# The map from a complex torus to the projective algebraic curve

• A

## Summary:

connectedness of the projective algebraic curve

## Main Question or Discussion Point

I am following the proof to show that the complex torus is the same as the projective algebraic curve.

First we consider the complex torus minus a point, punctured torus, and show there is a biholomorphic map or holomorphic isomorphism with the affine algebraic curve in ##\mathbb{C}^2##.

Second we make an extension and add the point removed above from the torus. But now the maps goes into ##\mathbb{P}^2## which is in ##\mathbb{C}^3##.

The extension sends the missing point of the torus coming from point #z=0# in the lattice to the point ##[0,1,0]## in ##\mathbb{P}^2##, the so called point of infinity.

But I don't see and cannot imagine why the map constructed this way is continuous after we add a point of infinity to a continuous curve in ##\mathbb{P}^2##. Also related, when we "compactify" and add a point of infinity to our connected affine algebraic curve in ##\mathbb{C}^2##, is the resulting projective algebraic curve in ##\mathbb{C}^3## connected too?

The proof which I am following doesn't check the continuity of the extended map and connectedness of the projective algebraic curve explicitly.

Do we make the conclusion about the connectedness of the projective algebraic curve from establishing a biholomorphic map with the complex torus ?

Am I missing something either intuitive or a proved earlier result?

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There are several ways to define a complex torus (of one complex dimension), and there are many projective algebraic curves (including many that are not complex tori).

So in order to know how to prove this, it's important to know exactly how each of these objects is defined in the first place.

Also: To show two objects are "the same" means different things depending on which category of object is being discussed. Tori can be thought of as purely topological surfaces. But most commonly a complex torus and a projective algebraic curve are thought of as Riemann surfaces, that is, complex analytic one-dimensional manifolds.

In the case of Riemann surfaces, the appropriate meaning for "the same" is conformal equivalence. Which indeed is what you are trying to show by finding a holomorphic isomorphism. But: There are many conformally inequivalent tori, so the proof would depend on just how a) your "complex torus" and b) your "projective algebraic curve" are each defined.

You say that ℙ2 is in ℂ3. But that's not correct. Rather, ℙ2 is the quotient space of ℂ3 - {0} obtained by setting

(u, v, w) ~ (αu, αv, αw)

for each (u, v, w) ∈ ℂ3 - {0} and each α ∈ ℂ - {0} (where ~ is the equivalence relation).

You say that ℙ2 is in ℂ3. But that's not correct. Rather, ℙ2 is the quotient space of ℂ3 - {0}
Isn't it true that the quotient space is isomorphic to a two dimensional complex manifold, i.e. a complex sphere, embedded in ##\mathbb{C}^3## and the projective algebraic curve lies on this sphere?

Infrared
Gold Member
I think a fact that you're missing is that if ##\phi:X\to Y## is a homeomorphism, then it extends to a homeomorphism of the one-point-compactifications. Does this answer your question?

You also seem to be asking why the one-point compactification of the plane algebraic curve ##C## is the corresponding projective curve. I guess this is the same thing as checking that the closure of ##C##, viewed as a subset of ##\mathbb{P}^2## (via the inclusion ##\mathbb{C}^2\subset\mathbb{P}^2##) is the projectivization of ##C##, which should be straightforward.

Isn't it true that the quotient space is isomorphic to a two dimensional complex manifold, i.e. a complex sphere, embedded in ##\mathbb{C}^3## and the projective algebraic curve lies on this sphere?
No, ##\mathbb{P}^2## is a compact complex manifold (of positive dimension), and therefore doesn't holomorphically embed into any ##\mathbb{C}^n##

mathwonk
Homework Helper
A nice general principle in these situations, is that if M-{p} is a punctured one dimensional complex manifold, and you have a map from this punctured manifold to P^n, which is defined by functions that are holomorphic on M-{p} and meromorphic at p, then the map always extends to a holomorphic map on all of M. E.g. if p has coordinate 0, this is because the functions (f/z, g/z, h/z) say, define the same map to projective space, as the functions (f,g,h). So if your map is defined by meromorphic functions (f,g,h) you just multiply them all by z^r, where r is the order of the biggest pole at 0 among the three functions.

As I recall, the usual map of a punctured torus to C^2, when the tous is described by a plane lattice, is defined by the associated Weierstrass "P-function" and its derivative, which have poles at zero of orders 2 and 3 respectively. The image of the map is a plane curve with equation in Weierstrass form y^2 = cubic in x.

Since the extended map is actually holomorphic, it is also continuous, and maps the closure of the punctured torus to the closure in P^2 of the affine cubic curve.

No, ##\mathbb{P}^2## is a compact complex manifold (of positive dimension), and therefore doesn't holomorphically embed into any ##\mathbb{C}^n##
What is the relationship between ##\mathbb{P}^2## and ##\mathbb{C}^3##?

Is ##\mathbb{P}^2## isomorphic to any subset of ##\mathbb{C}^3##?

Infrared
Gold Member
Is ##\mathbb{P}^2## isomorphic to any subset of ##\mathbb{C}^3##?
No, this is what is meant by "##\mathbb{P}^2## does not holomorphically embed into ##\mathbb{C}^3##."

If you want a proof: Suppose for contradiction that ##X## is a compact complex submanifold of ##\mathbb{C}^n## with positive dimension. Then by compactness the coordinate functions ##z_i## must attain maxima. The maximum principle of complex analysis tells us that the coordinate functions ##z_i## must be locally constant, so ##X## is a finite set of points.

No, this is what is meant by "##\mathbb{P}^2## does not holomorphically embed into ##\mathbb{C}^3##."
My key challenge is to imagine or visualize ##\mathbb{P}^2##!

So "locally" what does it look like?

And then visualize how a projective algebraic "curve" look like on it.

Infrared
Gold Member
Locally, it is the same as ##\mathbb{C}^2##. In projective coordinates ##[x:y:z]##, setting any of the coordinates equal to ##1## gives a subset isomorphic to ##\mathbb{C}^2##, e.g. ##U_x=\{[1:y:z]\}.## These coordinate charts cover ##\mathbb{P}^2## and the complement of, say, ##U_x## is the set of points ##\{[0:y:z]\},## which is a ##\mathbb{P}^1## with projective coordinates ##y,z##.

So you can think of the projective plane as the affine plane glued to a ##\mathbb{P}^1## (sometimes referred to as a line at infinity). You can then think of a projective curve as an affine curve in this affine plane, together with its intersections with the line at infinity.

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mathwonk
Homework Helper
notes from last time I taught this course, i.e. probably "more than you wanted to know":

Day two: Definition of the Riemann surface of a non singular affine plane curve

We start at the beginning with curves and Riemann surfaces which are not necessarily compact. We want to define holomorphic functions on a plane curve C:{ f(x,y) = 0}. Of course we want the coordinate functions x and y to be holomorphic on C, hence also all holomorphic combinations of them. This amounts to saying a function on X is holomorphic if it is the restriction of a holomorphic function of x and y, i.e. if it extends to a holomorphic function on the plane k^2. But if we want to be able to expand a function on C in a power series in a disc centered at a point of C, we need a single local coordinate near each point, instead of the two coordinates x and y. This we do next.

First example: C is the graph of a holomorphic function y = g(x), i.e. a plane curve of form 0 = f(x,y) = y – g(x) . As always, the graph of a function is isomorphic to its domain, which allows us to use the single coordinate x on the graph. I.e. since x and y are both holomorphic on the plane, their restrictions to the graph C should be holomorphic. In particular projection (x,y)---> x should be holomorphic on C. Conversely, since g is holomorphic by hypothesis, the inverse map x--->(x,g(x)) is holomorphic. Thus if we assume that both coordinates x and y are holomorphic on C, it follows that the graph C of a function g(x) is holomorphically isomorphic to the x axis. Hence it follows that a function on C will be holomorphic if and only if it is holomorphic as a function of x. Thus we will get a definition of holomorphic function on X in terms of only one coordinate, by letting x be the coordinate, and saying that a function h on C is holomorphic if and only if it is holomorphic as a function of x, via the composition map x--->(x,g(x))--->h(x,g(x)).

Thus we know how to put holomorphic coordinates on the graph of a holomorphic function. But holomorphicity is a local notion, so we should be able to do the same on a curve which is only locally a graph. Detecting such curves is exactly what the implicit function theorem is designed for. Draw a picture of a curve in the plane and project it onto the x axis. Look at the points of the curve where the projection is not a local isomorphism I.e. recall a curve is a graph if it passes the vertical line test, i.e. on some (complex) neighborhood, all vertical lines meet the curve exactly once. Look at points where this fails. These are points where the tangent line is vertical as well as all points where two branches of the curve cross. The implicit function theorem says these points on f(x,y) = 0 can be detected by the vanishing of the partial derivatives of the function f. I.e. f = 0 is a level set of f, and gradf is perpendicular to its level sets. Thus at points where two branches of the level set cross each other, at least if they cross transversely, grad f must be perpendicular to two independent tangent vectors, one to each branch. Thus gradf must be identically zero at such crossing points, i.e. both partials ∂f/∂x and ∂f/∂y are zero. At a point where the curve f = 0 is smooth, i.e. with only one tangent line but that tangent line is vertical, gradf must be horizontal, i.e. ∂f/∂y must be zero. At all points other than these two types, i.e. at all points where ∂f/∂y ≠ 0, projection to the x axis is a local isomorphism. So we need to exclude precisely those points where ∂f/∂y = 0, to have x be a local coordinate.

