(I don’t think that being educated dummy is something bad – in fact, I consider myself to be one).

Around month ago, I gave here, at Warsaw University, a very introductory talk about Gromov-Witten invariants. It was aimed at students who hadn’t heard about this topic but who knew standard things from university course: cohomology, vector bundles, Chern classes, Poincare duality, etc. There were also some local wise men and they didn’t say I totally screw it up.

It was based on first 9 chapters of a Sheldon Katz’s fantastic book: “Enumerative geometry and string theory” . This is suberbly written, it starts at the level of highschool students and it ends… well I don’t actually know where it ends, because I still have some chapters to read, but it surely introduces Gromov-Witten invariants.

Accordingly, the book has actually one disadvantage: Katz tries to explain really all, even things like what topology is. So, in case You know at least very roughly things I mentioned above, You may want to read notes I prepared.

Everything is in “algebraic geoemtry framework” – so that it’s meaningful to speak about for example cubics in P2. When I write Pn, I mean CPn. I use unicode for mathematical notation.

Any kind of feedback is welcomed!

BTW, I just found out that Sheldon is actually a name, not surname. When I was writing the notes, I didn’t check it and so I write “Sheldon’s book” instead of “Katz’s book”. Sorry for that.

BTW2, There’s a new edition of a book “J-holomorphic curves and symplectic topology” by Dusa McDuff and Dietmar Salamon. It has about 5 000 000 pages (I know what I’ve seen. However, Amazon says it has 669). First edition had 200 and I could at least dream of understanding it someday. Life’s so damn cruel.

BTW3, WOW!!! Maxim Kontsevich comes to Warsaw next week!!!! Believe it or not, but in the last post I chose Kontsevich as a random Fields medalist! :-).

*

We’re gonna try to find answers for 5 questions – from the very simple one to one without known answer.

(1)

We start with a question “Given two different lines in P2, in how many points they intersect?”. Obviously, in 1 point: we look at any affine part (it is isomorphic with C2) of P2; if given lines aren’t parallel there, then they intersect in this C2 and they don’t intersect in infinity; if they are parallel, then they don’t intersect in C2 but they do intersect in infinity.

 

(2)

Next question: “Given two different points in P2, how many lines go through them?”. Although it’s very easy to see that the answer is “1” we’ll come to this in a little bit strange way, one which will show the method we’ll use in the next problems.

What we are looking for? We’re looking for lines in P2 that fulfill some condition. What we’re going to do is to look at a space of all lines in P2 and try to express our condition in terms of this space.

Line in P2 is a set of solutions of equation

aX + bY + cZ = 0

where a, b, c are some complex coefficients not all equal to 0. Furthermore, two such equations with coefficients (a, b, c) and (a’, b’, c’) give rise to the same line iff a’/a = b’/b = c’/c. It follows that lines in P2 are in natural correspondence with points of another, dual, P2 (line <—-> coefficients of an equation) which, to avoid confusion, I’ll call P2*.

Now, suppose we have a point in P2 with coefficients (α, β, γ), and a line in P2 represented by a point in P2* with coefficients (a, b, c). Clearly, line (a,b,c) contains point (α, β, γ) iff

aα + bβ + cγ = 0.

So, if we fix our point (α, β, γ), we conclude that the set of all lines in P2 (we think that they are elements of P2*) that go through this point is a line defined by equation

Xα + Yβ + Zγ = 0.

Let’s translate our original problem of finding number of lines going trough 2 fixed points in P2 to this new situation. If points are (α, β, γ) and (α’, β’, γ’) than clearly what we want to know is in how many points two lines in P2*:

Xα + Yβ + Zγ = 0

Xα’ + Yβ’ + Zγ’ = 0

intersect. This was a first question; they intersect in 1 point.

 

(3)

Now something little bit harder. How many conics (curves of degree 2) in P2 contain given i points and are tangent to given (5-i) lines?

We’ll try to solve it by the same general scheme. We search for some conics in P2, so we find a manifold X of all conics in P2. For every of our 5 conditions, we search for a submanifold of X containing exactly these conics which fulfill particular condition. Finally, we investigate how many points are in an intersection of all 5 submanifolds – this is number we’re looking for.

