## The Nakayama automorphism of a Frobenius algebra

Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
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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 $x,y \in A.$

## A rank-nulity theorem for Frobenius algebras

Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
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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

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

## Frobenius algebras and linear functionals

Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
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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 $L=0. \ \Box$

## Definition of Frobenius algebras

Posted: May 4, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
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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 $B(xy,z)=\text{Tr}((xy)z)=\text{Tr}(x(yz))=B(x,yz). \ \Box$