Poincare polynomials of punctual Quot schemes

Note that since the matrix represents an ideal of colengthd, we must haved⁰+d⁰⁰ = d. Finally, by subtracting an appropriate multiple of the second column from the first column, we can further reducegto a polynomial,Íd⁰−1

0 aixⁱ, of degree less thand⁰. Hence, for each elementMofQuot_p(2,d), there exists a unique matrix of the form,

x^d⁰⁰ 0 Íd⁰−1

0 a_ixⁱ x^d⁰

! ,

where d⁰+ d⁰⁰ = d, that represents M. We saw that every element has such a representation, and two distinct matrices of the form above cannot differ by a series of column operations. We will calld⁰the length of the “sub-quotient" restricted to the second factor ofO. More precisely, consider the injectioni :C[[x]] →C[[x]]^⊕² into the second factor. Now consider the following cartesian square

C[[x]] ^M⁰ ^//

C[[x]]

C[[x]]^⊕² ^M ^//C[[x]]^⊕² .

Then, M⁰:C[[x]] → C[[x]]is precisely the ideal(x^d⁰).

Now, we can stratifyQuot_p(2,d)by the value ofd⁰in the (2,2)-entry of the matrixM in the reduced form. Alternatively, this stratification can be described as follows. Let i : O_C → O^⊕²

C be the second factor. Then, given(S → O^⊕²

C Q) ∈ Quot_p(2,d), we can consider the pullbackS⁰= S×_O⊕2

O_Cfrom the following cartesian square:

0 ^//S⁰ ^//

O_C

0 ^//S ^//O^⊕²

Then, we get a new exact sequence S⁰ → O_C Q⁰. SinceQ is a torsion sheaf supported on p, Q⁰ must also be a torsion sheaf supported on pof length d⁰ ≤ d. Then, (S⁰ → O_C Q⁰) is an element of Quot_p(1,d⁰). This gives us a map Quot_p(2,d) → Ý

d⁰≤dQuot_p(1,d⁰), which we can use to stratifyQuot_p(2,d). By the analysis above, we see that over Quot_p(1,d⁰), the fibers are isomorphic to A^d

02. The points of the stratum corresponding to d⁰ are uniquely determined by

2More precisely, the fibers are isomorphic toQuot_p(1,d −d⁰) ×A^d

since we get to choose x^d−d⁰ in the (1,1)-entry and a polynomial of degree at mostd⁰in the (2,1)-entry. However, since Quot_p(1,d−d⁰)is just a point, the fibers are isomorphic toA^d

matrices of the form

x^d−d⁰ 0 Íd⁰−1

0 a_ixⁱ x^d⁰

! .

In other words, these points are uniquely determined by the choice of coefficients a₀, . . . ,a_d⁰−1 ∈C. Therefore, we get the desired recursion

Quot_p(2,d)= Õ

d⁰≤d

L^d

0Quot_p(1,d⁰).

Now, we proceed by induction onn. Again, we take a matrixM ∈Quot_p(n,d). By column operation, we can reduceM to the form

M ⇒ x^d¹ 0 0 · · · 0

∗ ∗ ∗ · · · ∗

! ,

where * are the rest of the columns. In other words, we have a block matrix x^d¹ 0

∗ M⁰

! ,

where M⁰ is a (n − 1) × (n− 1) matrix with a nontrivial determinant. Hence, M⁰ ∈Quot_p(n−1,d⁰)for somed⁰. However, sinceM ∈Quot_p(n,d)we must have d⁰= d−d₁.

Now, by induction or by further column operations, we can reduce M to an lower triangular matrix whose diagonal entries are all powers of x, say x^d^r. And by further column operations we can assume ∗’s are polynomials of degree at most d_r −1, wherer is the corresponding column number:

x^d¹ 0 0 · · · 0

∗ x^d² 0 · · · 0

∗ ∗ x^d³ · · · 0 ... ... ... · · · ...

