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Prof. Lee in p. 30 of his "Riemannian Manifolds" says:

"Given a pseudo-Riemannian metric $g$ and a point $p \in M$, by a simple extension of the Gram-Schmidt algorithm one can construct a basis $(E_1, \dotsc, E_n)$ for $T_p M$ in which $g$ has the expression $$ g = -(\varphi^1)^2 - \dotsb - (\varphi^r)^2 + (\varphi^{r+1})^2 + \dotsb + (\varphi^n)^2 $$ for some integer $0 \le r \le n$."

However, I don't think it is quite that simple. In the Riemannian case, the coordinate frame $\partial_i$ forms a local frame, and applying the Gram-Schmidt algorithm to that brings the metric to the above form. But in the semi-Riemannian case, the coordinate vectors $\partial_i$ can all be null, e.g.

$$g = 2 \, du \, dv\text{.}$$

It seems to me that one would need to construct some local frame first in order to apply Gram-Schmidt.

An alternate way would be to represent the metric $g$ by a matrix and then eigendecompose it, i.e. represent it by $A D A^{-1}$ where $D$ is $\operatorname{diag}(\lambda_1, \dotsc, \lambda_n)$, where $\lambda_i$ are the (real) eigenvalues in ascending order. From that we can construct a local frame, and then rescale to bring $g$ to the above form. But this way doesn't involve Gram-Schmidt at all, as far as I can tell.

Am I missing something?

Fred Akalin
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1 Answers1

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You're right that it's not quite as simple as I made it appear. The correct generalization of the Gram-Schmidt algorithm uses the idea of a "nondegenerate basis": If $V$ is a finite-dimensional vector space endowed with a nondegenerate scalar product $g$, an ordered basis $(E_1,\dots,E_n)$ for $V$ is said to be nondegenerate if for each $k=1,\dots, n$, the scalar product $g$ restricts to a nondegenerate scalar product on the subspace spanned by $(E_1,\dots,E_k)$.

The two facts you need to make this work are

  1. $V$ has a nondegenerate basis, and
  2. If $(v_1,\dots,v_n)$ is a nondegenerate basis, then the Gram-Schmidt algorithm applied to $(v_1,\dots,v_n)$ produces an orthonormal basis $(E_1,\dots,E_n)$ with the property that $\operatorname{span}(E_1,\dots,E_k) = \operatorname{span}(v_1,\dots,v_k)$ for each $k=1,\dots,n$.

Proving these is only a little more complicated than proving the Gram-Schmidt algorithm itself. The nondegeneracy hypothesis guarantees that the expressions that appear in the denominators in the Gram-Schmidt algorithm are always nonzero.

Jack Lee
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  • You’re welcome! – Jack Lee Jan 29 '18 at 00:56
  • does your definition of nondegenerate basis exclude anything but the first basis vectors being lightlike? what if only the last one is lightlike? then they span the entire space, the product is known to be non-degenerate on it, but in gram-schmidt you have to divide by zero. – peter Sep 28 '22 at 07:12
  • @peter: In a nondegenerate basis, any basis vector other than the first can be lightlike. If the last one is lightlike, note that in the Gram-Schmidt algorithm you don't divide by the norm until after you've subtracted off the projections onto the other vectors. Once you've done that, the last vector will no longer be lightlike. (To see how it works, try applying the agorithm to the basis ${(1,0),(1,1)}$ on $\mathbb R^2$ with the Lorentz metric $dx^2 - dy^2$.) – Jack Lee Sep 28 '22 at 19:14
  • ic! thanks. makes perfect sense now why this is exactly the right condition. we ever only divide by norms of things spanned by the first k vectors. – peter Sep 29 '22 at 20:13