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## What do Weierstrass points look like?

As somebody who mostly works with smooth, real manifolds, I've always been a little uncomfortable with Weierstrass points. Smooth manifolds are totally homogeneous, but in the complex category you are just walking around, minding your own business, when suddenly you say to yourself: "this point feels different from all the others!"

I think I'd be happier if I could see a geometric feature which corresponds to the Weierstrass points. So here is a more precise sub-question: take a compact complex curve $X$ of genus $g \geq 2$, equipped with the hyperbolic metric. Is there something special happening for this metric at the Weierstrass points? Some kind of geodesic focusing, maybe?

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Oh, and again in genus two, every nonseparating simple closed geodesic (in any hyperbolic metric) goes through exactly two of the Weierstrass points. – Sam Nead Dec 17 2009 at 22:05

I think another interpretation is in terms of Euclidean geometry instead of hyperbolic geometry. If $\omega$ is a holomorphic 1-form on a Riemann surface $\Sigma$, then $|\omega|$ defines a Euclidean metric on $\Sigma$ with singularities at the zeroes of $\omega$. Think of the 1-form as locally defining a map to $\mathbb{C}$, and pull back the Euclidean metric. The metric near the zero is a cone of angle $2\pi k$, where $k$ is the order of the zero, since this is modeled locally on the map $z\to z^k$. The number of zeroes, counted with multiplicity, must be $2g-2$. A point $P\in \Sigma$ is a Weierstrass point if and only if there is a holomorphic 1-form with a zero at $P$ of order at least $g$. Equivalently, there is a flat cone metric (conformally equivalent to the underlying Riemann surface and with all cone angles of the form $2\pi k$) with a cone angle of $2\pi g$ at $P$ and trivial holonomy. I found this in some course notes of McMullen, but they seem to be no longer available online.

Addendum: I've added a link to McMullen's course notes on Riemann surfaces. Theorem 10.19 and the preceding discussion connect the Weierstrass points with holomorphic 1-forms. See Section 3 of another of McMullen's papers for a discussion of holomorphic 1-forms and branched Euclidean metrics,

Correction: I forgot that a holomorphic 1-form is not quite equivalent to a flat $2\pi k$-cone metric. The problem is that if one develops the surface into the plane around a closed loop, there might be a rotational part - that is why I added that there needs to be trivial holonomy. So a flat $2\pi k$ cone metric corresponds to a holomorphic 1-form twisted by a $U(1)$ character. It also corresponds to a special sort of holomorphic quadratic differential. It's still possible that high order zeroes of such twisted 1-forms correspond to Weierstrass points, but it's not equivalent to McMullen's characterization. As pointed out below by Dmitri, Troyanov has shown that there is a Euclidean structure on a surface with a cone point at any chosen point.

Addendum: Schmutz-Schaller stated this problem, which maybe is akin to what you are searching for (and as far as I know has not been answered). He gave a characterization of hyperelliptic Riemann surfaces in terms of closed geodesics.

Problem. Find a geometric characterization (based on closed geodesics) of Weierstrass points.

On a closed hyperbolic surface, there are points which have an open neighborhood which misses every simple closed geodesic. So there is certain "focussing" of simple closed geodesics.

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That explains a great deal. Thanks! – Greg Kuperberg Dec 18 2009 at 6:20
It is indeed a very nice answer. By the way, Schmutz-Schaller paper is freely available at ams.org/bull/1998-35-03/S0273-0979-98-00750-2/…. – jvp Dec 18 2009 at 10:07
Ian, I am not sure if I understand correctly your phrase "It's still possible that high order zeroes of such twisted 1-forms correspond to Weierstrass points". For every point of a Riemann surface of genus g there is a unique area 1 conformal flat metric on it with the cone angle $2\pi g$ at the point. This is a partial case of a theorem of Troyanov. – Dmitri Dec 22 2009 at 19:08
@ Dmitri: thanks, I didn't know of this result. I'll add a link to Troyanov's survey paper. – Ian Agol Dec 23 2009 at 2:58

Edit. After writing the answer below I realized it is essentially the same as the one by David Lehavi. As the different phrasing may improve understanding I will, reluctantly, not erase it.

If we consider the Poincaré metric (constant curvature $-1$) on an hyperbolic curve all the points are locally the same. I am afraid this implies that there is no local characterization of Weierstrass points. Of course this does not exclude the geodesic focusing property Matt suggests.

