Archive for the ‘Frobenius Algebras’ Category

Throughout k is a field and A is a Frobenius k-algebra. So there exists a bilinear form B : A \times A \longrightarrow k which is non-degenerate and B(xy,z)=B(x,yz), for all x,y,z \in A.

Theorem. There exists a k-algebra automorphism \sigma of A such that B(x,y)=B(y,\sigma(x)), for all x,y \in A.

Proof.  Let x \in A and define g : A \longrightarrow k by g(y)=B(x,y), for all y \in A. Clearly g \in A^* and so by this lemma, there exists a unique \theta_x \in A such that g(y)=B(y,\theta_x), for all y \in A. So B(x,y)=B(y,\theta_x), for all y \in A. Define the map \sigma : A \longrightarrow A by \sigma(x)=\theta_x, for all x \in A. So we need to prove that \sigma is a k-algebra automorphism. First we show that \sigma is k-linear. If \alpha \in k and x,y,z \in A, then

B(z, \alpha \sigma(x) + \sigma(y))=\alpha B(z, \sigma(x)) + B(z, \sigma(y))=\alpha B(x,z)+B(y,z)

=B(\alpha x + y,z)=B(z, \sigma(\alpha x + y)).

Therefore, since B is non-degenerate, \sigma(\alpha x + y)=\alpha \sigma(x) + \sigma(y) and so \sigma is k-linear. It is easy to see that \sigma is injective: if \sigma(x)=0, then B(x,y)=B(y,\sigma(x))=B(y,0)=0, for all y \in A and hence x = 0 because B is non-degenerate (see the Remark in this post). Hence \sigma is surjective as well because A is finite dimensional. So we only need to prove that \sigma is multiplicative. Recall that B(xy,z)=B(x,yz), for all x,y,z \in A and thus

B(z, \sigma(xy))=B(xy, z)=B(x,yz)=B(yz, \sigma(x))=B(y, z \sigma(x))=B(z \sigma(x), \sigma(y))

=B(z, \sigma(x)\sigma(y)).

 Therefore \sigma(xy)=\sigma(x)\sigma(y) because B is non-degenerate. \Box

Definition. The k-algebra automorphism \sigma in the above theorem is called the Nakayama automorphism.

We proved in here that there exists f \in A^* such that \ker f does not contain any non-zero left ideal of A. In fact we saw that B(x,y)=f(xy), for all x,y \in A. So if \sigma is the Nakayama automorphism of A, then f(xy)=f(y \sigma(x)), for all x,y \in A.

As usual k is a field. In this post I will prove a rank-nulity theorem for Frobenius algebras. First we need a lemma also known as the Riesz representation theorem.

Lemma. Let V be a finite dimensional k-vector space and let B be a non-degenerate bilinear form on V. If f \in V^*, then there exists a unique v_0 \in V such that f(v)=B(v,v_0), for all v \in V.

Proof. The uniqueness is obvious because B is non-degenerate. Now let \{v_1, \ldots , v_n \} be a basis for V and suppose that [B] is the matrix of B with respect to this basis, i.e. the (i,j)-entry of [B] is B(v_i,v_j), for all 1 \leq i,j \leq n.  Let \bold{y} \in k^n be a vector whose i-th coordinate is f(v_i). By the theorem in this post, [B] is invertible because B is non-degenerate. So the system of equations [B] \bold{x} = \bold{y} has a solution \bold{x} \in k^n. Let x_i be the i-th coordinate of \bold{x}. So

\sum_{j=1}^n x_j B(v_i,v_j)=f(v_i),

for all i. Let v_0=\sum_{i=1}^n x_iv_i. Then for any v=\sum_{j=1}^n \alpha_i v_i \in V we have

f(v)=\sum_i \alpha_i f(v_i)=\sum_{i,j} x_j \alpha_i B(v_i,v_j)=\sum_i \alpha_i B(v_i,v_0)=B(v,v_0). \ \Box

Notation. Let V be a k-vector space, W a k-vector subspace of V and B a bilinear form on V. We let W^{\perp}=\{v \in V : \ B(w,v)=0, \ \forall w \in W \}.

