Symmetric algebra (Frobenius algebra)

A finite-dimensional algebra $A$ is symmetric when there is a nondegenerate symmetric bilinear form $B:A\times A\to k$ satisfying $B(ab,c)=B(a,bc)$. Equivalently, $A\cong A^{\ast }$ as $A$-bimodules.

For a finite-dimensional symmetric algebra over any field, an indecomposable projective module $P=Ae$ has isomorphic simple head of a module and socle. A symmetrizing functional pairs $eA$ with $Ae$ nondegenerately, identifying $Ae$ with the left dual module $D(eA)$. Taking its socle gives $D(eA/eAJ(A))$. This dual of the right simple top and the left simple top $Ae/J(A)e$ belong to the same matrix block of the semisimple algebra $A/J(A)$, so they are isomorphic. The result does not require an algebraically closed ground field.

For a finite group $G$, the coefficient of the identity in $xy$ defines a nondegenerate symmetric associative bilinear form on $kG$. Hence $kG\cong (kG{)}^{\ast }$ as bimodules and the group algebra is a symmetric algebra.

For finite-dimensional modules over $kG$, projectivity and injectivity are equivalent. Indeed, $(kG{)}^{\ast }$ is injective because ${\mathrm{Hom}}_{kG}(-,(kG{)}^{\ast })\cong {\mathrm{Hom}}_k(-,k)$ is exact, while symmetry gives $(kG{)}^{\ast }\cong kG$.

If $P$ is an indecomposable projective $kG$-module, then its head and socle are simple and naturally isomorphic:This is the identity Nakayama permutation of the symmetric algebra $kG$.

For every simple finite-dimensional $kG$-module $S$,

Let $M$ be an indecomposable finite-dimensional $kG$-module and let $P_k$ be the projective cover of the trivial module. Then