國立成功大學

應用數學研究所

碩士論文

格拉斯曼流形、五次三維流形以及相交理論

Grassmannians, quintic threefolds, and intersection theory

研 究 生：宋政蒲

指導教授：章源慶

中 華 民 國 一 ○ 二 年 六 月

摘要

在這篇論文裡，我們將會探討一些格拉斯曼流形與五次三維流形的

相交理論。我們也將介紹格羅莫夫-威騰不變量，以及利用大量子積的結

合性，推導康采維奇的著名公式：計算出通過(3d-1)個一般點的 d 次平面

有理曲線的數目。

關鍵字：格拉斯曼流形、五次三維流形、相交理論、穩定映射、

格羅莫夫-威騰不變量、康采維奇公式。 ABSTRACT

.

In this thesis, we will investigate the intersection theory of some Grassmannians and the quintic threefolds. We will also introduce the notion of Gromov-Witten invariants, and derive Kontsevich’s celebrated formula for the number of plane rational curves of degree d passing through 3d 1 general points from the associativity of the big quantum product. −

Keywords: Grassmannian, quintic threefold, intersection theory, stable map, Gromov-Witten invariant, Kontsevich’s formula. 誌謝

首先要感謝家人，老爸不斷督促我，老媽也持續地給我鼓勵，讓我度過不少難

關，兩個哥哥雖然因不認同我的想法和做法，而給了我很多壓力，但我仍很感謝他們

讓我很有家的感覺，且因他們讓家裡有穩定的收入，我才能無後顧之憂地做研究。

還有興大保育社、成大野鳥社、墾丁解說志工認識的朋友，讓我在這三年可以到

處遊歷賞鳥，讓我在研究和學習遇到瓶頸時，能夠藉著大自然來抒發壓力和尋找靈感，

讓我的碩士生涯不是苦悶的三年，而是最充實的三年。特別感謝傻剛、婉瑄和國權，

陪我到處賞鳥到處遊玩，還常常討論許多我們關心的時事，也給了我很多觀點。

接下來是在應數所上認識的夥伴，416 的大家、學長們和學弟，常常舉辦賽局理

論研討會，讓我在學習和研究之間得以喘息。而他們也常常給我很多鼓勵和意見。特

別感謝孫維良，總是可以從他那邊聽到很多數學上有趣的事情，也是少數會和我談論

數學的人，同時在學術上，他也是我的楷模。另外還有黛玉，在我趕著論文，神經緊

繃而沮喪的時候，用簡短的一句話化解了我的緊張。以及呂庭維，在口試當天也幫我

跑東跑西的，還幫我煮了一壺讓夏杼教授讚譽有加的咖啡。

再來要感謝夏杼老師和陳若淳老師，同時也是我的口試委員。在口試前總是用玩

笑的口吻，跟我交談，減輕了我口試前的緊張。而在口試結束後也給予我許多鼓勵和

肯定，讓我能下更多的決心走向研究路線。

最後便是我的指導教授－章源慶老師，很感謝他在我一開始想請他當指導教授時，

沒多說甚麼就收了我當學生，也時常給予我指導，而不論我問甚麼樣的問題，總是會

告訴我線索或者相關書籍，讓我不至於沒有方向。在研究上也沒有給我太大的壓力，

且常常會鼓勵我或者是在我因瓶頸而沮喪時安慰我。非常感謝老師這兩年多的照顧！

最後，要謝謝的人太多了，只能謝天又謝地了。 CONTENTS

1. The Grassmannians G 2, n 1 ...... 7

( + ) 2. Quintic Threefold ...... 12

3. Lines On the Quintic Threefold ...... 13

4. Rational Curves of Degree d 2 On the Quintic Threefold ...... 16

≥ 5. Gromov-Witten Invariant ...... 24

6. Kontsevich’s Formula for Genus Zero Curves on P2 ...... 27

Bibliography ...... 31 1. The Grassmannians G(2, n + 1) 7

1. THE GRASSMANNIANS G 2, n 1

( + ) We would like to study the enumerative geometry of G 2, n 1 , the set of 2-dimensional subspaces

Cn+1 Pn of , which is also the set of lines of the n dimensional( projection+ ) space . A typical chart of G 2, n 1 is

( + ) 1 0 a11 a1 n 1 ( − ) . 0 1 a a ⎛ 21 ⋯ 2(n−1) ⎞ ⎜ ⎟ So the dimension of G 2, n 1 is 2 ⎝n 1 . ⋯ ⎠

( + ) ( − ) Cn+1 Now we fix a basis e1, . . . , en+1 for , and let Vi be the linear space of e1, . . . , ei .

Then we have a collection{ of subspaces} { }

−−⇀ Cn+1 0 V0 V1 Vn+1 , which is called a flag V . { } = ⊂ ⊂ ⋯ ⊂ =

Definition 1. For a flag V , we define the Schubert cycle σa,b n 1 a b 0 by

P ( P− ≥ ≥ ≥ ) σa,b V ` G 2, n 1 ` Vn−a , ` Vn+1−b .

( ) = { ∈ ( + ) T ∩ ( ) ≠ ∅ ⊂ ( )} When b is zero, we simply write σa,0 as σa.

A chart of σa,b is given by

1 a11 a12 a1 n a 1 0 0 ( − − ) , 0 1 a a 0 0 ⎛ 21 ⋯ 1(n+⋯1−b−2) ⋯ ⋯ ⎞ ⎜ ⎟ so the dimension of σ⎝a,b is n a 1⋯n ⋯1 b ⋯2 2n a b 2. ⋯ ⎠ ( − − ) + ( + − − ) = − − − 1. The Grassmannians G(2, n + 1) 8

The σa,b V is the closure of cellular decomposition , has codimension 2 n 1 2n a b 2 a b,

2 a b and σa,b V( ) H ( + ) G 2, n 1 . ( − )−( − − − ) = + The[ total( )] cohomology∈ ( (H∗ G+ 2)), n 1 is spanned by σa,b V n 1 a b 0 . Furthermore, if V ′ is another( ( flag,+ there)) exists a deforming{[ ( map)]TV− V≤ ′,≤ we≤ have}

′ → σa,b V σa,b V σa,b.

[ ( )] = [ ( )] = Now we are ready to calculate σa,b σc,d. G(2,n+1) Because the dimension of G 2, nS 1 is 2 n ⋅ 1 , σa,b σc,d is not zero unless G(2,n+1) ( + ) ( − ) ⋅ c d 2 n 1 S a b .

1 ,+ if= c ( n− 1) −b( , d+ )n 1 a. Theorem 1. σa,b σc,d G(2,n+1) ⎧ 0 , otherwise. ⎪ = − − = − − S ⋅ = ⎨ Proof. ⎪ ⎩⎪ The number of σa,b σc,d is equal to G(2,n+1) S ⋅ ′ σa,b V σc,d V , G(2,n+1)

′ S [ ( ) ∩ ′ ( )] where Vi intersect Vj transversely. (i.e. if i j n 2,Vi Vj .)

