Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi I had asked this already on math.stackexchange.com but got no answers.

I was wondering if there was any sort of (natural) analog of the residue of a meromorphic one form that made sense for a meromorphic quadratic differential. Ideally, one would take a differential $Q$ express it in local coordinates around a singular point and get such a residue as the factor in front of the $\frac{1}{z^2}$ term. However, taking $$Q=\left(\frac{1}{z^3}+\frac{1}{z^2}\right) dz^2$$ on $\mathbb{D}\backslash \lbrace 0 \rbrace$ shows that this is not well defined independent of coordinate changes.

One idea I had was the following. Let $U$ be a neighborhood of a singular point $p$ of $Q$. Let $\hat{U}^*$be the double cover of $U^*=U\backslash \lbrace p \rbrace$ and $\pi:\hat{U}^*\to U^*$ the (conformal) double cover. If I am not mistaken there is then a meromorphic one form $\omega$ on $\hat{U}^*$ so that $\pi^*Q=\omega \cdot \omega $ (i.e. in the double cover the square root of $Q$ should exist). If we define $\alpha$ to be the square of the residue of $\omega$ at the point over $p$ then $\alpha$ should be defined independent of choices and so is a (not necessarily natural) candidate.

Perhaps this is well known? (or incorrect...)

share|improve this question
    
McMullen (see math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf , page 15) thinks your definition DOES work... –  Igor Rivin Nov 8 '11 at 20:55
    
The problem with the square of (...) as a replacement for the residue is that it's nonlinear. –  Victor Protsak Nov 8 '11 at 21:22
add comment

2 Answers

up vote 9 down vote accepted

Let me point out a somewhat different answer. Rbega explicitly mentions $$ Q = \left(\frac1{z^3} + \frac1{z^2}\right)\ (dz)^2 $$ as an example of the sort of meromorphic quadratic differential that is of interest, and this is not at all covered by dalakov's answer.

Consider the more general problem of asking when there is a normal form (possibly with parameters) for differentials of the form $$ Q = \frac{h(z)\ (dz)^l}{z^k}\ , $$ where $k$ and $l$ are fixed positive integers and $h$ is holomorphic near $z=0$ with $h(0)\not=0$.

Now, when $k$ is not a multiple of $l$, there exists a local coordinate $w$ centered on $z=0$ such that $$ Q = \frac{h(z)\ (dz)^l}{z^k} = \frac{(dw)^l}{w^k}\ , $$ so all of these differentials are locally equivalent.

However, when $k=ml$ for some integer $m>0$, the story is quite different. For example, it is clear that the 'residue' $$ R_{l,l}\left(\frac{h(z)\ (dz)^l}{z^l}\right) = h(0) $$ is well-defined, independent of the choice of 0-centered local coordinate $z$. Moreover, in this case, there exists a coordinate $w$ centered on $z=0$ such that $$ \frac{h(z)\ (dz)^l}{z^l} = \frac{h(0)\ (dw)^l}{w^l}, $$ so the 'residue' $h(0)$ is the only invariant in the case $k=l$.

Now, in all the cases in which $k=ml$, there is such an invariant. For example, when $k=2l$, one has the 'nonlinear residue' $$ R_{l,2l}\left(\frac{h(z)\ (dz)^l}{z^{2l}}\right) = h(0)\left(\frac{h'(0)}{h(0)}\right)^l, $$ which is independent of choice of coordinates. One also has $$ R_{l,3l}\left(\frac{h(z)\ (dz)^l}{z^{3l}}\right) = h(0)\left(l\ \frac{h''(0)}{h(0)}-(l{-}1)\ \left(\frac{h'(0)}{h(0)}\right)^2\right)^l, $$ and $$ R_{l,4l}\left(\frac{h(z)\ (dz)^l}{z^{4l}}\right) = h(0)\left(l^2\ \frac{h'''(0)}{h(0)}-3l(l{-}1)\frac{h'(0)h''(0)}{h(0)^2} +(2l{-}1)(l{-}1)\ \frac{h'(0)^3}{h(0)^3}\right)^l, $$ and so forth.

