0
$\begingroup$

If one define the universal abelian covering $M_0$ of a manifold $M$ as the abelian covering (i.e. normal covering with abelian group of deck transformations) that covers any other abelian covering, then what can one say about $H_1(M_0)$ ? Note that Hurewicz Theorem gives us a group isomorphism between $H_1(M_0)$ and the abelianization of $[\pi_1(M),\pi_1(M)]$.

In particular, I would like to understand why is the following integral independent of the choice of a $C^1$ curve $\tau$ in $M_0$ with fixed endpoints: $$ \int_\tau \overline{\omega}, $$ where $\overline{\omega}$ is the lift of a closed 1-form $\omega$ on $M$.

$\endgroup$
4
$\begingroup$

This has not much to do with $H_1(M_0)$. If $\pi :M_0\rightarrow M$ is your abelian covering, we have $\int_{\tau }\overline{\omega} =\int_{\pi _*\tau }\omega $. But the exact sequence $0\rightarrow \pi _1(M_{0})\rightarrow \pi _1(M)\rightarrow H_1(M)\rightarrow 0$ shows that $\pi _*\tau $ is zero in $H_1(M)$, hence the integral is zero.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.