Note that + < + ∩ = ∅ P P σa,b V ` G 2, n 1 ` Vn−a , ` Vn+1−b ,

σ (V ′) = {`∈ G(2, n+ 1) T `∩ P(V ′ ) ≠ ∅ , ` ⊂ P(V ′ )}. c,d n−c n+1−d

For a line L σ V ( σ) = {V ′∈ , L( must+ contain) T ∩ ( points) ≠p∅ P ⊂V ( and )}q P V ′ . a,b c,d n−a n−c Because L is in P V , the point q contained in P V ′ has to be in P V ∈ ( n)+∩1−b ( ) n−∈c ( ) ∈ n(+1−b ) Therefore, q P V ′ P V . Similarly, p P V P V ′ . (n−c ) n+1−b n(−a ) n+1−d ( ) Then P V ′ P V and P V P V ′ are not empty n−c∈ ( n)+1∩−b ( ) n−a n+1∈−d ( ) ∩ ( ) ( ) ∩ ( ) ( ) ∩ ( ) P V ′ P Vn 1 b n c n 1 b n 2 n 1 b c n−c + − . P V P V ′ n a n 1 d n 2 n 1 a d ( n−a) ∩ ( n+1−d) ≠ ∅ ⇒ − + + − ≥ + − − ≥ ⇒ ⇒ However, c d(must) ∩ equal( to 2 )n≠ ∅1 ⇒ a −b .+ Therefore,+ − ≥c +n 1 b and− −d ≥n 1 a.

+ ( − ) − ( + ) = − − = − − 1. The Grassmannians G(2, n + 1) 9

So we know that σa,b , σc,d 0 unless c is equal to n 1 b and d is equal to n 1 a.

How many lines are⟨ in σa,b ⟩V= σc,d V ′ ? The question− is− easy, there is a unique− line− containing both p and q. ( ) ∩ ( ) ∗ The dual basis to σa,b , the basis of H G 2, n 1 , is σn−b−1,n−a−1 . For a cohomology g H2d G 2, n 1 , g a σ , if we want to find a , we can calculate { } ( ( + ))i d,d−i{ } i d 0≤i≤ 2 [ ] ∈ ( ( + )) [ ] = Q g σn−1−(d−i),n−1−d. G(2,n+1) STheorem[ ] 2.⋅ Pieri’s formula for G 2, n 1 ,

σa σb,c( + ) σd,e . {d+e=a+b+c,n−1≥d≥b≥e≥c} ⋅ = Q Proof. Because the codimension of σa σb,c is a b c, then σa σb,c can be written as

⋅ +m+d,e σd,e . ⋅ d+e=a+b+c Furthermore, Q

md,e σa σb,c , σn−1−d,n−1−b σa σb,c σn−1−e,n−1−d G(2,n+1) = ⟨ ⋅ ⟩ = ( ⋅ ) ⋅ ′ ∗ S σa V σb,c V σn−1−e,n−1−d V , G(2,n+1) ′ ∗ = [ ( ) ∩ ( ) ∩ ( )] where Vi ,Vj ,Vk intersect transversely. S

Suppose md,e is not zero,

σ V ′ ` G 2, n 1 ` P V ′ , ` P V ′ b,c n−b n+1−c

σ ( ) =V{∗ ∈ (` G+2,) nT ∩1 (` P)V≠∗∅ ⊂ , `( P V)}∗ n−1−e,n−1−d 1+e 2+d ` P V ′ ,L P V ∗ T n−b (2+d ) = { ∈ ( + ) ∩ ( ) ≠ ∅ ⊂ ( )} P V ′ P V ∗ n b 2 d n 2 d b (1). ∵ ∩ n(−b ) ≠ ∅2+d ⊂ ( ) ` P V ∗ ,L P V ′ ⇒ ( 1)+e∩ ( ) ≠ ∅ ⇒n+1−c− + + ≥ + ⇒ ≥ P V ∗ P V ′ 1 e n 1 c n 2 e c (2). ∵ ∩ 1(+e ) ≠ ∅n+1−⊂c ( ) Now we consider σ V , if line L is also contained to σ V , L must contain point s in P V . ⇒ ( ) ∩ ( a ) ≠ ∅ ⇒ + + + − ≥ + ⇒a ≥ n−a Here P V and P V ′ P V ∗ intersect transversely. n−a n(+1−)c 2+d ( ) ( ) If σ V σ V ′ σ V ∗ is nonempty set, then there exists a line L, L P V a( ) b,c ( n−1)−e,n∩ −1(−d ) n−a and L P V ′ P V ∗ . ( ) ∩n+1−c( ) ∩ 2+d ( ) ∩ ( ) ≠ ∅ ⊂ ( ) ∩ ( ) 1. The Grassmannians G(2, n + 1) 10

Therefore, the intersection P V and P V ′ P V ∗ , the dimension is n−a n+1−c 2+d n( 1 )c 2( d )n∩ 1( d) c 2, is nonempty. ( + − ) + ( + ) − ( + ) = − + So, d c 2 n a is greater than n 1. d c a 0 d c d e b c 0 b e (3).

In conclusion,− + + from− (1)(2)(3) we can+ see⇒ that−b, c,− d,≥ e are⇒ related− − ( by+ − − ) ≥ ⇒ ≥ n 1 d b e c 0 .

Thus we get a necessary condition of m−d,e ≥0,≥ now≥ we≥ calculate≥ md,e.

≠ Since P V ′ and P V ∗ intersect transversely, we can choose a local coordinate x , x , . . . , x n+1−c 2+d 1 2 n+1 such that the points of P V ′ are represented by ( ) ( n+1)−c ( ) ( ) x1, x2,...,...,...,xn+1−c, 0,..., 0 and the points of P V ∗ are represented( by ) 2+d ( ) 0,..., 0, yn−d,...,...,...,yn, yn+1 .

The points of P V ′ P (V ∗ can be represented by ) n+1−c 2+d ( ) ∩ ( ) 0,..., 0, zn−d, . . . , zn+1−c, 0,..., 0 .

A line L in σ V ′ σ ( V ∗ must contain a point p in) P V ′ and a point q in P V ∗ . b,c n−1−e,n−1−d n−b 1+e Therefore, the( line)∩L can be constructed( ) by ( ) ( )

a11 a1 n b 0 0 ... 0 0 ( − ) . 0 0 ... 0 0 a a ⎛ ⋯ 2(n+1−e) ⋯ 2(n+1) ⎞ ⎜ ⎟ But L is in P V ′ P V ∗ which is locally represented by n+1−c ⎝ 2+⋯d ⋯ ⎠ ( ) ∩ ( ) 0,..., 0, zn−d, . . . , zn+1−c, 0,..., 0 .

