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I'm trying to prove that all closed (holomorphic) $1$-forms on a riemann surface are exact by using zorn's lemma. That is I consider the set of $T$ of $(U,f)$ where $f$ is a $0$-form on $U$ such that $df=w$. Now I consider the relation $\leq$ on $T$ where $(U,f)\leq(V,g)$ if $U\subseteq V$ and the restriction of $g$ to $U$ is $f$. I now need the following topological theorem (which I hope is true) to expand these forms from the maximal element.

The theorem:

If $M$ is simply connected (riemann surface), $U_\alpha$ (the charts on $M$) an open cover for $M$, and $U\subset M$ an open connected set not equal to $M$. Then there exists an open set $V$ such that $V\not\subseteq U$, $U\cap V$ is connected (and non-empty) and $V\subseteq U_\alpha$ for some $\alpha$.

What I'm then able to do is expand $f$ to $U\cup V$ which is a larger set than $U$.

My intuition tells me that this should be true. If you consider for example the unit disk $D$ in $\mathbb{R}^2$ and $D\setminus\{(0,0)\}$, and $U$ is the set $D\setminus[-1,0]$, then you can add such a $V$ in the first case, but not in the second case.

If it is needed for the proof then we can of course assume that $M$ has all the nice properties of manifolds, for example that path-connectedness is equivalent to connectedness etc.

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  • $\begingroup$ sounds like you want something like the Lebesgue covering Lemma $\endgroup$
    – Dan Rust
    Dec 3, 2021 at 12:05
  • $\begingroup$ Are you requiring also that $V$ is connected? $\endgroup$ Dec 5, 2021 at 17:58
  • $\begingroup$ @MoisheKohan No I do not require that. $\endgroup$ Dec 5, 2021 at 18:25

1 Answer 1

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  1. If $cl(U)\ne X$ then such $V$ exists. Indeed, consider the open subset $W:= X\setminus cl(U)$; by our assumption, $W\ne \emptyset$. Since $X$ is connected, $cl(W)\cap cl(U)\ne \emptyset$, hence, choose $x\in cl(W)\cap cl(U)$. There exists a chart $U_\alpha$ containing $x$ and $U_\alpha\cap U\ne \emptyset$ and $U_\alpha\cap W\ne \emptyset$. Take an open set $V$ which is the disjoint union of a small open disk $V_1$ contained in $U$ and a small open disk $V_1$ contained in $W$. This subset $V$ is, of course, disconnected, but it satisfies all your requirements.

At the same time, I do not see how such subsets can be of any use for your purposes.

  1. In general, I think, the theorem you are after is simply false. A potential example is a pseudo-arc $A\subset S^2$. One can show that $A$ is connected and satisfies $\check{H}_1(A)=0$. It follows that $U=S^2\setminus A$ is simply-connected. I think, when the open subsets $U_\alpha$ are sufficiently small disks in $S^2$, the subset $V$ you are asking for does not exist. However, verifying this would be a serious task and I will not do this.

Moreover, proof of the result about holomorphic 1-forms that I know is much simpler than what you are trying to do and does not require Zorn's lemma. One argument uses sheaf cohomology. Another argument (lower tech) goes as follows. Firstly, you can assume that the cover $\{U_\alpha: \alpha\in J\}$ is a good cover of $X$, i.e. it is locally finite and nonempty intersections of its members are simply-connected. Next, for each $\alpha$ you find a holomorphic function $f_\alpha$ on $U_\alpha$ satisfying $df_\alpha=\omega$. Patching these functions together results in a multivalued holomorphic function $F$ on $X$. (You pick some $\alpha_0$ and then $F: U_\alpha\to {\mathbb C}$ inductively so that $F|U_\alpha$ differs from $f_\alpha$ by a constant. Since $X$ is simply-connected, you will find a branch $f$ of $F$. This will be your function.

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  • $\begingroup$ I will upvote your answer but not accept it. I have found an example of a riemann surface where we cannot find such a set. It is also the case that $cl(U)=X$. $\endgroup$ Dec 7, 2021 at 23:17
  • $\begingroup$ I just realized you also mention this. I will accept your answer. $\endgroup$ Dec 7, 2021 at 23:18

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