Implicit function theorem: If f(x,y) is a polynomial in two complex variables x,y, and if p = (x0,y0) is a point where f(x0,y0) = 0 and also ∂f/∂y ≠ 0, then on some neighborhood of p, the level set f = 0 is the graph of a holomorphic function y = g(x).

As a corollary, if f(x,y) is a polynomial and C is the level set: {f=0}, then x is a local coordinate for C at every point p where ∂f/∂y ≠ 0. Similarly, y is a local coordinate for C near every point q where ∂f/∂x ≠ 0. And if p is a point where both partials ∂f/x and ∂f/∂y are ≠ 0, then by the implicit function theorem, y and x are holomorphic functions of each other, so they give equivalent holomorphic local coordinates. I.e. at such points, a function on C, is holomorphic as a function of x if and only if it is holomorphic as a function of y. The points where both partials are zero are not “Riemann surface” points. They can be modified to become such as we discuss later, but now we want to formalize the properties of these good points, in the definition of a Riemann surface.

Definition: A Riemann surface X is a Hausdorff topological space equipped with a cover of X by open sets Uj, and for all j, homeomorphisms ƒj:Uj--->Vj where Vj is an open subset of the complex numbers, such that for all i,j, the composition ƒjƒi^-1 is holomorphic where defined, i.e. on the open set ƒi(UimeetUj) in the complex numbers.

Corollary of implicit function theorem: If f(x,y) is a complex polynomial, then the open subset of its level set f = 0 consisting of those points in k^2, k = complex field, where the partial derivatives ∂f/∂x and ∂f/∂y are not both zero, is a Riemann surface with coordinate functions given by restrictions of the functions x and y on appropriate sets.

A function g:X--->k on a Riemann surface is holomorphic if all compositions gƒ^-1 of g with coordinate functions are holomorphic where defined. The function g is holomorphic near p if this holds for one coordinate function defined at p. The derivative of such a g is zero at p if the derivative of gƒ^-1 is zero for one and hence for all charts ƒ near p.

Definition: A level set f(x,y) = 0 in k^2 of a polynomial f(x,y) is called an affine plane curve. A point of f = 0 where both partials ∂f/∂x and ∂f/∂y are zero is called a singular point, and other points are called non singular points. A curve with no singular points is called a non singular curve. Thus a non singular plane curve, with an appropriate open cover, is a Riemann surface.

Remark: The functions x and y are everywhere holomorphic functions on the Riemann surface of a non singular plane curve, but neither one is necessarily everywhere a local coordinate. The function x is a coordinate near all points where ∂f/∂y ≠ 0, and y is a coordinate near points where ∂f/∂x ≠ 0.

Example: If g(x) is a polynomial in one complex variable x with no repeated roots, then f(x,y) = y^n – g(x) = 0 defines a non singular plane curve in k^2 for n ≥ 1. If n = 1, then ∂f/∂y =1 is never zero, and if n ≥ 2, then if ∂f/∂y = ny^(n-1) = 0 at (x,y), then y = 0, so g(x) = 0, and ∂f/∂x = g’(x) is non zero by hypothesis on g, since then x is a root but not a multiple root of g.

Remark: Every Riemann surface is a smooth real 2 dimensional surface, and since multiplication by a ≠ 0 complex number, namely the derivative of a composition ƒjƒi^-1 of coordinate functions, preserves orientation of the complex number line, the local coordinates induce on the Riemann surface an orientation consistent with the usual one on the complex numbers, via the local coordinate neighborhoods. Thus every Riemann surface is an orientable 2 dimensional real surface with a preferred orientation consistent with that given by the ordered real basis {1,i} for the complex line k.

In fact every non singular affine plane curve f(x,y) = 0 is topologically a compact oriented 2 dimensional real surface with a finite set of points removed. Further, the non singular polynomial f is irreducible if and only if the compact Riemann surface is connected. We want to determine the genus of this compact surface, which will also have a complex structure of a Riemann surface.

An abstract example (not a plane curve): a “complex torus”
If a “lattice” is a Z submodule of the complex numbers k generated by two complex numbers which are linearly independent over the reals, then the quotient group T = k/lattice, with its natural quotient topology, is called a “torus”.

Ex: There is a unique structure of Riemann surface on such a torus T, so that the natural quotient map π:k––> T is a local holomorphic isomorphism. (A function on an open subset of T is holomorphic iff its composition with π is holomorphic on an open subset of k.)

Ex: Every such torus is holomorphically isomorphic to one whose lattice has generators {1,t} where Im(t) > 0.

Ex: All such tori are diffeomorphic.

*Ex: Not all such tori are holomorphically isomorphic.

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• lavinia
mathwonk
Homework Helper
Day three: Visualizing examples of Riemann surfaces of affine plane curves

Since every compact orientable surface has a genus, it is instructive to try to compute the genus in some examples. Even without compactifying our affine curves, the genus is determined and may be visible, since the affine curve only differs from the compact Riemann surface by the removal of a finite number of points.

1) C:{y^2 = x}, where x and y are complex coordinates on k^2, k = complex numbers. If we project onto the y axis, then over each y there is a unique x, namely x = y^2. So this projection is a bijection and this curve is isomorphic to the y axis, with inverse the map from the y axis to the curve taking y--->(y^2,y).

It is useful also to try to see the structure of the surface by projecting on the x axis. We have a map C--->k which is 2:1 over every x except x=0, where there is only the one point y = 0 over 0. If we think of the x axis as a large disc or rectangle, it is not easy to visualize how a smooth surface can lie directly over a rectangle, with two points over every point except the center, over which there is only one point. We know however since this is a holomorphic map, that locally it has the same structure as some power map z^k, which must be squaring, i.e. k must = 2.

Indeed we know the curve C is isomorphic to the y axis, and the composition of the isomorphism y--->(y^2,y) with projection on x, is just the squaring map y--->y^2. If we recall the behavior of the squaring map, it doubles angles about 0, so this map sends the top half of the y axis onto the whole x axis, with the horizontal line dividing the top half of the y axis from the bottom, going doubly onto the positive real line in the complex x axis. Similarly, the bottom half of the complex y axis also maps onto the whole complex x axis with the real line of the complex y axis mapping again doubly onto the positive real x axis. Thus we can visualize the way the curve C lies over the complex x axis as follows.

If we slit the complex x axis along the positive real line, there are two copies of this slit rectangle lying over it on C. These two copies are isomorphic to the upper and lower halves of the complex y axis. Now the upper and lower halves of the complex y axis are joined along their common border, and crossing the real line takes us from one half to the other half. So these two slit rectangles up on C are joined cross ways along their two slits, so that crossing the slit takes one from the upper copy to the lower one and vice versa. Thus we can visualize constructing the curve C from its image the complex x axis, without knowing in advance that the curve C was isomorphic to the complex y axis. I.e. we slit the x axis along its positive real line, take two copies of this slit rectangle, and glue the edges of their slits crossways. This is impossible in 3 space while keeping each point over its image without passing the glued slits through each other of course.

Thus if we want to look at the curve C as a rectangle, we have to visualize the branched 2:1 map defined on that rectangle as wrapping around the center and doubling angles. If we want to have the branched 2:1 map to be just vertical projection from the curve C down onto the x axis, then we have more work to do to visualize the way the curve C lies over the x axis. Viewed in 3 space it would then intersect itelf along the preimage of the positive x axis. Of course C actually lies in k^2 or real 4 space where this self intersection does not occur. So either we have a simple view of the curve C and a somewhat complicated view of the map from C (squaring), or we have a simple view of the map from C as just vertical projection, but then we have a harder task to visualize how the curve lies over its image. Anyway, since the affine curve C is isomorphic to the complex numbers, i.e. a copy of the real plane, the only possible compact surface that can be obtained by adding in a finite set of points is the sphere, so the genus is zero.

2) Let C: {y^2 = x^2 -1}. Now we do not know at once what the curve C looks like by using the first method above, since neither projection onto x, nor projection onto y is an isomorphism. So we adopt the second method above, choosing as map the vertical projection from C onto the x axis and trying to visualize how the curve C is constructed from gluing copies of the image x axis. Here again the map C--->k is usually 2:1 but there is only one point over x = -1 and one over x = 1. Moreover we know that locally near each of these points, the map looks like squaring, by the local structure theorem for holomorphic functions. So if we cut the complex x line by a vertical line separating the two points 0 and 1, i.e. if we cut the complex x line into two rectangles, then by example 1, the preimage of each rectangle is again a rectangle up in C. Further since every point of the line we cut along in the x plane has two disjoint preimages in C, we are joining our two rectangles in C by gluing two disjoint intervals in each rectangle.

