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I am reading A. Zorich's article "Square Tiled Surfaces and Teichmuller Volumes of the Moduli Spaces of Abelian Differentials" and there a main object is the space $$H^1(X,\{P_1,\dotsc,P_n\};\mathbb{Z}\oplus i\mathbb{Z}),$$ where $X$ is a compact genus $g$ Riemann space, and $\{P_1,\dotsc,P_n\}\subset X$.

First of all, we should understand what the author means by $H^1(X,\{P_1,\dotsc,P_n\};\mathbb{C})$. Since $X$ is a Riemann space, I should always see $H^1$ as the De Rham's cohomology, right? So in this case $H^1(X,\{P_1,\dotsc,P_n\};\mathbb{C})$ would be the relative De Rham cohomology.

If that is right, then what would be $H^1(X,\{P_1,\dotsc,P_n\};\mathbb{Z}\oplus i\mathbb{Z})$? The quotient module?

William
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Gabriel
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    Maybe singular cohomology with coeficients in that abelian group? – Eduardo Longa Apr 03 '19 at 23:28
  • Why would be anything but a special case of $H^k(X,A; G)$ where $G$ is an abelian group, $X$ a topological space and $A$ is a subset of $X$? – Moishe Kohan Apr 03 '19 at 23:51
  • The article says that its elements are differential forms. That what made me thought that it should be some sort of De Rham's cohomology – Gabriel Apr 04 '19 at 06:33
  • Seems like it could be complex differential forms with values in $\mathbb{Z}\oplus i\mathbb{Z}$? I'm not sure if that makes sense. I also don't know if this helps, but the inclusion $\mathbb{Z}\oplus i\mathbb{Z} \to \mathbb{C}$ gives a long exact sequence $\dots \to H^0(-; T^2) \to H^1(-;\mathbb{Z}\oplus i\mathbb{Z}) \to H^1(-;\mathbb{C}) \to H^1(-;T^2) \to \dots$ since $T^2$ is the quotient of $\mathbb{C}$ by the integral lattice. – William Apr 04 '19 at 12:36
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    I think the map $H^1(-;\mathbb{Z}\oplus i\mathbb{Z}) \to H^1(-;\mathbb{C})$ is actually the inclusion of a lattice in the complex vector space $H^1(-;\mathbb{C})$. – William Apr 04 '19 at 12:42
  • I actually don't know a lot about the singular cohomology, so I would prefer to understant this object in terms of the De Rham's cohomology. I also think that we could understand it as a $\mathbb{Z}\oplus i\mathbb{Z}$-module. – Gabriel Apr 04 '19 at 14:20

1 Answers1

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If $X$ is a compact Riemann surface of genus $g$ and $A= \{P_1,\dots,P_n\}\subset X$ is a finite set of points then there is an embedding $H^1(X,A;\mathbb{Z}\oplus i\mathbb{Z})\hookrightarrow H^1_{dR}(X,A;\mathbb{C})$ whose image is those cohomology classes (in the de Rham sense) of complex-valued $1$-forms which vanish on each $P_i$ and whose integral over any embedded circle $C\subset X$ is in $\mathbb{Z}\oplus i\mathbb{Z}$. I give some justification below.

Edit: It's not enough to consider just circles $C\subset X$, we also need to consider integrals over curves from $P_i \rightsquigarrow P_j$ since these are relative $1$-cycles.

I managed to find an old question that seems to cover an analogous general question in the real case. The inclusion $\mathbb{Z}\hookrightarrow \mathbb{R}$ induces a (not necessarily injective) homomorphism $$H^k(X;\mathbb{Z}) \to H^k(X;\mathbb{R}) \cong H^k_{dR}(X;\mathbb{R})$$ whose image consists of the cohomology classes of differential $k$-forms $\omega$ such that $\int_Z \omega\in \mathbb{Z}$ for any smooth $k$-cycle $Z\in H_k(X;\mathbb{R})$. Sometimes, as in the case $k=1$ and $dim(X)>1$, cycles can actually be represented by smooth submanifolds. Similarly in the complex case the image of $H^k(X;\mathbb{Z}\oplus i \mathbb{Z})\to H^k_{dR}(X;\mathbb{C})$ is represented by forms whose integral over $k$-cycles in $H_k(X;\mathbb{C})$ is in $\mathbb{Z}\oplus i\mathbb{Z}$. For example, Chern-Weil theory constructs chern classes $c_k\in H^{2k}_{dR}(X;\mathbb{C})$, but integrals of chern classes are always integers so they end up having a natural incarnation in $H^{2k}(X;\mathbb{Z})$.

There's also the twist that we're dealing with a relative cohomology group $H^k_{dR}(X,A)$ for some closed submanifold of $A\subset X$. As per this mathoverflow question this cohomology theory is defined by considering $k$-forms on $X$ whose restriction to $A$ vanish.

So I believe the general answer to your question is that if $G\subset \mathbb{C}$ is a subgroup and $A\subset X$ is a closed submanifold, then the image of $H^k(X,A;G) \to H^k_{dR}(X,A;\mathbb{C})$ consists of differential $k$-forms $\omega$ on $X$ which vanish on $A$ and such that $\int_Z \omega \in G$ for any smooth relative $k$-cycle $Z\in H_k(X,A;G)$. If this homomorphism is injective then we can faithfully identify $H^k(X,A;G)$ with this subgroup of forms.

In your particular case $X$ is a compact genus $g$ Riemann surface, and $A = \{P_1,\dots,P_n\}$ is $n$ points. Given a short exact sequence of groups $G\to H\to Q$, for any pair of spaces $(X,A)$ there is a long exact sequence in singular cohomology that looks like

$$\dots \to H^k(X,A;G)\to H^k(X,A;H) \to H^k(X,A;Q) \to H^{k+1}(X,A;G)\to \dots $$

In degree $0$ this sequence is just $G\to H\to Q$, so exactness implies that $H^1(X,A;G) \to H^1(X,A;H)$ is injective. Therefore we have an embedding $H^1(X,A; \mathbb{Z}\oplus i\mathbb{Z}) \hookrightarrow H^1_{dR}(X,A;\mathbb{C})$, and so we get our interpretation.

William
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