# Shifting the group homology of a topological group?

Let $$G$$ be a topological group. It has a classifying space $$BG$$, which has homology groups $$H_{*}BG$$. Changing the topology of $$G$$ affects the space $$BG$$ and hence its homology groups.

For example the group $$\mathbb{R}$$ with its usual topology has $$H_{*}B\mathbb{R}\simeq H_{*}pt$$. Changing the topology to be much finer, namely the discrete topology, results in a topological group $$\mathbb{R}^{\delta}$$ that has nontrivial $$H_{1}$$.

Question: Is it possible to go the other way? I.e. does there exist a topological group $$G$$, which has a second topology, coarser than the first, that makes the group homology larger?

There are many ways to make "larger" precise--I'm most interested in the case where initial topological group is acyclic. Then the meaning of "larger" is clear.

Motivation: It would help answer this question.

• Have you tried the translation-invariant topology on $\Bbb R$ (a set is open iff it's the inverse image of an open set of $S^1$)? – Tyler Lawson Feb 4 '20 at 5:04
• I hadn't, but doesn't that act freely and properly on $\mathbb{R}$? – John Greenwood Feb 4 '20 at 5:41
• I don't think that action is continuous. – Tyler Lawson Feb 4 '20 at 13:53
• I was hoping that because this topology gives the subgroup Z of integers the indiscrete topology (making them contractible), the group homology would be closer to that of $S^1$. – Tyler Lawson Feb 4 '20 at 13:53

## 1 Answer

Let $$p$$ be the projection map $$\Bbb R \to S^1$$. Let $$T$$ be $$\Bbb R$$, where we say that a subset $$U$$ is open if and only if $$U = p^{-1} V$$ for some open set $$V$$ of $$S^1$$. This makes $$T$$ a topological group, and the maps $$\Bbb R \to T \to S^1$$ are homomorphisms of topological groups. The space $$T$$ also has the property that a map $$f: X \to T$$ is continuous if and only if the composite $$p \circ f$$ is continuous.

The homotopy type of the space $$T$$ can also be understood: the projection $$p: T \to S^1$$ is a homotopy equivalence. To show this, we define $$i(x)$$ to be the unique element of $$[0,1)$$ such that $$p(i(x)) = x$$; then $$p \circ i = id$$ by definition, so $$i$$ is continuous. The map $$i \circ p$$ is also homotopic to the identity: we need to construct a homotopy $$H: T \times [0,1] \to T$$ which is continuous with various properties, but to verify continuity it suffices for $$p \circ H$$ to be continuous, and so defining $$H(t,s) = \begin{cases} t &\text{if }s = 0,\\i(p(t)) &\text{if }s > 0\end{cases}$$ works perfectly well.

Therefore, $$T$$ is a coarser topology on $$\Bbb R$$ with the same homotopy and homology groups as $$S^1$$, and (for definitions of $$B$$ in terms of principal bundles with contractible total space) $$BT$$ has the same homotopy and homology groups as $$BS^1$$, larger than the group homology of $$\Bbb R$$.