Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here I mean the version with all but finitely many components zero.

share|improve this question
2  
(Added for posterity): See also <mathoverflow.net/questions/38763/…; –  Loop Space Sep 15 '10 at 8:06
    
Maybe this should be closed? –  Sean Tilson Mar 26 '13 at 2:20
    
The same question was asked on math.stackexchange: math.stackexchange.com/questions/282268/… –  Seirios Jun 27 at 16:15

6 Answers 6

up vote 25 down vote accepted

This is the swindle, isn't it?

There's an elegant way to phrase this with lots of sines and cosines, but working it all out is too much like hard work. Here's the quick and dirty way.

Let $T: S^\infty \to S^\infty$ be the "shift everything down by 1" map.

Then for any point $x \in S^\infty$, $T(x)$ is not a multiple of $x$ and so the line between them does not go through the origin. We can therefore define a homotopy from the identity on $S^\infty$ to $T$ by taking the homotopy $t x + (1 - t)T(x)$ and renormalising so that it is always on the sphere (incidentally, although you are working in $\ell^0$, by talking about a sphere you implicitly have a norm).

Then we simply contract the image of $T$, which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...)$. Again, we can use 'orrible sines and cosines, but renormalising the direct path will do.

(Incidentally, there's nothing special about which space you are taking the sphere in. So long as your space is stable in the sense that $X \oplus \mathbb{R} \cong X$ then this works)

Added a bit later: Incidentally, if you want to work in a space that doesn't support a norm (such as an infinite product of copies of $\mathbb{R}$) you can still define the sphere as the quotient of $X$ without the origin by the action of $\mathbb{R}^+$. The argument above still works in this case.

Added even later: Revisiting this in the light of the duplicate: Is $L^p(\mathbb{R})$ minus the zero function contractible?, the key property on $T$ is that it be continuous, injective, have no eigenvectors, and be not surjective. These conditions imply the following:

  1. injective ⟹ the end-point of the homotopy is not the origin
  2. no eigenvalues ⟹ the homotopy does not pass through the origin en route
  3. not surjective ⟹ there is a point not in the image to which the image can be contracted
  4. continuous ⟹ the homotopy is jointly continuous

Finally, there's no difference between the sphere and the space minus a point (indeed, without a norm the "space minus a point" is easier to deal with). Indeed, the homotopy described here actually works on the "space minus a point" and is just renormalised to work on the sphere.

share|improve this answer

Another nice solution to a similar question is at http://katlas.math.toronto.edu/drorbn/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible:

Let $H=L^2([0,1])$ and define $S^\infty = \{x \in H : \|x\|=1\}$.

Claim. $S^\infty$ is contractible.

Proof. For any $t \in [0,1]$ and any $f \in H$ define $f_t(x)= f$ for $0<x<t$ and $f_t(x)=1$ for $t<x<1$. Observe that $t \mapsto f_t/\|f_t\|$ is continuous and gives the desired retraction to the point $f=1$.

share|improve this answer
    
Neat, even though this is contracting a different S^\infty. The argument can also be modified to contract the unit sphere in L^2(N) (where N is the set of natural numbers), a third version of S^\infty. –  Anton Geraschenko Oct 9 '09 at 6:52
    
Should that f_t(x) = f should be f_t(x) = f(x)? –  Loop Space Oct 9 '09 at 7:05
5  
One has to be careful in infinite dimensions with topology. This contraction does work, but proving that it is continuous takes a smidgen more work than simply saying that t -> f_t/||f_t|| is continuous. It needs to be jointly continuous in both f and t, which needs a line or two to show. It is not uniformly continuous in both f and t, though, which is why the obvious adaptation of this for the general linear group group only proves that that group is contractible in the weak topology, not the strong topology. Compare the length of Kuiper's theorem with that of Atiyah and Segal. –  Loop Space Oct 9 '09 at 7:40

Kind of late to the party, but the (weak) contractibility follows from $\pi_i(S^\infty) = 0$ for $i>0$.

share|improve this answer
3  
Indeed. This follows from the fact that every compact subset is contained in some finite S^n, which then can be contracted in S^{n+1}. This sort of argument is more generally useful to show the contractibility of this sort of infinite-dimensional object when it may be nonobvious how to write down a contraction explicitly. –  Eric Wofsey Oct 11 '09 at 23:20

3 proofs on Wikipedia - basically the same arguments as above. The Hilbert space part is superfluous.

share|improve this answer

Are follow up questions allowed? What does "version with all but finitely many components zero" mean? Is this different from taking the limit of n-spheres?

share|improve this answer
2  
I believe he means to take vectors of norm one in the infinite dimensional Euclidean space, where points are given by countably infinitely many real numbers such that almost all are zero. This is the same as the colimit (i.e., union) of n-spheres along equatorial embeddings. –  S. Carnahan Oct 14 '09 at 22:17

Here are my thoughts on the matter. However, this is not too much more than what is done above. I think...

$\quad$ We seek to show that a homotopy from the identity map of $S^{\infty}$ ($id_{S^{\infty}}$) to a constant map can be constructed and thus it must be null-homotopic, $i.e.$, contractible. Let $T: S^\infty \to S^\infty$ be the "shift everything 'down' by 1" map given by $(x_1, x_2, x_3,...) \mapsto (0, x_1, x_2,...)$. Then for any point, $x$ and its image $T(x)$, the line between them does not go through the origin.

$\quad$ We can therefore define a homotopy from the identity on $S^\infty$ to T by taking a homotopy and renormalizing, so that it is always on the sphere, as follows. Let $f_t: \mathbb{R}^{\infty} \setminus 0 \rightarrow \mathbb{R}^{\infty} \setminus 0$ be given by
$$f_t(x_1,x_2,...) = (1-t)(x_1, x_2,...) + tT(x_1, x_2,...).$$ (Note that the vector $f_0 = id_{\mathbb{R}^{\infty}}$ and that $f_t$ takes nonzero vectors to nonzero vectors $\forall t \in \left[ 0, 1\right]$.) Then we can renormalize it to ensure everything is still on $S^{\infty}$, $i.e.$, $$\frac{f_t}{\left| f_t\right|} = F(x,t) : S^{\infty} \times I \rightarrow S^{\infty}.$$ Thus we now have that $id_{S^{\infty}} \simeq T$. Or, in other words, $F$ gives a homotopy from the identity map of $S^{\infty}$ to the map $(x_1, x_2, x_3,...) \mapsto (0, x_1, x_2,...)$.

$\quad$ Then we simply contract the image of $T$ , which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...) = N$ (north pole). So let $g_{t} : \mathbb{R}^{\infty} \setminus 0 \rightarrow \mathbb{R}^{\infty} \setminus 0$ be given by $$g_t(x_1, x_2, ...) = (1-t)(0, x_1, x_2,....) + t(1, 0, 0, ...).$$ Now observe that $g_0 = f_1$, $f_0 = id_{S^{\infty}}$, and $g_1 = N$ (a constant). Again we can renormalize to guarantee everything is still on $S^{\infty}$, $i.e.$, $$\frac{g_t}{\left| g_t\right|} = G(x,t) : S^{\infty} \times I \rightarrow S^{\infty}.$$ Furthermore, we have that $\frac{g_0}{\left| g_0\right|} = \frac{f_1}{\left| f_1\right|}$. Therefore, it follows that $T \simeq N$ ($G$ gives a homotopy from
$T$ to the north pole) and since the composition of homotopies is again a homotopy we have that $id_{S^{\infty}} \simeq N$. Hence the desired result follows and we conclude that $S^{\infty}$ is contractible.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.