So the line L can be constructed( by )

0 0 zn−d zn−b 0 0 0 0 0 0 0 0 z z 0 0 ⎛ ⋯ ⋯ ⋯ n⋯+1−e ⋯n ⋯+1−c ⋯ ⎞ ⎜ ⎟ ⎝ n⋯ d 1⋯zeros ⋯ ⋯ ⋯b e zeros ⋯ c⋯zeros⎠

↑ ( − − ) ↑ ( − ) ↑ 1. The Grassmannians G(2, n + 1) 11

Look at the above matrix, the space it constructs has n d 1 b e c n a 1 zeros, and intersects P V only in one dimension n−a ( − − ) + ( − ) + = − − That means the point s contained in P V and the line L is unique. ( ) n−a Turning back to P V , which transversely intersect the space consisting of z , . . . , z and n−a ( ) n−d n−b containing the point p(. ) Because n a n b n d 1 n d e b c d b 1 n 1 e c n 2, therefore the point s is not contained in P V ′ . Similarly, it is not contained in P V ∗ . − + (( − ) − ( − ) + ) = n−b( + − − ) + ( − + ) = + − ( 1−+e ) < + So the line L contained in σ V σ V ′ σ V ∗ can be represented by a b,c ( ) n−1−e,n−1−d ( ) ( ) ∩ ( ) ∩ ( ) 0 0 zn d zn b 0 0 0 0 − − . 0 0 0 0 z z 0 0 ⎛ ⋯ ⋯ ⋯ n⋯+1−e ⋯n ⋯+1−c ⋯ ⎞ ⎜ ⎟ Because the unique⎝ point⋯ s is⋯ fixed, ⋯zn ⋯−d, . . . , z⋯n−b, zn+1−e, . . . , z⋯n+1−c are also⋯ fixed.⎠

Conclusively, this line L is unique, so md,e 1.

=

Now we can calculate σa, σb,c , but it is not so easy to determine σa,b, σc,d .

Let us see the table below.⟨ ⟩ ⟨ ⟩ σ1 σ2 σ3 σa σn−1 σ σ σ σ σ σ σ σ σ σ ⋅1 2 1,1 3 2,1 4 3,1 ⋯ a+1 a,1 ⋯ n−1,1 σ σ σ σ σ σ σ σ σ σ σ 2 + 4 +3,1 2,2 5 +4,1 3,2 ⋯ a+2 a+1,1 a,2 ⋯ n−1,2 σ σ σ σ σ σ 3 ⋯ + + 6+ + 3,3 ⋯ a++3 +a,3 ⋯ n−1,3 ⋯ ⋯ + ⋯ + ⋯ + ⋯ + ⋯ σ σ σ σ σ σ σ σ σ σ σ ⋮b b+1⋯ b,1 b+2 b⋯+1,1 b,2 b+3 ⋯ b,3 ⋯ a+b ⋯ a,b ⋯ n⋯−1,b

If we+ see this table+ row by+ row, σa,b is+ ⋯the + ba-entry⋯ of the table,+ ⋯ + which is⋯ simply σa σb (Where a b). ⋅ So σ can be represented by σ σ σ σ . Also, σ , σ , , σ can be ≥ a,b a b a+b a+1,b−1 a+b a+b−1,1 a+1,b−1 represented. Then the σ can be constructed by σ , , σ . a,b − ( + ⋯ + 1 n−)1 ⋯ That means H∗ G 2, n 1 can be algebraically constructed by σ , , σ . { ⋯ } 1 n−1 Now we can calculate( ( σ+a,b)), σc,d by using Pieri’s formula even if a,{ b, c,⋯ d are not} zero. ⟨ ⟩ 2. Quintic Threefold 12

2. QUINTIC THREEFOLD

A quintic threefold is a hypersurface in P4 defined by a degree 5 homogeneous polynomial. Now we are interested in enumerating rational curves on a general quintic threefold. Let X be the general quintic threefold defined by

I I I0 I1 I2 I3 I4 F x0, x1, x2, x3, x4 aI x 0, where x x0 x1 x2 x3 x4 , I0+I1+I2+I3+I4=5 and a degree d rational( curve C )is= parametrizedQ by = =

f y0, y1 f0 y0, y1 , f1 y0, y1 , f2 y0, y1 , f3 y0, y1 , f4 y0, y1 , d k d−k where fi y0, y1 ( aiky0)y1= ( ( ) ( ) ( ) ( ) ( )) k=0 ( ) = Q If C X, then

⊂ F f0 y0, y1 , f1 y0, y1 , f2 y0, y1 , f3 y0, y1 , f4 y0, y1 1 has to be zero for any y(0, y(1 P). ( ) ( ) ( ) ( )) 5d j 5d−j 0 F f0 y0(, y1 , f)1 ∈y0, y1 , f2 y0, y1 , f3 y0, y1 , f4 y0, y1 hj aik y0y1 j=0

For the fixed= ( quintic( threefold) ( X), this( is a equation) ( with) ( 5 d 1)) unknowns= Q { (} aik ), but for any

y0, y1 , F f y0, y1 is always zero. ( + ) { } That means the coefficients of yjy5d−j written simply by h a are all zero. Therefore, we have ( ) ( ( )) 0 1 j ik 5 d 1 unknown variables and 5d 1 equations, so the solutions{ } may be 4-dimensional. ( But+ ) for any reparametrization( of+ C) (i.e. f z0, z1 f g0 z0, z1 , g1 z0, z1 ), where g0, g1 are linear homogeneous polynomials. In addition, these( reparametrizations) = ( ( ) ( have 4)) dimensions. Also, these 4-dimensional reparametrizations will not change the consequence of

F f1 g1, g2 , f2 g1, g2 0

They may let the solutions be zero-dimensional,( ( ) ( but we)) still= can not sure the solutions are finite. 3. Lines On the Quintic Threefold 13

3. LINES ON THE QUINTIC THREEFOLD

Now we see how many lines (i.e. rational curve with degree 1) on a general quintic threefold. First of all, we construct a 2-dimensional vector bundle E over G 2, 5 by setting the fiber over a line ` G 2, 5 to be a 2-dimensional vector space E` of homogeneous( ) linear polynomials on `. For example,∈ ( the) fiber over ` defined by f x0, x1 x0, x1, 0, , 0 is spanned by x0, x1 . Then we set a section s of this vector( bundle) = be( a linear⋯ polynomial,) for example{ s} x0. For each line `, s ` x E . If s ` 0, the line ` is contained in Z s . In addition, the set of lines 0 ` ` = contained in (Z )s= isT a∈ Schubert( cycle) = σ1,1 Z s . So the second Chern( ) class of E is ( ) ( ( )) c2 E σ1,1.

( ) = And total Chern class of E is c E 1 σ1 σ1,1.

The first Chern class of E is( σ1),= see+ [3]. + For a general quintic threefold X defined by Z F , if there exists a vector bundle B whose section in general position is F , we can use the Chern( class) to find what cr B Z F X is. 5 Let B be Sym E , which is a 6-dimensional vector bundle of G 2, 5 , then( ) =F[ is( a)] section= [ ] of B in general position.( The) fiber over ` is a 6-dimensional vector space of( homogeneous) polynomials of degree 5. Now we return to see the 2-dimensional vector bundles E. If E is a direct sum of two line bundles

L1 and L2, then it is very convenient to calculate the Chern class by c Q c L1 c L2 , where denotes the cup product. ( ) = ( ) ⌣ ( ) But⌣ not all vector bundle can be written as the direct sum of line bundles, so we need to use “splitting principle”, see [1]. 3. Lines On the Quintic Threefold 14

Theorem 3 (Splitting Principle). Let p E X be a C∞ complex vector bundle of rank r over a manifold B. Then there exists a space ∶Y →F ` E , the flag bundle associated to E, and a map f Y X such that = ( ) ∶ (1)→ the induced cohomology homomorphism f ∗ H∗ X H∗ Y is injective, and (2) the pullback bundle f ∗ξ f ∗E Y splits into∶ a( direct) → sum( of) line bundles:

∗ ∶ f→ E L1 L2 Lr . where Li and their first characteristic class( ) = are⊕ called⊕ Chern ⋯ ⊕ roots.

