# Why are homeomorphism groups important?

For a compact metric space $$X$$ let $$\mathcal H(X)$$ denote the set of homeomorphisms in the compact-open topology (also generated by sup metric). It is known that $$\mathcal H(X)$$ is a Polish topological group under the composition operation. Many of the classical results in this area concern the homeomorphism type of $$\mathcal H(X)$$, for instance:

(0) $$\mathcal H(2^\omega)\simeq \mathbb R \setminus \mathbb Q$$;

(1) $$\mathcal H_\partial([0,1])\simeq \ell^2$$ (here $$\mathcal H_\partial$$ is the space of homeomorphisms which fix the boundary pointwise);

(2) $$\mathcal H_\partial([0,1] ^2)\simeq \ell^2$$;

(3) $$\mathcal H_\partial([0,1]^n)$$ is a mystery for $$3\leq n<\omega$$;

(4) $$\mathcal H([0,1]^\omega)\simeq \ell^2$$;

(5) $$\mathcal H(\text{Sierpinski carpet})$$ and $$\mathcal H(\text{Menger curve})$$ have dimension $$1$$. It is conjectured that these homeomorphism groups and others are homeomorphic to the $$\omega$$-power of $$\{x\in \ell^2:x_i\notin \mathbb Q\text{ for all }i<\omega\}$$;

(6) $$\mathcal H(\text{Pseudo-arc})$$ contains no continuum. It is unknown whether this group is zero-dimensional, connected, or something in-between.

(7) It is unknown if there is a compact space $$X$$ with $$1<\dim(\mathcal H(X))<\infty$$.

Question 1. What are some other major results and questions along these lines?

Question 2. I suppose the dimension/connectedness of $$\mathcal H(X)$$ says something about generalized homotopy between homeomorphisms. But specifically why is knowing the homeomorphism type of $$\mathcal H(X)$$ important/useful? Also are there applications and/or interpretations of $$\mathcal H(X)$$ which lead to a greater appreciation?

• Your claim that the homeomorphism group of $\mathbf{R}^2$ is connected sounds suspicious. I'd bet it has at least 2 connected components. Also, most of your examples are with $X$ locally compact, not always compact. – YCor Jan 9 at 3:47
• @YCor I will double-check my sources. It may be that the spaces should be products of $[0,1]$ and not $\mathbb R$. – D.S. Lipham Jan 9 at 4:10
• What is $\ell^2$? – ARG Jan 9 at 9:32
• @YCor Indeed $\mathcal H(\Bbb R^2)$ is homotopy equivalent to $O(2)$, combining a result of Smale with some fiber-sequence fiddling. Perhaps OP meant the group of homeomorphisms of $[0,1]^2$ which fix the boundary pointwise; Smale's result is that this is contractible. – Mike Miller Jan 9 at 12:48
• My memory had briefly failed me: Smale's result is on the diffeomorphism group. That the homeomorphism group is contractible is older and is called the Alexander trick. – Mike Miller Jan 10 at 6:48

## 1 Answer

Of course the automorphism group of any mathematical object is interesting if the object itself is interesting. But that doesn't single out the homeomorphism group . Let me give some examples.

Let $$X$$ be a space and $$H=H(X)$$ the homeomorphism group.

The homeomorphism group is a space. Knowing its homotopy type is much weaker than knowing its homeomorphism type, but still can give interesting results. Let me give two examples.

Fiber bundles over the sphere $$S^n$$ with fiber $$X$$ are in 1 to 1 correspondence with elements of $$\pi_{n-1}(H)$$. To classify the fiber bundles of any reasonable space $$Y$$, you need to understand all homotopy classes $$[Y,BH]$$, where $$BH$$ is the classifying space of $$H$$. Sometimes it is possible to partially compute these sets. This allows you to construct new bundles, with interesting properties.

Here is another nice theorem of Reeb, which depends on the fact that the group $$H_\partial(D^n)$$ is connected.

Let $$f:M\rightarrow \mathbb{R}$$ be a smooth function on a smooth compact manifold $$M$$. Suppose that $$f$$ has only two critical points. Then $$M$$ is homeomorphic to a sphere.

This theorem is false if we want to construct a diffeomorphism. Indeed, the group of diffeomorphisms $$\mathrm{Diff}_{\partial}(D^n)$$ is not always connected. Milnor famously found the first examples of manifolds that are homeomorphic but not diffeomorphic.