Timeline for multivalued holomorphic function on Riemann surfaces
Current License: CC BY-SA 4.0
18 events
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| Feb 7, 2020 at 16:49 | comment | added | Dmitri Panov | The point is that the flow is defined outside of measure $0$ subset $E$ in both directions of time. So if you through away this subset $E$ of measure $0$ the flow is well defined for all time. But then you will have an invertible map $M\setminus E\to M\setminus E$ that contracts area. Such thing can not exist. | |
| Feb 7, 2020 at 15:34 | comment | added | Dmitri Panov | Yes, indeed, this is because the volume of $M$ is finite. By the assumptions, $M$ is a compact surface with punctures. At each puncture the hyperbolic metric has a cusp. Assume that the number of cusps is $n$. Moreover, the surface has a finite number of conical points of angle $2\pi n_j$. Then Gauss-Bonnet theorem tells us that the area of $M$ is $-2\pi\chi (M)+2\pi n+\sum_j(1-n_j)$, where the sum is over all points with conical angles $2\pi n_j$. This is bounded by $-2\pi\chi (M)+2\pi n$. | |
| Feb 7, 2020 at 9:39 | comment | added | Yu Feng | I have understood the sentence “The field is contracting the area form". But I don't understand how to get the contradiction. Is it because the above sentence contradicts “the volume of $M$ is finite"? Could you please explain why? Thank you. | |
| Feb 3, 2020 at 9:03 | history | edited | Dmitri Panov | CC BY-SA 4.0 |
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| Feb 3, 2020 at 9:02 | comment | added | Dmitri Panov | Thanks for the comments. I should have said the following: "The flow of the field on $\mathbb H$ is contracting the area form". I have updated the answer the make it a bit more clear. | |
| Feb 3, 2020 at 1:57 | comment | added | Yu Feng | [1] R. C. Gunning, \textit{Special coordinate coverings of Riemann surfaces.} [2] R. Mandelbaum, \textit{Branched structures on Riemann surfaces.} | |
| Feb 3, 2020 at 1:57 | comment | added | Yu Feng | We can construct an affine connection (which can be found in [1] or [2]) on $\Sigma$, then we get a contradiction by a property of affine connection on a compact Riemann surface ([2] Lemma 2). We proved that $h:=\{h_{\alpha}:=\frac{f''}{f'}:U_{\alpha}\rightarrow\overline{\mathbb{C}}\}$ is an affine connection on $\Sigma$, where $\{U_{\alpha},z_{\alpha}\}$ is a complex atlas on $\Sigma$. If $\Sigma$ is $\mathbb{C}$, $\mathbb{C}-\{0\}$ or the punctured tori $\mathbb{T}\setminus\{q\}$, we also can obtain the conclusion. Moreover, we proved special cases for subgroup $B$ or $C$. | |
| Feb 3, 2020 at 1:49 | comment | added | Yu Feng | Thanks a lot. What do you mean by "this field is contracting the metric"? Let $\Sigma$ be a Riemann surface and $\{p_{i}\}_{i=1}^\infty\subset\Sigma$ is a closed discrete subset. $\mathrm{d} s^2$ is a (singular) hyperbolic metric with cusp or conical singularities at $p_{i}$. $f:\Sigma-\{p_{i}\}_{i=1}^\infty\longrightarrow\mathbb{H}$ is a developing map of $\mathrm{d} s^2$ and the monodromy of $f$ lies in $A$. So your answer proves $\Sigma$ is not compact. We proved it in another way, then we conjectured that $\Sigma$ must be a hyperbolic surface. The following are partial results. | |
| Jan 31, 2020 at 16:05 | history | edited | Dmitri Panov | CC BY-SA 4.0 |
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| Jan 30, 2020 at 21:15 | comment | added | Yu Feng | I already understand what you mean. Sorry for the misunderstanding. I think I should edit my question again to make it more clear and complete. | |
| Jan 30, 2020 at 12:58 | comment | added | Dmitri Panov | Ok, I finally got it. You mean that the monodromy is a subgroup of this $3$-dimensional solvable group. I thought that the monodromi is generated by one matrix from this group, and then the action is of course discrete. Of course with this new understanding obviously my post is not correct. I'll update it soon | |
| Jan 30, 2020 at 5:13 | comment | added | Yu Feng | If $\Gamma\subset {PSL}(2,\mathbb{R})$ is a discrete subgroup acting freely on $\mathbb{H}^{2}$, then $\mathbb{H}^{2}/\Gamma$ is a Riemann surface. However, the monodromy group $G$ is not necessarily discrete. $A$ and $B$ are not discrete subgroup. So $\mathbb{H}^{2}/A$ is not a Riemann surface. | |
| Jan 29, 2020 at 23:07 | comment | added | Dmitri Panov | So you are objecting the following statement: Suppose that $f$ is a multivalued function on $M$ with values in $H^2$ and with monodromy group $G$, then $f$ defines to you a uni-valued function on $M$ with values in $H^2/G$. Could you please explain why this is wrong? | |
| Jan 29, 2020 at 17:15 | comment | added | Yu Feng | In other words, the monodromy of $f$ is the image of its monodromy representation. So at each point $x$ of $M$, there is a one-to-one correspondence between the values of $f$ at $x$ and its monodromy group. Actually I don't quite understand your proof. I think the function $\widetilde{f}$ you construct is not necessarily single-valued. | |
| Jan 29, 2020 at 17:15 | comment | added | Yu Feng | Let $M$ be a Riemann surface and $x_{0}$ a base point on $M$. Suppose that the monodromy of $f$ lies in the group $A=\left\{\begin{pmatrix} a & b \\ 0 & \frac{1}{a}\end{pmatrix}:a>0, b\in\mathbb{R}\right\},$ i.e. its monodromy representation is a group homomorphism $\mathcal{M}_{f}: \pi_{1}(M, x_{0})\longrightarrow A.$ For each single-valued analytic branch $\mathfrak{f}$ of $f$, $\mathcal{M}_{\gamma}(\mathfrak{f})=\frac{a_{\gamma}\mathfrak{f}+b_{\gamma}}{a_{\gamma}^{-1}}$, where $\begin{pmatrix} a_{\gamma} & b_{\gamma} \\ 0 & a_{\gamma}^{-1}\end{pmatrix}\in A$, $\gamma\in \pi_{1}(M, x_{0})$. | |
| Jan 28, 2020 at 8:30 | comment | added | Dmitri Panov | Maybe I have misunderstood what you mean by monodromy. But I think that from what you write in your question it follows that at each point $x$ of $M$ the values of $f$ lies in one orbit of the action of $\mathbb Z$ on $\mathbb H$. Correct? What I say is that the space of such orbits is the quotient $\mathbb H/\mathbb Z$. In this quotient space to each point $x\in M$ corresponds only one point - the corresponding orbit. So the map $\tilde f$ is the composition of $f$ with the quotient map $\mathbb H \to \mathbb D^*$. Does this make sense? | |
| Jan 28, 2020 at 8:10 | comment | added | Yu Feng | Thanks for your comment. I don't understand why $\widetilde{f}$ is single-valued? For the case $a=1, b\neq0$, the quotient group is the additive integer group. If I consider the universal covering from $\mathbb{H}$ to $\mathbb{D}^{*}$, $z\mapsto e^{iz}$, where $\mathbb{D}^{*}$ is the punctured disk, how can we get $\widetilde{f}$? | |
| Jan 25, 2020 at 18:03 | history | answered | Dmitri Panov | CC BY-SA 4.0 |