So, we can pretend that E L1 L2, then c Q c L1 c L2 .

Let = ⊕ ( ) = ( ) ⌣ ( )

c1 L1 α, c1 L2 β, then c E c L1 L2 1 α 1 β 1 α β αβ .

But we( know) = the total( ) = Chern class( of) E= is( ⊕ ) = ( + )( + ) = + ( + ) + ( )

1 σ1 σ1,1, so α β σ1, αβ σ1,1.

And + + ( + ) = =

5 ⊕5 ⊕4 ⊕3 ⊕2 ⊕2 ⊕3 ⊕4 ⊕4 ⊕5 Sym V1 V2 V1 V1 V2 V1 V2 V1 V2 V1 V2 V2 .

So ( ⊕ ) ≅ ⊕ ( ⊗ ) ⊕ ( ⊗ ) ⊕ ( ⊗ ) ⊕ ( ⊗ ) ⊕

5 5 ⊕5 ⊕4 ⊕3 ⊕2 ⊕2 ⊕3 ⊕4 ⊕4 ⊕5 Sym B Sym L1 L2 L1 L1 L2 L1 L2 L1 L2 L1 L2 L2 .

Then( the) ≅ total( Chern⊕ class) ≅ of Sym⊕ ( 5 B⊗is ) ⊕ ( ⊗ ) ⊕ ( ⊗ ) ⊕ ( ⊗ ) ⊕

1 5α 1 4α β 1 ( 3)α 2β 1 2α 3β 1 α 4β 1 5β .

And the 6-th( Chern+ )( class+ of + )( + + )( + + )( + + )( + )

5 c6 Sym B 5α 4α β 3α 2β 2α 3β α 4β 5β

4 3 2 2 3 4 ( ( )) = 25αβ( 24+ α)( 154+α β)( 269+α β)( +154αβ)( )24β

2 2 2 4 = 25αβ(9α β + 58αβ α+ β 24+ α β + )

2 2 4 = 25σ1,1( 9σ1,1 +58σ1,1(σ1 +24)σ1+ ( + ) )

3 2 2 4 = 25 9σ1(,1 58+σ1,1σ1 24+σ1,1σ1).

= ( + + ) 3. Lines On the Quintic Threefold 15

2 2 And σ1 σ1,1 σ2 σ1,1 σ1 σ2, then we can use the Pieri’s formula and Table 1 to calculate σ3 , σ2 σ2 , σ σ4, the answer are 1,1 1,1 1 = 1,1 +1 ⇔ = −

3 2 2 4 σ1,1 σ3,3 , σ1,1σ1 σ3,3 , σ1,1σ1 2σ3,3.

= = = Therefore,

3 2 2 4 25 9σ1,1 58σ1,1σ1 24σ1,1σ1 25 9 58 48 σ3,3 2875σ3,3 .

So, ( + + ) = ( + + ) = 5 Z F c6 Sym E 2875σ3,3 2875 . G(2,5) G(2,5) G(2,5) S [ ( )] = S ( ( )) = S = There are 2875 lines on a general quintic threefold. 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 16

4. RATIONAL CURVES OF DEGREE d 2 ON THE QUINTIC THREEFOLD

≥ There is an unsolved conjecture on the rational curves in a quintic threeflod.

Clemens Conjecture Let X be a general quintic threefold, and let d N. Then there are finitely many rational curves of degree d on X. ∈

To study rational curves on a quintic threefold, we consider the moduli space M Pn, d of stable maps firstly. In what follows, I follow the exposition of [4]. ( )

1 1 Definition 2. Let C Ci be a tree of P ’s with parametrizations φi P Ci. A morphism f C Pn is a mapping such that each f f φ P1 Pn is a parametrized rational curve or a = ⋃ i i ∶ → constant∶ → map. The degree of f is the sum of= the○ degree∶ → of fi’s.

Definition 3. A genus 0 stable map to Pn is a morphism f C Pn from a tree of P1’s such that if f is constant on a component Ci of C, then Ci is required∶ to contain→ at least 3 nodes.

Definition 4. Let f C Pn and f ′ C′ Pn be genus 0 stable maps with components parametrized by φ P1 C and φ′ P1 C′. An isomorphism from f C Pn to f ′ C′ Pn is i ∶ →i j ∶ j → a map g C C′ such∶ that→ the following∶ → hold. ∶ → ∶ → f ′ ∶ g →f. For each C we have g C C′ for some j, and given any j, there is a unique C ● ○ = i i j i with g C C′. ● i j ( ) = φ′ −1 g φ P1 P1 is a degree 1 parametrized rational curve whenever g C C′. j ( ) =i i j ●An( isomorphism) ○ ○ ∶ from→f to itself is called an automorphism. ( ) ⊂ 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 17

Definition 5. The moduli space of stable maps is the set of all isomorphism classes of degree d genus 0 stable maps to Pn. This space is denoted by M Pn, d .

Now let us concentrate on M P4, d . If M P4, d is a( manifold,) that is easy to see its geometric.

Although it is not a manifold, we( pretent) that( it is.) Let f C P1 P4 be a stable map of degree 2. We will focus on the Aut f instead of Iso f .

The reason∶ is≅ in the→ following. ( ) ( )

Let f x0, x1 y0, y1, y2, y3, y4 , where each yi is a degree 2 homogeneous polynomial of x0, x1 .

Without( loss of) generality,= ( let y0 )and y1 be nonzero polynomials which are not multiples( of each) other.

Then there exists an isomorphism g such that f g h, where h x0, x1 z0, z1, z2, z3, z4 and

2 2 z0 x0 ○a1=x1 ( ) = ( )

2 2 z1 = a2x+0 + x1

2 2 z2 = a3x0 + a4x0x1 a5x1

2 2 z3 = a6x0 + a7x0x1 + a8x1

2 2 z4 = a9x0 + a10x0x1+ a11x1.

Since g is an isomorphism, then f h = M P+4, 2 . Therefore,+ for any stable map class f , there exists a stable map h, such that f h and h has x2 a x2 and a x2 x2 as its entries. [ ] = [ ] ∈ ( ) 0 1 1 2 0 1 [ ] Let us see the above stable map f as[ an] = example.[ ] We replace+ f by h. There+ is a good reason to replace f by h. The reason is that the automorphism set Aut f has only two possible forms, just the trivial automorphism or the trivial automorphism and g x(0,) x1 x0, x1 . C11 C11 Let q q1, q2, . . . , q11 , and q in a neighborhood U( of 0) = (− . We) can regard q as the P4 stable map= (fq approaching) ∈f M , 2 , where fq zq0, zq1, zq2, zq3∈, zq4 and

2 2 ∈ z(q0 x0) q1 a1 x=1( )

2 2 zq1 = q2+ (a2 +x0 )x1

2 2 zq2 = (q3 + a3)x0 + q4 a4 x0x1 q5 a5 x1

2 2 zq3 = (q6 + a6)x0 + (q7 + a7)x0x1 + (q8 + a8)x1

2 2 zq4 = (q9 + a9)x0 + (q10+ a10) x0x+1 ( q+11 )a11 x1.