Theorem. Let A be a Frobenius k-algebra. Let the bilinear form B and f \in V^* be as stated in Definition 3 and the theorem in this post, respectively. Then

1) L^{\perp} = \{x \in A : \ f(yx)=0, \ \forall y \in L \},

2) \dim_k L + \dim_k L^{\perp} = \dim_k A.

Proof.  Part 1) is clear because B(y,x)=f(yx), for all x,y \in A. So we only need to prove the second part of the theorem. Define the map \varphi : A \longrightarrow L^* by \varphi(x)(y)=f(yx), for all x \in A and y \in L. Obviously \varphi is a k-linear map and \ker \varphi = L^{\perp}, by part 1). We are now going to prove that \varphi is onto. So let g \in L^*. Let B_1 be the restriction of B to L. By the above lemma, there exists y_0 \in L such that

g(y)=B_1(y,y_0)=B(y,y_0)=f(yy_0),

for all y \in L. Hence g = \varphi(y_0) and so \varphi is onto. Thus, by the rank-nulity theorem for vector spaces, we have

\dim_k A = \dim_k \ker \varphi + \dim_k L^*=\dim_k L^{\perp} + \dim_k L. \ \Box

Throughout this post k is a field. Recall that if V is a k-vector space, then the dual space of V is the vetor spcae V^* consisting of all k-linear maps from V to k. If V is finite dimensiona, say with the basis \{v_1, \ldots , v_n \}, then for every 1 \leq i \leq n we define e_i \in V^* by e_i(v_i)=1 and e_i(v_j)=0 for all j \neq i. It is easy to see that \{e_1, \ldots , e_n \} is a basis for V^*. In particular, \dim V = \dim V^*.  

Theorem. Let A be a finite dimensional k-algebra. Then A is a Frobenius k-algebra if and only if there exists f \in A^* such that \ker f does not contain any non-zero left ideal of A.

Proof. Suppose first that A is Frobenius. So there exists a non-degenerate bilinear form B of A such that B(xy,z)=B(x,yz), for all x,y,z \in A. Define f : A \longrightarrow k by f(x)=B(x,1), for all x \in A. Obviously f \in A^* because B is linear. Now suppose that L is a left ideal of A and L \subseteq \ker f. Let y \in L and x \in A. Then xy \in L and so 0=f(xy)=B(xy,1)=B(x,y). Therefore y=0 because B is non-degenerate. So we have proved that L=0. Conversely, suppose that there exists f \in A^* such that \ker f does not contain any non-zero left ideal of A. Define B: A \times A \longrightarrow k by B(x,y)=f(xy), for all x,y \in A. Clearly B is bilinear because f is linear. Also, clearly B(xy,z)=f((xy)z)=f(x(yz))=B(x,yz), for all x,y,z \in A. To complete the proof of the theorem we only need to show that B is non-degenerate. So suppose that B(x,y)=0 for some y \in A and all x \in A. Then f(xy)=0 and thus xy \in \ker f for all x \in A, i.e. Ay \subseteq \ker f. Hence Ay=0, because Ay is a left ideal of A and \ker f does not contain any non-zero left ideal of A. So y=0 and we are done. \Box

Example 1. If K/k is a finite field extension, then K is a Frobenius k-algebra.

Proof. Let \{x_1, \ldots , x_n \} be a k-basis for K. Define f : K \longrightarrow k by f(\alpha_1 x_1 + \ldots + \alpha_n x_n)=\alpha_1, for all \alpha_i \in k. Clearly f is a non-zero k-linear map and \ker f does not contain any non-zero ideal of K because K is a field and so it has no non-zero proper ideal.

Example 2. Let G be a finite group. Then the group algebra k[G] is a Frobenius k-algebra.

