Triple intersections

Addendum 4.14. DivisorF from lemma 4.13 furthermore satisfies:

(ζ_n⁻¹)^β∗(ζ_n^α∗F)−(ζ_n^α∗F) =π^γ∗(D_d^γ)−π^α∗(D_d^α).

With different choices ofpα,pβandpγ, there exists a (different) divisor,Fon Σ˜ of degree^dn(q−1)₂ onΣ, such that:˜

(ζn)^α∗F−F =π^β∗(D^β_d)−π^γ∗(D_d^γ) and

(ζn)^β∗F−F =π^α∗(D^α_d)−π^γ∗(D^γ_d).

Proof. For the first claim, simply observe that:

π^γ∗(D_d^γ)−π^α∗(D_d^α) = π^β∗(D^β_d)−π^α∗(D^α_d)−(π^β∗(D^β_d)−π^γ∗(D_d^γ))

= ζ_n^γ∗F−F−ζ_n^α∗F+F

= (ζ_n⁻¹)^β∗(ζ_n^α∗F)−(ζ_n^α∗F)

For the second claim, proceed in the proof of lemma 4.13, only this time lettingpα=π^α(˜p),pβ=π^β(˜p)andpγ=π^γ(˜p). Then put

F=













0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 1 0













Forneven,a∈ ^α₂,b∈ ^β₂ andc∈ ^γ₂, and forζ, ζ^′, ζ^′′∈µr,r= (n, d), denote:

|M(n,∆d)|^ζ,ζ_a,b,c^′,ζ′′=|M(n,∆d)|^ζ_a∩ |M(n,∆d)|^ζ_b^′∩ |M(n,∆d)|^ζ_c^′′

Lemma 4.16. Assume that|M(n,∆d)|α,β 6=∅. Letγ∈J⁽ⁿ⁾. We then have:

|M(n,∆d)|α,β,γ6=∅ ⇔γ∈ hα, βi.

In the affirmative case,|M(n,∆d)|α,β,γ=|M(n,∆d)|α,β.

Proof. Suppose that|M(n,∆d)|α∩|M(n,∆d)|β∩|M(n,∆d)|γ 6=∅. The first claim follows becauseJ⁽ⁿ⁾/hα, βiacts freely on|M(n,∆d)|α,β according to proposi- tion 4.6. The second claim is trivial.

The next obvious question is: Whenγ =α+β, how do the components of

|M(n,∆d)|α,|M(n,∆d)|βand|M(n,∆d)|γintersect?

Notice that the results about|M(n,∆d)|αapply to|M(n,∆d)|βand|M(n,∆d)|γ

as well. Moreover,λn(α, β) =λn(α, γ) =λn(γ, β). Hence, the above description of the pairwise intersections applies to each of the three pairs.

4.3.1 Component intersections, odd rank

Whennis odd, there is a complete answer to the question of which triple com- ponent intersection are non-empty.

Theorem 4.17. Assumenis odd. Letr= (n, d)andq=ⁿ_r. Supposeα, β∈J⁽ⁿ⁾are primitive elements withhαi ∩ hβi= 0andord(λn(α, β)) =q. Letγ =α+β. For each tripleζ, ζ^′, ζ^′′∈µrwe have:

|M(n,∆d)|^ζ,ζ_α,β,γ^′,ζ′′6=∅ ⇔ ζ^′′

ζζ^′ = 1.

Proof. Givenζ, ζ^′, ζ^′′ ∈ µr, we may chooseδ ∈ J⁽ⁿ⁾such thatλn(α, δ)^q = ζ⁻¹ andλn(β, δ)^q = (ζ^′)⁻¹. Then, by proposition 3.42 we have:

E∈ |M(n,∆d)|^ζ,ζ_α,β,γ^′,ζ′′⇔Lδ⊗E∈ |M(n,∆d)|^1,1,

ζ′′

ζζ′

α,β,γ .

Hence it suffices to prove the seemingly weaker statement that for eachζ∈µr,

|M(n,∆d)|^1,1,ζ_α,β,γ6=∅ ⇔ζ= 1.