= ( + ) + ( + ) + ( + ) 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 18

Here, we set f0 f.

So, we can pretend= that fq is a 11-dimensional manifold at f, denoted by M. P4 Suppose that fq g { M} , 2 can be verified by an isomorphism (but not an automor- P1 phism) which is a[ linear] = [ reparametrization] ∈ ( ) of . The form (zq1, zq2) will change over the course of reparametrizing, then the stable map g which is identified with f M P4, 2 can not be defined in

M. ∈ ( ) 4 If fq g M P , 2 can be verified by an automorphism, then the form will not change C11 during[ reparametrizing.] = [ ] ∈ ( Then) r U such that g fr, so g can be defined in M. ∃ ∈ ⊂ = Unfortunately, if there exists a nontrivial automorphism in Aut f , the M we defined will not C11 be a manifold. Because fq is identified with fr, and q may not be( equal) to r , M is not a manifold and locally looks like ∈ U Aut f , the quotient of U by Aut f . It is not a manifold€ but( ) an orbifold. P4 Now, let us construct( the) vector bundle E5 over M , d which has F as a section in general position. (The quintic threefold X is defined by F .) ( ) Firstly, we need to study the vector bundle over M P4, d . We know that M P4, d can be locally seen as a quotient space of some open neighborhood( U over) Aut f . Therefore,( we) have the quotient map ρ, ( ) U ρ U Aut f . ∶ → € U Let EU be a vector bundle over U, we can induce the( vector) bundle E over Aut f by EU .

€ ( ) For each g U, we can identify Eρ(g), the fiber of E at ρ g , with EU g, the fiber of EU at g.

And the zero∈ loci of section s of E also can be induced by( the) section( of) EU . Let sU is a section of EU , the zero loci of the section s induced by sU can be regarded as a quotient space

1 ρ Z s . Aut f U ( ( )) SS ( )SS 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 19

P4 Now, we construct the vector bundle EU5 over U one of whose section is F . Let f C be a stable map of degree d, then we defined the fiber of E at f by U5 ∶ →

0 ∗ 4 EU5 f H f P 5 ,

( ) = ( (O ( ))) the vector space of global sections of the pullback bundle of P4 5 , where P4 5 is a line bundle P4 over . O ( ) O ( ) 4 1 If f M P , d is defined on P of degree d, the sections of f ∗ P4 5 can be regarded as the

P1 ∗ degree 5∈d homogeneous( ) polynomials defined on . (i.e. f P4(O5 ( ))P1 5d .) Therefore, the space of global sections of f ∗ P4 5 is 5d 1 dimensional, and we can see that the fiber E (O ( )) ≅ O ( ) U5 f also has dimension 5d 1. (O ( )) + ( ) P4 Now consider the degree+ d stable map f M , d defined on n ∈ ( ) 1 C Ci, where Ci P . i=1 = ≅ The pullback bundle f ∗ P4 5 is not isomorphic to P1 5d , but its section space is still a 5d 1 dimensional space. (O ( )) O ( ) +

Lemma 1. Given a degree d stable map f M P4, d definded on

n ∈ ( ) 1 C Ci where Ci P . i=1 = ( ≅ ) The space of global sections of the pullback bundle f ∗ P4 5 is 5d 1 -dimensional.

Firstly, let us see an example. (O ( )) ( + )

P4 Example 1. Check that dim EU5 f 41 for the degree 8 stable map f C described as n follows. The curve C is i 0 Ci, and fi are denoted by = ( ) =Ci ∶ → ⋃ T f0 x00, x01 x00, x01, 0, 0, 0 and ( ) = ( )

fi xi0, xi1 xi0, ixi0, xi1, ixi1, 0 , for i 1, 2,..., 7 .

1 Here Ci is identified with(P . The) = curve( C has 7 nodes )pi, and∈ each{ pi is formed} by gluing the point

1, i of C0 to the point 1, 0 of Ci.

( ) ( ) 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 20

1 4 Proof. Each fi is a degree 1 stable map from P to P , so for each Ci, its pullback bundle can be

∗ regarded as fi P4 5 P1 40 . Its space of sections is a 41 dimensional vector space spanned by (O ( )) ≅ O ( ) 5 4 3 2 2 3 4 5 ai0xi0, ai1xi0xi1, ai2xi0xi1, ai3xi0xi1, ai4xi0xi1, ai5xi1 .

Now we can regard { }

5 5 5 5 a00x00, . . . , a05x01, . . . , a70x70, . . . , a75x41 as the section s of the vector( bundle f ∗ P4 5 . It seems that the) space of sections is a vector space of dimension 8 5 1 48 instead of(O dimension( )) 5 8 1 41. We claim that each node reduce the dimension by 1. ( + ) = ⋅ + =

For any p Ci, we define

5 ∈ 5−k k s p aikxi0 xi1, where p is written as xi0, xi1 in Ci. k=0 ( ) = Q ( ) Then for the point p C0 Ci, we have

5 ∈ k ∩ s p a0ki ai0, where p is written as 1, i in C0 and as 1, 0 in Ci. k=0 ( ) = = ( ) ( ) Then we get 7 linearQ equations

a00 a01 a02 a03 a04 a05 a10 0

2 3 4 5 a00 + 2a01 + 2 a02 + 2 a03 + 2 a04 + 2 a05 − a20 = 0

2 3 4 5 a00 + 3a01 + 3 a02 + 3 a03 + 3 a04 + 3 a05 − a30 = 0

+ + + + + − =

2 3 4 5 a00 7a01 7 a02 7 a03⋮ 7 a04 7 a05 a70 0.

+ + + + + − = The matrix of coefficients is a Vandermonde matrix, so they are linearly independent.

5 5 5 5 5 Therefore, span a00x00, . . . , a05x01, a10x10, a20x20, . . . , a70x70 has only 5 dimensions instead of 0 12 dimensions. Then({ the dimension of H f ∗ P4 5 is (48-7)=41}) because of the above equations. ( (O ( ))) 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 21

1 0 Proof of lemma 1. We already know that if C P , then H f ∗ P4 5 is a 5d 1 dimensional vector space. n ≅ ( (O ( ))) + 1 Now let us consider the curve C Ci which is a tree of P ′s. The tree C has exctly n 1 i=1 nodes. = −

Let the restriction fi of f to Ci is a degree di parametrized rational curve or a constant map (if n di=0), and di d. i=1 0 So for each C , its pullback bundle can be regarded as f ∗ P4 5 P1 5d , and H f ∗ P4 5 Q i= i i i can be spanned by (O ( )) ≅ O ( ) ( (O ( ))) 5di 5di−1 5di xi0 , xi0 xi1, . . . , xi1 .

