Contractibility of unit sphere in Hilbert space

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In topology, it is a surprising fact that the unit sphere in (infinite-dimensional) Hilbert space is a contractible space, since no finite-dimensional spheres are contractible.

This can be demonstrated in several different ways.

1 Topological proof
2 Constructive proof via shift operator
3 Second constructive proof
4 References

[edit] 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 $π k (S)$ . Any element in this group is represented by a map $\gamma:S^k \to S$ where $S k$ is the standard $k$ -sphere. Since $S k$ is compact, so is its image under $γ$ . Hence the image is contained in some $n$ -dimensional subspace of $H$ . Since the intersection of this subspace with $S$ is an $n$ -sphere, by choosing $n$ large enough (i.e. larger than $k$ ), one sees that the map $γ$ is null-homotopic when considered as a map $S^k \to S^n$ , hence it is also null-homotopic as a map $S^k \to S$ . Thus, all of the homotopy groups $π k (S)$ vanish. Furthermore, since the inclusion $\lbrace point \rbrace \to S$ induces isomorphisms on all homotopy groups, and all spaces under consideration are connected CW-complexes, the conditions of the Whitehead theorem apply and we can conclude that $S$ is homotopy-equivalent to a point, i.e. contractible.

[edit] Constructive proof via shift operator

Let H be a separable, infinite-dimensional Hilbert space. One can assume H = l²(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 the contractibility 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:

$\gamma_x'(t) = \frac{\gamma_x(t)}{\| \gamma_x(t)\|}.$

The homotopy defined by

$I_t (x) = \gamma'_x(t), \; t \in [0,1]$

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

$I'_t (x) = \phi_x(t) \,$

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

[edit] Second constructive proof

Another manifestation of a separable infinite dimensional Hilbert space is, the function space L²[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

$f_t(x) = \begin{cases} f(x/(1-t)) & \mbox{if }x < (1-t) \\ 1 & \mbox{if } x \ge (1-t) \end{cases}.$

This is a "sped-up" version of f for the interval [0,(1 - t)], and the 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

$H(t, f) (x) = f_t(x)\,$