It is possible that on a sphere $S^n$ there is a natural Riemannian metric in $R^(n+1)$. But it is not always possible for pseudo Riemann metric since the sum of two symmetric matrix which are not positive definite but may have rank different from the two matrix. So I wonder what is the sufficient and necessary condition for the dimension of a sphere which can be endowed with a Lorentz metric.



For which $b$ it is possible that $S^n$ can have a Lorentz metric? Why?
**A compact simply connected manifold carriez a Lorenz metric iff its Euler characteristic vanishes.** 


Proof: If $\chi(M)=0$, $M$ carries a nowhere vanishing vector field $X$. Pick up a Riemannian metric $g$ on $M$ (using a partition of unity argument) and denote by $\eta$ the 1-form dual to $X$: $\eta(Y):=g(X,Y)$ for all $Y\in TM$. Then  

$$g-2\frac{\eta\otimes\eta}{g(X,X)}$$
is a Lorenz metric.


Conversely, if $M$ has a Lorenz metric $h$ of signature $(n-1,1)$, pick again a Riemannian metric $g$ and consider the symmetric endomorphism $A$ of $TM$ defined by $h(.,.)=g(A.,.)$. The eigenspaces of $A$ corresponding to the unique negative eigenvalue define a line sub-bundle of $TM$ which is trivial if $M$ is simply connected, so $\chi(M)=0$.


Therefore, the answer to your question is: $S^n$ carries a Lorenz metric iff $n$ is odd. 


