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

Let [math]\displaystyle{ H=\{(x_1,x_2,...)| \sum x_n^2\lt \infty\} }[/math] and define [math]\displaystyle{ S^{\infty}=\{x\in H| ||x||=1\} }[/math]

Claim

[math]\displaystyle{ S^{\infty} }[/math] is contractible

Proof

Suppose [math]\displaystyle{ x=(x_1,x_2,...)\in S^{\infty} }[/math] then [math]\displaystyle{ \sum x_n^2=1 }[/math]

Define [math]\displaystyle{ F_1:S^{\infty}\times I\rightarrow S^{\infty} }[/math] by [math]\displaystyle{ 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,...)|| }[/math]

[math]\displaystyle{ F_2:S^{\infty}\times I\rightarrow S^{\infty} }[/math] by [math]\displaystyle{ 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,...)|| }[/math]

[math]\displaystyle{ F_3:S^{\infty}\times I\rightarrow S^{\infty} }[/math] by [math]\displaystyle{ 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,...)|| }[/math]

and so on ...

applying the homotopy [math]\displaystyle{ F_1 }[/math] in the time interval [math]\displaystyle{ [0,1/2] }[/math], [math]\displaystyle{ F_2 }[/math] in the interval [math]\displaystyle{ [1/2,3/4] }[/math], [math]\displaystyle{ F_3 }[/math] in [math]\displaystyle{ [3/4,5/6] }[/math] etc...

we get the desired contraction to the point [math]\displaystyle{ (1,0,0,...) }[/math].

A different way to see this is via the cellular structure of [math]\displaystyle{ S^{\infty} }[/math]. If [math]\displaystyle{ S^{\infy}=C_0C_1... }[/math] you can always contract [math]\displaystyle{ C_k }[/math] along [math]\displaystyle{ C_{k+1} }[/math] like moving contracting the equator along the surface of the earth.