The general rule is that, in these cases, one can write $$ Q = \frac{h(z)\ (dz)^l}{z^{ml}} = \left(\frac{\bigl(h(z)\bigr)^{1/l}\ dz}{z^m}\right)^l = \omega^l\ , $$ where $\omega$ is a meromorphic $1$-form, well defined near $z=0$ up to multiplication by an $l$-th root of unity. The $l$-th power of the usual residue of $\omega$ at $z=0$ then provides an invariant that works out to be a rational expression in the coefficients of the power series of $h$, as in the cases noted above.

Finally, when $m>1$ and $h(0)\not=0$, it can be shown that there exists a local coordinate $w$ centered on $z=0$ such that $$ \frac{h(z)\ (dz)^l}{z^{ml}} = \left(\frac{(1 + a\ w^{m-1})\ dw }{w^m}\right)^l\ . $$ In this case, the 'residue' is (up to a universal constant multiple) simply $a^l$. An alternative normal form is $$ \frac{h(z)\ (dz)^l}{z^{ml}} = \frac{(1 + b\ \zeta^{m-1})\ (d\zeta)^l}{\zeta^{ml}} $$ for some constant $b$, and the nonlinear residue in this case is (a universal constant multiple of) $b^l$.

share|improve this answer
    
Thanks Robert. Just to clarify, for the example $Q$ I gave what would the 'residue' be? Since $k=3$ and $l=2$ am I right to take it to be 0? (this agrees with the speculative definition). –  Rbega Nov 9 '11 at 22:51
    
@Rbega: You could say that the residue' is zero in this case ($k=3$ and $l=2$, but what I prefer to say is that there is no meaningful notion of residue in this case because all germs of meromorphic quadratic differentials that have a third order pole are equivalent up to change of variables, i.e., there is no such invariant as a residue' to distinguish them. –  Robert Bryant Nov 10 '11 at 0:00
    
@Robert Bryant: Mea culpa! –  Peter Dalakov Nov 11 '11 at 13:44
    
@Peter Dalakov: ?? Your answer wasn't at fault, so I don't think that a 'mea culpa' is needed. In fact, your answer and your reference are particularly valuable because, in fact, quadratic differentials with poles of order at most 2 are particularly important in the theory of moduli of Riemann surfaces, and consequently, as you pointed out, there's a well-developed theory for them. It's just that Rbega was asking a somewhat different question, and I thought that deserved an answer, too. –  Robert Bryant Nov 11 '11 at 15:36
add comment

Your definition definitely works when the quadratic differential has poles of order at most two. Let $X$ be a Riemann surface and $z$ a local coordinate centered at $x\in X$. The principal part of order at most 2 of a meromorphic differential $\omega$ is of the form $$ R_2(\omega) = \left( \frac{a}{z^2} + \frac{b}{z}+ \ldots \right) dz^{\otimes 2} $$ and $Res^2_x (\omega):=a$. Principal parts of order at most $n$ form a rank $n$ vector bundle $R_n$, and there is a filtration $R_n\supset R_{n-1}\supset\ldots$ . In particular, we have a short exact sequence $$ 0 \to R_1\to R_2 \to \mathcal{O}_X\to 0, $$ and the last map is $Res^2:R_2\to \mathcal{O}_X$.

Stalkwise, this can be identified with $$ 0\to \mu_x/\mu_x^2\to \mathcal{O}_{X,x}/\mu_x^2\to \mathcal{O}_{X,x}/\mu_x\to 0. $$ A more familiar form of the first sequence is $$ 0\to \Omega_X^1\to J\to \mathcal{O}_X\to 0 $$ where $J$ is the unique non-trivial extension of $\mathcal{O}_X$ by $\Omega_X^1$ (corresponding to the Kaehler class). Or, if you want, $J= J^1(K^{1/2}_X)\otimes K^{-1/2}_X$.

The theory of residues of quadratic differentials is quite classical. A great overview is A.N.Tyurin, "On periods of quadratic differentials", Russ.Math.Surveys, 33 (169), 1978.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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