∗ ∗ ∗ · · · x^dⁿ ª

¬ ,

whereÍ

d_i = d. Analogous to the Quot_p(2,d) case, we get a mapQuot_p(n,d) → Ý

d⁰≤dQuot_p(n−1,d⁰) by restricting to the last (n− 1) factors of O^⊕n, which is precisely represented by M⁰ from above3. Fixing d⁰ = d₂+· · ·+d_n, the fibers of the above map overQuotp(n−1,d⁰)are again isomorphic toA^d

coming from the

3As before, this is equivalent to taking the fiber product of the kernel,S→ O^⊕n

C Q, with the injection of the last(n−1)factors,i:O_C^⊕n−1→ O_C.

choice of polynomialsÍdr−1

0 a_ixⁱfor allr = 2, . . . ,non the first column. Therefore, we get the desired relation inK(V ar/C):

[Quotp(n,d)]= Õ

d⁰≤d

L^d

0[Quotp(n−1,d⁰)].

With the recursion in Lemma 2.2, we can compute the class of Quot_p(n,d) in K(V ar/C).

Proposition 2.3. For all n ≥ 1and d ≥ 0, the class of Quot_p(n,d) in K(V ar/C) can be written in the following form4

[Quot_p(n,d)]= (Lⁿ−1)(Lⁿ⁺¹−1) · · · (L^n+d⁻¹−1)

(L−1)(L²−1) · · · (L^d−1) . (2.1) Proof. First, note that when n = 1, Quot_p(1,d) SpecC. Points of Quot_p(1,d) parametrize colengthdideals inC[[x]]. However, sinceC[[x]]is a PID,(x^d)is the unique ideal of colengthdinC[[x]]. Hence,Quot_p(1,d) SpecC.

Now, the boundary conditions and the recursive relation in Lemma 2.2 uniquely determine the class of Quot_p(n,d) in K(V ar/C). Hence, it suffices to show that equation 2.1 satisfy the conditions.

Forn= 1, equation 2.1 becomes

[Quot_p(1,d)]= (L−1)(L²−1) · · · (L^d−1) (L−1)(L²−1) · · · (L^d−1) =1.

Hence, we only have left to show that the equation satisfies the recursion. Suppose the equation 2.1 holds for alln⁰< nand alld⁰. We want to show that the recursion in Lemma 2.2 gives us equation 2.1. By the recursion we have

[Quot_p(n,d)] = Õ

d⁰≤d

L^d

0[Quot_p(n−1,d⁰)] (2.2)

= Õ

d⁰≤d

L^d

0(Lⁿ⁻¹−1)(Lⁿ−1) · · · (L^n+d

0−2−1)

(L−1)(L²−1) · · · (L^d⁰ −1) . (2.3) However, (2.3) is equal to

(Lⁿ−1) · · · (L^n+d⁻¹−1) (L−1) · · · (L^d−1)

d⁰≤d

L^d

0(Lⁿ⁻¹−1) (L^d

0+1−1) · · · (L^d−1) (Lⁿ⁺^d⁰⁻¹−1) · · · (Lⁿ⁺^d−¹−1).

4Whend =0, the formula gives[Quot_p(n,0)]=1.

Hence, it suffices to show that Õ

d⁰≤d

L^d

0(Lⁿ⁻¹−1) (L^d

0+1−1) · · · (L^d−1)

(L^n+d⁰⁻¹−1) · · · (L^n+d⁻¹−1) = 1. We will prove this by induction ond. Ford = 0, we have

(Lⁿ⁻¹−1) 1

Lⁿ⁻¹−1 =1.

Now letd ≥ 1 and assume the equality holds for alld⁰< d. Then, we get L^d(Lⁿ⁻¹−1)

L^n+d−¹−1 + Õ

d⁰≤d−1

L^d

0(Lⁿ⁻¹−1) (L^d

0+1−1) · · · (L^d−1) (L^n+d⁰⁻¹−1) · · · (L^n+d−¹−1)

= L^d(Lⁿ⁻¹−1)

L^n+d⁻¹−1 + L^d−1 L^n+d⁻¹−1

d⁰≤d−1

L^d

0(Lⁿ⁻¹−1)(L^d

0+1−1) · · · (L^d⁻¹−1) (L^n+d⁰⁻¹−1) · · · (L^n+d⁻²−1)

= L^d(Lⁿ⁻¹−1)

L^n+d⁻¹−1 + L^d−1 L^n+d⁻¹−1

= L^d(Lⁿ⁻¹−1)+L^d−1 L^n+d−¹−1

= L^n+d⁻¹−1 Lⁿ⁺^d−¹−1

= 1.