Thus we are choosing two disjoint edges of each rectangle and gluing them together, each edge from one rectangle to one edge of the other rectangle. This gives either a cylinder or a twisted cylinder (Mobius strip), but since C is orientable it must be a cylinder. Now the only compact surface obtained by adding a finite number of points to a cylinder is a sphere, so this curve again has genus zero. Note that last time we only added one point to the complex line k to get a sphere, and this time we add two points to a cylinder, which is homeomorphic to k-{0}, to get a sphere. So the compact Riemann surfaces are the same, but the affine ones we started from were different.

To see this construction using a different cut in the x plane, make a slit along the interval joining the two branch points -1 and 1. A closed loop in the plane which goes around both branch points deforms to the union of two closed circles, one going around 1 and the other going around -1, with the circles having one common point at 0. Up on the curve C, this means the curve lying over these loops has passed from one preimage of 0 to the other and back again, so after going around both branch points the preimage of 0 is the same at both ends of the full closed curve. This means every closed loop in the complex x line which goes around both branch points once has as preimage a curve that is closed on C. Thus every closed loop in the x line that misses the interval between the two branch points must have a closed preimage on C. Hence if we take two disjoint copies of the x line, both cut along the interval from -1 to 1, then C is obtained from these two copies by cross gluing them along opposite edges of the cuts. Since the preimage on C of the slit betwen -1 and 1 is a circle, we thus form C by gluing in a circle to join up two slit planes, obtaining a cylinder. The gluing must be done across the slits to retain the orientation of C.

In fact the first method above also works if we first change coordinates, setting s = x+y and t = x-y. Since C has equation y^2 = x^2 - 1, or equivalently 1 = x^2-y^2, this gives as equation just 1 = st, or equivalently t = 1/s, for s≠0. Thus projection of this curve onto the s - axis, is an isomorphism between C and the set of non zero complex numbers, which is again seen to be topologically a cylinder. In the original coordinates this map takes (x,y) to x+y = s. This is an isomorphism from the curve to the non zero complex s line. I.e. 0 ≠ s = x+y implies that on C, 1/s = t = x-y, so we can recover x as [s+(1/s)]/2, and y from [s-(1/s)]/2. I.e. s--->([s + 1/s]/2, [s – (1/s)]/2) = (x,y) is inverse to the map (x,y)--->x+y = s, between the curve C and the set {s≠0} on the s line.

3) C: {y^2 = x(x^2-1)}. Here there are three branch points, so we can view C as obtained by gluing three rectangles. We get C topologically by gluing two opposite sides of the center rectangle to two opposite side of the left rectangle, and then gluing the remaining two opposing sides of the center rectangle to two opposite sides of the right rectangle. This gives a cylinder with a strip glued from part of the border of one hole to part of the border of the other hole, like two links in a paper chain. I claim the compact surface has genus one, after adding one additional point. (What is the boundary curve?)

4) C: {y^2 = (x^2 - 1)(x^2 - 4)}. Show this has genus one, after adding two points.

5) C:{y^2 = x(x^2 - 1)(x^2 - 4)}. Show the genus is two, after adding one point.

6) C:{y^2 = (x^2 - 1)(x^2 - 4)(x^2 - 9)}. ????

7) C: {y^2 = (x-a1)....(x-an)}, all aj distinct. ????

8) C: { y^3 = 1 – x^2}. This one can be done by projecting on the y axis as above, since this is a 2:1 map branched over the three cube roots of 1, by using the equation in the form x^2 = 1-y^3. But if we project onto x, we have a triple cover of the x axis, branched at the 2 square roots of 1, so by separating again those 2 branch points by the imaginary line in the x plane, we have two disjoint rectangles, each covered by a triple cover with on branch point. Since that map is equivalent to y--->y^3, its domain up in C is a disc, or rectangle, or let’s say a hexagon. Thus we form C by gluing two disjoint hexagons along the preimage curves of the imaginary line in the x plane. These preimage curve in each hexagon, are three disjoint edges of the hexagon, but when we join these images from one hexagon to the other, the result is a single closed curve in C, because of the way the edges are joined cyclically over each rectangle. If one chooses three disjoint edges in each hexagon and glues in the simplest way, one obtains a sphere with three disjoint discs removed, not a single connected boundary curve. This puzzled me for some time, as this would give genus zero, the wrong answer. But it is possible to visualize how to join the edges of the hexagons to get a connected boundary curve as follows. Take two hexagons and label their edges as if looking at a clock face, noon, 2 o’clock, 4 o’clock, 6 o’clock, 8 o’clock, and 10 o’clock. We want to identify three of these pairs of edges, the ones at 2 o’clock, 6 o’clock, and 10 o’clock on the first hexagon, with those at 4 o’clock, 8 o’clock, and noon on the second hexagon. So first identify 6 o’clock on the first hexagon with noon on the second, and then identify 2 o’clock on the first with 8 o’clock on the second, and finally identify 2 o’clock on the first with 4 o’clock on the second. This system of cross identifications leaves a simple connected boundary curve and is homeomorphic to a genus one surface with one point (or disc) removed.

9) C: { y^3 = 1 – x^3}. ???? I have not visualized this one or the next one, but this one should be obtainable by identifying edges on three dodecagons, since for the center hexagon we need two pairs of disjoint edges and another unidentified edge in between each of those 6 edges.

10) C:{ y^4 = 1 – x^4}. ???? Maybe one should begin on { y^4 = 1 – x^2} where we know the answer, as we did in problem 8.

Obviously these are getting harder to visualize, so we need some more powerful tools to even compute the topological genus of a plane curve. Soon we will introduce the method of “degeneration”, or variation of curves in families, to help us make more computations.

Preview: We will define the compact Riemann surface associated to a possibly singular affine plane curve, in two steps. First we will compactify the affine plane curve by enlarging it to a projective plane curve. Then we will desingularize the projective curve by removing its singular points and replacing each singular point by a finite number of new points which will become non singular on the Riemann surface. I.e. at a non singular point, projection on a suitably chosen axis is a local isomorphism from a neighborhood of the non singular point to a disc. It can be shown that projection of a neighborhood of a singular point onto a suitably chosen line, gives a finite covering space from a punctured neighborhood of the singular point to a punctured disc.

Such a finite covering space is always a finite disjoint union of punctured discs, on each of which the covering map is isomorphic to z^k for some k. On the curve, these punctured discs all share the same center point, the singular point of the curve. Then the Riemann surface is obtained by giving a different center to each of these punctured discs, so that we get a disjoint union of discs. The number of disjoint discs in a neighborhood of a singular point is called the number of local analytic branches at that singularity. Thus a singular point is replaced by as many non singular points as there are branches at the singularity. However, even when there is only one branch at a singularity, and the Riemann surface has only one point corresponding to the singularity, the map from the disc neighborhood of that non singular point on the Riemann surface, to a neighborhood of the singular point of the curve, is never a local isomorphism even if it may be a local homeomorphism. I.e. at a singular point with one branch, although there is a neighborhood of the singularity which is homeomorphic to a disc, that disc is pinched, and there is no projection to any axis which is a local coordinate. There is always a unique holomorphic map from the Riemann surface of the projective curve to the projective curve, but it is either not one to one over a singularity, or has derivative zero at some preimage of the singularity, as a holomorphic map to the plane. E.g. k--->k^2 defined by (x,y) = (t^2,t^3) mapping the t line onto the curve y^2 = x^3, is a homeomorphism but has derivative zero at t = 0. Then k is the (non compact) Riemann surface of the affine plane curve y^2 = x^3 with a singularity at (0,0).

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mathwonk
Homework Helper
8320 sp2010 day 4. Projective line P^1 and projective plane P^2

P^1: We want to compactify the complex line k and complex plane k^2 in a natural way, obtaining compact complex manifolds P^1, P^2 of complex dimension one and two respectively. The complex line is compactified by adding just one point “at infinity”, to form P^1 = k + {infinity}. This is done by taking another copy of the complex line and gluing the two copies together on the complement of their origins by the reciprocal map. I.e. let s be an affine coordinate for one copy of the complex line and let t be an affine coordinate for another copy of the complex line. Then identify the open sets {s≠0} and {t≠0} by mapping s to t = 1/s. This is a holomorphic isomorphism, and if we form the identification space from the disjoint union of these two copies of the complex line, joined along those two isomorphic open sets, we get a compact Hausdorff space called P^1, the complex projective line. Moreover since the map s--->1/s is an invertible holomorphic map, this space P^1 has an open cover by compatible holomorphic coordinate charts making it into a complex one - manifold, or compact Riemann surface. To see P^1 is compact it suffices to note it is the union of the two compact subsets {|s|≤1} and {|t|≤1|}. To see it is Hausdorff, since the s line k is Hausdorff, it suffices to note that the additional point t=0 in P^1, can be separated from any point s0 by the open sets, |s| < 2|s0|, and |s| > 2|s0|, i.e. |t| < 1/|2s0|.

