Let $X$ be a compact connected manifold. Since $\mathbb T^1$ is an Eilenberg-MacLane space $K(\mathbb Z,1)$, it follows that for every morphism $\varphi\colon\pi_1(X)\to\pi_1(\mathbb T^1)$ there is a continuous map $f\colon X\to\mathbb T^1$ such that $\varphi$ coincides with the induced morphism $f_*$.

Now assume that $G$ is a compact connected Lie group and $\varphi\colon\pi_1(G)\to\pi_1(\mathbb T^1)$ is a morphism. Does there exist a morphism of Lie groups $h\colon G\to\mathbb T^1$ with $h_*=\varphi$? If not, are there any necessary or sufficient conditions for this to be the case? What about non-compact Lie groups $G$?

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    $\begingroup$ It's wrong for non-compact groups, $SL_2({\mathbb R})$ being a counter example. $\endgroup$ – user1688 Aug 31 '16 at 13:32
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    $\begingroup$ Since the circle is abelian, every morphism $G \to \mathbb T^1$ factors through the abelianization of $G$, so if $G$ is simple every such morphism is constant! $\endgroup$ – Jens Reinhold Aug 31 '16 at 14:12
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    $\begingroup$ But if $G$ is simple in addition to being compact and connected then $\pi_1(G)$ is finite, so there also aren't any morphisms $\pi_1(G) \to \mathbb{Z}$ either... $\endgroup$ – Qiaochu Yuan Aug 31 '16 at 18:19

There is no need to assume that $X$ is compact, connected, or a manifold in your first claim. Anyway, here is some context in which to put this question. A morphism $G \to S^1$ of Lie groups gives a map $BG \to BS^1 \cong B^2 \mathbb{Z}$ of classifying spaces, and hence a cohomology class in $H^2(BG, \mathbb{Z})$. If $G$ is connected, $BG$ is simply connected, and hence

$$\pi_1(G) \cong \pi_2(BG) \cong H_2(BG, \mathbb{Z}).$$

Now, by universal coefficients, we have an isomorphism

$$H^2(BG, \mathbb{Z}) \cong \text{Hom}(H_2(BG, \mathbb{Z}), \mathbb{Z})$$

from which it follows that every morphism $\pi_1(G) \to \mathbb{Z}$ corresponds to a map $BG \to B^2 \mathbb{Z}$ of classifying spaces, and hence to a map $G \to S^1$ of $\infty$-groups (meaning, loosely, a group homomorphism up to coherent homotopy). The remaining question is when we can strictify a map $G \to S^1$ of $\infty$-groups to a genuine Lie group homomorphism.

As Anton says in the comments, this is not possible if $G$ is noncompact, and $SL_2(\mathbb{R})$ is an explicit counterexample (the point being that the inclusion $SO(2) \to SL_2(\mathbb{R})$ of the maximal compact is a homotopy equivalence). It's less clear to me what happens when $G$ is compact. Certainly if $G$ is itself a torus it's always possible. On the other hand, if $G$ is semisimple then $\pi_1(G)$ is finite, so admits no nontrivial homomorphisms to $\mathbb{Z}$. So it might be possible to construct a more exotic counterexample by taking something like $SU(2) \times S^1$ mod $(-1, -1)$? I haven't thought much about these groups.


Finally I managed to work out a solution for compact Lie groups $G$.

By the structure theory of compact groups, $G$ can be expressed as a topological semi-direct product $G=\mathbb T^n\ltimes K$ of a torus $\mathbb T^n$ and a semi-simple Lie group $K$. Let $p\colon G\to\mathbb T^n$ be the projection morphism and $p_*\colon\pi_1(G)\to\pi_1(\mathbb T^n)$ be the induced morphism. It is known that $\pi_1(K)$ is a finite group. Also, there is the usual identification $\pi_1(G)=\pi_1(\mathbb T^n)\times\pi_1(K)$.

Now let $\varphi\colon \pi_1(G)\to\pi_1(\mathbb T^1)$ be a morphism. Since $\pi_1(\mathbb T^1)$ is torsion-free and $\pi_1(K)$ is a torsion group, $\varphi$ annihilates $\pi_1(K)$ and so $\varphi=\psi\circ p_*$ for some $\psi\in\text{Hom}(\pi_1(\mathbb T^n),\pi_1(\mathbb T^1))$. Clearly, we have $\psi=k_*$ for some $k\in\text{Hom}(\mathbb T^n,\mathbb T^1)$. Thus, $\varphi=k_*\circ p_*=(k\circ p)_*$ and so $k\circ p\in\text{Hom}(G,\mathbb T^1)$ is the desired morphism.


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