Now we can replace the section s {of f ∗ P4 5 by }

(O ( )) a10, a11, . . . , a1(5d1), a20, . . . , an(5dn)

n ( ) which has 5di 1 5d n entries. i=1 For any point p C C, p can be written as x , x in C , and we define s p by Q( + ) =i + i0 i1 i ∈ ⊂ 5di ( ) ( ) 5di−k k s p aikxi0 xi1. k=0 ( ) = Q Because of well-definedness, for the point p Ci Cj, where p is written as xi0, xi1 in Ci and as xj0, xj1 in Cj, we have a linear equation ∈ ∩ ( )

5d ( ) 5di j 5di−k k 5dj −k k aikxi0 xi1 ajkxj0 xj1. k=0 k=0 Q = Q This equation restricts one dimension of ai0, . . . , ai(5di), aj0, . . . , aj(5dj ) which has 5di 5dj 2 entries. ( ) ( + + ) In addition, for each node, we also have such equation as above. There are n 1 linear equations which restrict n 1 dimension of a , a , . . . , a , and these equations are linearly independent. 01 02 n(5dn) − (Otherwise, the− curve C has cycles,( then C is not of) genus zero.) 0 Therefore, the vector space H f ∗ P4 5 has dimension 5d n n 1 5d 1.

( (O ( ))) ( + ) − ( − ) = +

Then we know that EU5 and E5 are rank (5d+1) bundles. 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 22

Set sUF is the section of EU5 induced by F defined by sUF f F f EU5 f . If sUF f 0, that means the image of the stable map f is contained in Z F (, where) = ○Z ∈F( is) the quintic( ) threefold.= 1 Then we can calculate Z sF defined by ρ Z sU . So the number of stable maps of degree SSAut(f)SS F ( ) ( ) d contained in X can be( seen) as ( ( ))

Nd c5d+1 E5 . M(P4,d)

The Nd is a Gromov-Witten Invariant= .S It counts not( ) only degree d rational curves but lots of lower degree rational curves reparametrized by higher degree reparametrizations. And by the definition of Z sF , we see that Nd may be not an integer but a rational number.

Firstly, let( us consider) the higher degree reparametrizations of lower degree rational curves. In what follows, I follow the exposition of [4].

Theorem 4. For any line L P1 on a general quintic threefold, the component M P1, d contributes

1 3 to the invariant Nd 4 c5d 1 E5 . d M(P≅ ,d) + ( )

See [2] for the general= ∫ version and( the) proof.

Definition 6. A curve C on X is called infinitesimally rigid if the normal bundle NC~X has no nonzero sections.

If k in an integer dividing d, it is said that the space of degree k multiple covers of an infinitesi-

d 1 mally rigid smooth rational curve C of degree k contributes k3 to the Gromov-Witten invariant.

Let nd be the number of disjoint rational curves of degree d on a generic quintic threefold, it is conjectured that n N d~k . d k3 kSd And we get the d d lower triangular system= Q of linear equations

× 1 0 0 0 0 n1 N1 1 1 0 0 0 n N ⎛ 23 ⋯ ⎞ ⎛ 2 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 1 0 1 0 0 ⎟ ⎜ n ⎟ ⎜ N ⎟ ⎜ 33 ⋯ ⎟ ⎜ 3 ⎟ ⎜ 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ . ⎜ 1 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 3 3 0 1 0 ⎟ ⎜ n4 ⎟ ⎜ N4 ⎟ ⎜ 4 2 ⋯ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⋯ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 3 1 ⎟ ⎜ nd ⎟ ⎜ Nd ⎟ ⎜ d⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⎟ ⎜ ⋮ ⎟ ⎜ ⋮ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⋯ ⋯ ⋯ ⋯ ⎠ ⎝ ⎠ ⎝ ⎠ 4. Rational Curves of Degree d ≥ 2 On the Quintic Threefold 23

From this, we can solve nd by

N N n N , n N 1 , n N 1 ,.... 1 1 2 2 23 3 3 33 = = − = − Conjecture (Gopakumar-Vafa).

The nd, defined above, are integers.

nd~k However, this conjecture and the inference Nd k3 are based on the validity of Clemens Conjecture which is still an open problem. = ∑ 5. Gromov-Witten Invariant 24

5. GROMOV-WITTEN INVARIANT

Let X be a complex projective manifold. In this section, we will define n-pointed stable maps to X and their moduli. In what follows, I follow the exposition of [4].

1 Definition 7. An n-pointed map f C, p1, p2, . . . , pn X, where C is a tree of P ’s and the marked points pi are distinct smooth∶ points,( is stable) if→ it satisfies the property: If Ci C is a component such that f Ci is a point, then the Ci contains at least three special points (a⊂ node or a marked point). ( )

Definition 8. The moduli space of stable maps to X of the homology class β H2 X is the set of isomorphism classes of n-pointed stable map f, defined above, satisfying f C β. This space is ∗ ∈ ( ) denoted by M 0,n X, β . [ ] =

The space M(0,n X,) β admits a virtual fundamental class

vir ( ) M 0,n X, β H2D M 0,n X, β , Q . where D c1 TX dim X
3 n( is the) expected∈ ( dimesion.( ) We) need rational coefficients due to β the automorphisms of stable maps. = S ( ) + − + 1 Note that the term D c1 TX dim X is the dimension of the space of maps f P X. β Because of the automorphisms of P1, we need to subtract 3. In addition, for each marked point = S ( ) + ∶ → 1 pi P , pi can be arbitrarily chosen. Therefore, each pi contributes dimension one to D. PN PN ∈Fortunately, for X , the dimension of M 0,n , d is equal to D, where β d line . And

N = D dim M 0,n P , d (Nd N) d n 3. = [ ]

Definition 9. The evaluation map= evi M( 0,n X,) β= X+ is+ defined+ − by

evi f C,∶ p1, .( . . , pn) → X f pi , for any f M 0,n X, β . [ ∶ ( ) → )] = ( )

[ ] ∈ ( ) 5. Gromov-Witten Invariant 25

Definition 10. For ω1, . . . , ωn H∗ X , the (genus zero) n-point Gromov-Witten invariant is de-

fined by ∈ ( ) ∗ ∗ ω1, . . . , ωn β vir ev1 ω1 evn ωn .
M 0,n(X,β) 4 The Nd defined in Section⟨ 4 is a Gromov-Witten⟩ = S invariant( ) of⋯ a quintic( ) threefold on P , where β is a rational curve class of degree d.

∗ Geometrically, let ωi Zi , where Zi are submanifolds of X in general position, then evi ωi ev −1 Z and ev∗ ω ev∗ ω ev −1 Z ... ev −1 Z is equal to i i 1 1 = [ n] n 1 1 n n = [( ) ( )] ( )⋯ ( ) = [( ) ( ) ∩ ∩ ( ) ( )] f C, p1, . . . , pn X f pi Zi, i 1, . . . , n .

