I am look for a way to determine the definiteness of a quadratic form on the orthogonal complement of a line. More specifically, I have the quadratic form $Q^-(x) = -x_0^2 + x_1^2 + \cdots + x_n^2$, which I know is indefinite. The text I am reading makes the claim that if you have a vector $x$ such that $Q^-(x) >0 $, then it must be true that $Q^-$ is indefinite on the orthogonal complement of $x$. I am not sure how one would prove this. Any suggestions on a general strategy to prove something like this would be appreciated as I also would like to then determine what would be true about the orthogonal complement when $Q^-(x) = 0$ or $Q^-(x) < 0$.

Note that the inner product being used here is $\langle x,y\rangle = -x_0y_0 + x_1y_1 + \cdots + x_ny_n$.

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    $\begingroup$ The text is Three-dimensional geometry and topology, by Thurston $\endgroup$ – luthien Jun 13 '18 at 20:07
  • $\begingroup$ false for $n=1.$ $\endgroup$ – Will Jagy Jun 13 '18 at 20:10
  • $\begingroup$ Ah yes, you are correct. I think perhaps Thurston was assuming $n>1$. $\endgroup$ – luthien Jun 13 '18 at 20:21
  • $\begingroup$ I'm looking at a pdf from an early edition/version. I'm not sure you have correctly interpreted what he wrote $\endgroup$ – Will Jagy Jun 13 '18 at 20:22
  • $\begingroup$ Here is one instance I am looking at. He's talking about the relationship between the hyperboloid model and projective model of hyperbolic space. He says "If $x\in \Bbb{R}\Bbb{P}^n$ is such a point, [a point not representing a point on the hyperboloid], $Q^-$ is positive on the associated line $X\subset \Bbb{E}^{n,1}$. This means $Q^-$ is indefinite on the orthogonal complement $X^{\bot}$ of $X$, and that the corresponding hyperspace $x^{\bot}\subset \Bbb{R}\Bbb{P}^n$ intersects hyperbolic space." (page 71) $\endgroup$ – luthien Jun 13 '18 at 20:34

Hint The standard basis of $\Bbb R^{n, 1}$ is orthogonal with respect to $Q^-$, and we can read off that the signature of $Q^-$ is $(n, 1)$

On the other hand, if $Q^- \vert_{x^{\perp}}$ is definite, we can choose an orthogonal basis $(E_1, \ldots, E_n)$ for whichever of $\pm Q^-$ is positive definite, so that $(x, E_1, \ldots, E_n)$ is again an orthogonal basis. What, then, is the signature of $Q^-$?

Note that this approach makes plain the exception Will Jagy pointed out in the comments.

  • $\begingroup$ I know very little about quadratic forms, so I'm not entirely sure how to determine the signature. Do you have a suggestion of something I could read to understand this? $\endgroup$ – luthien Jun 13 '18 at 22:50
  • $\begingroup$ You can just replace all of the references in my answer to $Q^-$ with references to the bilinear form $\langle\,\cdot\,,\,\cdot\,\rangle$. $\endgroup$ – Travis Jun 14 '18 at 2:38
  • $\begingroup$ I just don't know a good way in general to determine a signature. How does the basis relate to the signature? $\endgroup$ – luthien Jun 14 '18 at 5:22
  • $\begingroup$ A basis of a vector space $(E_a)$ is orthogonal with respect to an inner product $B$ iff $B(E_a, E_b) = 0$ for all $a \neq b$; in particular, this forces $B(E_a, E_a) \neq 0$ for all $a$. Given an orthogonal basis $(E_a)$, the signature of $B$ is $(p, q)$, where $p$ is the number of positive elements $B(E_a, E_a)$, and $q$ is the number of negative elements. Critically, the numbers $p, q$ do not dependent on the choice of basis. $\endgroup$ – Travis Jun 15 '18 at 3:10

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