1
$\begingroup$

Assume that $V$ is a finite dimensional real or complex normed linear space. Let $Iso(V)\subset GL(V)\subset L(V)$ be the space of linear isometric endomorphisms, invertible endomorphism and linear endomorphisms on $V$, respectively. These space have their natural and unique topologies arising from the operator norm on $L(V)$.

Is $Iso(V)$ a deformation retract of $GL(V)$?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ For a generic real normed space $Iso(V)$ is trivial, isn't it? $\endgroup$
    – aglearner
    Commented Jun 22, 2018 at 22:59
  • 2
    $\begingroup$ The answer to the question is yes iff the norm is Euclidean up to a linear automorphism. $\endgroup$
    – YCor
    Commented Jun 22, 2018 at 23:02
  • $\begingroup$ @aglearner what is an argument? $\endgroup$
    – Ali Taghavi
    Commented Jun 22, 2018 at 23:04
  • $\begingroup$ @YCor may you give a reference or an sketch of proof? $\endgroup$
    – Ali Taghavi
    Commented Jun 22, 2018 at 23:04
  • 3
    $\begingroup$ Write $K=\mathrm{Iso}(V)$ and $L$ a maximal compact subgroup of $G=\mathrm{GL}(V)$ containing $K$, so $L\subset G$ is a homotopy equivalence. If $K$ is a deformation retract of $G$, then $K\subset G$ is a homotopy equivalence, and hence so is $K\subset L$. So $K$ and $L$ have the same number of components, and $K^0\subset L^0$ is a homotopy equivalence; homology in highest degree then says that $K^0$ and $L^0$ have the same dimension and hence $K^0=L^0$, so $K=L$. Since $L$ is conjugate to $\mathrm{O}(V)$, which only preserves, among norms, scalar multiples of the standard one, we conclude. $\endgroup$
    – YCor
    Commented Jun 22, 2018 at 23:14

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .