Let $k$ be a field, $V,W$ be two vector spaces over $k$ and $\beta : V \times W \to k$ be a bilinear form.

I have always seen non-degeneracy for $\beta$ over $V$ stated in the following way : $$\forall \,v \in V\setminus\{0\}\,\,,\,\, \exists w \in W \,\,,\,\, \beta(v,w)\neq 0. $$

While reading Kock's book "Frobenius algebras and 2D topological quantum field theories" I read the following definition.

If we write $\bar\beta : V \otimes W \to k$ for the corresponding linear map then $\bar\beta$ is non-degenerate over $V$ if there exists some linear map $\gamma : k \to W \otimes V$ such that the map $$ V = V\otimes k \xrightarrow{\text{$id_V\otimes\gamma$}} V \otimes W \otimes V \xrightarrow{\beta\otimes id_V} k \otimes V = V $$ is the identity of $V$.

My question is : are this two definitions equivalent ?

I have already shown that the second implies the first but I'm completely stuck when I try to do the converse.


No, they are not equivalent. The second definition is equivalent to the condition that the map $V \ni v \mapsto \beta(v, -) \in W^{\ast}$ is an isomorphism (equivalently, that the map $W \ni w \mapsto \beta(-, w) \in V^{\ast}$ is an isomorphism). The first definition only states that the first map is injective; in particular, it does not imply that $V, W$ are finite-dimensional.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.