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Let $(M , g)$ a $n$-dimensional pseudo-riemannian manifold. As $g(p)$ is a bilinear mapping from $T_pM \times T_pM$ to $\mathbb{R}$, we can get a basis $$ {\left\{{\left(\frac{\partial}{\partial x_i}\right)}_p\right\}}_{i = 1}^n $$ on $T_pM$ such that $$ g(p)\left({\left(\frac{\partial}{\partial x_i}\right)}_p , {\left(\frac{\partial}{\partial x_j}\right)}_p\right) = 0 $$ for each $i , j \in \{1 , \ldots , n\}$ such that $i \neq j$. It allows us to define the index (or signature) of $g(p)$ because it is independent of the choice of basis in $T_pM$, a fact known classically as Sylvester's law of inertia.

Well, many texts talk about index of $g$ strightly, and define for instance riemannian metrics as pseudo-riemannian metrics which index coincide with the dimension of the manifold (or signature ($n , 0$) in this case).

How can I show that it is independent of the point $p$ in $M$? As it can be observed, I have not used neither $g(p)$ is non-degenerate or symetric nor the fact $T_pM \cong {\mathbb{R}}^n \cong T_qM$, as isomorphism of vector spaces ($p , q \in M$), for defining signature of $g(p)$ (for $p \in M$ fixed). Must I use one of these statements to show that the index of $g(p)$ and $g(q)$ coincide for each $p , q \in M$?

joseabp91
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1 Answers1

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Let $v_1, \dots, v_n$ be a basis for $T_pM$ such that $g(p)(v_i, v_i) > 0$ for $i = 1, \dots, k$, and $g(p)(v_i, v_i) < 0$ for $i = k + 1, \dots, n$. We can extend this basis to a basis of local smooth vector fields $V_1, \dots, V_n$ for $TM|_U$ where $U$ is an open neighbourhood of $p$.

Now consider the functions $f_i : U \to \mathbb{R}$ given by $f_i(q) = g(q)(V_i|_q, V_i|_q)$. Note that $f_i$ is smooth and $f_i(p) = g(p)(V_i|_p, V_i|_p) = g(p)(v_i, v_i)$ which has a sign. For $i = 1, \dots, k$, let $U_i = f_i^{-1}((0, \infty))$ and for $i = k + 1, \dots, n$ let $U_i = f_i^{-1}((-\infty, 0))$. Then on $U' = U_1\cap\dots\cap U_k\cap U_{k+1}\cap\dots\cap U_n$ each $f_i$ has a sign. For any $q \in U'$, $f_i(q) > 0$ for $i = 1, \dots, k$ and $f_i(q) < 0$ for $i = k + 1, \dots, n$, so $g(q)$ has the same index of $g(p)$. That is, the index of $g$ is a locally constant integer-valued function and is therefore constant on each connected component of $M$.

Michael Albanese
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