Furthermore, since the assumptions ensure that|M(n,∆d)|^1,1_α,β 6=∅, it is enough to show that anyE∈ |M(n,∆d)|^1,1_α,βmust lie in|M(n,∆d)|¹_γ.

LetE ∈ |M(n,∆d)|^1,1_α,β. SupposeE ∈ |M(n,∆d)|^ζ_γ. According to section 3.6 we may find line bundlesL1,L2andL3onΣ^α,Σ^βandΣ^γrespectively, such that E ∼= πα∗(L1) ∼= πβ∗(L2) ∼= πγ∗(L3). By proposition 3.31 and definition 3.36, there exist line bundlesK1onΣ^αandK2onΣ^β, both of degree zero, such that:

L1∼= ∆^αd ⊗(ζ_n⁻¹)^∗_α(K1)⊗K₁⁻¹ L2∼= ∆^β_d ⊗(ζn)^∗_β(K2)⊗K₂⁻¹ Now define:

Ke3=π^α∗(K1)⁻¹⊗π^β∗(K2)⊗[F] whereFis the divisor from lemma 4.13. Notice that:

(ζn)^γ∗(Ke3) ∼= π^α∗((ζ_n⁻¹)^∗_α(K1))⁻¹⊗π^β∗((ζn)^∗_β(K2))⊗[(ζn)^γ∗F]

∼= Ke3⊗π^α∗(L⁻¹₁ ⊗∆^α_d)⊗π^β∗(L2⊗(∆^β_d)⁻¹)⊗[(ζn)^γ∗F−F]

∼= Ke3

–Where the last step used lemma 4.12 as well as the first property ofF.

The above calculation implies thatKe3 ∼= π^γ∗(K3)for some line bundleK3

onΣ^γ. Notice thatdeg(K3) = _n¹deg(Ke3) =_n¹deg(F) =^d(q−1)₂ . Now letL= ∆^γ_d⊗(ζ_n⁻¹)^∗_γ(K3)⊗K₃⁻¹. We have:

π^γ∗(L) ∼= π^γ∗(∆^γ_d)⊗(ζn)^α∗(Ke3)⊗Ke₃⁻¹

∼= π^γ∗(∆^γ_d)⊗(ζn)^α∗π^β∗(K2)⊗π^β∗(K2)⁻¹⊗[(ζn)^α∗F−F]

∼= π^β∗(∆^β_d ⊗(ζn)^∗_β(K2)⊗K₂⁻¹)

∼= π^β∗(L2)

∼= π^γ∗(L3)

–where we used the second property ofFas well as lemma 4.12 in the last two steps.

SinceKer(π^γ∗) =hπ^∗_γ(Lβ)i, we get for somej∈ {0,1, . . . n−1}:

L∼=L3⊗π^∗_γ(Lβ)^⊗j.

Hence, sinceE∈ |M(n,∆d)|α,β,γ:

πγ∗(L)∼=πγ∗(L3⊗π^∗_γ(Lβ)^⊗j)∼=E⊗L^⊗j_β ∼=E.

But by proposition 3.31, L ∈ ∆^γ_d ⊗Pγ^{ζn−deg(K3 )}, and hence: E ∼= πγ∗(L) ∈

|M(n,∆d)|^ζ_γ, where

ζ= (ζ_n^−deg(K3))^q =ζ⁻

dq(q−1)

n 2 = 1

– sincenand therebyqis odd, andn|dq.

4.3.2 Component intersections, even rank

Whennis even, the situation is somewhat more complicated. In the general case we only show a partial result, which is complete in the case where(n, d) = 2, thus generalising the important casen= 2, d= 0treated in [1].

Subsequently we propose a method for treating values ofnthat are twice an odd number. Unfortunately, it is not clear how to generalise to greater powers of2, leaving the casen= 4still a partial mystery.

Proposition 4.18. Assume that bothnanddare even. Letr = (n, d)andq = ⁿ_r. Supposeα, β∈J⁽ⁿ⁾are primitive elements withhαi∩hβi= 0andord(λn(α, β)) =q.

