The index of a bilinear form on a vector space $V$ is the dimension of a maximal negative definite subspace of $V$. Let $M$ be a manifold. A fundamental bilinear form $b$ on M is a $C^\infty$ symmetric tensor field of type $(0,2)$ defined on all points of $M$ and which is nondegenerate at every point.
I want to proof that the index of $b$ is constant on each connected component of $M$.
First approach: assume $m$ and $n$ are points in the same connected component of $M$. Then there is a $C^\infty$ curve connecting the two points. I just need to show the index is constant/continuous through the curve. Then the problem reduces to showing that the index is a continuous function of the components of b in the basis induced by the coordinates in each coordinate domain.
I'm not sure how to complete the proof.
A useful trick to use for such questions is the argument principle. Namely, let $\gamma \colon [0,1] \rightarrow M$ be a continuous curve and choose a frame $(v_1,\dots,v_n)$ of $TM$ along $\gamma$. Let $a_{ij}(t) := b(v_i(t),v_j(t))$ be the matrix representing $b(\gamma(t))$ with respect to the frame $(v_1(t),\dots,v_n(t))$. The matrix $A(t)$ is continuous and you want to show that the number of positive eigenvalues of $A(t)$ is constant along $\gamma$.
Let $p_t(z) = \det(A(t) - zI)$ be the characteristic polynomial of $A(t)$ and set $N = \sup_{t \in [0,1]} \| A(t) \|$. Let $\beta \colon I \rightarrow \mathbb{C}$ be the boundary of the disc $B(N,N)$. Consider the function
$$ t \mapsto \frac{1}{2\pi i} \int_{\beta} \frac{p_t'(z)}{p_t(z)} \, dz. $$
Since $A(t)$ has only real eigenvalues in $[-N,N] \setminus \{ 0 \}$, this function is well-defined and by the argument principle gives you the number of zeroes of $p_t(z)$ in the disc $B(N,N)$ which is precisely the number of positive eigenvalues of $A(t)$. This function is continuous (the integrand is a continuous function of $t$) and integer-valued and so must be constant.