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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.

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  • $\begingroup$ 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

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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 $\Bbb R$.

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