Leta∈ ^α₂,b∈ ^β₂, and defineγ=α+β andc=a+b. For each tripleζ, ζ^′, ζ^′′∈µr, we have:

|M(n,∆d)|^ζ,ζ_a,b,c^′,ζ′′6=∅ ⇒ ζ^′′

ζζ^{′ r}₂

= (−1)^d(q−1)2 λ2n(a, b)ⁿ².

Proof. Suppose E ∼= πα∗(L1) ∼= πβ∗(L2) ∼= πγ∗(L3) ∈ |M(n,∆d)|^ζ,ζ_a,b,c^′,ζ′′. By proposition 3.31 and definition 3.37, we may pick line bundlesK1,K2 andK3

onΣ^α,Σ^βandΣ^γrespectively, such that:

L1∼= ∆^αd ⊗π^∗_α(La)⊗(ζn)^∗_αK1⊗K₁⁻¹ , (ζn)^deg(K1)·q =ζ L2∼= ∆^β_d ⊗π_β^∗(Lb)⊗(ζn)^∗_βK2⊗K₂⁻¹ , (ζn)^deg(K2)·q =ζ^′ L3∼= ∆^γ_d⊗π^∗_γ(Lc)⊗(ζn)^∗_γK3⊗K₃⁻¹ , (ζn)^deg(K3)·q=ζ^′′.

Pick divisorsEiwith[Ei] =Ki,(i= 1,2,3). Furthermore, pick divisorsDa, DbandDconΣ, representingLa,LbandLc, respectively. Define:

D1=D_d^α+π^∗_α(Da) + (ζn)^∗_αE1−E1

D2=D^β_d+π^∗_β(Db) + (ζn)^∗_βE2−E2

D3=D^γ_d+π^∗_γ(Dc) + (ζn)^∗_γE3−E3

I.e.Li∼= [Di]. Now, lemma 4.12 implies the existence of meromorphic func- tionsh1andh2onΣ, such that:˜

π^α∗(D1) + (h1) =π^β∗(D2) + (h2) =π^γ∗(D3).

For the sake of readability as well as later reference, the rest of the proof is divided into lemmas.

Lemma 4.19. Letgα,gβandgγdenote the meromorphic functions onΣ^α,Σ^βandΣ^γ, given by proposition 3.25.

We may chooseDa, Db and Dc so thatπ^∗_α(2Da) = (gα), π_β^∗(2Db) = (gβ)and π^∗_γ(2Dc) = (gγ).

Proof. LetΣ^adenote the2n-sheeted covering associated toa. Letgabe the mero- morphic function, given by lemma 3.25 applied toa. Consider the diagram:

Σ^a

πa

A

AA AA AA A

π //Σ^α

πα

~~||||||||

Σ

LetNmdenote the norm map associated toπ, and notice that forζ ∈ µ2n: (ζ²)^∗_α(Nm(ga)) = Nm(ζ_a^∗(ga)) = Nm(ζ⁻¹·ga) = (ζ²)⁻¹·Nm(ga). This shows that we may useNm(ga)asgα. Thus, choosingDasuch thatπ^∗_a(Da) = (ga), we get:

π^∗_α(2Da) = Nm(π^∗(π_α^∗(Da))) = Nm((ga)) = (gα).

Similarly forbandc.

Lemma 4.20. We may assume without loss of generality that Nm^α(h2) = (gα)ⁿ², Nm^β(h1) = (gβ)ⁿ² andNm^γ(^h_h¹₂) = (gγ)ⁿ².

Furthermore, forζ∈µnlettingkA(ζ) = ^ζα∗_h^h1

1 ,kB(ζ) = ^ζβ∗_h^h2

2 andk(ζ) = ^k_k^A(ζ)

B(ζ), thenk(ζ)is constant, andk(ζ)ⁿ=ζ⁻ⁿ² ∈ {±1}

Proof. For the first claim, notice that

(π^α∗(Nm^α(h2))) = π^α∗(Nm^α(π^γ∗(D3)−π^β∗(D2))

= π˜^∗(Nmγ(D3)−Nmβ(D2))

= π˜^∗(Nmγ(D^γ_d) +nDc−Nmβ(D^β_d)−nDb)

= π^α∗(π^∗_α(nDa))

= π^α∗(n 2(gα))

-where we used lemma 4.19 in the last step. This shows thatNm^α(h2)and(gα)ⁿ² have the same divisors. Hence, after scalingh2, we may assume thatNm^α(h2) = (gα)ⁿ². Similarly in the other two cases.

