# About the cohomology of $BG^\delta$. Making a Lie group discrete

Let $$G$$ be a connected Lie group. Recall that the topological group $$G^\delta$$ is $$G$$ endowed with the discrete topology. The inclusion $$G^\delta \to G$$ induces a map between the classifying spaces $$\eta: BG^\delta\to BG$$.

Question 1

Let $$\eta^*:H^*(BG,\mathbb{Z})\to H^*(BG^\delta,\mathbb{Z})$$ be the induced map in integral cohomology. By Corollary 1 in Milnor, On the homology of Lie groups made discrete, we get that $$\eta^*$$ is injective.

On the other hand, by Lemma 10 in the same paper, we learn that the kernel of $$\eta_{\mathbb{Q}}^*:H^*(BG,\mathbb{Q})\to H^*(BG^\delta,\mathbb{Q})$$ (notice the rational coefficients here) is equal to the kernel of $$\eta^*_{\mathbb{Q}}:H^*(BG,\mathbb{Q})\to H^*(B\Gamma,\mathbb{Q})$$, where $$\Gamma is a discrete cocompact group.

Consider $$G=U(n)$$. Then $$H^*(BG,\mathbb{Z})= \mathbb{Z}[c_1,c_2,\dots,c_n]$$ injects in $$H^*(BG^\delta, \mathbb{Z})$$, in particular $$\eta^*(c_1)\neq 0$$. However, we can take $$\Gamma= \{\mathbb{1}\}$$, which implies that $$H^*(B\Gamma,\mathbb{Q})=H^*(\mathbb{R},\mathbb{Q})$$, hence $$\eta^*_{\mathbb{Q}}$$ is trivial and in particular $$\eta^*_{\mathbb{Q}}(c_1) = 0$$.

Why doesn't this give a contradiction?

I will only attempt to answer your first question. The reason there is no contradiction is that it is not true for arbitrary spaces that $$H^{\ast}(X;\mathbb Q) = H^{\ast}(X;\mathbb Z) \otimes \mathbb Q$$.
For instance, take $$X = B\mathbb Q$$. Then $$H^2(X;\mathbb Z) = \text{Ext}(\mathbb Q,\mathbb Z)$$ is a $$\mathbb Q$$ vector space of uncountable dimension. In particular, we can find a map $$B\mathbb Q \to K(\mathbb Z,2)$$ that is injective on second cohomology with $$\mathbb Z$$ coefficients. But clearly the map induced on rational cohomology is zero.
• Probably it's obvious but why the map on rational cohomology is zero? I tried to compute $H^2(B\mathbb{Q},\mathbb{Q}) = H^2(\mathbb{Q},\mathbb{Q}) = Ext^2_{\mathbb{Z}[\mathbb{Q}]}(\mathbb{Z},\mathbb{Q})$ but I can't see why it should be trivial. – Warlock of Firetop Mountain Mar 27 at 18:04
• I would compute it differently, $H^2(B\mathbb Q;\mathbb Q) = \text{Hom}(H_2(B\mathbb Q;\mathbb Q),\mathbb Q)$, and $H_2(B\mathbb Q;\mathbb Q) = \text{Tor}_1(\mathbb Q,\mathbb Q) = 0$ since $\mathbb Q$ is flat. – Jens Reinhold Mar 27 at 19:17