Let $M$ be some real smooth manifold (non-compact in particular). In literature, I encounter a statement that the space $L$ of symmetric non-degenerate (0,2) tensors with some fixed signature, let's say $(n,m)$ ($n$ minuses, $m$ pluses) is a subbundle of the second symmetric power $\mathrm{Sym}^2(T^{*}M)$. I don't expect it to be a vector subbundle, but I would like to show that it is a fibre bundle, as I wasn't able to find the proof anywhere.
Maybe people consider it trivial, so this is my first naive attempt. Let $V \cong T^{\ast}_{x}M$ be the cotangent space. Both bundles under consideration would have the same base and projection maps (up to restriction), so the only question is whether $L$ is a submanifold of $\mathrm{Sym}^2(T^{*}M)$. Let's move to local trivialisation. Let $O$ be a trivialising element of cover on $M$. We topologise the vector space of symmetric bilinear forms $\mathrm{Sym}^2(V)$ by means of induced standard topology from $\mathbb{R}^N$. The topology on $\mathrm{Sym}^2(T^{*}M)$ is given by means of local homeomorphism \begin{align*} \phi_O \colon O \times \mathrm{Sym}^2(V) \to \pi^{-1}(O) \subset \mathrm{Sym}^2(T^{*}M) \end{align*} Now, since topological space $S$ of symmetric bilinear forms with signature $(n,m)$ is open in $\mathrm{Sym}^2(V)$, $\phi_O(O \times S)$ will be open in $\mathrm{Sym}^2(T^{*}M)$. Since the same works for all of local trivialisations and any union of open sets is open, $L$ should be an open submanifold of $\mathrm{Sym}^2(T^{*}M)$, and hence a subbundle.
Is this reasoning right? Thank you for help.