Hence equation 2.1 satisfies the recursive relation from Lemma 2.2 and the initial

conditions.

In particular, we have

[Quotp(n,1)]= Lⁿ−1 L−1 =

n−1

L^k = [Pⁿ⁻¹],

for alln.5 Moreover, by Proposition 2.2, we obtain the following corollary.

Corollary 2.1. LetPn,d(t)denote the Poincare polynomial ofQuotp(n,d). Then, we have

Pn,d(t)= (t²ⁿ−1)(t²ⁿ⁺²−1) · · · (t²ⁿ⁺²^d−²−1) (t²−1)(t⁴−1) · · · (t²^d−1) .

5In fact,Quot_p(n,1)Pⁿ⁻¹for alln.

2.6 Generating series for Poincare polynomials ofQuot_C(n,d)

In this section, we will find the Poincare polynomial ofQuot_C(n,d). Then, we will show that the generating series for the Poincare polynomials ofQuot_C(n,d) is, in fact, a rational function.

We first start by examining the class of[Quot_C(n,d)]inK(V ar/C)using the same stratification as in the proof of Lemma 2.2.

Lemma 2.3. LetC be a smooth curve overC, and letn,d ≥ 1. Then, we have the following relation inK(V ar/C):

[Quot_C(n,d)]= Õ

(d₁,...,dn)∈P(d)

Ön

i=1

[Hilb_C^dⁱ]L^idⁱ,

where Hilb_C^dⁱ Quot_C(1,d_i) is the Hilbert scheme over C of colength d_i ideal sheaves, and P(d) = {(d₁, . . . ,d_n) ∈ Zⁿ | d_i ≥ 0,Í

d_i = d}is the set of all ordered partitions ofd.

Proof. Recall that points ofQuot_C(n,d)parametrize exact sequence(S → O^⊕n

Q)such thatQis a torsion sheaf of lengthd. Leti: O^⊕(n−¹⁾

C → O^⊕n

C be the inclusion of the last(n−1)factors ofO_C. Now, consider the following cartesian diagram:

S⁰ ^//

O^⊕(n−¹⁾

S ^//O^⊕ⁿ

The diagram above gives us another exact sequence(S⁰→ O^⊕(ⁿ⁻¹⁾

C Q⁰), whereQ is a torsion sheaf of lengthd⁰ ≤ d. In other words, we have(S⁰→ O^⊕(ⁿ⁻¹⁾

C Q⁰) ∈

Quot_C(n−1,d⁰).

Therefore, we have a mapQuot_C(n,d) →Ý

d⁰≤dQuot_C(n−1,d⁰). Now, the fibers overQuot_C(n−1,d⁰)can be further stratified. Consider the expanded diagram

0 ^//S⁰ ^//

O^⊕(ⁿ⁻¹⁾

C //

Q⁰ ^//

0 ^//S ^//

O^⊕ⁿ

C //

Q ^//

0 ^//S⁰⁰ ^//O_C ^//Q⁰⁰ ^//0 .

obtained from the cartesian diagram above. Now,(S⁰⁰→ O_C → Q⁰⁰) ∈Quot_C(1,d− d⁰)and thus we can stratify the fibers ofQuot_C(n,d)overQuot_C(n−1,d⁰)by where each point maps to Quot_C(1,d − d⁰). Now, the fibers over Quot_C(1,d − d⁰) are isomorphic toA^d

by exactly the same argument as in the proof of Lemma 2.2.