Conic in P2 is given by an equation

aX2 + bY2 + cZ2 + dXY + eXZ + fYZ = 0.

Just as before, we see that conics in P2 are simply points of some P5, which I’ll call P5*. Given point in P2, we see (again as before) that conics that go trough this point form a hyperplane in P5*.

What about condition for a conic (point of P5*) to be tangent to given line? We want to express this condition algebraically (so that we have a submanifold in P5*). To do it we modify our problem slightly: instead of looking for conics tangent to given line, we’ll look for conics that intersect with given line in precisely 1 point. Although these two problems are very similar, they are not exactly the same – but let’s pretend for a moment that they are.

Let’s take a line in P2 given by equation

Z = 0.

(This is without a loss of generality – every line is of such form after suitable change of coordinates.) Conic (a, b, c, d, e, f)∊P5* has exactly 1 common point with this line iff

aX2 + bY2 + dXY = 0 has exactly one solution.

It’s straightforward to see that this equation has exactly one solution iff

Δ = b2-4ad = 0.

So, submanifold of P5* for condition of tangency (or rather of having one point in common with given line) is a conic in P5*.

Accordingly, to obtain an answer for a question (3) we need to know how many points are in an intersection of i hyperplanes and 5-i conics in P5*.

(Let’s observe that if we were intersecting 3 planes in R3 then we’d obtain one common point, unless these 3 planes would be chosen in a mischievous way, and we’d end with a whole line as an intersection. The same is in P5*: we have five 4-dimensional (I mean complex dimension) things in 5-dimensional space – we expect to get something 0-dimensional (i.e. points) as an intersection. However, if the input data (5 points and 5-i lines) were chosen mischievously we’ll end with something unexpected. To prevent this, one formulates question (3) like this: Given general i points and general 5-i lines, how many conics…? Suppose I answered “58” to this question. Word “general” in this context means that I proved that there is dense open subset of {all i-tuples of points and (5-1)-tuples of lines} = (P2)i x (P5)5-i such that for all input data in this subset one gets answer “58”. However, we won’t worry about really proving existence of such dense open set – we’ll rely on our intuition (and 58i is not the right answer).)

Problems like this can be solved with Bezout’s theorem:

Theorem (Bezout): Suppose we’re given n hypersurfaces C1,…, Cn in Pn of degrees c1,…cn respectively. Suppose also that their intersection is finite. Then we have

ΣIp = Πci ,

where sum is over points p in intersection, product is over i, and Ip denotes intersection factor (“how tangently” things are intersecting in p).

Sheldon says that in our case, for general input data, all Ip=1. Thus, we have, provided that intersection is indeed finite, an answer: 25-i.

Unfortunately, this is not precisely the case. For i = 3, 4, 5 the intersection is finite and additionally conics in intersection are (for general input data) smooth (not like 2 lines intersecting: (aX+bY)(cX+dY)=0). For i = 2 the intersection is finite, but some of conics in it are pairs of intersecting lines. For i = 1, 0 intersection isn’t finite – for i=0 whatever 5 lines we choose, any double line (aX+bY)2 has exactly one point in common with any of chosen lines. This also shows the difference between real tangency and “having one point in common” – this double line doesn’t really look like it was tangent to other 5 lines.

However, Bezout’s theorem produces some kind of answer even for infinite intersection. What one can do is identify in this intersection submanifolds that one doesn’t need (like submanifold consisting of double lines) and learn how much they add to answer Bezout’s theorem provides. This is called “excess intersection theory” and some ideas of it are described nicely in Sheldon’s book.

Alternatively, one could try to compactify space of smooth conics differently, obtaining space X P5 , express given conditions in terms of submanifolds of X and intersect them (hoping that this time there won’t be excess intersection or it’ll easer to handle). Unfortunately, on general X one lacks Bezout’s theorem which is of great help in studying intersection. (One could also try to deal just with noncompact manifold of smooth conics, but investigating intersections on noncompact manifolds is much harder)

However, it’d be naive to think that space parametrising objects we’re interested in will always be Pn (or that we’d be able to compactify this space always to Pn), so, even if we’re not interested particularly in problem (3), we should find some substitution for Bezout’s theorem. Fortunately, we don’t have to look very far – natural candidates (for compact X) are cohomology ring of X and Poincare duality.

