Consider a smooth manifold $M$ of arbitrary dimension. We have notions of psuedoRiemannian or Riemannian metrics on a manifold, and they differ in the slightest way of being positivedefinite or not. However, what happens if we drop positivedefinite AND symmetry? For example, if we had a nondegenerate bilinear form $g_p: T_p M \times T_p M \to \mathbb{R}$ that varied smoothly between points. Has this been explored in depth? It appears to me at surface level that one could still concoct connections, curvature, and possibly a notion of parallel transport in this flavor of 'smooth geometry'. A motivation for me to ask is as follows. Suppose $R$ is an $S$algebra where $\Omega_{R/S}$ is reflexive and the canonical isomorphism $\phi: \Omega_{R/S} \to \Theta_{R/S}$ is an isomorphism of $R$modules (i.e. nonsingular varieties). There exists a canonical map $\Omega_{R/S} \times \Theta_{R/S} \to R$, which is $R$bilinear and nondegenerate, and is given by $\langle \omega, V \rangle = l(\omega)$ where $l:\Omega_{R/S} \to k$ such that $l \circ d_{R/S} = V$. This induces a morphism $$\Theta_{R/S} \times \Theta_{R/S} \xrightarrow{\phi^{1} \times 1}\Omega_{R/S} \times \Theta_{R/S} \to R.$$ Natural questions that arise are is this composition $R$bilinear nondegenerate, and when is it symmetric? Which symmetric $R$bilinear forms factor through $\phi^{1} \times 1$? When we work with a manifold and have a metric tensor that is bilinear and nondegenerate, just how interesting is this flavor of curvature (whatever it is supposed to mean)?

8$\begingroup$ There is no functorial way to define a connection on $TX$ starting from a nonsymmetric nondegenerate form. As an extreme opposite case to a metric you can consider a symplectic manifold, which is equipped with a nondegenerate form. It is not the case that a symplectic manifold has a canonically definable connection on its tangent bundle: for example, if this were the case then every manifold would carry a connection on $TX \oplus T^*X$ which is invariant under autodiffeomorphisms, and this is not the case e.g. for $\mathbb{R}$. $\endgroup$– Dmitry VaintrobCommented May 26, 2020 at 3:17
2 Answers
Consider a bilinear form $b \in \mathcal{C}^\infty (T^*M\otimes T^*M, \mathbb{R})$ and an affine connection $\nabla \colon \mathcal{\Gamma}^\infty(TM) \to \mathcal{\Gamma}^\infty(T^*M\otimes TM)$ whose parallel transport preserves $b$. That can be expressed by the condition $b(\nabla_X Y, Z) + b(Y, \nabla_X Z) = 0$ for all $X, Y, Z \in \Gamma(TM).$
When $b$ is symmetric nondegenerate tensor (i.e. a pseudoRiemannian metric), then this $\nabla$ is caleld metric connection. It always exists, but it is not unique! To get a unique connection one has to impose also that $\nabla$ has zero torsion tensor.
When $b$ is antisymmetric nondegenerate tensor (i.e. a presymplectic form), then $\nabla$ is known as symplectic connection but this time the uniqueness is not saved by torsionfreeness.
The way to handle these problems in general is to consider one torsion free $b$connection $\nabla$ and study its modification $\nabla + A$ where $A \in \Gamma(\mathrm{End}(TM)).$ Such modified connection preserves $b$ if and only if $b(A(X)Y, Z) + b(Y, A(X)Z) = 0.$ Torsionfreeness is equivalent to $A(X)Y  A(Y)X = 0.$ From this it is obvious that the affine space of torsionfree $b$connections is governed by the representation theory of Lie algebra of the isotropy group of $b$. Sometimes you get just the trivial representation, sometimes you get bigger space.
The punchline here is that connection preserving some tensorial objects are generally not unique. You can either add some additional data to fix them, or you can try to construct invariants which do not depend on the possible choices.
A few remarks:
First, in a sense, (special cases of) this (are) is very commonly studied. Because a bilinear differential form $g$ as you have defined it can naturally be written as a sum $g = \sigma + \alpha$ where $\sigma$ is symmetric and $\alpha$ is skewsymmetric, you are, equivalently, asking about the geometry of the pair $(\sigma,\alpha)$. The most famous example is that of Kähler geometry, i.e., where $\sigma$ is positive definite and $\alpha$ is nondegenerate and parallel with respect to the LeviCivita connection of $\sigma$, but there are many variations on this (pseudoKähler, Hermitian, unitary...) that essentially have the same nature. In general, when $\sigma$ is nondegenerate, you know that there is the canonical LeviCivita connection of $\sigma$, but, of course, depending on the assumptions you make about $\alpha$, there could be other `canonical' connections.
Second, it is not clear how you ought to define 'nondegenerate' for a general $g$, because there are different notions, and it might well be that what you want to call 'nondegenerate' depends on the applications you have in mind. For example, if $M=\mathbb{R}^2$, and $g = \mathrm{d}x\otimes\mathrm{d}y$, then $g$ is 'degenerate' in the naïve sense (since it has tensor rank $1$ instead of $2$), but both $\sigma = \tfrac12(\mathrm{d}x\otimes\mathrm{d}y+\mathrm{d}y\otimes\mathrm{d}x)$ and $\alpha=\tfrac12(\mathrm{d}x\otimes\mathrm{d}y\mathrm{d}y\otimes\mathrm{d}x)$ are nondegenerate in the usual senses for symmetric and antisymmetric quadratic differential forms. In particular, since $\sigma$ has a 'functorial' connection (being nondegenerate), $g$ does also. You'd want to count such a $g$ as 'nondegenerate', no?
Finally, as Dmitry has already remarked, you do need some hypotheses beyond simple algebraic 'nondegeneracy' in order to have a 'functorial' (i.e., invariant under all diffeomorphisms)construction of a connection. Such considerations will enter into any discussion of the right hypotheses, as Vit's discussion above makes clear. I'll just add that sometimes it's important to take higher order derivatives of $g$ into consideration in order to define a class of structures for which a 'functorial' connection exists. For example, as Dmitry pointed out, a closed nondegenerate $2$form on $M$ does not determine a 'functorial' affine connection on $M$. However, if you drop the assumption 'closed' and replace it with an appropriate higher order nondegeneracy condition, there sometimes is an associated functorial connection. For example, there is an open set $\mathcal{F}$ of germs of $2$forms on $4$manifolds that is perserved by (local) diffeomorphisms that does have the property that any $2$form $\alpha$ on $M^4$ whose germ at every point belongs to $\mathcal{F}$ possesses a functorial torsionfree connection $\nabla^\alpha$, but the definition of $\mathcal{F}$ (and of $\nabla^\alpha$) depends on higher derivatives of $\alpha$ than just first derivatives. (In particular, $\mathrm{d}\alpha$ will be nowhere vanishing if $\alpha$ has its germs belonging to $\mathcal{F}$ at every point.)
Remark: I was asked about how the above $\mathcal{F}$ is defined and how it works. Here's a quick sketch of the result of applying Cartan's method of equivalence to this question:
Start with a nondegenerate $2$form $\alpha$ on a $4$manifold $M$. First, there exists a unique $1$form $\beta$ on $M$ such that $\mathrm{d}\alpha = \beta\wedge\alpha$. Let $\gamma = \mathrm{d}\beta$. Then $0 = \mathrm{d}(\beta\wedge\alpha) = \gamma\wedge\alpha$. Second, there exists a uniqe function $F$ on $M$ such that $\gamma^2 = F\,\alpha^2$, and the first condition defining $\mathcal{F}$ is that $F$ should be nowhere vanishing. For simplicity, I'm going to continue the analysis under the assumption that $F<0$ (there is a similar branch when $F>0$, but I'll leave that to the interested reader). Set $F = f^2$ where $f>0$. Then, using $\gamma\wedge\alpha= 0$ and $\gamma^2 = f^2\,\alpha^2$, we see that $(\gamma\pm f\alpha)^2 = 0$, but $(\gamma+f\alpha)\wedge(\gammaf\alpha) = 2f^2\,\alpha^2\not=0$, so, setting $\gamma\pm f\,\alpha = \pm 2f\,\alpha_{\pm}$, we have $$ \alpha = \alpha_+ + \alpha_\quad\text{and}\quad \gamma = f\,(\alpha_+  \alpha_), $$ where $\alpha_\pm$ are a pair of (nonvanishing) decomposable $2$forms whose wedge product is nonvanishing. Finally, there are unique decompositions $$ \beta = \beta_+ + \beta_ \quad\text{and}\quad \mathrm{d}f = \phi_+ + \phi_ $$ where $\beta_\pm \wedge\alpha_{\pm} = \phi_\pm \wedge\alpha_{\pm} = 0$, and so there will be unique functions $g_\pm$ such that $$ \beta_\pm\wedge\phi_\pm = g_\pm\,\alpha_\pm\,. $$ The final 'open' conditions on $\alpha$ needed to define $\mathcal{F}$ are that $g_+$ and $g_$ be nonvanishing.
In this case, the $1$forms $\beta_+$, $\beta_$, $\phi_+$, and $\phi_$ define a coframing on $M$ that is functorially associated to $\alpha$. Once one has such a 'canonical' coframing, it is easy to define a connection, in fact, a large family of connections, on $M^4$ (for example, one of these connections (with torsion) will make the given coframing parallel) including some that are torsionfree.

$\begingroup$ Concerning nondegeneracy: is not just an isomorphism between the tangent and the cotangent bundle the most natural and uniquely determined notion of a nondegenerate "nonsymmetric"? $\endgroup$ Commented May 26, 2020 at 13:09

3$\begingroup$ @მამუკაჯიბლაძე: Ah, but the question is which map are you going to ask to be an isomorphism? If you have a bilinear form $g:V\times V\to\mathbb{R}$ that is neither symmetric nor antisymmetric, there are two possibilities: $\alpha(v)(w) = g(v,w)$ or $\beta(v)(w) = g(w,v)$. When $g$ is neither symmetric nor antisymmetric, one of $\alpha:V\to V^*$ or $\beta:V\to V^*$ could be an isomorphism without the other being an isomorphism. Which one are you going to use? $\endgroup$ Commented May 26, 2020 at 13:48

$\begingroup$ Ooooh. Life is tough :) Thank you for the explanation. Indeed $V\to W^*$ and $W\to V^*$ are safely distinguishable, but when $V=W$, there is a problem, yes. $\endgroup$ Commented May 26, 2020 at 14:01

1$\begingroup$ @RobertBryant Could you please add some references to that higher order symplectic example? $\endgroup$ Commented May 27, 2020 at 15:57

1$\begingroup$ @VítTuček: Well, I'm not sure of the best place to see a reference to Cartan's equivalence problem applied to 'generic' nondegenerate $2$forms on $4$manifolds, but this is a wellknown exercise in applying Cartan's method of equivalence. I'll add a paragraph at the end, that sketches how this goes. $\endgroup$ Commented May 27, 2020 at 18:34