Note that Quot_C(1,d) is isomorphic to Hilb_C^d, the Hilbert scheme parametrizing ideal sheaves on C of colength d, by definition. Hence, by Proposition 2.2, we conclude that the following relation holds inK(V ar/C):

[Quot_C(n,d)]= Õ

d⁰≤d

[Quot_C(n−1,d⁰)][Hilb^d⁻^d

C ]L^d

By induction onn, we get [Quot_C(n,d)] = Õ

d⁰≤d

[Quot_C(n−1,d⁰)][Hilb^d⁻^d

C ]L^d

= Õ

(d₁,...,dn)∈P(d)

[Hilb_C^d¹]L^d²⁺^···⁺^dⁿ[Hilb_C^d²]L^d³⁺^···⁺^dⁿ· · · [Hilb^d_Cⁿ]

= Õ

(d₁,...,d_n)∈P(d)

[Hilb_C^d¹][Hilb_C^d²]L^d²[Hilb^d_C³]L²^d³· · · [Hilb^d_Cⁿ]L⁽ⁿ⁻¹⁾^dⁿ

= Õ

(d₁,...,dn)∈P(d)

Ön

i=1

[Hilb^d_Cⁱ]L^idⁱ,

as desired.

Now, letQC,n,d(t) denote the Poincare polynomial of Quot_C(n,d). We will show that the generating series forQC,n,d(t)is a rational function.

Theorem 2.4. Let C be a projective curve and let QC,n,d(t) denote the Poincare polynomial ofQuot_C(n,d). Then, the generating series for QC,n,d(t) is a rational function. More precisely, let

QC,n(t,x):=Õ

QC,n,d(t)x^d ∈Z[[x,t]]

be the generating series forQC,n,d(t). Then, we can write

QC,n(t,x)= [(1+t x)(1+t³x) · · · (1+t²ⁿ⁻¹x)]²^g

(1−x)(1−t²ⁿx)[(1−t²x)(1−t⁴x) · · · (1−t²ⁿ⁻²x)]².

Proof. First, we find the generating series for the classes ofQuot_C(n,d)in the power series Grothendieck ring,K(V ar/C)[[x]], using Lemma 2.3.

[Quot_C(n,d)]x^d = Õ

(d₁,...,dn)∈P(d)

[Hilb^d_C¹][Hilb^d_C²]L^d²· · · [Hilb_C^dⁿ]L⁽ⁿ⁻¹^)dⁿx^dª

= Õ

(d₁,...,d_n)∈P(d)

[Hilb^d_C¹]x^d¹[Hilb^d_C²]L^d²x^d²· · · [Hilb_C^dⁿ]L⁽ⁿ⁻¹⁾^dⁿx^dⁿª

= Õ

[Hilb^d_C]x^d

! Õ

[Hilb_C^d]L^dx^d

· · · Õ

[Hilb_C^d]L⁽ⁿ⁻¹⁾^dx^d

n−1

i=0

[Hilb^d_C]L^idx^d

! .

Since C is a projective curve, we know by [7] that the generating series for the Poincare polynomials,QC,1,d(t), of Hilb^d_Cis

QC,1(t,x)=Õ

QC,1,d(t)x^d = (1+t x)²^g (1−x)(1−t²x), wheregis the genus ofC. Hence, we have

QC,1,d(t)t²^{k d}x^d = (1+t²^k+1x)²^g (1−t²^kx)(1−t²^k⁺²x). By Lemma 2.3 and Proposition 2.2, we know that

Q_C,n,d(t)= Õ

(d₁,...,d_n)∈P(d)

Q_C,₁_,d

1(t)Q_C,₁_,d

2(t)t²^d²Q_C,₁_,d

3(t)t⁴^d³· · ·Q_C,₁_,d_n(t)t²⁽ⁿ⁻¹^)dⁿ. Now, we are ready to find the generating seriesQ_n(t,x). We use the same manip- ulation as we did in finding the generating series for the classes of Quot_C(n,d)in K(V ar/C)[[x]]to findQC,n(t,x)inZ[[x,t]].

