Consider $GL^+(n) \supset SL(n) \supset SO(n)$ the groups of matrices $n \times n$ with positive determinant, determinant $1$ and orthogonal with positive determinant, respectively. Show that these three spaces have the same homotopy type.

Attempt: Two spaces are said to have the same homotopy type when for a continuous map $f : X \to Y$ there exists a continuous $g : Y \to X$ such that $f \circ g \simeq id_Y$ and $g \circ f \simeq id_X$.

(Main idea): Now if we show that $GL^+(n)$ is homeomorphic to $\mathbb R^+ \times SL(n)$ and $P \times SO(n)$, where $P$ is the set of all positive definite matrices in $GL^+(n)$, then we are done.

For the second homeomorphism we notice that for any matrix $A \in GL^+(n)$ if we take the positive definite matrix $A^TA$ then we may find a unique positive definite matrix $B$ such that $B^2 = A^T A$. Then $AB^{-1}$ is orthogonal and we have the (unique) decomposition $A = AB^{-1}\cdot B$. We consider the map $$\begin{align} \psi: P \times SO(n) &\to GL^+(n)\\(X,A) &\mapsto A\exp X\end{align}$$

is a homeomorphism (still have to show that is surjective).

As for the first homeomorphism I still haven't figured out. I thought maybe considering

$$\begin{align} \varphi: \mathbb R^+ \times SL(n) &\to GL^+(n)\\(t,A) &\mapsto \exp tA\end{align}$$

or maybe something like

$$\begin{align} \varphi: Gl^+(n) &\to\mathbb R^+ \times SL(n) \\A &\mapsto (\det A, \,\,\,?\,\,\,)\end{align}$$

Any ideas are welcome.

Note: I am not interested in any different approaches other than the one presented.

  • $\begingroup$ Is $P$ simply connected? $\endgroup$ – Anon May 11 '16 at 18:23
  • $\begingroup$ I know it is connected. I'm not sure it is simply connected P though. $\endgroup$ – Aaron Maroja May 11 '16 at 18:24
  • $\begingroup$ This is almost a duplicate of math.stackexchange.com/questions/214784/… $\endgroup$ – Travis Willse May 11 '16 at 20:02
  • $\begingroup$ I'm looking for the homeomorphism mentioned. $\endgroup$ – Aaron Maroja May 11 '16 at 20:11
  • 1
    $\begingroup$ I don't believe that the tag homotopy-type-theory is warranted, unless you are looking for a solution in the new foundational framework of homotopy type theory. It sure would be an interesting question in this framework, although a question of a vastly different spirit. $\endgroup$ – Ingo Blechschmidt May 11 '16 at 21:45

Your approach for the first question will work, but you need to note that the matrix $B$ you construct necessarily has distinct eigenvalues: since $A^TA$ is symmetric and positive definite, it has distinct eigenvalues, and thus so does its square root. (This is how you contract your space $P$ by the way - send the eigenvalues - which are positive reals - continuously to 1. The same argument does not work over the complex numbers, by the way!)

You should also be a little careful with matrix square roots and continuity. But I'll leave you to solve that problem.

For the second decomposition things are much simpler. Remember that $$\det(tA) = t^n\det(A)$$ for any $t \in \mathbb R$ and $A \in GL(n)$. So let's construct the maps $$f: GL^+(n) \to \mathbb R^+ \times SL(n), \qquad f(A) = \left(\sqrt[n]{\det A},\frac{1}{\sqrt[n]{\det A}}A\right)$$ and $$g: \mathbb R^+ \times SL(n) \to GL(n), \qquad g(t, B) = tB.$$ Then clearly $f$ and $g$ are inverses; they're both continuous and hence give the desired homeomorphisms.

  • $\begingroup$ Why do you say "distinct" eigenvalues? Plenty of symmetric positive definite matrices have repeated eigenvalues, but the point is that there are $n$ of them with multiplicity, no? $\endgroup$ – Kevin Carlson May 13 '16 at 14:22
  • $\begingroup$ Oh, hmm, you don't actually need diagonalizability to contract, do you - I think you can just get away with a nice enough decomposition. I had required distinct eigenvalues because I wanted to guarantee diagonalizability. $\endgroup$ – Thurmond May 13 '16 at 23:27
  • $\begingroup$ But, as I said, not all symmetric positive definite matrices do have distinct eigenvalues. Anyway, you can indeed contract the whole space of symmetric positive definition matrices, for instance using the map $A^t$, which is continuous on the whole space and contracts everything to the identity between time $1$ and $0$. Regardless, it's very well known that an arbitrary real symmetric matrix is diagonalizable. $\endgroup$ – Kevin Carlson May 14 '16 at 0:46

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