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I am trying to understand congruent matrices and Sylvester's Law of Inertia, and came upon the following space. Let $X$ be the space $M_n(\mathbb{R})$ mod congruence of matrices.

Question 1.: Is this space studied in any part of mathematics (if so, does it have a name)? I assume it must be important since it is the space of quadratic forms classified by their indices.

Question 2. Is it a manifold? Are there notable topological properties about this space? (Anything interesting about its homology, cohomology, fundamental group etc.)?

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  • $\begingroup$ There are finitely many elements in this quotient, so it is only a manifold in the trivial sense: the induced topology on this quotient would be discrete. A more interesting question (topologically) might be what the orbits under matrix congruence look like $\endgroup$ Aug 30, 2020 at 10:13
  • $\begingroup$ @BenGrossmann: No: The number of orbits is finite only if you restrict to the action on symmetric matrices, otherwise, the orbit space has the cardinality of continuum. $\endgroup$ Aug 30, 2020 at 15:24

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There is a very good question behind this post, but the true answer lies beyond linear algebra and in the realm of algebraic or symplectic geometry. Since you are only learning Sylvester's Law, I will only give a rather superficial answer.

  1. The quotient space $X$ is non-Hausdorff, not even $T_1$ (for all $n\ge 1$), hence, it cannot be a manifold. In order to see this, take any nonzero matrix $A\in Y=M_n({\mathbb R})$, a sequence of matrices $B_i\in G=GL_n({\mathbb R})$ converging to the zero matrix and observe that $$ \lim_{i\to\infty} B^T_i A B_i = {\mathbf 0}. $$
    From this, you conclude that the corresponding congruence classes $[A]\in X$ and $[{\mathbf 0}]\in X$ have the property that every neighborhood of $[{\mathbf 0}]$ contains $[A]$. One can verify that the space $X$ is contractible, hence, very boring from the viewpoint of algebraic topology. However, this is not the end of the story.

  2. The above observation suggests that taking the naive quotient is simply a wrong thing to use in this setting. There are several areas of math where it is defined how to make the "right" quotient in this setting, one is GIT = Geometric Invariant Theory (pioneered by David Mumford, although many early ideas go back to David Hilbert), the other area is Symplectic Geometry. I will not attempt to describe what the "right" quotient means in this situation. Briefly, there are two open and dense subsets $Y_{sst}$ (semistable matrices) and $Y_{st}$ (stable matrices) in $Y$, which are both invariant under the action of the group $G$ and satisfy $Y_{st}\subset Y_{sst}$. In order to form the "right" quotient, one takes the naive quotient of $Y_{st}$ by the action of $G$, while for the quotient of $Y_{sst}$ one uses the extended orbit equivalence: $A\sim B$ if and only if the closures of their orbits in $Y_{sst}$ have nonempty intersection. Then the "right" quotient space (also called "Mumford quotient" or GIT quotient) is $$ Z=Y//G= Y_{sst}/\sim. $$

I did not try hard to compute $Y_{sst}$ and $Y_{st}$ in this example, but I am pretty sure that $Y_{st}$ consists of invertible matrices. It is also follows from the definition that zero matrix does not belong to $Y_{sst}$.

I did not try to analyze the geometry/topology of the quotient space $Z$: This usually requires very serious work.

  1. Lastly, if you are just interested in exhibiting representatives of congruence classes of matrices, you can find it in

Lee, Jeffrey M.; Weinberg, David A., A note on canonical forms for matrix congruence, Linear Algebra Appl. 249, 207-215 (1996). ZBL0886.15008.

as well as in

De Terán, Fernando, Canonical forms for congruence of matrices and (T)-palindromic matrix pencils: a tribute to H. W. Turnbull and A. C. Aitken, SeMA J. 73, No. 1, 7-16 (2016). ZBL1338.15030.

where more references are given.

In particular, you will see that the quotient space $X$ has the cardinality of continuum. However, the authors did not think in terms of algebraic geometry, so they did not address the question of geometry/topology of the "right" quotient space. I am not sure if algebraic geometers ever looked at the problem. When I have more time, I will post this question on Mathoverflow.

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  • $\begingroup$ Hi Moishe. What is the relationship between your space and the $\mathbb{R}$-points of the GIT quotient $\mathrm{GL}_n//\mathrm{GL}_n$? I only ask since this quotient space is fairly simple. Namely, it's just $\mathbb{G}_{m,\mathbb{R}}^n/S_n$ as follows from Chevalley's theorem. $\endgroup$ Sep 1, 2020 at 22:51
  • $\begingroup$ @AlexYoucis: Most likely, it will be the set of real points of Mumford quotient. But in some cases (different quotient problem), the set of real points is larger. Maybe you should add an answer using Chevalley's theorem. $\endgroup$ Sep 2, 2020 at 17:08

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