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.


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