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Let $G=(\mathbb C^\times)^n$ the algebraic torus, I am trying to understand the isomorphism between its character group and the second cohomology of its classifying space $M(G)\cong H^2(BG,\mathbb Z)$. The construction is as follows, given a character $\rho\in M(G)$, we have an associated 1 dimensional representation $\mathbb C_\rho$, when taken as $G$-space, we can form the space $$(\mathbb C_\rho)_G=\mathbb C_\rho\times_G E_G$$ Which is a line bundle over $BG$, the the negative of its first Chern class defines a map $M(G)\to H^2(BG,\mathbb Z)$. I can show that this is indeed an isomorphism by explicitly writing down a set of generators.

I realized there is another construction. An element $\rho\in M(G)=\operatorname{Hom}((\mathbb C^\times)^n,\mathbb C^\times)$ induces a map $BG\to B\mathbb C^\times$, whose homotopy class can be regarded as element of $[BG,B\mathbb C^\times]=H^2(BG,\mathbb Z)$. It seems natural that the two maps $M(G)\to H^2(BG,\mathbb Z)$ should be the same. Is it true? If it is then how to prove it? I cannot even show that the second construction is an isomorphism, in particular I don't know how to explicitly describe the induced map between the classifying spaces. Thanks for your help.

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  • $\begingroup$ And perhaps a follow-up question: does the same result holds if we replace $G$ by a topological abelian group? I would like to read a bit more on classifying spaces as well so I'd appreciate if someone can recommend some references to look into. Thanks! $\endgroup$
    – lEm
    Jul 21, 2021 at 13:54

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Yes. Upto sign issues, this is the same map.

The cohomology ring of $BU(1)=\mathbf{CP}^{\infty}$ is given by $$H^*(\mathbf{CP}^\infty,\mathbf{Z})=\mathbf{Z}[c_1]$$ where $c_1$ is a 'universal Chern class' in cohomological degree $2$. It is the Chern class of the universal line bundle $\gamma$ on $\mathbf{CP}^{\infty}$ ( this is the colimit of the usual tautological line bundles $\gamma_n$ on $\mathbf{CP}^n$).

Now for any space $M$, the data of a line bundle $\mathcal{L}\in \operatorname{Pic}(M)$, is equivalent (essentially by the definition of $BU(1)$) upto homotopy to the data of a map $f:M\to \mathbf{CP}^{\infty}$. The line bundle $\mathcal{L}$ is isomorphic to the pullback of $\gamma$ under $f:M\to \mathbf{CP}^{\infty}$.

This map induces a pullback $$f^*:H^2(\mathbf{CP}^{\infty},\mathbf{Z})\to H^2(M,\mathbf{Z})$$ and the pullback of the universal Chern class $f^*c_1$ is equal to the Chern class $c_1(\mathcal{L})$ of $\mathcal{L}$ which also lives in $H^2(M,\mathbf{Z})$.

Thus for a topological group, a character $\rho:G\to \mathbf{C}^\times\simeq U(1)$ induces a pullback $$\rho^*:H^2(\mathbf{CP}^\infty,\mathbf{Z})\to H^2(BG,\mathbf{Z})$$ and the image is the Chern class of the line bundle associated to $\rho$. You should be able to show that the pullback of the tautological bundle $\rho^*\gamma$ and $\mathbf{C}_{\rho}\times_G EG$ are in fact isomorphic.

The question whether $M(G)\to H^2(BG,\mathbf{Z})$ is always an isomorphism should also be true, but I am not sure.

In the case that $G$ is finite, then $$H^2(BG,\mathbf{Z})\cong H^2(G,\mathbf{Z})\cong H^1(G,\mathbf{C}^\times)=MG$$ where the first identification follows from identifying the singular cohomology of $BG$ with $\mathbf{Z}$ coefficients with the group cohomology of $G$ with trivial action on $\mathbf{Z}$, and the second arises from the connecting map of the sheaf sequence

$$0\to \mathbf{Z}\to \mathbf{C}\xrightarrow{\operatorname{exp}2\pi i}\mathbf{C}^\times\to 0$$ on $BG$ and once again identifying group cohomology of $G$ with singular cohomology of $BG$ and the last isomorphism is standard.

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  • $\begingroup$ Thanks for your answer. I understand more now. I have some follow-up questions if you don't mind. I would like to know how does $G\to\mathbb C^\times$ induces the map between classifying spaces? Because I am not sure how would $\rho^*$ work. Also may I ask for a reference where I can read up more related stuff (classifying spaces, group cohomology, etc). Thanks a lot! $\endgroup$
    – lEm
    Jul 22, 2021 at 1:36
  • $\begingroup$ Also another question I have, it seems like in the topological settings $\operatorname{Hom}(\mathbb C^\times,\mathbb C^\times)=\mathbb R\times\mathbb Z$ because we have maps of the form $z\mapsto |z|^t z^n$. But in the isomorphism $M(\mathbb C^\times)\cong H^2(\mathbb C\mathbb P^\infty, \mathbb Z)=\mathbb Z$, it seems like we are taking $M(G)$ in the algebraic category or lie group category? I am unsure of what is happening here since we seem to be taking $\mathbb C^\times$ as in different settings on both sides of the isomorphism. $\endgroup$
    – lEm
    Jul 22, 2021 at 7:33
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    $\begingroup$ For your first question. There is a construction known as the Milnor join construction taken as $EG=\varinjlim *_n G$ i.e. the colimit over the $\mathbf{N}$-indexed topological joins of $G$. Since the join construction is functorial and so is taking colimits, this gives you a morphism $EG\to EH$ for any morphism of topological groups $G\to H$. Note $EG$ is contractible. $\endgroup$
    – S.S.
    Jul 22, 2021 at 9:04
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    $\begingroup$ The group $G$ acts on the right of $EG$ by multiplication and so quotienting by $G$ gives you $BG$ and so a morphism $BG\to BH$ for a morphism $f:G\to H$. Note that this construction also works for finite groups with the usual discrete topology. $\endgroup$
    – S.S.
    Jul 22, 2021 at 9:06
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    $\begingroup$ For your second question, it depends on what category your representations $\rho: (\mathbf{C}^\times)^n\to \mathbf{C}^\times$ live in. Usually, in this setting, you will have holomorphic or algebraic representations. What I wrote about above works in the topological category, so in both these settings because holomorphic and algebraic functions are continuous in the Euclidean topology. $\endgroup$
    – S.S.
    Jul 22, 2021 at 9:14

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