QC,n(t,x) = Õ

QC,n,d(t)x^d

= Õ

QC,1,d(t)x^d

! Õ

QC,1,d(t)t^2dx^d

· · · Õ

QC,1,d(t)t²⁽ⁿ⁻¹⁾^dx^d

= (1+t x)²^g (1− x)(1−t²x)

(1+t³x)²^g (1−t²x)(1−t⁴x)

· · ·

(1+t²ⁿ⁻¹x)²^g (1−t²ⁿ⁻²x)(1−t²ⁿx)

= [(1+t x)(1+t³x) · · · (1+t²ⁿ⁻¹x)]²^g

(1− x)(1−t²ⁿx)[(1−t²x)(1−t⁴x) · · · (1−t²ⁿ⁻²x)]².

Therefore, we conclude that the generating series, QC,n(t,x), is in fact a rational

function.

When, C = P¹ we get a particularly nice expression for the generating series for Quot_P1(n,d)in the power series Grothendieck ring.

Corollary 2.2. The generating series for[Quot_P1(n,d)]in K(V ar/C)[[x]]is a ra- tional function inx. More precisely, we have

Quot_P1(n,d)x^d = 1

(1−x)(1−Lx)²· · · (1−Lⁿ⁻¹x)²(1−Lⁿx). Proof. From the proof of Theorem 2.4, we saw

Quot_P1(n,d)x^d =

n−1

i=0

[Hilb^d_P1]L^idx^d

! . We know that Hilb^d

P¹ P^d 6. Since[P^d]= Í_d

i=0Lⁱ, we have Õ

[Hilb^d

P¹]x^d =Õ

d d

i=0

Lⁱ

x^d = 1

(1−x)(1−Lx). Similarly, for generalk, we have

[Hilb^d

P¹]L^{k d}x^d =Õ

Õd

i=0

Lⁱ

L^{k d}x^d = 1

(1−L^kx)(1−L^k+1x). Hence, we have

Quot_P1(n,d)x^d =

n−1

i=0

[Hilb_P^d1]L^idx^d

n−1

i=0

(1−Lⁱx)(1−Lⁱ⁺¹x)

= 1

(1−x)(1−Lx)²· · · (1−Lⁿ⁻¹x)²(1−Lⁿx).

2.7 The generating series for Poincare polynomials ofQ₀_,₂(G(n,n),d) Lemma 2 of [8] computes the Poincare polynomial ofQ₀_,₂(G(1,1),d)ford > 0.

Lemma 2.4. [8] Let P_d(t) be the Poincare polynomial of Q₀_,₂(G(1,1),d), where d >0. Then,

Pd(t)=(1+t²)^d−¹.

6See for example [4].

In the proof of this lemma, they show thatQ₀_,₂(G(n,n),d)can be stratified as follows.

Lemma 2.5. [8] Q₀_,₂(G(n,n),d) can be written as a disjoint union of quasi- projective strata,S(d₁,...,d_k), indexed by ordered partitions,(d₁, . . . ,d_k), ofd7. More- over, for each ordered partition,(d₁, . . . ,d_k), we have

S_(d

1,...,dk) Ök

i=1

Quot_C^×(n,d_i)/C^×.

In fact,S_(d

1,...,dk) ⊂Q₀_,₂(G(n,n),d)is the stratum of points parametrizing(C,p₁,p₂,Q), whereC is a nodal curve with k rational components,C₁, . . . ,C_k, such thatQ has lengthd_i onC_i. Note that since there are only two marked points onC, p₁ and p₂ must lie on the two extremal components ofC. Moreover, all components ofCare rational with two special points and thus, by the stability condition, Q must have positive length on eachC_i.

To simplify notation, we will write

Qn,d(t):=Q_C^×,n,d, Q_n(t,x):=Q_C^×,n(t,x)=Õ

Q_C^×,n,d(t)x^d,

Q_n,d(t):=P_Q

0,2(G(n,n),d)(t), S_n,d(t):= P_S_n,d(t),

where S_n_,d ⊂ Q₀_,₂(G(n,n),d) is the stratum associated to the ordered partition d = (d₁, . . . ,d_k).

Now, we try to find the generating function for the Poincare polynomials ofQ₀_,₂(G(n,n),d) withnfixed.

