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In my search to understand how $S^2$ has no Lorentz metric, I found this Pseudo-Riemannian Metric on Manifold which helps me a lot because the hairy ball theorem concludes.

But I don't understand how to prove that

A necessary and sufficient condition for a smooth $n$-manifold $X$ to admit a metric of signature $(r,s)$ is that the tangent bundle admits a direct sum decomposition into a bundle of rank $r$ and a bundle of rank $s$.

Of course, I will be statisfied with a proof for the case $r$ or $s$ equal to 1 but I think the structure of the proof will stay the same.

References are appreciated.

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Let me outline how you can prove directly that if a manifold admits an $(r,s)$ metric then the tangent bundle admits a direct sum decomposition into a bundle of rank $r$ and a bundle or rank $s$.

Start with the linear algebra. Given a vector space $V$ and an $(r,s)$ metric $h$ on it, we would like somehow to decompose $V$ into a direct sum of two subspaces $W_1,W_2$ of dimensions $r$ and $s$ respectively. A natural thing to do is to try and take $W_1$ to be "the" subspace on which $h$ is positive-definite and $W_2$ to be "the" subspace on which $h$ is negative-definite. The problem is that there isn't a unique subspace $W_1$ of dimension $r$ on which $h$ is positive-definite so we need some way to choose such a subspace. We don't want to do it arbitrary because we want to carry this construction on the tangent bundle on each fiber and there we will want our choices "to vary" smoothly so that we actually get subbundles.

Choose an arbitrary auxiliary inner product $g$ on $V$ and represent $h$ as a self-adjoint operator on $V$. Namely, there exists a unique $T_h \colon V \rightarrow V$ such that $g(T_h v, w) = h(v, w)$ for all $v, w \in V$. Now we can take $W_1$ to be the subspace of $V$ spanned by the eigenvectors of $T$ associated to the positive eigenvalues and $W_2$ to be the subspace of $V$ spanned by the eigenvectors of $T_h$ associated to the negative eigenvalues. The choice of $g$ eliminates the arbitrary choice of subspaces and gives us a unique direct sum decomposition.

Now apply the above argument for $TM$. Choose a metric $g$ and represent $h$ as a self-adjoint operator $T$ (so that $T$ is a $(1,1)$-tensor). Denote by $W_1$ the subset of $TM$ such that $W_1|_{p}$ consists of the span of the eigenvectors of $T|_p$ associated to positive eigenvalues and similarly for $W_2$. Since $h$ has signature $(r,s)$, each $W_1|_{p}$ is of dimension $r$, each $W_2|_{p}$ is of dimension $s$ and we have a direct sum decomposition $W_1 \oplus W_2 = TM$. Argue (this is the subtle part) that $W_i$ are in fact smooth subbundles of $TM$.

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  • $\begingroup$ How do you know that you can represent $h$ with a self-adjoint operator? $\endgroup$ – Sov Sep 4 '17 at 0:02
  • $\begingroup$ The formula $g(Tv,w) = h(v,w)$ implicitly defines $Tv$ by defining the inner product of $Tv$ with any vector $w$. This is the same argument that shows any operator has and adjoint. $\endgroup$ – levap Sep 4 '17 at 3:02
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Sufficiency follows easily from the fact that any vector bundle can be endowed with a positive definite bundle metric. Thus if you have $TX=E\oplus F$, then you choose positive definite bundle metrics $g_E$ and $g_F$ and the orthogonal direct sum $g_e\oplus -g_F$ will be a pseudo-Riemannian metric of signature $(r,s)$, where $r$ and $s$ are the ranks of $E$ and $F$.

For the other direction, there are direct arguments using local orthonormal frames, but they are a bit tedious. If you are familiar with the description of pseudo-Riemannian metrics in terms of reductions of structure group, then you can bypass these arguments as follows: Choosing a splitting of $\mathbb R^{(r,s)}$ into a positive definite and a negative definite part defines an inclusion $O(r)\times O(s)\hookrightarrow O(r,s)$ as a maximal compact subgroup. Either from general results or from a direct description of $O(r,s)$ you can then deduce that this inclusion is a homomotopy equivalence, so the homogeneous space $O(r,s)/(O(r)\times O(s))$ is contractible. For a manifold $X$ of dimension $r+s$, a pseudo-Riemannian metric $g$ of signature $(r,s)$ is equivalent to reduction of structure group to $O(r,s)\subset GL(r+s,\mathbb R)$. By the contractibility observed above, such a reduction always gives you a further reduction to $O(r)\times O(s)$, which is equivalent to a decomposition $TX=E\oplus F$ together with positive definite bundle metrics on the summands such that $g=g_E\oplus -g_F$.

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  • $\begingroup$ Thank you, but I'm not familiar with the description of pseudo-riemannian metrics in terms of reductions of structure groups. Anyway, I will give a look to it and try to understand this second part. $\endgroup$ – Sov Aug 29 '17 at 13:54

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