# 0708-1300/the unit sphere in a Hilbert space is contractible

Let $H=\{(x_1,x_2,...)| \sum x_n^2<\infty\}$ and define $S^{\infty}=\{x\in H| ||x||=1\}$

Claim

$S^{\infty}$ is contractible

Proof

Suppose $x=(x_1,x_2,...)\in S^{\infty}$ then $\sum x_n^2=1$

Define $F_1:S^{\infty}\times I\rightarrow S^{\infty}$ by $F_1(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)||$

$F_2:S^{\infty}\times I\rightarrow S^{\infty}$ by $F_2(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)||$

$F_3:S^{\infty}\times I\rightarrow S^{\infty}$ by $F_3(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)||$

and so on ...

applying the homotopy $F_1$ in the time interval $[0,1/2]$, $F_2$ in the interval $[1/2,3/4]$, $F_3$ in $[3/4,5/6]$ etc...

we get the desired contraction to the point $(1,0,0,...)$.

A different way to see this is via the cellular structure of $S^{\infty}$. If $S^{\infty}=C_0 C_1 ...$ you can always contract $C_k$ along $C_{k+1}$ like moving contracting the equator along the surface of the earth.

This proof only works in separable Hilbert spaces. Is the unit ball in a non-separable Hilbert space contractible?
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