4
$\begingroup$

How can one see that the Gram Schmidt process is a homotopy equivalence $ SL(m,\mathbb{R})\rightarrow SO(m)$? Intuitively it seems plausibel to me, but I can't find an explicit homotopy between the Gram Schmidt map and the identity, since this homotopy always has to "stay inside $SL(m,\mathbb{R}))$". I am thankful for any help.

$\endgroup$

1 Answer 1

5
$\begingroup$

Do it step by step. Pick the first vector $v_1$ of your frame. Rescaling it to unit is obviously a homotopy equivalence. Rotating the span of $v_2, \dotsc, v_n$ to the orthogonal complement of $v_1$ is also. Lather, rinse, repeat.

$\endgroup$
1
  • $\begingroup$ But it is not a homotopy equivalence which stays inside $ SL(m,\mathbb{R})$. I think one would have to modify the Gram Schmidt process quite a bit. $\endgroup$ Jul 13 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.