[{ ∶ ( ) → S ( ) ∈ = }] Definition 11. For a fixed basis Ti for H∗ X , let γ H∗ X , the Gromov-Witten potential of X is the formal function defined{ by} ( ) ∈ ( ) 1 Φ γ γ, . . . , γ qβ. n! β β,n. If β=0, n≥3

T ( ) = Q ⟨ ⟩ β ∫β i q is a shorthand of qi . i An n-pointed stableM map of class β 0 is a constant map, they are constant maps. So each component of C must contain at least 3 special= points. That is the reason why n can not less than 3 if β 0.

i ∗ Let= Ti be a basis of H∗ X with dual basis T and γ tiTi H X . Then for fixed β i and n, γ, . . . , γ consists of t t . . . t T ,...,T . { } β (i1 i)2 in i1 in{ β } = Q ∈ ( ) Lemma⟨ 2. Let⟩β 0, then the Gromov-Witten⟨ invariant⟩

=

Ti1 Ti2 Ti3 if n 3 X Ti1 ,...,Tin 0 . ⎧ 0 otherwise ⎪ S ⋅ ⋅ = ⟨ ⟩ = ⎨ Proof of lemma 2 with X PN . We already⎪ know that there are no n-pointed stable maps with ⎩⎪ n 2 if β 0. Let’s consider= n 3. Let T Z , where Z is a submanifold of PN , then we have ev∗T ev −1 Z ≤ i=j j j ≥ j ij j j M 0,n Zj, 0 .= The[ dimension] of M 0,n Zj, 0 is 0 dim Zj n 3 and its codimension= [( is ) ( )] = ( ) ( ) + + − 0 dim X n 3 0 dim Zj n 3 dim X dim Zj.

+ + − − ( + + − ) = − 5. Gromov-Witten Invariant 26

∗ The invariant Ti1 ,...,Tin 0 is nonzero only if the sum of the codimensions of evi Ti is equal to the dimension of M⟨ 0,n X, 0 .⟩ This means that n ( ) n dim X dim Zj 0 dim X n 3 n 1 dim X n 3 dim Zi 1 . j=1 j=1 Q( − ) = + + − ⇒ ( − ) = − + Q ( )

And we know that

∗ ∗ ev1Ti1 evnTin f C, p1, . . . , pn X f pj q Zj, j 1, . . . , n .

⋯ = [{ ∶ ( n ) → S ( ) = ∈ = }] Suppose Ti1 ,...,Tin 0 0. Then Zj can not be empty. Because Z1,...,Zn intersect trans- j=1 n n ⟨ ⟩ ≠ versely, the dimension of Zj is dim Zj n 1 dim X. j=1 j=1 Therefore, − ( − )  n Q dim Zj n 1 dim X 0 2 . j=1 Q − ( − ) ≥ ( ) In conclusion, from (1),(2) and the restriction that n 3,, we have n 3.

For n 3, the space of constant maps is X itself. Also, ev∗T ev −1 Z is just Z . ≥ j ij j= j j Therefore,= = [( ) ( )] [ ]

∗ ∗ ∗ ev1Ti1 ev2Ti2 ev3Ti3 Z1 Z2 Z3 Ti1 Ti2 Ti3 . M 0,3(X,0) X X S = S [ ] ⋅ [ ] ⋅ [ ] = S ⋅ ⋅ 6. Kontsevich’s Formula for Genus Zero Curves on P2 27

6. KONTSEVICH’S FORMULA FOR GENUS ZERO CURVES ON P2

2 2 2 Now consider the Gromov-Witten potential of P . Let T0,T1,T2 1,H,H be a basis of H∗ P .

∗ P2 For any γ H , γ takes the form t0T0 t1T1 t2T2. So( we can) write= ( the Gromov-Witten) potential( ) Φ γ as Φ∈ t0, t(1, t2). + + ( In) what( follows,) I follow the exposition of [4]. 2 2 2 For H ,H ,...,H d 0, it must satisfy

⟨ ⟩ ≠ 2n Nd N d n 3 2d 2 d n 3 n 3d 1.

2 2 = 2+ + + − = + + + − ⇒ = − Now set Kd H ,H ,...,H d with n 3d 1. Note that Kd is the number of rational curves of P2 degree d on = ⟨passing through⟩ 3d 1 general= − points. For example, the number of lines passing two distinct points is 1, so K1 1. And− there is a unique conic passing through five general points, so K2 1. (See Chapter 2 of [4])= Next, let us look at the Gromov-Witten invariant H2,H2,...,H2,H,...,H with c copies of = d 2 H . If it is nonzero, c must satisfy ⟨ ⟩

2c n c 2d 2 d n 3d n 1 c 3d 1.

+ ( − ) = + + + = + − 2⇒ = − Consider the n-point stable map f C, p1, . . . , pn P in the underlying moduli space of H2,H2,...,H2,H,...,H . The 3d 1 copies of H2 are 3d 1 general points q and f p q , d ∶ ( ) → i i i where⟨ i 3d 1. This means⟩ that f−passes through 3d 1 general− points. So{ the} image( of )f =is a P2 rational≤ curve− of degree d in . And it is one of the Kd−rational curves of degree d.

The remaining n 3d 1 copies of H are n 3d 1 general lines Lj and f pj Lj, where

3d 1 j n. And−pj must+ be in the intersection− of+ Lj and the image of f.( By) the∈ B´ezout’s theorem,− ≤ the≤ f pj has only d choices. Therefore, the meaning of H2,...,H2,H,...,H with 3d 1 copies of H2 is the K rational ( ) d d ⟨ ⟩ − 6. Kontsevich’s Formula for Genus Zero Curves on P2 28

curve of degree d with each f pj having d choice. Then

( ) H2,...,H2,H,...,H d n−3d+1K . d d

a f =2 2 Now we can compute the Gromov-Witten potential of P . Let γ t01 t1H t2H .

1 = + + Φ γ γ, . . . , γ qd n! d d,n≥0. if d=0, n≥3 ( ) = 1 ⟨ ⟩ 1 Qγ, . . . , γ γ, . . . , γ qd n! 0 n! d n=3 d,n≥0. β≠0 = 1 ⟨ ⟩ + 1 ⟨ ⟩ Q γ, γ, γ Q γ, . . . , γ qd. 3! 0 n! d d,n≥0. β≠0 = ⟨ ⟩ + Q ⟨ ⟩ 1 1 3 2 2 1 3 2 1 2 1 2 2 And 3! γ, γ, γ 0 3! C2 H , 1, 1 0 t0t2 3! C1 H,H, 1 0 t0t1 2 t0t2 2 t0t1 because P2 H 1 and H H 1. P2 ⟨ ⟩ = ⟨ ⟩ + ⟨ ⟩ = + ∫ = And for fixed d 0, ∫ ⋅ = 1 > 1 γ, . . . , γ qd γ, . . . , γ qd n! d n! d n=0 n=3d+1 1 Q ⟨ ⟩ = Q C⟨ n H2⟩,...,H2,H,...,H tn−3d+1 t3d−1qd n! 3d−1 d 1 2 n=3d+1 1 n! = Q a d n−3d+1fK tn−3d+1 t3d−1qd n! 3d 1 ! n 3d 1 ! d 1 2 n=3d+1 1 1 = Q K t3d−1qd t d n−3d+1 3d 1 !( d −2 ) ( − +n) 3d 1 ! 1 n=3d+1 t3d−1 = 2 t1d d Q ( ) K( d − ) e q . ( − + ) 3d 1 !