Proof.  Let G=\{g_1, \ldots , g_n \}, where g_1 is the identity element of G. Define the map f : k[G] \longrightarrow k by f(\alpha_1g_1 + \ldots + \alpha_n g_n)=\alpha_1, for all \alpha_i \in k. Clearly f is a k-linear map and \ker f =kg_2 + \ldots + kg_n. Suppose that L \subseteq \ker f and L is a non-zero left ideal of k[G]. Let 0 \neq x= \sum_{i=2}^n \alpha_ig_i \in L. Then \alpha_j \neq 0, for some j \geq 2. Since L is a left ideal of k[G], we have g_j^{-1}L \subseteq L \subseteq \ker f. But the coefficient of g_1 in g_j^{-1}x is \alpha_j \neq 0 and so g_j^{-1}x \notin \ker f, which is a contradiction. Thus L=0. \ \Box     

Throughout k is a field and all vetor spcaes are over k.

Definition 1. Let V be a k-vector space. A bilinear form on V is a map B : V \times V \longrightarrow k such that for every \alpha_1, \alpha_2 \in k and v_1,v_2,v \in V we have

1) B(\alpha_1v_1+\alpha_2v_2, v)=\alpha_1B(v_1,v)+ \alpha_2B(v_2,v),

2) B(v, \alpha_1v_1+\alpha_2v_2)=\alpha_1B(v,v_1)+\alpha_2B(v,v_2).

Definition 2. Let B be a bilinear form on a vector space V. we say that B is non-degenerate if B(v,w)=0, \ \forall v \in V, implies w=0. If \{v_1, \ldots , v_n \} is a basis for V, then the n \times n matrix whose (i,j)-entry is B(v_i,v_j) is called the matrix of B with respect to that basis of V.

Theorem. Let V be a finite dimensional vector space with a bilinear form B. Then B is non-degenerate if and only if  the matrix of B is invertible.

Proof.  Choose a basis \{v_1, \ldots , v_n \} for V. Let [B] be the matrix of B. Then [B] \bold{x}=\bold{0}, for some \bold{x} \in k^n, if and only if \sum_{j=1}^n B(v_i,v_j)x_j=0, for all 1 \leq i \leq n, where x_j is the j-th coordinate of \bold{x}. Thus [B] \bold{x}=\bold{0} if and only if B(v_i, \sum_{j=1}^n x_jv_j)=0, for all 1 \leq i \leq n, if and only if B(v, \sum_{j=1}^n x_j v_j)=0, for all v \in V. Hence [B] \bold{x}=\bold{0} has a non-zero solution for \bold{x} if and only if there exists 0 \neq w \in V such that B(v,w)=0 for all v \in V. Thus [B] is not invertible if and only if B is not non-degenerate. \Box

Remark. Since [B] is invertible if and only if [B]^T, the transpose of [B], is invertible, a similar argument to the above theorem gives us that B is non-degenerate if and only if B(w,v)=0, \ \forall v \in V, implies w=0.

Definition 3. A finite dimensional k-algebra A is called a Frobenius algebra if A, as a k-vector space, has a non-degenerate bilinear form B such that B(xy,z)=B(x,yz), for all x,y,z \in A.

Example. Let A=M_n(k), the algebra of n \times n matrices with entries in k. Then A is a Frobenius k-algebra.

Proof. Define B : A \times A \longrightarrow k by B(x,y)=\text{Tr}(xy), the trace of xy. It is clear that B is a bilinear form. To prove that B is non-degenerate, suppose that B(x,y)=0 for all x \in A. So \text{Tr}(xy)=0 for all x \in A. Let e_{ij} be the element of A with (i,j)-entry 1 and 0 anywhere else. Let y_{ij} be the (i,j)-entry of y. Let 1 \leq r,s \leq n and let x = e_{rs}. Then 0=\text{Tr}(xy)=y_{sr} and so y=0. This proves that B is non-degenerate. Finally, for any x,y,z \in A, we have B(xy,z)=\text{Tr}((xy)z)=\text{Tr}(x(yz))=B(x,yz). \ \Box