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

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


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

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

<math>F_3:S^{\infty}\times I\rightarrow S^{\infty}</math> by <math>F_3(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_4^2},0,0,(1-t)x_4,x_5,x_6,...)/||((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>F_1</math> in the time interval <math>[0,1/2]</math>, <math>F_2</math> in the interval <math>[1/2,3/4]</math>, <math>F_3</math> in <math>[3/4,5/6]</math> etc...

we get the desired contraction to the point <math>(1,0,0,...)</math>.

Revision as of 10:46, 2 November 2007

Let and define

Claim

is contractible

Proof

Suppose then

Define by

by

by

and so on ...

applying the homotopy in the time interval , in the interval , in etc...

we get the desired contraction to the point .