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.

  • $\begingroup$ 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 $S^1$)? $\endgroup$ Feb 4, 2020 at 5:04
  • $\begingroup$ I hadn't, but doesn't that act freely and properly on $\mathbb{R}$? $\endgroup$ Feb 4, 2020 at 5:41
  • $\begingroup$ I don't think that action is continuous. $\endgroup$ Feb 4, 2020 at 13:53
  • $\begingroup$ 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 $S^1$. $\endgroup$ Feb 4, 2020 at 13:53

1 Answer 1


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 $\Bbb R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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