# Index of fundamental bilinear form is constant in each connected component of a manifold

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.

• You just need to prove $\{ y\in M : \text{Index at }x = \text{Index at }x_0\}$ are both open and closed. – user99914 Jan 29 '17 at 4:03

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.
• $M$ are used twice. And what is $B(M, M)$? – user99914 Jan 29 '17 at 4:30
• Sorry :| I changed the letters. $N$ is a bound on the eigenvalues of $A(t)$ and $B(N,N)$ is a disc of radius $N$ around $N$ (so $B(N,N)$ contains all the positive eigenvalues of $A(t)$ for all $t \in [0,1]$ and non of the negative eigenvalues). – levap Jan 29 '17 at 4:42