# Existence theorem for symmetric nondegenerate forms over a ring

There exists a rich theory for inner product spaces (i.e. vector spaces with a symmetric nondegenerate bilinear form) over fields, and it can be discussed in the context of local rings and free modules. Furthermore, there exists a very nice result using cohomology in a paper of Jardine

https://www.uwo.ca/math/faculty/jardine/preprints/preprint-coh-inv.pdf

that talks about these objects up to isomorphisms and relates it to the absolute Galois group of a field. This has made me wonder if there exists a “criterion” or existence theorem for such structures on a module over a commutative Unitas ring. For instance, if one has a Noetherian algebra over a field of characteristic prime to two and the module of Kahler differentials is finitely generated projective, then when does it have a symmetric nondegenerate bilinear form? Such conditions ensure that its isomorphic to its dual, so when one works with say a smooth variety, its canonical bundle has 2-torsion, does there exists geometric conditions that ensure there exists such a form? Ideally, it would be nice to have a statement that says under certain assumptions that there exists a nondegenerate symmetric bilinear form on a finitely generated module iff some condition. This might be too general, but even something in a particular context would be interesting as it seems this theory seems to become difficult to study when one steps away from certain finiteness conditions like free modules.

• A nondegenerate bilinear form (no symmetry) on a f.g. projective module exists iff it’s isomorphic to its dual. In the topological setting a necessary condition for this is that all the odd Chern classes vanish and I imagine something similar is true in the AG setting for the Chern classes in the Chow ring. Commented Sep 25, 2020 at 16:24
• Also for line bundles a necessary and sufficient condition is that its order in the Picard group divides $2$. Commented Sep 25, 2020 at 16:30
• So if I understand correctly, nondegenracy can be equivalently stated that the order of the canonical line bundle has order two in the Picard group.. I get that, but any idea on how symmetry could be understood in this way? I will read up on the Chern classes and Chow ring setting for symmetry. Commented Sep 25, 2020 at 16:50
• Symmetry is complicated. Passing to a different setting, for representations of groups it depends on the value of the Frobenius-Schur indicator. If $V$ is isomorphic to its dual then the isomorphism lives in $V^{\otimes 2}$ and symmetry is the question of whether it lives in $S^2(V)$ (and skew-symmetry corresponds to $\wedge^2(V)$). Again topologically there's an obstruction in the Chern classes. (Also something funny might happen in characteristic $2$.) Commented Sep 25, 2020 at 16:56
• Yeah I figured symmetry is complicated, I have been working on a problem and was hoping there was a case for which symmetry could dwindle down to checking if a particular criterion was met. Passing to the setting of a smooth variety $V$ over a filed $k$ of char. not $2$, this amount to checking if there exists a section $g$ of $\textrm{Sym}^2 \Omega_{V/k}$ being nondegenerate at all points. One could choose finite set of global sections and show that $g$ has a symmetric matrix whose determinant is a unit in the stalk, but this approach has too much going on. I wanted to try and simplify it. Commented Sep 25, 2020 at 17:03