# Construct Torsion element in $H^2(X,\mathbb{Z})$ with Ambrose–Singer theorem

Let $$X$$ be a Kahler manifold. Using the exponential sequence one obtains a homomorphism $$c_1:H^1(X,\mathcal{O}_X^*)\rightarrow H^2(X,\mathbb{Z})$$. This is associating to a holomorphic line bundle $$L$$ its Chern class $$c_1(L)$$.

Take any connection $$D$$ on a holomorphic line bundle $$L$$, if we do $$D_L\circ D_L = \Omega_L$$ we obtain a curvature and then take its class in $$H^2(X,\mathbb{C})$$ we get a curvature form independent of the choice of the connection. After multiplying by $$-i/(2\pi)$$ we run into $$c_1(L)$$ when identifying $$H^2(X,\mathbb{Z})$$ as a subspace of $$H^2(X,\mathbb{C})$$, if we know that $$\phi: H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$$ is injective. This is not always true but if $$H^2(X,\mathbb{Z})$$ has no torsion using universal coefficient thm we see $$H^2(X,\Bbb C) \cong H^2(X,\Bbb Z)\otimes_{\Bbb Z}\Bbb C$$ and thus $$\phi$$ is in that case injective.

Looking for examples where $$\phi$$ is not injective, so where $$H^2(X,\mathbb{Z})$$ has torsion I encountered the class of flat bundles, which provide by definition connections with zero curvature.

What I still not understand is how to see explicitely that these flat bundles provide a curvature what induces in $$H^2(X,\mathbb{Z})$$ a torsion class. Technically, one can simply say that because the class of cuvatures of flat bundles are zero in $$H^2(X,\mathbb{C})$$, it must be torsion.

The point is that I wan't to have a better intuitive inside look what really is going on in the torsion. That is I aim to consider a representative of the curvature of a flat line bundle "alive" in $$H^2(X,\mathbb{Z})$$ instead simply kill it by tensoring with $$C$$ and say "thus it was torsion".

The promissing tool for this seems to lie in the Ambrose–Singer theorem which allows to "measure" the infinitesimal Holonomy by curvature. Especially that allows to see that for prinzipal $$G$$-bundles with flat curvature the infinitesimal holonomy has to be trivial (not globally in general) . That is homotopical paths induce same holonomy action and we obtain well defined representation

$$\pi _{1}M\rightarrow G$$

More precisely a consequence of Ambrose–Singer is that we obtain a bijection between flat $$G$$-Bundles ($$G$$ certain group) over connected manifolds and representations $$\rho : \pi_1(M) \to G$$

More concretly we associate to a given representation a flat bundle using monodromy action of $$\pi_1(M)$$ on the universal cover $$\tilde{M}$$ of $$M$$ and set

$$E_{\rho }:=\widetilde {M}\times G/\sim$$

with equivalence relation $$(\gamma x,g)\sim (x,\rho (\gamma )g)$$ for $${\displaystyle \gamma \in \pi _{1}M,x\in {\widetilde {M}},g\in G}$$.

If $$\pi(M)$$ is for exaple finite that it looks promising since this gives a non trivial action of a finite group on the line bundle.

Problem/Question: can this action be "transfered" in certain way to the curvature induced by concerned principal line line bundle or it's class in $$H^2(X,\mathbb{Z})$$. My goal is if I find out that this action is still non trivial and $$\pi_1(M)$$ is finite group, then the class of the curvature is torsion and we have constructed a desired example of a torsion element explicitly. What I still not understand is how to expand the action of fundamnetal group via $$\rho$$ to the curvature form of $$E_{\rho}$$ and why it stays non trivial?

• I do not understand most of your question (for a flat bundle the action of the fundamental group on the curvature form is zero, because the curvature form is zero...) but you can see how torsion classes arise explicitly. One obtains the isomorphism $H^1(X;U(1)) \cong \text{Hom}(\pi_1 X, U(1))$ in the usual fashion. This splits as a sum of $U(1)^{b_1(X)}$ and $\text{Hom}(H_1(X;\Bbb Z)_{\text{tors}}, U(1))$ (the free part and the torsion part). The free part is what dies in the boundary map $H^1(X;U(1)) \to H^2(X;\Bbb Z)$ from the exponential exact sequence. – Mike Miller Jan 5 at 23:51
• So the desired classes in $H^2$ are just those coming from the boundary map in the exact sequence associated to $\Bbb Z \to \Bbb R \to U(1)$, and those that survive come from homomorphisms $\pi_1(X) \to U(1)$ with finite image. – Mike Miller Jan 5 at 23:52
• @MikeMiller: two questions: how do you obtain the isomorphism $H^1(X;U(1)) \cong \text{Hom}(\pi_1 X, U(1))$? And secondly: this concerns the case if a $G$ bundle $P$ is not flat, ie. the the curvature $\Omega_P$ not vanish, is it possible to extend the action of $G$ to the curvature in canonical way? – user7391733 Jan 6 at 0:39
• That isomorphism holds for any abelian group in the place of $U(1)$ and follows from the universal coefficient theorem. I thought you were talking about the action of $\pi_1$ earlier. There is certainly an action of $G$ on the bundle $\Lambda^2 T^* M \otimes T\text{Aut}(G)$, which is one way to describe where curvature lives. This gives you an action of $G$ on the curvature form at each point, but not globally. I still don't know why you would want this anyway. – Mike Miller Jan 6 at 0:55
• the background arosed in this discussion math.stackexchange.com/questions/2073846/…. Primary I wanted to understand the torsion part in $H^2(X,\mathbb{Z})$ better. Probably I misunderstood a hint from that thread in order to approach them by flat line bundles. – user7391733 Jan 6 at 1:11