= 2 Then the Gromov-Witten potential( Φ− γ) Φ t0, t1, t2 of P is given by

1 1 ( ) = ( t3d)−1 t2t t t2 K 2 et1dqd. 2 0 2 2 0 1 d 3d 1 ! d=1

+ + Q i Definition 12. Let Ti be a basis of H∗ X and (T −be) the corresponding dual basis. The big quantum product is defined as { } ( ) {3 } ∂ Φ k Ti Tj T . k ∂ti∂tj∂tk ∗ = Q The determination of Kd’s is a long-standing enumerative problem. In [5], Kontsevich solves the problem by using the associativity of the big quantum product for P2. 6. Kontsevich’s Formula for Genus Zero Curves on P2 29

3 2 2 2 1 0 ∂ Φ Let X P and T0,T1,T2 1,H,H T ,T ,T and Φijk , we have ∂ti∂tj∂tk = ( ) = ( ) = ( ) = Φ000 Φ001 Φ012 Φ022 0, Φ002 Φ011 1,

= t3d−1 = = = = t3d−=2 Φ K 2 d3et1dqd, Φ K 2 d2et1dqd, 111 d 3d 1 ! 112 d 3d 2 ! d=1 d=1 = Q t3d−3 = Q t3d−4 Φ K ( 2 − ) det1dqd and Φ K( −2 ) et1dqd. 122 d 3d 3 ! 222 d 3d 4 ! d=1 d=1 = Q = Q ( − ) ( − ) By the associativity, we have T1 T1 T2 T1 T1 T2 .

And ( ∗ ) ∗ = ∗ ( ∗ ) 2 i T1 T1 Φ11iT i=0 ∗ = Q 0 1 2 Φ110T Φ111T Φ112T

= Φ011T2 +Φ111T1 +Φ112T2

= T2 Φ111+ T1 Φ112+ T0.

= + + It follows that

T1 T1 T2 T2 Φ111T1 Φ112T0 T2

( ∗ ) ∗ = T( 2 +T2 Φ111+T1 T2 ) Φ∗ 112T0 T2

= Φ220∗ T2+ Φ221T1∗ Φ222+ T0 Φ∗111 Φ120T2 Φ121T1 Φ122T0 Φ112 Φ020T2 Φ021T1 Φ022T0

= (Φ122T1 + Φ222T0 + Φ111 Φ) 112+ T1 (Φ122T0 + Φ112T2+ ) + ( + + )

= (Φ222 Φ+111Φ122 )T+0 Φ(122 Φ111+Φ112 T1) +Φ112T2.

= ( + ) + ( + ) + And

T1 T1 T2 T1 Φ112T1 Φ122T0

∗ ( ∗ ) = Φ112∗T(1 T1 +Φ122T1 )T0

= Φ112 T2∗ Φ+111T1 Φ∗112T0 Φ122T1

2 = Φ112(T0 + Φ111Φ112+ Φ122 )T+1 Φ112T2.

= + ( + ) + 6. Kontsevich’s Formula for Genus Zero Curves on P2 30

2 By comparing the coefficients of T0 in T1 T1 T2 and T1 T1 T2 , we have Φ222 Φ111Φ122 Φ112.

( ∗ )∗ ∗( ∗ ) + = 2 t3d−2 t3d1−2 t3d2−2 Φ2 K 2 d2et1dqd K 2 d2 et1d1 qd K 2 d2 et1d2 qd . 112 d 3d 2 ! d1 3d 2 ! 1 1 d2 3d 2 ! 2 2 d=1 d1=1 1 d2=1 2

= ŒQ d 2 ‘ = Œ Q ‘ Œ Q ‘ The coefficient( of−q )of Φ112 is ( − ) ( − )

t3d−4 K K 2 d2d2 et1d. d1 d2 3d 2 ! 3d 2 ! 1 2 d1+d2=d 1 2 Q And ( − ) ( − )

t3d1−1 t3d2−3 Φ Φ K 2 d3 et1d1 qd K 2 d et1d2 qd . 111 122 d1 3d 1 ! 1 1 d2 3d 3 ! 2 2 d1=1 1 d2=1 2 d = Œ Q ‘ Œ Q ‘ The coefficient of q of Φ111Φ122( Φ222− )is ( − ) t3d−4 + t3d−4 K 2 et1d K K 2 d3d et1d. d 3d 4 ! d1 d2 3d 1 ! 3d 3 ! 1 2 d1+d2=d 1 2 + Q Thus we have ( − ) ( − ) ( − )

t3d−4 t3d−4 t3d−4 K 2 et1d K K 2 d2d2 et1d K K 2 d3d et1d d 3d 4 ! d1 d2 3d 2 ! 3d 2 ! 1 2 d1 d2 3d 1 ! 3d 3 ! 1 2 d1+d2=d 1 2 1 2 = Q − ( − ) ( − ) ( − ) ( − ) ( − ) 1 1 K K t3d−4 et1d d2d2 d3d d1 d2 2 3d 2 ! 3d 2 ! 1 2 3d 1 ! 3d 3 ! 1 2 d1+d2=d 1 2 1 2 = Q Œ − ‘ So, ( − ) ( − ) ( − ) ( − )

3d 4 ! 2 2 3d 4 ! 3 Kd Kd1 Kd2 d1d2 d1d2 d d d 3d1 2 ! 3d2 2 ! 3d1 1 ! 3d2 3 ! 1+ 2= ( − ) ( − ) = Q Œ − ‘ ( − ) ( − ) ( − ) ( − ) 3d 4 3d 4 2 2 3 Kd1 Kd2 d1d2 d1d2 . d1+d2=d ⎛⎛ 3d1− 2 ⎞ ⎛ 3d1− 1 ⎞ ⎞ = Q ⎜⎜ ⎟ − ⎜ ⎟ ⎟ 2 This is the Kontsevich’s formula⎝⎝ for genus− ⎠ zero curves⎝ on− P⎠. From⎠ K1 1 and K2 1, we can calculate all Kd with d 0. = = ≥ Bibliography 31

BIBLIOGRAPHY

[1] R. Bott and L. W. Tu, Differential forms in algebraic topology, Springer-Verlag, New York, 1982.

[2] D. A. Cox and S. Katz, Mirror symmetry and algebraic geometry, American Mathematical Society, 1999.

[3] P. Griffiths and J. Harris, Principles of algebraic geometry, John Wiley Sons, New York, 1994.

[4] S. Katz, Enumerative geometry and string theory, American Mathematical Society, Institute for Advanced Study, 2006.

[5] M. Kontsevich and Yu. Manin, Gromov-Witten classes, quantum cohomology, and enumerative geometry, Comm. Math. Phys 164 (1994), no. 3, 525–562.