There is another construction of P^1 by starting from the complex affine plane k^2, and considering P^1 to be the set of lines through the origin. I.e. take the punctured affine plane k^2 –(0,0), and define an action of k* = non zero complex numbers on it, by saying that c in k* sends (a,b) in k^2 – (0,0) to (ca,cb). Then the orbits of this action are exactly the punctured lines through the origin of k^2. The quotient space [k^2-(0,0)]/k* is also P^1. I.e. if we take coordinates (z,w) on k^2, and use [z:w] to represent the equivalence class of (z,w) in P^1, then we can cover this manifestation of P^1 by the two open sets {z≠0} and {w≠0}. The first open set {z≠0} is isomorphic to the t line by the map s--->[1:t] = [z:w] from k to P^1, with inverse map [z:w]--->(w/z) = t. The second open set {w≠0} is isomorphic to the s line by the map s--->[s:1] = [z:w], with inverse map [z:w]---> (z/w) = s. In the overlap zw≠0, a point [z:w] of P^1 has s and t coordinates related by s = (z/w) = 1/(w/z) = 1/t. Hence this is isomorphic to the identification space version of P^1 constructed above from identifying the s line and t line by reciprocation.

This gives another way to see P^1 is compact since the 3 sphere in k^2 ≈ R^4, real 4 space, surjects continuously onto it. I.e. every complex line in k^2 is a real plane in R^4 and meets the 3 sphere in a circle. Thus the map S^3--->P^1 is a circle bundle over the sphere P^1, called the “Hopf map”. This map represents a non zero generator of the higher homotopy group π3(S^2), a rather fascinating object since it seems strange for a 2 sphere to admit a homotopically non trivial map from a 3 sphere. There are no non trivial maps from higher spheres to S^1 for instance because the universal cover of S^1 is contractible. The classification of all homotopically non - trivial maps from higher dimensional spheres to lower dimensional spheres is still an open problem.

Since P^1 is homeomorphic to a sphere, it has genus zero. Moreover, this is the only compact topological surface that can be constructed from the complex line k by adding a finite number of points, hence the Riemann surface of any affine curve homeomorphic to k also has genus zero. E.g. any affine line {ax+by = 0}, a,b not both zero, must have a Riemann surface of genus zero. Of course the proof of this must await the definition of the Riemann surface associated to an affine plane curve.

We can also give the construction of P^1 in a coordinate - free way, by starting from any 2 dimensional complex vector space V and considering the set P(V) of all lines through the origin. Note that if (a,b) and (c,d) are non parallel vectors in k^2, then the map taking (z,w) --->(az+bw, cz+dw) = (u,v) is an isomorphism of k^2, and takes the set{az+bw≠0} to the set {u≠0}, and takes the set {cz+dw≠0} to the set {v≠0}. Moreover in local holomorphic coordinates, as a map say from an open subset of {z≠0} to {v≠0}, in the local coordinates t = w/z, and s = u/v, this map has form s = [a+bt]/[c+dt] where c+dt≠0, hence defines a holomorphic bijection thus an isomorphism of P^1. It follows that we could recoordinatize P^1 as the union of any two affine open sets u = az+bw ≠ 0, and v = cz+dw ≠ 0. In particular any choice of vector basis for V yields the same holomorphic structure on P(V) ≈ P^1.

P^2: Now we want to compactify the complex plane k^2 to obtain a compact complex two-manifold P^2. First of all we may define a holomorphic function on a open subset of k^2 to be a continuous function which is holomorphic in both variables separately. [In fact continuity is guaranteed by the separate holomorphicity.] Now we will compactify k^2 by adding in a copy of P^1 to form P^2. Thus if we think of k^0 as one point, then P^2 = k^2 + k^1 + k^0, as disjoint union of sets. The coordinate definition is to start from k^3 and consider all lines through the origin. I.e. on k^3 – (0,0,0) define the k* action of scaling as before, with a non zero scalar c sending (z0,z1,z2) to (cz0,cz1,cz2). The quotient space by this action [k^3 – (0,0,0)]/k* = P^2. It has an open cover by three affine open sets {z0≠0}, {z1≠0}, {z2≠0}, with local holomorphic coordinates on {z0≠0} given by (z1/z0, z2/z0), which defines a homeomorphism from P^2 with its quotient topology, to k^2. Composition of these local coordinates gives holomorphic rational functions in the affine coordinates with non zero denominators, hence the local coordinates are holomorphically compatible and we have a complex 2 manifold, analogous to our definition of a complex Riemann surface or complex one manifold.

Since complex 3 space k^3 ≈ R^6 = real 6 space, and every complex line through the origin of complex 3 space, is a real plane through the origin of R^6, which meets the 5-sphere in a circle, the 5-sphere maps surjectively onto the quotient space P^2, displaying S^5 as a circle bundle over P^2. Since the sphere is compact, so is its continuous image P^2.

Homogenizing an affine equation: Now we can take the first step toward defining the Riemann surface of a plane curve, by compactifying the curve. I.e. we will take the closure in P^2 of an affine plane curve in k^2. Suppose {f(X,Y) = 0} is a plane curve in the (X,Y) plane k^2. Think of X,Y as affine coordinates X = Z1/Z0, Y = Z2/Z0 in P^2, and write the equation for the curve as f(Z1/Z0, Z2/Z0) = 0. Then multiply by just a high enough power of Z0 to clear denominators; e.g. if f has degree d, we obtain Z0^d f(Z1/Z0, Z2/Z0) = F(Z0,Z1,Z2). Then this polynomial F will be homogeneous of degree d in the three variables Z0,Z1,Z2. Then the zero locus of F in P^2, which makes sense because a homogeneous equation vanishes on [a:b:c] if and only if it vanishes on [ta:tb:tc], is just the closure in P^2 of the zero locus of the affine equation f. Moreover F is irreducible if and only if f is.

Example: Start from f(X,Y) = Y^2 – X^4 – 1 of degree 4. This is essentially the same topologically as problem #4, day 3, hence is a non singular affine curve whose Riemann surface is obtained by adding two points to this affine curve, and has genus one. But is this visible from the compact projective version? Homogenizing gives F(Z0,Z1,Z2) = Z0^4[Z2^2/Z0^2 – Z1^4/Z0^4 – 1] = Z0^2Z2^2 – Z1^4 – Z0^4, which is homogeneous of degree 4. Now the original affine curve was the part of this projective curve lying in the open set {Z0≠0}. Thus the added points lie on the line Z0=0. Setting Z0=0 in this new equation implies Z1 = 0, and hence Z2 ≠ 0, since in projective space some coordinate must be non zero. Since coordinates may be scaled in P^2, there is only one added point, [0:0:1]. To see whether this point is non singular we can use a different set of affine coordinates. I.e. since Z2 ≠ 0 at this point, we can look in the affine open set {Z2 ≠ 0}, where affine coordinates are given by u = Z0/Z2, v = Z1/Z2. Thus in these coordinates, the equation becomes (1/Z2^4)[ Z0^2Z2^2 – Z1^4 – Z0^4] = 0, or [(Z0^2/Z2^2) – (Z1^4/Z2^4) – (Z0^4/Z2^4)] = 0, or u^2 – v^4 – u^4 = 0. Since the point [Z0:Z1:Z2] = [0:0:1] has affine coordinates (u,v) = (0,0), this is a singular point.

To compute the number of local analytic branches we can proceed as follows. This equation can be written as u^2 – u^4 – v^4 = u^2(1-u^2) – v^2 = 0. Since near (0,0), 1-u^2 is an analytic unit, we can take a square root e of it and take as new local analytic coordinate t = u.e. In this coordinate, the equation becomes t^2 – v^4 = 0, which factors locally as (t-v^2)(t+v^2) = 0. Hence there are two local analytic branches t = v^2 and t = -v^2, both tangent to each other, but distinct. Hence the Riemann surface has two distinct points in place of this singular point. This is what we “saw” from our explicit pictures earlier for similar examples, when we concluded the genus was one. Thus the projectivization only helped us find the number of missing points, but we will soon have a way to compute the genus of this example as well, using Milnor numbers of isolated singular points.

Remark: Sometimes we are careless with homogeneous coordinates and their affine counterparts, using the same letter for both. This can be confusing but is often quicker for some calculations. For instance in the previous example, starting from affine equation y^2 - x^4 -1 = 0, we could have “homogenized” just by boosting all monomials up to degree 4, adding one new letter, getting y^2 z^2 –x^4 – z^4 = 0. Then z=0 implies x = 0, so y = 1, and we have the point [x:y:z] = [0:1:0]. Here y=1, so we set y =1 and take affine coordinates x,z, getting equation z^2 –z^4 – x^4 = 0. This is essentially equivalent to the equation above, or at least it gives the same local analysis of the point (x,z) = (0,0). Of course x means something different in all three equations here. First it represents Z1/Z0, then it represents Z1, and finally it represents Z1/Z2. So be careful with this.

