Contractibility of unit sphere in Hilbert space

In topology, it is a surprising fact that the unit sphere in (infinite-dimensional) Hilbert space is a contractible space, sinceno finite-dimensional spheres are contractible.

This can be demonstrated in several different ways.

Topological proof

First, one can use a purely topological argument, appealing to the Whitehead theorem.This proof proceeds as follows: Let $H$ denote the Hilbert space, and $S$ the unit sphere in $H$. Consider the $k$-th homotopy group $pi\_k(S)$. Any element in this group is represented by a map $gamma:S^k\; o\; S$where $S^k$ is the standard $k$-sphere.Since $S^k$ is compact, so is itsimage under $gamma$. Hence the imageis contained in some $n$-dimensional subspaceof $H$. Since the intersection ofthis subspace with $S$ is an $n$-sphere,by choosing $n$ large enough (i.e. largerthan $k$), one sees that the map$gamma$ is null-homotopic when consideredas a map $S^k\; o\; S^n$, hence it is alsonull-homotopic as a map $S^k\; o\; S$. Thus, all of the homotopy groups $pi\_k(S)$ vanish. Furthermore, since the inclusion $lbrace\; point\; brace\; o\; S$induces isomorphisms on all homotopy groups, andall spaces under consideration are connected CW-complexes,the conditions of the Whitehead theorem apply and we canconclude that $S$ is homotopy-equivalent toa point, i.e. contractible.

Constructive proof via shift operator

Let "H" be a separable, infinite-dimensional Hilbert space. One can assume "H" = "l"²(N), the space of square-summable complex sequences. Let "S" be the unit ball in "H".

The topological proof given in the preceding section is non-constructive. It is possible to prove thecontractibility of "S" by explicitly showing that "S" deformation retracts to a single point, that is, writing down a homotopy between the identity map on "S" and a constant map

:$e(\; (z\_i)\_\{i\; geq\; 1\}\; )\; =\; (1,0,0,cdots)$

on "S". This can be done in two stages. First we show that the identity map on "S" is homotopic to the shift operator "T" restricted to "S". Second, we show that "T" restricted to "S" is homotopic to "e". The desired result then follows from the transitivity of homotopy equivalence.

The shift operator "T" on "H" is defined by

:$T\; (z\_1,\; z\_2,\; ...)\; =\; (0,\; z\_1,\; z\_2,\; ...).\; ,$

"T" is an isometry (which is not the case if "H" were finite dimensional), and therefore leaves "S" invariant. The image "T(S)" is the equator of "S", which is homeomorphic to the sphere "S" itself (which clearly does not hold in the finite-dimensional case).

Consider the line segment "γ" from "x"∈ "S" to "Tx"∈ "S":

:$gamma\_x\; (t)\; =\; (1-t)x\; +\; tT(x).\; ,$

Now "T" has no eigenvalues. So "γ"("t") is never the zero vector in "H". Thus one can normalize "γ" to obtain a path in "S":

:$gamma\_x\text{'}(t)\; =\; frac\{gamma\_x(t)\}\{|\; gamma\_x(t).$

The homotopy defined by

:$I\_t\; (x)\; =\; gamma\text{'}\_x(t),\; ;\; t\; in\; [0,1]$

takes the identity to the shift.

Next we show that "T" is null-homotopic. For each

:$x\; =(z\_1,\; z\_2,\; z\_3,\; ...)\; in\; S,$

the path in "S" defined by (here one again needs the fact that "T" is an isometry)

:$phi\_x(t)\; =\; sin\; frac\{pi\}\{2\}\; t\; ,\; (1,\; 0,\; cdots)+\; cos\; frac\{pi\}\{2\}\; t\; ,\; Tx$

goes from "Tx" to the element (1,0,0...). So the homotopy

:$I\text{'}\_t\; (x)\; =\; phi\_x(t)\; ,$

takes "T" to the constant map "e". This, together with the homotopy "I" from the above, proves the claim.

Second constructive proof

Another manifestation of a separable infinite dimensional Hilbert space is, the function space [Lp_space|"L"² [0,1] ] , complex-valued functions on the interval [0,1] square integrable with respect to the Lebesgue measure, modulo equality almost everywhere.

Given a function "f" in "S" and "t" ∈ [0,1] , define

:

This is a "sped-up" version of "f" for the interval [0,(1 - "t")] , andthe constant function 1 on [(1 - "t"),1] . Direct calculation shows that that "f_t" lies in "S" for all "t" ∈ [0,1] . The map "H" on [0, 1] × "S" given by

:$H(t,\; f)\; (x)\; =\; f\_t(x),$

is continuous and therefore a homotopy from the identity map on "S" to the constant map whose range is 1 ∈ "L"² [0,1] .

References

* John Baez, "This Week's Finds in Mathematical Physics, Week 151", [http://www.math.ucr.edu/home/baez/week151.html]
* Dave Rusin, newsgroup posting http://www.math.niu.edu/~rusin/known-math/93_back/s-infty