Difference between revisions of "0708-1300/the unit sphere in a Hilbert space is contractible"

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we get the desired contraction to the point <math>(1,0,0,...)</math>.
 
we get the desired contraction to the point <math>(1,0,0,...)</math>.
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A different way to see this is via the cellular structure of <math>S^{\infty}</math>. If <math>S^{\infy}=C_0C_1...</math> you can always contract <math>C_k</math> along <math>C_{k+1}</math> like moving contracting the equator along the surface of the earth.

Revision as of 11:04, 2 November 2007

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 Failed to parse (unknown function\infy): S^{\infy}=C_0C_1... 

you can always contract C_k along C_{k+1} like moving contracting the equator along the surface of the earth.
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