The homogeneous implicit function theorem

There is another way to check singularity of a point on a projective curve directly from the homogeneous equation, without using affine coordinates. I.e. a point on a projective curve with homogeneous equation F(Z0,Z1,Z2) = 0 is singular if and only if all partials ∂F/∂Zj = 0 at the point. In the previous example we could have thus used the equation F = Z0^2Z2^2 – Z1^4 – Z0^4, with partials ∂F/∂Z0 = 2Z0.Z2^2 – 4Z0^3, ∂F/∂Z1 = -4Z1^3, and ∂F/∂Z2 = 2Z0^2Z2. All these partials vanish at [0:0:1], so this is a singular point of the curve. In fact in the homogeneous case, it follows from the vanishing of the partials that the homogeneous polynopmial itself vanishes at the point, so we do not even have to check that the point is on the curve. I.e., if all ∂F/∂Zj = 0 at a point then F also vanishes at that point. This follows from “Euler’s theorem”, the easy result that d.F = Z0.∂F/∂Z0 + Z1.∂F/∂Z1 + Z2.∂F/∂Z2. This can be checked on monomials F = Zj^d since both sides are linear in F.

To see that the homogeneous criterion for singularity is equivalent to the affine one is not hard either using the chain rule and the basic fact that the gradient of a function is perpendicular to the level sets of the function. E.g. in the open set Z0 ≠ 0, we have affine coordinates X,Y and a parametrization defined by sending (X,Y)--->[1:X:Y] = {Z0,Z1,Z2]. If F(Z0,Z1,Z2) is the homogeneous equation of our curve, the affine equation of this piece of it is f(X,Y) = F(1,X,Y). Then by the chain rule, if all partials ∂F/∂Zj vanish at a point, then all partials ∂f/∂X, ∂f/∂Y vanish at the affine coordinates (X,Y) of that point. Conversely if both affine partials vanish at a point [1:x0:y0], then by the chain rule at least ∂F/∂Z1, and ∂F/∂Z2 vanish there. If ∂F/∂Z0 did not vanish at this point, then ∂F/∂Z0 has form [a,0,0] where a ≠ 0. But gradF = (∂F/∂Z0,∂F/∂Z1,∂F/∂Z2) is perpendicular to the level set of F which contains the line (t,x0,y0), for all t. Thus gradF should be perpendicular to the tangent vector (1,0,0) of this line, a contradiction, since (1,0,0) is not perpendicular to (a,0,0) unless a = 0.

Projective n space is defined exactly like P^2, as the quotient space of k^(n+1)-(0,...,0) by the k* action of scaling the coordinates. It is covered by n+1 holomorphically compatible affine open subsets each homeomorphic to k^n, and has thereby the structure of a complex manifold of dimension n. We can define curves in P^n for n > 2 also but it requires at least (n-1) equations to do so in P^n. For instance in P^3, each equation will define a 2 dimensional surface, and two of these surfaces will usually intersect in a curve.

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mathwonk
Homework Helper
8320 Day 5, Wed Jan 20, 2010 Milnor numbers & the genus of a Riemann surface

We are assuming a big theorem, the existence of the Riemann surface of an reduced projective plane curve. I.e. let F = F1....Fn be a product of distinct irreducible homogeneous polynomials in (x,y,z). The curve they define in P^2 is called a reduced projective plane curve. It is the union of the irreducible components Fj =0. The Riemann surface of F=0 is the disjoint union of the Riemann surfaces of each of the Fj, so it suffices to state the result about an irreducible curve C:{F=0}. If F is irreducible, there is a compact connected Riemann surface X and a holomorphic map X--->C which is an isomorphism over the open set of non singular points of C. For each singular point p of C, a small punctured neighborhhood of p is homeomorphic to a non empty disjoint union of a finite set of punctured discs. The number of these discs is called the number of local branches at p. There are exactly as many inverse images of p in X as there are local branches at p on C.

Thus constructing X from C is a matter of separating all the local branches at each singular point, and then smoothing the branches out if necessary. I.e. a singularity can arise because several non singular branches intersect at p, or because one singular branch occurs at p, or a combination of these phenomena. The map X--->C is bijective, and thus a homeomorphism, if and only if at each point there is only one branch, and is a holomorphic isomorphism if and only if all points of C are non singular. Of course if a curve C is non singular, then it equals its own Riemann surface, so we can say we have constructed the Riemann surface in these cases, using the implicit function theorem to define a coordinate cover. We have not actually constructed any Riemann surfaces from singular curves, but assuming they exist, we have been using the description above to study the genus of the Riemann surfaces of various singular and non singular curves.

I.e. we are trying to understand the Riemann surface associated to an irreducible complex projective plane curve, a compact connected orientable real surface with a holomorphic structure. So first of all we want to compute its topological genus. Without proof, we are going to state some powerful facts about the topology of complex plane curves that is classical, but has become known as the theory of Milnor numbers and Milnor fibers, because Milnor made it precise and generalized it to higher dimensional complex hypersurfaces in a highly recommended pamphlet, Singularities of complex hypersurfaces. Other good references include Algebraic plane curves, by Brieskorn and Knorrer, and Singularities of plane curves, by C.T.C.Wall.

We want to discuss how plane curves vary in families, especially as they acquire isolated singular points. It turns out that non singular plane curves in nice families do not vary at all topologically, but they can vary holomorphically. If a plane curve acquires an isolated singular point however, the topology of the curve will change, and the genus of the associated Riemann surface will go down. We want to compute how much it goes down. Then we can calculate the genus of our Riemann surface from knowing the genus of a non singular curve of the same degree, and then computing how much the genus of our curve is diminished by its singularities. This is determined by an invariant called the Milnor number µ, plus the number r of irreducible local analytic branches of the curve near the singularity.

In the other direction, it can be easier to understand the genus of a singular curve than a non singular one, so we can also use a sort of inductive procedure to go back from the genus of a singular curve to the genus of a non singular one, again by knowing how much the genus changes due to the singularities. E.g. a line, a curve of degree one, is isomorphic to P^1, which is topologically a sphere, and every curve of degree d can be thought of as obtained by "smoothing" the singularities of a set of d general lines. So if we know the genus of a line, and how the genus changes due to a simple singularity like two lines crossing, we can work backwards and learn the genus of any non singular curve of degree d.

Here is the answer. If we define the genus of a smooth surface as the unique g such that the Euler characteristic is 2-2g, then d disjoint lines have genus 1-d. Then smoothing an intersection of two lines is like forming a connected sum of two disjoint spheres, hence increases the genus by one. Since d lines cross pairwise at (1/2)d(d-1) points, smoothing all self intersections results in a genus for a smooth curve of degree d of (1-d) + (1/2)d(d-1) = (1/2)(d-1)(d-2).

In general the main result is that the genus of the Riemann surface of an irreducible curve is completely determined by the degree and the singularities of the curve as follows. First of all, every non singular plane curve of degree d has the same genus pa(d). Each singularity lowers this genus by a positive amount determined by the local analytic equation at the singularity. Thus our computation of the genus of a plane curve can be done by computing the genus of a non singular curve of degree d, and computing the amount of the decrease due to each singular point of our given curve.

Every singular point p on an an irreducible plane curve C has a neighborhood homeomorphic to a “one point union” of discs joined at their centers. The number of these discs is called the number of local analytic branches of C at p. We will see that a non singular complex projective plane curve of degree d has genus pa(d) = (1/2)(d-1)(d-2). For the decrease due to a given singular point, assume C has irreducible affine equation f(x,y) = 0, and (0,0) is the given singular point, i.e. f = ∂f/x = ∂f/∂y = 0 at (0,0). The Milnor number of the singularity µ = dim(C[[x,y]]/(∂f/∂x, ∂f/∂y), the dimension as (finite dimensional) complex vector space, where C[[x,y]] is the ring of formal power series in the two variables x,y. Then the decrease in the genus due to this singularity is ∂(p) = (1/2)(µ + r-1) where r is the number of local analytic branches of the curve at the singularity. In particular, the genus can drop quite a lot if either there are many branches at the singularity or if the Milnor number is large.

Thus the genus of the Riemann surface of an irreducible complex projective plane curve C of degree d equals pa(d) = (1/2)(d-1)(d-2), if C is non singular, and equals pa(d) - ∂, where ∂ = the sum of the ∂(p), summed over the finite set of singularities p of C. In particular the genus of the Riemann surface of an irreducible plane curve C can be computed from the degree of C, and the number of local analytic branches and the Milnor number for each singularity of C.

More precisely, a topological model of a non singular curve of degree d can be constructed from any irreducible curve C of degree d as follows. For each singularity p of the given curve C, remove an open neighborhood of the singularity. This neighborhood has a boundary homeomorphic to a disjoint union of r circles, where r is the number of local analytic branches at p. Then glue in place of that neighborhood of p, a compact connected orientable manifold H with boundary homeomorphic to r circles, and with 1st homology group of rank µ. H has the homotopy type of a wedge of µ circles.

For example, if we define an ordinary double point (odp) as a point p where f = ∂f/∂x = ∂f/∂y = 0, and the Hessian matrix of second partials evaluated at p is invertible, then by the inverse function theorem (∂f/dx, ∂f/∂y) is a local analytic coordinate system for k^2 at p, where k = the complex number field, so (∂f/∂x, ∂f/∂y) generates the maximal ideal of the ring of formal power series k[[x-x(p),y-y(p)]] at p, so the Milnor number µ(p) = dim(k[[x-x(p),y-y(p)]]/ (∂f/∂x, ∂f/∂y)) = 1. Then ∂(p) = (1/2)(µ+r-1) = (1/2)(1+2-1) = 1. I.e. for any ordinary double point p, ∂(p) = 1. Thus the genus of a plane curve C is reduced from that of a smooth curve of the same degree, by one for each odp on C.

