## 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 $B(xy,z)=\text{Tr}((xy)z)=\text{Tr}(x(yz))=B(x,yz). \ \Box$