Timeline for k-differentials and their residues
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2013 at 8:13 | comment | added | Adam Epstein | That coefficient will survive unchanged when the expression of the $k$-differential is transformed by a local holomorphic change of coordinate. See the discussion of "natural parameters" in Strebel's book "Quadatic Differentials" | |
| Nov 18, 2013 at 23:37 | comment | added | Jame Ake | Can you explain 'formal invariant', please? or Tell me some resource I can refer to. | |
| Oct 22, 2013 at 19:13 | comment | added | Adam Epstein | If two quadratic differentials have double poles at $\zeta$, and if the residues at $\zeta$ are the same, then the difference has at worst a simple pole at $\zeta$. | |
| Oct 22, 2013 at 15:33 | comment | added | sam | Still I would very much appreciate if you could somehow explain why this space is n+3 dimensional. You state that this is because the difference between any pair of elements has at most a simple pole. Could you elaborate on this? For instance, are the elements you mean the $q_{\alpha ,\beta}$'s? Their differences do have poles of order two. | |
| Oct 22, 2013 at 12:22 | comment | added | sam | All right thanks a lot for your help, I'll maybe post i new question about the subject. Anyhow, your answers helped me | |
| Oct 21, 2013 at 19:26 | comment | added | Adam Epstein | I have lost track of what you were asking for. I was trying to give an example of a quadratic differential with all residues equal to 1. I do not attempt to simultaneously control the zeros. | |
| Oct 21, 2013 at 9:11 | comment | added | sam | My factor of two comes from the fact that $q_{\alpha ,\beta}=q_{\beta ,\alpha}$. I am still not sure what you mean by the sum. It gives a big quadratic differential with double poles at n-points and some complicated polynomial in the numerator, however with no zeroes at one of the n points, I think. | |
| Oct 20, 2013 at 18:49 | comment | added | Adam Epstein | For each $\alpha$ there are $n-1$ different $q_{\alpha,\beta}$ each contributing residue 1. The expressions you give yield quadratic differentials with some negative residues. | |
| Oct 20, 2013 at 16:37 | comment | added | sam | Shouldn't the normalization be $\frac{1}{2(n-1)}$? What do you precisely mean with the sum of the $q_{\alpha ,\beta}$'s? For instance, would for $n=4$ $q_{\alpha ,\beta}-q_{\gamma ,\delta}$ be the 1 linearly independent vector in the hyperplane? Similarly, for $n=6$ ${q_{1,2}-q_{3,4},q_{1,2}-q_{5,6},q_{5,6}-q_{3,4}}$? | |
| Oct 20, 2013 at 15:08 | comment | added | Adam Epstein | The normalizing constant should be $\frac{1}{n-1}$. | |
| Oct 20, 2013 at 1:24 | comment | added | Adam Epstein | ...Then for any set $S$ with cardinality $n$, consider $\frac{1}{n(n−1)}∑_{(α,β)∈S×S}q_{α,β}$. | |
| Oct 19, 2013 at 7:47 | comment | added | Adam Epstein | The space is an $n-3$ dimensional hyperplane (but not a linear space) precisely because the difference between any pair of elements is a quadratic differential with at worst simple poles. Regardinng existence, for $\alpha,\beta\in\widehat{\mathbb{C}}$ let $q_{\alpha,\beta}$ be the unique quadratic differential having double poles with residue 1 at $\alpha$ and $\beta$, that is, $q_{\alpha,\beta}=\left(\frac{(\alpha-\beta)dz}{(z-\alpha)(z-\beta)}\right)^2$. Then for any finite set $S$ consider $\frac{1}{n(n-1)}\sum_{(\alpha,\beta)\in S\times S} q_{\alpha,\beta}$. | |
| Oct 14, 2013 at 17:04 | comment | added | sam | By the way, Adam, isn't the dimension of quadratic differentials with simple poles equal to $n-3$. Does this also hold for quadratics with double poles? | |
| Oct 14, 2013 at 13:27 | comment | added | sam | I don't really see why this space is n-3-dimensional. Doesn't there exist a quadratic differential for $n=2$ right? This would be the one you've mentioned, $dz^2/z^2$ which has a double pole at $z=0$ and a double pole at $z=\infty$. If not, could you explain for $n=4,5$ what would be the beasis vectors of the space? | |
| Oct 11, 2013 at 18:27 | comment | added | Adam Epstein | I'n not entirely sure what you are asking. But, for example, if you specify a finite set of $n\geq 2$ points then there will exists a quadratic differential which has a double pole with 'residue' 1 at each of those points, and which is holomorphic elsewhere. You can write down such a quadratic differential as a sum of Mbbius translates of $\frac{dz^2}{z^2}$,at least for $n$ even, and a modified recipe works for $n$ odd. Moreover, the space of such quadratic differentials will have dimension $\max(n-3,0)$. | |
| Oct 11, 2013 at 13:23 | comment | added | sam | I am asking about these 'free parameters' or freely chooseable zeroes because in the end I need to check that a certain one form, $\lambda = xdz$ with $x=\frac{G(F_1(z),F_2(z),...,F_k(z)))}{\prod (z-z_i)}$, differentiated wrt to one of the moduli is a holomorphic one-form on a curve defined by: $$x^k+F_1(z)x^{k-1}+...+F_k(z)=0$$. Hence, I want some sort of feeling or better yet concrete vision of how the $F_i$ depend on their moduli (and ofcourse how many moduli there are for given k and n). | |
| Oct 11, 2013 at 11:54 | comment | added | sam | Thanks for your answer. For 3 or more second order poles of a quadratic residue, $\frac{c(z-u_1)(z-u_2)}{(z-z_1)^2(z-z_2)^2(z-z_3)^2}dz^2$, there do not seem to be any relations anymore between the residues at the separate points $z_i$. Does this mean that when we fix the residues of a quadratic differential, we lose one more freely chooseable zero than in the case with a 1-differential (from which the residues do sum to zero and therefore we have one less constraint)? | |
| Oct 10, 2013 at 17:07 | history | answered | Adam Epstein | CC BY-SA 3.0 |