I like to think on Weierstrass points as a concept of projective differential geometry. It is an extension of the concept of inflection point of a plane curve.

Indeed if $C$ is a smooth plane quartic then its Weierstrass points are exactly its inflection points, i.e., points at which the tangent line has abnormal order of contact with the curve.

For a curve with very ample canonical bundle( the generic case when $g\ge 3$ ) essentially the same interpretation holds. More precisely we have that

• the canonical bundle $K_C$ determines an embedding of $C$ into the projective space $\mathbb P H^0(C,K_C)^{\ast}$;
• the Weierstrass points correspond to the points in $\mathbb P H^0(C,K_C)^{\ast}$ at which the image of $C$ and the corresponding osculating hyperplane have abnormal order of contact.
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 I like the connection between Weierstrass points and special points under the canonical embedding, but this isn't quite what I'm after. Ideally, I would like something more intrinsic, rather than extrinsic (even if it is canonically extrinsic). I realize this is a little tricky, since the notion of Weierstrass point isn't local... – Matt Noonan Dec 18 2009 at 2:25 Re: no local characterization: that's why I was hoping there was something like a focusing effect. That would be global, but still intrinsic and rooted in the surface's natural geometry. – Matt Noonan Dec 18 2009 at 3:22 I think I was editing that while you were writing your comment. Of course you are right. – jvp Dec 18 2009 at 3:30 Note that in the plane quartic case you have two types of Weierstrass points: inflections points (= flexes), and hyperflexes, where the tangent lines has contact 4 with the quartic. E.g the the point (0:0:1) on the quartic (x^2-yz)^2 + y^4 (tangent line y=0) – David Lehavi Dec 18 2009 at 6:53 Hyperflexes are special kind of flexes but are still flexes, at least if you define them as contact with tangent line of order greater than or equal to three. – jvp Dec 18 2009 at 9:59

In 1969, Joseph Lewittes wrote a paper in the Transactions of the AMS where he attempted to give a differential geometric interpretation of Weierstrass points. From his introduction: In particular, we show that the Weierstrass points of a compact Riemann surface can be characterized as points of zero curvature of certain "naturally defined" metrics on the surface"

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The way I got a feel for them was in a very elementary way by starting with hyperelliptic curves $y^2=P(x)$ and "manually" trying to construct rational functions of x and y with given poles and zeroes on the curve. (Ensuring, of course, that I had the correct zero or pole at infinity too). You quickly find inequalities bounding the sizes of the powers of x and y that you're allowed in the numerator or denominator and it'll become clear that for poles at some locations you have a bit more freedom. That way you can get a direct feeling for these things (at least in certain easy cases) without going through Riemann-Roch 'n' all that.

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Any time you get a statment on OC(D), you can translate the question via Geometric Riemann-Roch.

Genus 2: Weierstrass points the only points such that 2p = K

Genus 3 non hyperelliptic: there are no g12 s so the first interseting case is g13 with some point p such that g13 = |3p|. a g13 is a projection from a point on the canonical curve. So Weierstrass points are flexes under the canonical embedding (projecting from the point where the tangent line intersect). In curves which admit hyperflexes, the hyperflexes are another type of Weierstrass point, and that's all you can get in genus 3 (because 2g-2 = 4)

Higher genus hyperelliptic: canonical map is to a rational normal curve (which is the he system) so just like genus 2.

genus 4 and up non hyperelliptci: similar to genus 3.

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Associated to a dessin d'enfant (or hypermap) there is an associated complex analytic structure. (See Grothendieck's esquisse d'un programme or D. Singerman Automorphisms of maps, permutation groups and Riemann surfaces, Bull. London Math. Soc. 8 (1976), 65-68.) Grothendieck observed that Belyi's theorem implies that the corresponding algebraic curve is defined over the field of algebraic numbers. The most symmetric dessins are the regular maps, long studied by Coxeter and many others. For example, the famous Klein quartic corresponds to the well-known map of type {3,7} of 24 heptagons on a genus 3 surface. Here the 24 Weierstrass points are the face-centres. More generally we can ask whether vertices, face-centres or edge-centres are Weierstrass points. The answer is yes for genus 2,3 4 and with one possible exception for genus 5 too. See D. Singerman and P. Watson, Weierstrass points on regular maps, Geom. Dedicata, 66 (1997), 69-88.

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