Timeline for Is $Iso(V)$ a deformation retract of $GL(V)$ when $V$ is a finite dimensional linear normed space
Current License: CC BY-SA 4.0
8 events
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| Jun 23, 2018 at 12:05 | comment | added | aglearner | The argument is the following. 1. It is is easy to see that $Iso(V)$ is compact. Hence, by averaging we can find a Euclidean metric on $V$ invariant under $Iso(V)$. 2. Now, the unit sphere of the norm is a convex body in a Euclidean space and $Iso(V)$ is acting by isometries of this convex body (iso with respect to the Euclidean metric). 3. A generic convex body does not have non-trivial isometries | |
| Jun 22, 2018 at 23:14 | comment | added | YCor | 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. | |
| Jun 22, 2018 at 23:04 | comment | added | Ali Taghavi | @YCor may you give a reference or an sketch of proof? | |
| Jun 22, 2018 at 23:04 | comment | added | Ali Taghavi | @aglearner what is an argument? | |
| Jun 22, 2018 at 23:02 | comment | added | YCor | The answer to the question is yes iff the norm is Euclidean up to a linear automorphism. | |
| Jun 22, 2018 at 22:59 | comment | added | aglearner | For a generic real normed space $Iso(V)$ is trivial, isn't it? | |
| Jun 22, 2018 at 22:59 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 66 characters in body
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| Jun 22, 2018 at 22:52 | history | asked | Ali Taghavi | CC BY-SA 4.0 |