Non-degenerate $2$-form on a manifold gives tangent-cotangent bundle isomorphism

Question

I am trying to solve the following problem:

Let $M$ be a smooth $n$-manifold, and let $\omega\in \Omega^2(M)$ be such that $\omega_p\colon T_pM\times T_pM\to \Bbb R$ is non-degenerate for each $p\in M$. Then, we have a bundle isomorphism $\omega^\flat\colon TM\to T^*M$.

My attempt: As $\omega$ is non-degenerate, the musical isomorphism $\omega^\flat_p\colon T_pM\to T^*_pM$ at every point $p\in M$ is a linear isomorphism. For a vector field $X$ on $M$, define the $1$-form $\omega^\flat(X)$ as follows $\omega^\flat(X)_p(-):= \omega_p^\flat\left(X_p\right)=\omega_p\left(X_p,-\right)$ for each $p\in M$. The smoothness of $\omega^\flat(X)$ can be checked as follows.

Write $\omega=\sum_{i,j=1}^n c_{ij} dx^i\wedge dx^j$ and $X=\sum_{\ell=1}^n X_\ell\frac{\partial}{\partial x^\ell}$ in local coordinates. Then for $Y=\sum_{k=1}^n Y^k\frac{\partial }{\partial x^k}$, we have $$ \omega^\flat(X)(Y) =\omega\left(\sum_{\ell=1}^n X_\ell\frac{\partial}{\partial x^\ell},\sum_{k=1}^n Y^k\frac{\partial }{\partial x^k}\right)=\sum_{i,j=1}^n(c_{ij}-c_{ji}) X^i Y^j.$$ Thus, $\omega^\flat(X)=\sum_{k=1}^n \sum_{i=1}^n(c_{ik}-c_{ki}) X^i dy^k$ in local coordinates. Therefore, in local coordinates $\omega^\flat\colon TM\to T^*M$ can be written as $$\left(x^1,....,x^n, X_1,...,X_n\right)\longmapsto\left(x^1,....,x^n,\sum_{i=1}^n(c_{i1}-c_{1i}) X^i,...,\sum_{i=1}^n(c_{in}-c_{ni}) X^i\right),$$ which shows $\omega^\flat$ is a smooth bundle map.

If $\omega^\sharp_p\colon T_p^*M\to T_pM$ is the inverse of $\omega^\flat_p\colon T_pM\to T^*_pM$, then for a $1$-form $\theta$ on $M$, define the vector field $\omega^\sharp(\theta)$ as follows $\omega^\sharp(\theta)_p:=\omega^\sharp_p(\theta_p)$.

Difficulty: What is the local expression of $\omega^\sharp(\theta)$ if $\theta=\sum_{\ell=1}^n \theta_\ell dx^\ell$? Is it $\omega^\sharp(\theta)=\sum_{\ell=1}^n\widetilde{c^{i\ell}}\theta_\ell\frac{\partial}{\partial x^i}$, where $\left(\widetilde{c^{i\ell}}\right)$ is the inverse of $\left(c_{ij}\right)$?

Many thanks for reading.

Answer 1

You showed $\omega^{\flat}:TM\to T^*M$ is a smooth vector bundle map, and it is a fiberwise linear isomorphism (and since we’re on a finite-dimensional space, it means the restriction to each fiber is a diffeomorphism). Also, clearly $\omega^{\sharp}$ is fiberwise linear, so the only thing left is smoothness of $\omega^{\sharp}$. In your specific case, you can give an argument that the inverse map is smooth by a relatively direct argument using (a slightly dressed up phrasing of) Cramer’s formula for the inverse of a matrix.

But here’s a general purpose theorem which immediately allows you to conclude:

Let $(X,\pi,B)$ and $(X’,\pi’,B’)$ be fiber bundles and $(f,g)$ a smooth fiber bundle morphism, i.e $f,g$ are smooth maps such that the diagram below commutes: $\require{AMScd}$ \begin{CD} X@>{g}>> X’\\ @V{\pi}VV @VV{\pi’}V \\ B @>>{f}> B’. \end{CD} Then, $(f,g)$ is an isomorphism (in the sense of fiber-bundles) if and only if $f$ is a diffeomorphism and for each $b\in B$, the restriction $g_b:X_b\to X’_{f(b)}$ is a diffeomorphism.

As a corollary, if $(X,\pi,B),(X’,\pi’,B’)$ are vector bundles and $(f,g)$ is a smooth bundle map, then $(f,g)$ is a vector-bundle isomorphism if and only if $f$ is a diffeomorphism and for each $b\in B$, the restriction $g_b:X_b\to X’_{f(b)}$ is a linear isomorphism.

The corollary is immediate from the theorem. Next, the ‘only if’ direction of the theorem is clear. For the ‘if’ direction, note that our assumptions tell us that $g$ is smooth and bijective, so we only need to show that the inverse is smooth. Fix a point $b_0\in B$ and a point $x_0\in X_{b_0}$. Pick small enough neighborhoods of $b_0\in B$ and $f(b_0)\in B’$, which are diffeomorphic via $f$, and over which the bundles $X,X’$ are trivializable. The upshot of all of this is that we reduce to the case where $B=B’$ is open in some $\Bbb{R}^n$ and $X=B\times F$ and $X’=B\times F’$ and $g$ takes the form $g(b,x)=(b,\gamma(b,x))$ with $\gamma$ smooth and each $\gamma(b,\cdot):F\to F’$ being a diffeomorphism. Now, notice that the derivative of $g$ at a point $(b,x)$ has the form \begin{align} Dg_{(b,x)}&= \begin{pmatrix} I_n&0\\ *&\frac{\partial \gamma}{\partial x}(b,x) \end{pmatrix} \end{align} The lower right block is invertible since $\gamma(b,\cdot):F\to F’$ is a diffeomorphism. Hence, $Dg_{(b,x)}$ is invertible. So, by the inverse-function theorem, $g$ has a smooth local inverse about $(b,x)$. But of course, this smooth local inverse coincides with the restriction of the inverse of $g$ (which we already know exists). Hence, $g^{-1}$ is smooth, thereby proving the claim.