(Indeed, what is a Poincare duality for compact manifold X? One can see it roughly as a map from set of oriented submanifolds of X to cohomology ring of X which transforms intersection into cup-product. So, if we know everything about cohomology ring of X (and often we do) than we know really much about intersections of submanifolds.)

 

(4)

Now something really nice: How many lines are contained in given general quintic threefold (that is, in 3-dimensional hypersurface of degree 5 in P4)?

Scheme is the same: take space of all lines in P4, express condition for a line to be contained in a threefold in terms of submanifolds of this space, count all points in a suitable intersection. This time however, we won’t need to intersect anything – we have only one condition (being subspace of a given threefold) and so we’ll have only one (0-dimensional) submanifold.

Technologies we’re going to use are complex vector bundles and their Chern classes. Let me roughly remind what the (highest) Chern class is: suppose we have an n-dimensional manifold X and a k-dimensional vector bundle V on it. Chern class of V is an element of k-th cohomology group of X (with coefficients in Z (integers)). I won’t write how one exactly defines this element – see Milnor, Stasheff, Characteristic classes if you want to know. The property of Chern class we need is the following: suppose we are given transverse section s of V; the set of zeros of s ({x∊X: s(x)=0}) is an (n-k)-dimensional oriented submanifold of X, and its Poincare dual is a Chern class of V.

We are particularly interested in a case where X and V are both n-dimensional. Then, Chern class of V (denoted cnV, because there are also other, lower than n-th, Chern classes) is an element of Hn(X), and its Poincare dual is a 0-dimensional submanifold (i.e. set of points). Counting (oriented) points in this submanifold is equivalent to evaluating Chern class on fundamental class of X, and this is often denoted as

ʃX cnV

because if one represents cnV as a differential form than evaluating cnV on fundamental class of X really becomes above integral.

What one should keep in mind is that Chern classes are really computable. That is, if we’re given manifold which we know very well and some vector bundle on it, odds are good that we’re actually able to compute its Chern class.

Back to our problem (4). We’re interested in lines in P4 that fulfill some condition so we’ll work in a space of all lines in P4. That is, we’ll work in a complex Grassmannian G(2,5) of 2-dimensional complex linear subspaces of C5.

We’ll now define a bundle B on this G(2,5) and a section of this bundle whose zero-set consists exactly of lines we’re interested in. As we can assume that we know G(2,5) very well, in particular we know precisely the strucure of its cohomology ring (see for example Griffiths, Harris, Principles of algebraic geometry, chapter 1, paragraph 5, for multiplicative structure of cohomology in arbitrary Grassmannian G(k,n)), this will provide us with an answer.

Bundle B is as follows: Above fixed line k in P4 lie restrictions to this line of homogeneous polynomials of degree 5. Now, let’s suppose that our quintic threefold is defined as a set of solutions of equation

Q(X, Y, Z, V, W) = 0,

where Q is homogeneous form of degree 5. Q clearly defines a section of B. (on the point xͧ∊G(2,5) this section takes value “restriction of Q to line x”) and clearly lines that lie in the threefold defined by Q are precisely are precisely these points of G(2,5) on which this section takes value 0!

One proves that for general choice of Q this section is transverse and all orientations of points in a zero-set are positive. So, we can compute Chern class of B and give an answer

ʃc6(B),

where integral is over G(2,5). And we can treat it as a numerical answer: we know everything about G(2,5) and B defined explicitly, so we can easily compute (using standard tools) its Chern class and apropriate integral. The actual number is 2875 (and was first obtained by Schubert in XIX century).

(Sheldon in his book gives also simpler description of Q: as a 5th symmetic power of a well-known bundle – so that it becomes obvious that one can compute total Chern class of B using Cartan’s formule. However, I didn’t understand that completely so you have to check it on your own)

 

(5)

The climax: what is the number of rational curves of degree d on general quintic threefold? (rational curve is a curve that is an image of P1)

One should note first that answer to this question is not known. In fact, it is known only for d<10 that there is finite number of degree d curves on general quintic threefold. The statement that this number is always finite is known as Clemens conjecture. Actual number of degree d curves is known only for d=1,2,3 (and it is 2875 (Schubert, XIX century), 609250 (Katz, 1985), 317206375 (Ellingorud, Strome, 1991).