For the final claim, we calculate:

(k(ζ)) = (kA(ζ))−(kB(ζ))

= ζ^α∗(π^γ∗(D3)−π^α∗(D1))−(π^γ∗(D3)−π^α∗(D1))

−ζ^β∗(π^γ∗(D3)−π^β∗(D2)) + (π^γ∗(D3)−π^β∗(D2))

= ζ^α∗(π^γ∗(D3))−ζ^β∗(π^γ∗(D3))

= π^γ∗(ζ_γ^∗(D3)−ζ_γ^∗(D3))

= 0

This shows thatk(ζ)is constant. To see that it is a root of unity:

k(ζ)ⁿ= Nm^β(k(ζ)) = Nm^β(ζ^α∗(h1) h1

) =

ζ_β^∗(gβ) gβ

ⁿ2

=ζ⁻ⁿ² =±1

Lemma 4.21. λ2n(a, b)ⁿ² = (−1)^d(q−1)2

ζ^′′

ζζ^′

^r₂

Proof. Since by lemma 4.19, 2nDa = (Nmα(gα))and 2nDb = (Nmβ(gβ)), we may calculate: (Using lemma 4.20 in the second equality.)

λ2n(a, b)ⁿ² =

Nmα(gα)(Db) Nmβ(gβ)(Da)

ⁿ₂

= Nm(hg 2)(Db) Nm(hg 1)(Da)

= h2(˜π^∗Db) h1(˜π^∗Da)

= h2(˜π^∗Dc−˜π^∗Da) h1(˜π^∗Dc−π˜^∗Db)

= h2(π^γ∗(D3−D_d^γ−(ζn)^∗_γE3+E3)−π^α∗(D1−D^α_d −(ζn)^∗_αE1+E1)) h1(π^γ∗(D3−D^γ_d−(ζn)^∗_γE3+E3)−π^β∗(D2−D^β_d −(ζn)^∗_βE2+E2))

= X·Y ·Z

–Here (by Weil reciprocity):

X= h2(π^γ∗D3−π^α∗D1)

h1(π^γ∗D3−π^β∗D2) =h2(h1) h1(h2)= 1.

And:

Y = h2(π^α∗(D^α_d)−π^γ∗D_d^γ))

h1(π^β∗(D^α_d)−π^γ∗(D^γ_d))= h2((ζn)^β∗(F)−F) h1((ζn)^α∗(F)−F) =

(ζ_n⁻¹)^β∗h2

h2 (F)

(ζ⁻¹n )^α∗h1

h1 (F) –whereFis the divisor from addendum 4.14. So, by the last part of lemma 4.20, and using the assumption thatdis even:

Y =k(ζ_n⁻¹)(F) =k(ζ_n⁻¹)^deg(F)=k(ζ_n⁻¹)^nd(q−1)2 = (−1)^d(q−1)2 .

And finally:

Z = h2(π^α∗((ζn)^∗_αE1−E1)−π^γ∗((ζn)^∗_γE3−E3)) h1(π^β∗((ζn)^∗_βE2−E2)−π^γ∗((ζn)^∗_γE3−E3))

= Nm^α(h2)((ζn)^∗_αE1−E1)

Nm^β(h1)((ζn)^∗_βE2−E2)Nm^γ(h2

h1

)((ζn)^∗_γE3−E3)

= gα((ζn)^∗αE1−E1)

gβ((ζn)^∗_βE2−E2)·gγ((ζn)^∗_γE3−E3)

!ⁿ₂

= ζn^deg(E1)

ζn^deg(E2)ζn^deg(E3)

!ⁿ₂

= ζn^deg(K3)·q

ζn^deg(K1)·qζn^deg(K2·q)

!_2qⁿ

= ζ^′′

ζζ^{′ r}₂

–Where we used the assumption thatdis even, and henceq | ⁿ₂, again in the second but last equality.

