nLab infinite-dimensional sphere

Idea

General

There are several ways to make sense of the notion of the n-sphere in the limit (rather: colimit) that $n \to \infty$. Typically the resulting infinite-dimensional sphere $S^\infty$ has the remarkable property that it is contractible as a topological space – or at least weakly contractible, in that its map to the point is a weak homotopy equivalence.

In homotopy theory

Due to their weak homotopy equivalence to points, in homotopy theory infinite-dimensional spheres provide nothing new in themselves, but as a source of big contractible spaces they serve as a starting point for making concrete models of classifying spaces.

For example, the 2k-sphere is the total space of the tautological circle group-principal bundle over complex projective k-space, so that in the colimit $2k \to \infty$ the infinite-dimensional sphere emerges as a model for the universal principal bundle over the classifying space $B \mathrm{U}(1)$. This being contractible relates to important statements such as that the zero-section into the Thom space of the universal line bundle is a weak equivalence.

Realizations

An an infinite spherical cell complex

Realizing the n-sphere as a cell complex of sorts, such that $S^n \hookrightarrow S^{n+1}$ is a relative cell complex-inclusion for all $n \in \mathbb{N}$, the infinite-dimensional sphere may be taken to be the colimit over this sequence of inclusions. Since the n+1-sphere is n-connected it follows that $S^\infty$ is $\infty$-connected and hence weakly contractible:

Specifically in the category Top of topological spaces, the n-sphere has a standard CW-complex structure with exactly 2-cells in each dimension, obtained inductively by attaching two $n$-dimensional hemispheres to the $(n-1)$-sphere regarded as the equator in the $n$-sphere.

Since forming homotopy groups $\pi_k(-)$ commutes with taking the colimit over these relative cell complex inclusions (by the skeleton filtration), and since the n-sphere has trivial homotopy groups in dimension $k \lt n$, it follows at once that the homotopy groups of the infinite-dimensional sphere all vanish, and hence that it is weakly equivalent to the point in the classical model structure on topological spaces:

$$\begin{aligned} \pi_k \big( S^\infty \big) & =\; \pi_k \big( \underset{ \underset{n \in \mathbb{N}}{\longrightarrow} }{\lim} \; S^n \big) \\ & = \; \underset{ \underset{n \in \mathbb{N}}{\longrightarrow} }{\lim} \; \pi_k \big( S^n \big) \\ & = \; \underset{ \underset{n \in \mathbb{N}}{\longrightarrow} }{\lim} \; \underset{ = \, \ast }{ \underbrace{ \pi_k \big( S^{n+k+1} \big) } } \\ & \simeq \ast \end{aligned}$$

As the unit spheres in a topological vector space

One can also talk about a sphere in an arbitrary (possibly infinite-dimensional) normed vector space $V$:

$$S(V) = \{ x: V \;\text{such that}\; {\|x\|} = 1 \} .$$

If a locally convex topological vector space admits a continuous linear injection into a normed vector space, this can be used to define its sphere. If not, one can still define the sphere as a quotient of the space of non-zero vectors under the scalar action of $(0,\infty)$.

Homotopy theorists define $S^\infty$ to be the sphere in the (incomplete) normed vector space (traditionally with the $l^2$ norm) of infinite sequences almost all of whose values are $0$, which is the directed colimit of the $S^n$:

$$S^{-1} \hookrightarrow S^0 \hookrightarrow S^1 \hookrightarrow S^2 \hookrightarrow \cdots S^\infty \,.$$

If the vector space is a shift space, then contractibility is straightforward to prove.

Theorem

Let $V$ be a shift space of some order. Let $S V$ be its sphere (either via a norm or as the quotient of non-zero vectors). Then $S V$ is contractible.

Proof

Let $T \colon V \to V$ be a shift map. The idea is to homotop the sphere onto the image of $T$, and then down to a point.

It is simplest to start with the non-zero vectors, $V \setminus \{0\}$. As $T$ is injective, it restricts to a map from this space to itself which commutes with the scalar action of $(0,\infty)$. Define a homotopy $H \colon [0,1] \times (V \setminus \{0\}) \to V \setminus \{0\}$ by $H_t(v) = (1 - t)v + t T v$. It is clear that, assuming it is well-defined, it is a homotopy from the identity to $T$. To see that it is well-defined, we need to show that $H_t(v)$ is never zero. The only place where it could be zero would be on an eigenvector of $T$, but as $T$ is a shift map then it has none.

As $T$ is a shift map, it is not surjective and so we can pick some $v_0$ not in its image. Then we define a homotopy $G \colon [0,1] \times (V \setminus \{0\}) \to V \setminus \{0\}$ by $G_t(v) = (1 - t)T v + t v_0$. As $v_0$ is not in the image of $T$, this is well-defined on $V \setminus \{0\}$. Combining these two homotopies results in the desired contraction of $V \setminus \{0\}$.

If $V$ admits a suitable function defining a spherical subset (such as a norm) then we can modify the above to a contraction of the spherical subset simply by dividing out by this function. If not, as the homotopies above all commute with the scalar action of $(0,\infty)$, they descend to the definition of the sphere as the quotient of $V \setminus \{0\}$.

Related concepts

• Kuiper's theorem

• n-connected object