For example, we have seen by studying the topology of branched double covers that (the projective closure of) {y^2 = x(x^2-1)} is non singular of genus one. This confirms our formula since every non singular cubic has genus pa(3) = (1/2)(2)(1) = 1.

As another example, {y^2 - x^2(x+1) = 0} is singular at p = (0,0) and otherwise is non singular, including at infinity. The partials at the origin are (-2x-3x^3, 2y), and hence generate the maximal ideal (x,y) of the power series ring k[[x,y]] at the origin, (since -2-3x^2 is a unit in this ring). So this is an odp, µ = 1, and the Riemann surface has genus one less than a non singular cubic, i.e. 1-1 = 0.

Since a projective line is isomorphic as we shall see later, to P^1, hence diffeomorphic to a sphere, the union of two transverse lines looks topologically like two spheres joined at one point. There are two branches at the singularity and the Milnor number is 1, since we may take affine coordinates in which the two lines are defined by xy = 0, and the singularity is at (0,0). Thus the smooth curve of degree 2 is obtained topologically from two spheres by removing a small neighborhood of their common point, i.e. removing one disc from each sphere, and replacing them by a single connected manifold with boundary consisting of two circles and having the homotopy type of 1 circle. This is a tube, or cylinder, and results in a surface again topologically equivalent to a sphere, so a non singular degree two projective plane curve, i.e. a non singular plane conic, has genus 0.

For a non singular curve of degree 3 we know the genus is one, but we can see it this way also. Let us consider a non singular conic and a line transverse to the conic, meeting it twice again in two ordinary double points. Then both surfaces that meet have genus zero, and they meet at ordinary double points each of Milnor number 1, but there are two of them, so we join the two spheres at two points using two separate cyclinders, and we obtain a torus of genus one. Similarly each time we add a transverse line to a non singular curve of degree d-1 to get a singular curve of degree d, we obtain d-1 ordinary double points at the intersections, and the genus of a nearby smoothed non singular curve of degree d, is increased by d-2 from that of the non singular curve of degree d-1. Hence the genus of a curve of degree d equals 1+2+....+(d-2) = (1/2)(d-1)(d-2), and the formula holds for all d≥1. Now that we have used the genus change from curves with easy singularities to learn the genus of a non singular curve, we can work backwards to learn the genus of some curves with more complicated singularities.

To recap, on a singular curve a small neighborhood of an isolated singularity p with r local analytic branches looks topologically like a cone over a disjoint union of r circles, i.e. a one point union of r discs. [The boundary circles are linked and/or knotted however in the 3 sphere in C^2.] We know the Riemann surface is then obtained by replacing this neighborhood of the singularity by a disjoint union of r discs. Note that in some sense, these two sets have the same boundary curve, a disjoint union of r circles. Milnor's theory (which was already known classically in this dimension) tells us how to obtain a topological model of a nearby non singular curve in a related way. I.e. if our singular curve is f = 0, a nearby non singular curve will have form f = e for some small non zero complex number e. A small open ball centered at the original singularity p this time will meet the nearby non singular curve in a set of non singular points, hence an open Riemann surface. Milnor's theory gives us the topology of this Riemann surface. It says the intersection of the nearby non singular curve with a small closed ball centered at p, will be a compact connected real 2 manifold, a “handlebody” whose boundary is still r disjoint circles, and whose 1st betti number equals µ = dim(C[[x,y]]/(∂f/∂x, ∂f/∂y)), as a finite dimensional complex vector space. Indeed this manifold has the homotopy type of a wedge of µ circles. If we cap off each boundary circle in the handlebody by a disc we get a manifold of genus g' where g' + (r-1) is the difference in the genus of the Riemann surface of the original singular curve and the genus of the nearby non singular curve, assuming p is the only singularity of our singular curve. I.e. g'+(r-1) is the drop in the genus of the non singular curve attributed to that one isolated singular point p. As stated above this drop equals ∂(p) = (1/2)(µ + r-1) = g' + (r-1), i.e. µ = 2g’ + r-1.

Test this on a curve with affine equation y^2 = x^6-1. This curve is non singular except at infinity, and we find the homogeneous equation y^2z^4 = x^6-z^6, and set z = 0. Then we get 0 = x^6, a single point of the projective plane [0:1:0]. To find an affine equation at this point we set y = 1, and get the equation f(x,z) = z^4 - x^6+z^6 = 0. Then ∂f/∂x = -6x^5, and ∂f/∂z = 6z^5 + 4z^3 = z^3(6z^2 + 4). The ideal these two partials generate in the power series ring k[[x,z]] is generated by (x^5,z^3) since in the power series ring 4 + 6z^2 is a unit. Thus the dimension µ of the quotient vector space k[[x,z]]/( ∂f/∂x, ∂f/∂z) = k[[x,z]]/(x^5,z^3), equals the number of monomials of form x^iz^j with 0≤i≤4, 0≤j≤2, i.e. µ = 15.

Now we have studied hyperelliptic curves like this in pictures and we know this one has two local branches at infinity. Thus we obtain a smooth curve of degree 6 from this singular degree 6 curve by plumbing in a connected manifold with two boundary circles, and the homotopy type of a wedge of 15 circles. This is a manifold of genus 7 with 2 discs removed. Whether you can see that or not, our formulas show that plumbing this manifold into the hole left by removing the singularity, raises the genus of our singular curve by (1/2)(µ + r-1) = 8. In fact the formula for the genus of a non singular curve of degree 6 says it should be 10, and we know from our work in class that the singular curve {y^2 = x^6-1} above, has a Riemann surface which is a double cover of a sphere with 6 branch points, hence has genus 2. So it checks out.

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Now we will compactify k^2 by adding in a copy of P^1 to form P^2. Thus if we think of k^0 as one point, then P^2 = k^2 + k^1 + k^0, as disjoint union of sets.

Since complex 3 space k^3 ≈ R^6 = real 6 space, and every complex line through the origin of complex 3 space, is a real plane through the origin of R^6, which meets the 5-sphere in a circle, the 5-sphere maps surjectively onto the quotient space P^2. Since the sphere is compact, so is P^2.
I'm mostly interested to understand this part.

Can you explain it further how a real 5-sphere is essentially the same as k^2 + k^1 + k^0?

Infrared
Gold Member
I think you meant to quote @mathwonk there, but I showed in my post 9 how to write ##\mathbb{P}^n=k^n\cup\mathbb{P}^{n-1}.## Now just repeat this ##n## times.

But note that ##\mathbb{P}^2_{\mathbb{C}}## is a quotient of ##S^5##, not homeomorphic to it. For one, they have different dimensions.

• victorvmotti
##\mathbb{P}^n=k^n\cup\mathbb{P}^{n-1}.## Now just repeat this ##n## times.
When we say that two edges of a rectangular are glued together to make a cylinder I can imagine it.

But when we say that ##\mathbb{P}^2=\mathbb{C}^2 \cup\mathbb{C}^1\cup\mathbb{C}^0## a little challenge to visualize for me.

How ##\mathbb{C}^2## and ##\mathbb{C}^1## are glued? Which points go to which points?

Are we saying that all the points at the infinity in ##\mathbb{C}^2## are identified with ##\mathbb{C}^1## and also all the points at the infinity in ##\mathbb{C}^1## are identified with ##\mathbb{C}^0##? And after all these identifications we build ##\mathbb{P}^2##?

mathwonk
Homework Helper
since every complex "plane" (i.e. copy of C^1) through the origin of C^3 represents only one point of P^2, but meets the 5 sphere in a circle, the map from the 5 dimensional sphere to the (4 dimensional) P^2, has circles for fibers, hence represents the 5 sphere as a circle bundle over P^2.

• victorvmotti
since every complex "plane" (i.e. copy of C^1) through the origin of C^3 represents only one point of P^2, but meets the 5 sphere in a circle, the map from the 5 dimensional sphere to the (4 dimensional) P^2, has circles for fibers, hence represents the 5 sphere as a circle bundle over P^2.
The bundle language and picture was very helpful for me to make sense of this.

mathwonk
Homework Helper
yes, well by definition, P^2 is obtained from C^3 by identifying each linear copy of C^1 passing through the origin, to a point. If we remove the origin, this means we are collapsing each punctured real plane (since these are copies of C^1 we really should call them complex lines) through the origin, to a point. Thus simply by definition, each point of P^2 is the image of a punctured real plane, which is topologically similar to a circle, i.e. it retracts to a circle. So we have a map C^3 - {0}-->P^2, whose fibers are punctured planes, and if we restrict this map to the 5 sphere in C^3, we get a map S^5-->P^2 whose fibers are circles.

One can also add in extra origins, one for each punctured plane, i.e. one for each point of P^2, and then this defines a bundle over P^2 whose fibers are copies of C^1, no longer punctured ones, called the tautological complex "Line bundle" over P^2. The total space of this bundle is C^3 but modified to have a whole copy of P^2 where there used to be just the one point 0, the origin of C^3. This is called "blowing up" C^3 at a point. This is sort of the opposite of putting a copy of P^2 at "infinity" of C^3 to form P^3. This line bundle is called O(-1), bercause its chern class is -1, in particular it does not have a nice section.