However, in 1992 physicists said that they computed the answer to our question for all d, and they didn’t worry that the numbers they have are in fact not integers. It seems that numbers they came up with are good enough approximations for their purposes (which have to do with string theory and are described in further chapters of Sheldon’s book. I didn’t read it yet) .

Shortly after, Kontsevich made physicists’ ideas precise and it was understood what really Physicists computed (basically, difference between what physicist computed and the real answer to question (5) is analogous to difference between what we computed in problem (3) and real answer to problem (3): physicists solved problem similar to question (5), but not exactly question (5), and there is some excess intersection). We turn to that right now.

The main problem in a search for an answer to question (5) with d>1 (we’ve seen that d=1 it’s fairly easy) is a lack of simple compactification of space of degree d curves in P4 (in a case d=1 we didn’t even have to compactify – degree 1 curves in P4, i.e. lines, is a well understood Grassmannian G(2,5)).

Kontsevich’s idea is to study not the space of degree d curves, but rather the space of degree d parametrized curves. More precisely:

Tree of P1‘s is a result of following construction: we take disjoint sum of n P1‘s let’s call it T0. One can form n*(n-1)/2 pairs out of them – we choose k such pairs, give them labels 1,…,k and for every chosen pair we chose 1 point on each member of a pair. Now, let T1 be a T with both chosen points in a pair 1 identified, T2 be a T1 with both chosen points in a pair 2 identified and so on. If the result, Tk, has trivial fundamental group (“there is no cycles in a graph”) then we call it tree of P1‘s. Tree of P1‘s has a natural structure of algebraic variety.

Example of a tree is a subset of P2 defined by an equation xy = 0 (two P1‘s, {x=0} and {y=0}, joined in a one point (0,0,1)). Antiexample is a subset of P2 defined by an equation xy(x+y+z)=0.

Morphism of trees α: T –> T’ is a morphism of algebraic varieties that takes every P1 of T into some single P1 of T’. It follows that isomorphism of trees is an isomorphism of algebraic varieties which is an isomorphism when restricted to any P1.

Definition: stable map of degree d and genus 0 to Pn is a map

f: C –> Pn

where C=∪Pi1 is a tree of P1‘s, such that

i) if one denotes degree of f restricted to Pi1 by di then Σdi=d

ii) if f is constant when restricted to a Pi1 then this Pi1 must have at least 3 neighbours.

We consider maps from trees, instead of simple P1, because space of maps from P1 is not compact. Furthermore, we see that for example sequences of maps of degree 2 “want to” have a map from a tree as a limit: consider sequence fn of maps that parametrize curves {xy=z2/n}. This family of curves has {xy=0} as a limit and this one can be parametrize only by tree of two P1‘s.

Sheldon explains condition ii) of above definition by saying that maps that show up as limits of maps from P1 fulfill this condition. However, one of the wise guys who listened to my original talk, Adrian Langer, said that this is not really the case and that it’s a little bit more complicated. He actually mentioned other reasons for requiring condition ii) to hold, but they were to technical for me to understand.

Definition: Moduli space of stable maps to Pn of degree d and genus 0 is a set of isomorphism classes of stable maps to Pn of degree d and genus 0. We denote it by M‾(Pn, d). (The dash should be really over M).

Unfortunately, M‾(Pn, d) is not a manifold. The obstruction are so called nontrivial automorphism of stable maps. Example: let stable map f0 be defined as

f0 = (x2, y2, 0, 0, 0).

Consider also reparametrization α of a P1 (this will be nontrivial automorphism of f0) given by

α(x, y) = (-x, y).

We see that f0∘α = f0. Suppose now that we start the set of stable maps and perform all necessary identifications to obtain M‾(Pn, d) except those identifications that are coming from α.

One can prove that what we end with (in the neighbourhood of f0) is a set {fa, a∊C11}, where

fa(x,y) = (z0, z1, z2, z3, z4)

z0 = x2 + a1y2

z1 = a2x2 + y2

z2 = a3x2 + a4xy + a5y2

z3 = a6x2 + a7xy + a8x2

z4 = a9x2 + a10xy + a11y2.

