in an ideal Icontained in the kernel of f , then xy ∈ I implying thatf(xy) = 0 = η(x ⊗ y) and η is degenerate.


DefinitionA symmetric Frobenius algebra is a Frobenius algebra such that thenon-degenerate linear form η defined in theorem 2 is actually a trace mapwhere η(ab) = η(ba). If A is a symmetric Frobenius algebra, the linearform will be denoted θ.

Note: The term symmetric Frobenius algebra is not the universal termused in defining this structure.

RemarkClearly if A is a commutative algebra, ie. ab = ba for all a, b ∈ A then acommutative Frobenius algebra is a symmetric Frobenius algebra.


ExampleConsider the algebra Mn(K) where θ(ab) = tr(ab), the trace function.θ(a · b , c) = tr((ab)c) = tr(a(bc)) = θ(a, b · c)


ExampleThe field C is a Frobenius algebra over R with form η(a + bi) B a.

Another form could be based on a different mapping:Consider the form η(2 + 3i) B 7; η(1 − i) B 4.[?]


ExampleLet G be a finite group, R a ring and R[G] be the group ring.

θ(a · b) is the coefficient of the identity element of a · b.

This defines a Frobenius algebra:θ(a · b , c) is the coefficient of the identity element of(a · b) · c = a · b · c = a · (b · c). The coefficient of the identity element ofa · (b · c) is θ(a, b · c).

θ is non-degenerate:If θ(C[G]a) = 0 for a ∈ C[G]. Then θ(g−1a) = 0∀g ∈ G. Thus, θ(g−1a) isthe coefficient of g in a, implying that a = 0. Similarly, θ(aC[G]) = 0implies a = 0.


Theorem ((3) Abrams)If A is a Frobenius algebra over a field K with form f, then every otherFrobenius form on A is given by u · f for u an invertible element in A.


Proof.

Let u ∈ A be a unit. Then for any a ∈ A such that u · f(ax) = f(uax) = 0for all a ∈ A, ua = 0 and therefore a = 0.

Thus there are no non-trivial ideals in the kernel and u · f is a Frobeniusform.


Proof con’t.

Consider g ∈ A∗ a Frobenius form not equal to f .

Then by the proof in theorem 2, g = λ(u) = u · f for some u ∈ A.

Since g is a Frobenius form, the map λ′ B gβ is an isomorphismA → A∗, where β : A → End (A), the map which takes an elementa ∈ A to the map “multiplication by a”.

Thus, there is a v ∈ A such that f = λ′(v) = v · g = vu · f . Thenλ(1A) = f = uv · f = λ(vu) implies that 1A = vu. Since λ is anisomorphism, u is a unit in A.


Theorem (4)If A is a Frobenius algebra, then A∗ is a Frobenius algebra.

Proof.Let A be a Frobenius algebra with form f ∈ A∗. By theorem 2, allelements of A∗ are of the form a · f for a ∈ A.

As A and A∗ are isomorphic, define mutiplication in A∗ by(a · f)(b · f) B (ab · f).

Define τ : A∗ → K to be “evaluation at 1A.” Then the identityτ(ax · f) = f(ax) lets A∗ with structure τ be a Frobenius algebra.


DefinitionA coalgebra A over a field K is a vector space A with two K-linear maps:

α : A → A⊗A and δ : A → K

such that the following diagrams commute

Aα //

α

��

A⊗A

IA⊗α��

A⊗Aα⊗IA // A⊗A ⊗A

AIA

**UUUUUUUUUUUUUUUUUUUα //

α

��

A⊗A

IA⊗δ��

A⊗Aδ⊗IA // K ⊗A � A � A⊗ K

The map α is the comultiplication map, δ is the counit map with axiomscoassociativity [(a ⊗ (b ⊗ c) = (a ⊗ b) ⊗ c] and the counit condition[A⊗ K � K ⊗A is a natural isomorphism].


We can use the multiplication map µ on A:

µ : A⊗A → A∑ai ⊗ bj 7→

∑aibj

to define a comultiplication map β∗ on A∗:

β∗ : A∗ → A∗ ⊗A∗ � (A⊗A)∗∑f 7→ f ◦ µ


As a Frobenius algebra, A and A∗ are isomorphic, thus A is prescribeda coalgebra structure by defining α, the comultiplication mapA → A⊗Aas (λ−1 ⊗ λ−1) ◦ β∗ ◦ λ:

Aα //

λ

��

A⊗A

A∗β∗ // A∗

⊗A∗

λ−1⊗λ−1

OO

A is coassociative and cocommutative by the definition of α.


α can be also used to define the multiplication in A∗ since

(a · f)(b · f) = [a · f ⊗ b · f ] ◦ α = ab · f .

Let g : K→ A denote the unit map. The commutativity of

A∗

g∗

��>>>

>>>>

>>f ◦ β(a)

''PPPPPPPPPPPP

A

λ

OO

f // K a //

OO

f(a) = f ◦ β(a)(1A)

guarantees the commutativity of

A∗β∗ // A∗

⊗A∗

g∗⊗IA∗ // K⊗A∗

λ−1

��A

λ

OO

α // A⊗A

λ⊗λ

OO

f⊗IA // K⊗A

Since the top row is IA∗ , the bottom row is IA. Thus f is the counit in A.


TheoremEvery finite dimensional semi-simple algebra admits a symmetricFrobenius structure.


Proof.

Part 1 A matrix algebra over a Frobenius algebra is a Frobenius algebra:Let A be a Frobenius algebra over K with form η. Let Mn(A) be thealgebra of n × n matrices over A with the standard trace. Show that thecomposition

Mn(A)Tr // A

η // K

is a Frobenius form.

Recall that every finite dimensional simple algebra is a matrix algebraover a skew-field, showing that every finite-dimensional simple algebraadmits a Frobenius algebra structure.


Proof con’t.

Part 2 Every finite-dimensional simple algebra is a matrix algebra admitsa symmetric Frobenius structure.

Part 3 Frobenius algebras are compatible over direct products.

Part 4 By Wedderburn, every finite dimensional semi-simple algebra isisomorphic to a direct sum of matrix algebras over skew-fields.

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Frobenius Algebrasgeillan/research/tqft_frob.pdfPart 3 Frobenius algebras are compatible over direct products. Part 4 By Wedderburn, every ﬁnite dimensional semi-simple algebra is