So there are two ways to add a copy of P^2 to C^3-{0}. You can put it at "infinity" or you can put it at {0}. If you put it at infinity, thus getting a punctured copy of P^3, you also get a C^1 bundle over that copy of P^2. Namely draw any complex line through the hole where 0 was, and then it meets the P^2 in one point at infinity, so that adds back in the point of C^1 that was removed at 0. So the family of all complex lines through the missing 0, give you a disjoint family of complex lines, one over each point of the P^2 at infinity. Thism bundle, O(1), does have a nice section, namely choose any affine linear copy of C^2 in C^3 that misses the origin, and then it will meet every complex line through the origin exactly once, except for those that are "parallel" to it. But it meets those once each at infinity. Since this line bundle over P^2, has a nice section that meets the base space in a line, it's chern class is 1, so that gives it its name, O(1).

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WWGD
Gold Member
2019 Award
8320 Day 5, Wed Jan 20, 2010 Milnor numbers & the genus of a Riemann surface

We are assuming a big theorem, the existence of the Riemann surface of an reduced projective plane curve. I.e. let F = F1....Fn be a product of distinct irreducible homogeneous polynomials in (x,y,z). The curve they define in P^2 is called a reduced projective plane curve. It is the union of the irreducible components Fj =0. The Riemann surface of F=0 is the disjoint union of the Riemann surfaces of each of the Fj, so it suffices to state the result about an irreducible curve C:{F=0}. If F is irreducible, there is a compact connected Riemann surface X and a holomorphic map X--->C which is an isomorphism over the open set of non singular points of C. For each singular point p of C, a small punctured neighborhhood of p is homeomorphic to a non empty disjoint union of a finite set of punctured discs. The number of these discs is called the number of local branches at p. There are exactly as many inverse images of p in X as there are local branches at p on C.

Thus constructing X from C is a matter of separating all the local branches at each singular point, and then smoothing the branches out if necessary. I.e. a singularity can arise because several non singular branches intersect at p, or because one singular branch occurs at p, or a combination of these phenomena. The map X--->C is bijective if and only if at each point there is only one branch, and is an isomorphism if and only if all points of C are non singular. Of course if a curve C is non singular, then it equals its own Riemann surface, so we can say we have constructed the Riemann surface in these cases, using the implicit function theorem to define a coordinate cover. We have not actually constructed any Riemann surfaces from singular curves, but assuming they exist, we have been using the description above to study the genus of the Riemann surfaces of various singular and non singular curves.

I.e. we are trying to understand the Riemann surface associated to an irreducible complex projective plane curve, a compact connected orientable real surface with a holomorphic structure. So first of all we want to compute its topological genus. Without proof, we are going to state some powerful facts about the topology of complex plane curves that is classical, but has become known as the theory of Milnor numbers and Milnor fibers, because Milnor made it precise and generalized it to higher dimensional complex hypersurfaces in a highly recommended pamphlet, Singularities of complex hypersurfaces. Other good references include Algebraic plane curves, by Brieskorn and Knorrer, and Singularities of plane curves, by C.T.C.Wall.

We want to discuss how plane curves vary in families, especially as they acquire isolated singular points. It turns out that non singular plane curves in nice families do not vary at all topologically, but they can vary holomorphically. If a plane curve acquires an isolated singular point however, the topology of the curve will change, and the genus of the associated Riemann surface will go down. We want to compute how much it goes down. Then we can calculate the genus of our Riemann surface from knowing the genus of a non singular curve of the same degree, and then computing how much the genus of our curve is diminished by its singularities. This is determined by an invariant called the Milnor number µ, plus the number r of irreducible local analytic branches of the curve near the singularity. In the other direction, it can be easier to understand the genus of a singular curve than a non singular one, so we can also use a sort of inductive procedure to go back from the genus of a singular curve to the genus of a non singular one, again by knowing how much the genus changes due to the singularities. E.g. a line, a curve of degree one, is isomorphic to P^1, which is topologically a sphere, and every curve of degree d can be thought of as obtained by removing the singularities of a set of d general lines. So if we know the genus of a line, and how the genus changes due to a simple singularity like two lines crossing, we can work backwards and learn the genus of any non singular curve of degree d. Here is the answer. If we define the genus of a smooth surface as the unique g such that the Euler characteristic is 2-2g, then d disjoint lines have genus 1-d. Then smoothing an intersection of two lines is like forming a connected sum of two disjoint spheres, hence increases the genus by one. Since d lines cross pairwise at (1/2)d(d-1) points, smoothing all self intersections results in a genus for a smooth curve of degree d of (1-d) + (1/2)d(d-1) = (1/2)(d-1)(d-2).

In general the main result is that the genus of the Riemann surface of an irreducible curve is completely determined by the degree and the singularities of the curve as follows. First of all, every non singular plane curve of degree d has the same genus pa(d). Each singularity lowers this genus by a positive amount determined by the local analytic equation at the singularity. Thus our computation of the genus of a plane curve can be done by computing the genus of a non singular curve of degree d, and computing the amount of the decrease due to each singular point of our given curve.

Every singular point p on an an irreducible plane curve C has a neighborhood homeomorphic to a “one point union” of discs joined at their centers. The number of these discs is called the number of local analytic branches of C at p. We will see that a non singular complex projective plane curve of degree d has genus pa(d) = (1/2)(d-1)(d-2). For the decrease due to a given singular point, assume C has irreducible affine equation f(x,y) = 0, and (0,0) is the given singular point, i.e. f = ∂f/x = ∂f/∂y = 0 at (0,0). The Milnor number of the singularity µ = dim(C[[x,y]]/(∂f/∂x, ∂f/∂y), the dimension as (finite dimensional) complex vector space. Then the decrease in the genus due to this singularity is ∂(p) = (1/2)(µ + r-1) where r is the number of local analytic branches of the curve at the singularity.

Thus the genus of the Riemann surface of an irreducible complex projective plane curve C of degree d equals pa(d) = (1/2)(d-1)(d-2), if C is non singular, and equals pa(d) - ∂, where ∂ = the sum of the ∂(p), summed over the finite set of singularities p of C. In particular the genus of the Riemann surface of an irreducible plane curve C can be computed from the degree of C, and the number of local analytic branches and the Milnor number for each singularity of C.

More precisely, a topological model of a non singular curve of degree d can be constructed from any irreducible curve C of degree d as follows. For each singularity p of the given curve C, remove an open neighborhood of the singularity. This neighborhood has a boundary homeomorphic to a disjoint union of r circles, where r is the number of local analytic branches at p. Then glue in place of that neighborhood of p, a compact connected orientable manifold H with boundary homeomorphic to r circles, and with 1st homology group of rank µ. H has the homotopy type of a wedge of µ circles.

For example, if we define an ordinary double point (odp) as a point p where f = ∂f/∂x = ∂f/∂y = 0, and the Hessian matrix of second partials evaluated at p is invertible, then by the inverse function theorem (∂f/dx, ∂f/∂y) is a local analytic coordinate system for k^2 at p, where k = the complex number field, so (∂f/∂x, ∂f/∂y) generates the maximal ideal of the ring of formal power series k[[x-x(p),y-y(p)]] at p, so the Milnor number µ(p) = dim(k[[x-x(p),y-y(p)]]/ (∂f/∂x, ∂f/∂y)) = 1. Then ∂(p) = (1/2)(µ+r-1) = (1/2)(1+2-1) = 1. I.e. for any ordinary double point p, ∂(p) = 1. Thus the genus of a plane curve C is reduced by one from that of a smooth curve of the same degree, by each odp on C.

For example, we have seen by studying the topology of branched double covers that (the projective closure of) {y^2 = x(x^2-1)} is non singular of genus one. This confirms our formula since every non singular cubic has genus pa(3) = (1/2)(2)(1) = 1.

As another example, {y^2 - x^2(x+1) = 0} is singular at p = (0,0) and otherwise is non singular, including at infinity. The partials at the origin are (-2x-3x^3, 2y) hence they generate the ideal of the power series ring k[[x,y]] at the origin, so this is an odp, µ = 1, and the Riemann surface has genus one less than a non singular cubic, i.e. 1-1 = 0.

Since a line is isomorphic as we shall see later, to P^1, hence diffeomorphic to a sphere, the union of two transverse lines looks topologically like two spheres joined at one point. There are two branches at the singularity and the Milnor number is 1, since we may take affine coordinates in which the two lines are defined by xy = 0, and the singularity is at (0,0). Thus the smooth curve of degree 2 is obtained topologically from two spheres by removing a small neighborhood of their common point, i.e. removing one disc from each sphere, and replacing them by a single connected manifold with boundary consisting of two circles and having the homotopy type of 1 circle. This is a tube, or cylinder, and results in a surface again topologically equivalent to a sphere, so a non singular degree two projective plane curve, i.e. a non singular plane conic, has genus 0.