Now, the problem with alfa is that it doesn’t move f0, but it does move some maps in any neighbourhood of f0 (because fa∘α=fσ(a), σ(a)=(a1, a2, a3, -a4, a5, a6, -a7, a8, a9, -a10, a11)). So, M‾ in a neighbourhood of f0 is not really a C11, but a quotient of C11 by an action of Z/2.

Such quotients generally don’t want to be manifolds. However, there are some good news:

i) Aut(f) is finite for all f.

ii) There are no other problems: every point in M‾ has a neighbourhood isomorphic with U/Aut(f), U being a subset of C5d+1. Objects with this property are called orbifolds, so we have that M‾ is an orbifold.

iii) On orbifolds we also have technology of bundles, Chern classes and even Poincare duality. However, I know very little about suitable definitions. For example, bundles we work with on orbifolds don’t have to be locally trivial -instead, they should be glued from pieces that “come” from locally trivial bundles on thing we work with before we divide by action of Aut(f). Accordingly, Chern classes have to be made rational. Sheldon writes a little bit more, but no really much.

So, let’s pretend that on M‾ everything works just as on ordinary manifolds! :-) We’ll define a bundle (only point-wisely, because indeed it is not locally trivial) and a section of this bundle whose zero-set is what we’re looking for.

On P4 we have a bundle O(5).

∘8728

‾ 8254

Let’s suppose again that our quintic threefold is defined as a zero-set of an equation

Q(X, Y, Z, V, W) = 0

On P4 we have a bundle O(5). What we need to know about it is that this is a bundle whose sections are homogeneous polynomials of degree 5. For example, Q defines a section of O(5), which we’ll call q.

(Precise definition of O(5) can be found… One fast way to precisely define O(5) is to give a sheaf of sections – homogeneous polynomials of degree d. However, if you’re familiar with sheaves then you probably don’t need definition of O(5)).

We define bundle on M‾ in a following way: over point represented by stable map f: C –> P4 we have a vector space “Global sections of a bundle f*(O(5))”, where f*(O(5)) is push-back of O(5) (of course, we mean algebraic sections, so this vector space is finite-dimensional. This bundle is called E5.

Now let’s define a section qd of E5: over point represented by a stable map f our section takes value q∘f.

(In addition to problems I wrote about earlier, one has to prove that this definitions are somehow independent from choice of representation of point in M‾. Sheldon doesn’t say anything about it so neither will I.)

It’s clear now that this section takes value 0 precisely on these stable maps whose image is contained in a zero-set of Q. This is what we wanted – we can define, as in a solution of problem (4), suitable integral over M‾(P4, d):

Nd := ʃc5d+1(E5).

This number is an example of a Gromov-Witten invariant (I don’t know of what). Physicists were able to compute Nd for all d, using ideas from string theory. I hope I’ll find out more (and make a suitable report) soon.

Similarly to situation with problem (3), we should ponder over what this number really means (it’s not always integer, so it’s definitely not a number of degree d curves on a general quintic threefold). Zero-set of qd consists of stable maps, not degree d curves, and accordingly it has things we wouldn’t want to count: higher dimensional reparametrizations of lower dimensional curves (if one has a stable map g: P1 –> P4 of degree d/k, k>1, and reparametrization β: P1 –> P1 of degree k then g∘β is a stable map of degree d, but its image is a curve of degree ≤d/k. Actually, Sheldon says that such reparametrization occur in zero-set of qd in positive-dimensional families, so it seems that qd isn’t even a transverse section (M‾ has the same dimension as E5). However, Sheldon doesn’t mention anything like that.

What I think is most important from our discussion of problem (5) is that although technically it’s probably very difficult, conceptually it is natural extension of ideas we used when dealing with problem (4) (and previous ones). What I like about Sheldon’s book very much: it shows that mathematical ideas are simple. When I read it I’m in this cool state of not being spiritually crushed by technical difficulties (most of the time I’m in this state) but understanding ideas well enough to be happy because of this understnading.

It’s all for now. You can find some more details in Sheldon’s very enjoyable book. Most notably, there’s a chapter on excess intersection which I didn’t really need up to chapter 9, but which can come in handy later.

As soon as I read all the book, I’ll write a report. Unless resistance of matter holds me from doing this, of course (or rather, unless I happen to loose this hopeless battle we all have to fight against that bastard, resistance of matter).