Theorem 2.5. Let

Q_n(t,x):=Õ

Q_n,d(t)x^d ∈Z[[t,x]]

be the generating series for the Poincare polynomials of Q_n_,d(G(n,n),d). Then, Q_n(t,x)is a rational function intand x. More precisely, we have

Q_n(t,x)= (1−t²ⁿx)(t²−1) t²−t²ⁿ⁺²x+x−1.

7i.e. tuples(d₁, . . . ,d_k)such thatd_i >0 for alliandÍd_i =d.

Proof. We first computeQ_n(t,x). The computation is similar to the computation in the proof of Theorem 2.4. MacDonald’s generating function forQ₁,d(t)does not hold in our case sinceC^×is not projective.

From the proof of Lemma 2 in [8], we know that the Poincare polynomial of Hilb^d(C^×)is

Q₁,d(t)=t^2d−t^2d⁻². Hence, we get

Q₁(t,x) = Õ

d≥0

Q₁,d(t)x^d

= 1+Õ

d>0

(t²^d−t²^d⁻²)x^d

= 1+(t²−1)Õ

d>0

t²^d−²x^d

= 1+(t²−1)xÕ

d≥0

t²^dx^d

= 1+ x(t²−1) 1−t²x

= 1−x 1−t²x. It follows that for general kwe have

P(Hilb^d(C^×))(t)t²^{k d}x^d = 1−t²^kx 1−t^2k+2x. Now, by Lemma 2.3, we obtain

Q_n(x,t) = Õ

Q₁,d(t)x^d

! Õ

Q₁,d(t)t²^dx^d

· · · Õ

Q₁,d(t)t²⁽ⁿ⁻¹⁾^dx^d

= 1−x

1−t²x · 1−t²x

1−t⁴x · · ·1−t²ⁿ⁻²x 1−t²ⁿx

= 1− x 1−t²ⁿx.

Now, we computeQ_n,d(t,x). By Lemma 2.5 we have [Q₀_,₂(G(n,n),d)]= Õ

d∈P(d)

[S_n,d],

where P(d) = {(d₁, . . . ,d_k) | d_i > 0∀i and Í

d_i = d} is the set of all ordered partitions ofd. Therefore,

Q_n,d(t)= Õ

d∈P(d)

S_n,d(t).

Recall that

S_n,(_d

1,...,d_k) Ök

i=1

Quot_C^×(n,d_i)/C^×.

From [2], we know that the Poincare polynomial ofQuot_C^×(n,d_i)/C^× is equal to Q_n_,d_i(t)

t²−1 . Hence,

Sn,(d₁,...,d_k)(t)= Ök

i=1

Qn,d_i(t) t²−1 . We are now ready to computeQ_n(t,x):

Q_n(t,x) = Õ

d≥0

Q_n_,d(t)x^d

= Õ

d≥0

« Õ

d∈P(d)

S_n_,d(t)x^dª

= Õ

k≥0

d₁,...,dk≥1

S_n,(d

1,...,d_k)(t)x^d¹⁺^···^+d^k

= Õ

k≥0

d₁,...,d_k≥1

Ök

i=1

Qn,d_i(t) t²−1

x^d¹⁺^···^+d^k

= Õ

k≥0

d₁,...,d_k≥1

Ök

i=1

Qn,d_i(t)x^dⁱ t²−1

! !

= Õ

k≥0

d₁≥1

Q_n,d

1(t)x^d¹ (t²−1)

· · · Õ

d_k≥1

Qn,d_k(t)x^d^k (t²−1)

! !

= Õ

k≥0

Q_n(x,t) −1 t²−1

= 1

1− (Q_n(x,t) −1) t²−1−1

1− x(t²ⁿ−1) (1−t²ⁿx)(t²−1)

⁻1

= (1−t²ⁿx)(t²−1) (1−t²ⁿx)(t²−1) −x(t²ⁿ−1)

= (1−t²ⁿx)(t²−1) t²−t²ⁿ⁺²x+x−1

Hence,Q_n(t,x)is a rational function int andx.

Dalam dokumen The Geometry of Moduli Spaces of Maps from Curves (Halaman 72-84)