For a non singular curve of degree 3 we know the genus is one, but we can see it this way also. Let us consider a non singular conic and a line transverse to the conic, meeting it twice again in two ordinary double points. Then both surfaces that meet have genus zero, and they meet at ordinary double points each of Milnor number 1, but there are two of them, so we join the two spheres at two points using two separate cyclinders, and we obtain a torus of genus one. Similarly each time we add a transverse line to a non singular curve of degree d-1 to get a singular curve of degree d, we obtain d-1 ordinary double points at the intersections, and the genus of a nearby smoothed non singular curve of degree d, is increased by d-2 from that of the non singular curve of degree d-1. Hence the genus of a curve of degree d equals 1+2+....+(d-2) = (1/2)(d-1)(d-2), and the formula holds for all d≥1. Now that we have used the genus change from curves with easy singularities to learn the genus of a non singular curve, we can work backwards to learn the genus of some curves with more complicated singularities.

To recap, on a singular curve a small neighborhood of an isolated singularity p with r local analytic branches looks topologically like a cone over a disjoint union of r circles, i.e. a one point union of r discs. [The boundary circles are linked and/or knotted however in the 3 sphere in C^2.] We know the Riemann surface is then obtained by replacing this neighborhood of the singularity by a disjoint union of r discs. Note that in some sense, these two sets have the same boundary curve, a disjoint union of r circles. Milnor's theory (which is classical in this dimension) tells us how to obtain a topological model of a nearby non singular curve in a related way. I.e. if our singular curve is f = 0, a nearby non singular curve will have form f = e for some small non zero complex number e. A small open ball centered at the original singularity p this time will meet the nearby non singular curve in a set of non singular points, hence an open Riemann surface. Milnor's theory gives us the topology of this Riemann surface. It says the intersection of the nearby non singular curve with a small closed ball centered at p, will be a compact connected real 2 manifold, a “handlebody” whose boundary is still r disjoint circles, and whose 1st betti number equals µ = dim(C[[x,y]]/(∂f/∂x, ∂f/∂y)), as a finite dimensional complex vector space. Indeed this manifold has the homotopy type of a wedge of µ circles. If we cap off each boundary circle in the handlebody by a disc we get a manifold of genus g' where g' + (r-1) is the difference in the genus of the Riemann surface of the original singular curve and the genus of the nearby non singular curve, assuming p is the only singularity of our singular curve. I.e. g'+(r-1) is the drop in the genus of the non singular curve attributed to that one isolated singular point p. As stated above this drop equals ∂(p) = (1/2)(µ + r-1) = g' + (r-1), i.e. µ = 2g’ + r-1.

Test this on a curve with affine equation y^2 = x^6-1. This curve is non singular except at infinity, and we find the homogeneous equation y^2z^4 = x^6-z^6, and set z = 0. Then we get 0 = x^6, a single point of the projective plane [0:1:0]. To find an affine equation at this point we set y = 1, and get the equation f(x,z) = z^4 - x^6+z^6 = 0. Then ∂f/∂x = -6x^5, and ∂f/∂z = 6z^5 + 4z^3 = z^3(6z^2 + 4). The ideal these two partials generate in the power series ring k[[x,z]] is generated by (x^5,z^3) since in the power series ring 4 + 6z^2 is a unit. Thus the dimension µ of the quotient vector space

k[[x,z]]/( ∂f/∂x, ∂f/∂z) = k[[x,z]]/(x^5,z^3), equals the number of monomials of form x^iz^j with 0≤i≤4, 0≤j≤2, i.e. µ = 15. Now we have studied hyperelliptic curves like this in pictures and we know this one has two local branches at infinity. Thus we obtain a smooth curve of degree 6 from this singular degree 6 curve by plumbing in a connected manifold with two boundary circles, and the homotopy type of a wedge of 15 circles. This is a manifold of genus 7 with 2 discs removed. Whether you can see that or not, our formulas show that plumbing this manifold into the hole left by removing the singularity, raises the genus of our singular curve by (1/2)(µ + r-1) = 8. In fact the formula for the genus of a non singular curve of degree 6 says it should be 10, and we know from our work in class that the singular curve {y^2 = x^6-1} above, has a Riemann surface which is a double cover of a sphere with 6 branch points, hence has genus 2. So it checks out.
Why dont you Latex these entries into a book?

mathwonk
Homework Helper
I was trying to write up these notes back in 2010 when I retired, and never finished the work. I also have yet to learn Latex, but that is a good excuse to do so. A lot of my course was also cribbed from the lovely book of Rick Miranda, Riemann surfaces and algebraic curves. When you teach a course you use the best available source for every topic, mostly by other people, and then when you want to publish your own book, I thought you sort of need to only publish your own version, which requires changing everything you borrowed. Of course in reality many books borrow from each other. Sometimes it also seems like cheating, in math books, to use big results you do not prove, but in this topic it really helps to have Milnor numbers available, to understand the topology of plane curves. I like to give my students some power to calculate things, even if they do not always see all the proofs of those powerful techniques.

In other real life difficulties, I can no longer open most of the lengthy and detailed notes I wrote some 20 years ago, because the word processing software has changed so much those files are not compatible with my current system. Indeed this should be motivation to learn to use Latex.

WWGD
Gold Member
2019 Award
I was trying to write up these notes back in 2010 when I retired, and never finished the work. I also have yet to learn Latex, but that is a good excuse to do so. A lot of my course was also cribbed from the lovely book of Rick Miranda, Riemann surfaces and algebraic curves. When you teach a course you use the best available source for every topic, mostly by other people, and then when you want to publish your own book, I thought you sort of need to only publish your own version, which requires changing everything you borrowed. Of course in reality many books borrow from each other. Sometimes it also seems like cheating, in math books, to use big results you do not prove, but in this topic it really helps to have Milnor numbers available, to understand the topology of plane curves. I like to give my students some power to calculate things, even if they do not always see all the proofs of those powerful techniques.

In other real life difficulties, I can no longer open most of the lengthy and detailed notes I wrote some 20 years ago, because the word processing software has changed so much those files are not compatible with my current system. Indeed this should be motivation to learn to use Latex.
I am willing to help with a bit of it, maybe others here can too. Maybe you can pose this as a question? I used the software Scientific Workplace a while back , which had a Latex drop menu . Maybe you can check it out? Id offer a link but Im on my phone, hard to do.

"But when we say that P2=C2∪C1∪C0 a little challenge to visualize for me."

You know complex projective n-space Pn is defined as Cn+1 - {0} factored out by the equivalence relation (u, v, ..., w) ~ (cu, cv, ..., cw) for any (u, v, ..., w) ∈ Cn+1 - {0} and any c ∈ C - {0}. Now, every equivalence class (i.e., every 1-dimensional complex subspace of Cn+1) intersects the unit sphere S2n+1 in Cn+1 = R2n+2. So, Pn can perhaps be even more simply expressed as the quotient of the unit sphere S2n+1 in Cn+1 factored out by the equivalence relation

(u, v, ..., w) ~ (cu, cv, ..., cw)

for any (u, v, ..., w) ∈ S2n+1 and c ∈ C with |c| = 1.

Using this, you can check that for n = 1, the quotient map S3 → P1 is what's called the Hopf fibration. And then in general if you have Pn , you can make a Pn+1 by taking the unit ball

{(z_0, ...,. z_n) ∈ Cn | |z_0|2 + ... + |z_n|2 <= 1}

(which is a real (2n+2)-dimensional closed ball) and mapping its boundary sphere, which is a (2n+1)-dimensional sphere, onto Pn by the above quotient map ... thereby glueing the boundary of the ball onto the Pn by that map, and creating a new space. This space will be a copy of Pn+1.

Doing this process inductively, you can see that each time you glue a ball to the previous stuff, you are adding new material only from the interior of the ball. (The boundary of the ball got identified to stuff you already had.) So by induction, the final result is the union of a 0-ball and a 2-ball, and ... and a (2n+2)-ball.

• victorvmotti and mathwonk
mathwonk
Homework Helper
@WWGD: my advisor said he used that software too, but when i looked into it they had stopped producing the version that worked with a mac i think.

well their current page says the newest version 6 is again ok for macOS except not 10.15 catalina. I have 10.13 high sierra, so apparently that is good, and their home office is apparently about 15 miles away from me in poulsbo, so maybe someday i can drop by. Oops, pricey software, now that I no longer have an NSF grant for such things. But thanks for the tip.

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write ##\mathbb{P}^n=k^n\cup\mathbb{P}^{n-1}.## Now just repeat this ##n## times.
After noting the above, can we say there is a bijective map or isomorphism between ##\mathbb{P}^2## which is ##\mathbb{C}^2\cup\mathbb{P}^1## and a unit closed disc defined in ##\mathbb{C}^2##? Then conclude that because such a disc is compact then ##\mathbb{P}^2## is also compact? Can we say that by defining or constructing ##\mathbb{P}^2## we are actually taking ##\mathbb{C}^2## and then compactify all the points of infinity in ##\mathbb{C}^2## into a a boundary and in this way we arrive at a compact space ##\mathbb{P}^2## which is locally isomorphic to a subset of ##\mathbb{C}^2##?

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