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Consider the $\ell^2$ complex Hilbert space.

Let $m\in \mathbb{N}^*$ be a fixed number, and set $$ S=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^m \frac{|x_n|^2}{n^2}=1\right\}.$$

I want to show that $S$ is not homeomorphic to $$ S(0,1)=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^\infty |x_n|^2=1\right\}.$$

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  • $\begingroup$ What do you mean by $(x_n)_n$? $\endgroup$
    – Amir Sagiv
    Commented Apr 16, 2018 at 20:05
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    $\begingroup$ @AmirSagiv $(x_n)_n$ is a standard notation to mean the sequence $n\mapsto x_n$. Does this answer your question? $\endgroup$
    – YCor
    Commented Apr 16, 2018 at 20:06
  • $\begingroup$ Sure, now I get it $\endgroup$
    – Amir Sagiv
    Commented Apr 16, 2018 at 20:10

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$S$ contains the ellipsoid $\{x\in \mathbb C^m: \sum \frac{|x_n|^2}{n^2}=1\}$ (which is homeomorphic to $S^{2m-1}$) as a strong deformation retract via $A_t(x_1,\dots,x_m,x_{m+1},\dots) = (x_1,x_2,\dots,x_m,(1-t)x_{m+1},(1-t)x_{m+2},\dots)$ for $t\in[0,1]$, thus the homotopy group $\pi_{2m-1}(S) = \mathbb Z$.

$S(0,1)$, however, has $\pi_{2m-1}(S(0,1)) = 0$ since it is contractible: shift first all coordinates one step to the right and then contract in the stereographic chart opposite $(1,0,0,\dots)$. Or use the contraction described on page 513 of here.

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    $\begingroup$ I feel that you should modify your deformation formula so that it would indeed work (a minor thing). $\endgroup$
    – Wlod AA
    Commented Apr 17, 2018 at 19:39
  • $\begingroup$ It is true that $S$ is homeomorphic to $S_m\times \ell^2$ with $$S_m:=\{y\in \mathbb{C}^m; \|y\|=1\}?$$ $\endgroup$
    – Schüler
    Commented Apr 18, 2018 at 14:50
  • $\begingroup$ @Schüler: yes, it is: $$S\ni (x_1,\dots)\mapsto (x_1,2x_2,\dots,mx_m,x_{m+1},\dots)\in S_m \times \ell^2;$$ Note $S_m = S^{2m-1}$. $\endgroup$
    – Peter Michor
    Commented Apr 18, 2018 at 17:59
  • $\begingroup$ @PeterMichor I think that $S_m\times \ell^2$ cannot be homeomorphic to the $1$-shere of $\ell^2$. And thus $S$ is not homeomorphic to the $1$-shere of $\ell^2$. Do you agree with me? Thanks $\endgroup$
    – Schüler
    Commented Apr 21, 2018 at 15:48
  • $\begingroup$ @Schüler: Yes, a proof of that is the content of my answer. $\endgroup$
    – Peter Michor
    Commented Apr 21, 2018 at 15:57

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