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I am looking, without luck so far, for references for an equivariant Gram-Schmidt.

Specifically, Let $P$ be the permutation matrices, and $D$ the diagonal matrices with entries in $\mathbb R^\times$. Then $P,D$ have the obvious right actions on $GL(n,\mathbb R)$, and the coset spaces are (respectively) the space of unordered bases and what we might call the projective frames, i.e., $n$ maximally non-degenerate lines. We can take the coset space of the closure of these two groups acting on the right, too, to get unordered projective bases.

Each of the above has an orthogonal version as a subset, which is the quotient of $O(n)$ by the given right action.

One can argue, fairly straightforwardly, that inclusion of the orthogonal version in the general version is a homotopy equivalence cofibration and hence a deformation retraction exists in the other direction. One should also be able to prove this by arguing that $\sum_{i\ne j}|v_i\cdot v_j|/|v_i||v_j|$ is an equivariant Morse-Bott function, with critical points exactly the orthogonal version.

Anyway, this deformation retraction (perhaps even explicitly given) seems like something which should be well-known. Might anyone have a reference?

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BTW the scaling by real value on the diagonal of $D$ obviously doesn't matter, I can take D to just be $\pm 1$ on the diagonal. –  Andrew Marshall Feb 19 '12 at 19:35
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1 Answer 1

I don't know a reference, but here is a partial answer:

If you keep track of the matrix $b$ that effects Gram-Schmidt on $m$ to produce $g$, you get a diffeomorphism $GL\rightarrow O\times B$ given by $m\to (g,b^{-1})$ (this is "QR factorization"). Here, $B$ is the upper triangular matrices with positive diagonal entries. The inverse is multiplication. A deformation retraction of $B$ to a point induces the usual deformation retraction of $GL$ onto $O$.

If $c$ is positive, multiplication of a column of $m$ by $c$ corresponds to dividing a row of $b$ by $c$, which corresponds to multiplication of a column of $b^{-1}$ by $c$ (and no change on the $O$ factor). That is, our diffeomorphism is equivariant for the right action of $H=\mathbb{R}^{+}\times ...\mathbb{R}^{+}$ on $GL$ and the action on $O\times B$ induced by that on $B$. $B/H$ is still contractible, giving a deformation retraction of $GL/H$ onto $O$.

The rest of your group $D$, the group $K=\mathbb{Z}_2\times...$ acts as expected on $O$. But the element $(0,0,..,1,...)$ (in the $i$th slot) acts on the $b$ as well, by negating the non-diagonal entries of row $i$ and column $i$. It's easy to believe that, passing to $b^{-1}$, we get still a deformation retraction of $(B/H)/K=B/D$ to a point, and hence a deformation retraction of $GL/D$ onto $O/K$.

So you need to prove that the discrete quotient $(B/D)/P$ is still contractible, giving the desired deformation retraction $(GL/D)/P$ onto $(O/D)/P$.

*I should point out that I wouldn't be surprised if it is not true...

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So, as you have put it, our right action is the same as conjugation on the upper triangular and right multiplication on the orthogonal components, resp. E.g., $q\cdot b \cdot k=q\cdot k \cdot k^{-1} \cdot b\cdot k$, and this decomposition of the action is necessary since $K$ doesn't act by right multiplication on $B$. This action isn't free, of course, either, so the resulting orbifold may well be connected even though the space is connected and each orbit is discrete. But we have a problem, because $P$ doesn't even act by conjugation on $B$...hmmm... –  Andrew Marshall Feb 20 '12 at 0:17
    
oops, meant the resulting orbifold may indeed be contractible. It is, of course. –  Andrew Marshall Feb 20 '12 at 0:55
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