Corollary 4.22. With the assumptions in proposition 4.18, if(n, d) = 2, we have:

|M(n,∆d)|^ζ,ζ_a,b,c^′,ζ′′6=∅ ⇔ ζ^′′

ζζ^′ = (−1)^d(q−1)2 λ2n(a, b)ⁿ².

Proof. The arrow to the right comes from proposition 4.18. Conversely, given ζ, ζ^′, ζ^′′ such that the right hand statement is true, the assumptions ensure by proposition 4.9 and lemma 4.16 that

|M(n,∆d)|^ζ,ζ_a,b,c^′,ξ6=∅

for someξ∈µr, but then the implication to the right gives that ξ=ζζ^′(−1)^d(q−1)2 λ2n(a, b)ⁿ² =ζ^′′

Finally, we seek to strengthen the result of proposition 4.18 in the special case whered= 0andnis twice an odd number. Assume throughout the investiga- tion thatn= 2˜n, wheren˜is odd, letα∈J⁽ⁿ⁾be primitive, and denote:

α1= 2α∈J^(˜n) α2= ˜nα∈J⁽²⁾

Lemma 4.23. IfE1∈ |M(˜n,O)|α1 andE2∈ |M(2,O)|α2, then:

E1⊗E2∈ |M(n,O)|α

Proof. Observe that

Lα∼=L^−˜_αⁿ⊗(L²_α)^n+1˜2 ∼=L⁻¹_α1 ⊗L

˜ n+1

α22 . Hence, ifE1∈ |M(˜n,O)|α1 andE2∈ |M(2,O)|α2, then:

(E1⊗E2)⊗Lα∼= (E1⊗L⁻¹_α1)⊗(E2⊗L

˜ n+1

α22 )∼=E1⊗E2.

Next, introduce the coveringsΣ^α1,Σ^α2andΣ^α, and notice that we have the following commutative diagram:

Σ^α

π^α2

""

EE EE EE EE

π^α1

||yyyyyyyy

πα

Σ^α1

πα1

""

EE EE EE

EE Σ^α2

πα2

||yyyyyyyy

Σ

–whereπ^α1 is the2-sheeted covering associated to π_α^∗₁(Lα)and π^α2 is then-˜ sheeted covering associated toπ^∗_α2(Lα).

Assume thatβ ∈ J⁽ⁿ⁾ is primitive, with hαi ∩ hβi = 0 andλn(α, β) = 1.

Denote:

β1= 2β∈J^(˜n) β2= ˜nβ∈J⁽²⁾

Lemma 4.24. Suppose that

E1∼=πα1∗(L1)∈ |M(˜n,O)|α1,β1

and

E2∼=πα2∗(L2)∈ |M(2,O)|α2,β2

Then:

E1⊗E2∼=πα∗(π^α1∗(L1)⊗π^α2∗(L2))

Proof. By lemma 4.23,E1⊗E2 ∈ |M(n,O)|α,β. Hence, by proposition 4.6, we may find some line bundleLonΣ^αwithπα∗(L) ∼= E1⊗E2. PullingE1⊗E2

back toΣ^αthen yields:

π_α^∗(E1⊗E2)∼=π^∗_α(πα∗(L))∼= M

ζ∈µn

ζ^∗L∼= Mn i=1

L⊗π^∗_α(Lβ)^⊗i (4.3) Where the last isomorphism is becauseL∈Iζ for some primitiveζ∈µnby proposition 4.6.

On the other hand,

π_α^∗(E1⊗E2) ∼= π^∗_α(E1)⊗π_α^∗(E2)

∼= π^α1∗(M

ζ∈µn˜

ζ^∗(L1))⊗π^α2∗(M

ζ∈µ2

ζ^∗(L2))

∼= π^α1∗(

˜

Mn i=1

L1⊗π_α^∗₁(Lβ1)^⊗i)⊗π^α2∗( M2 i=1

L2⊗π_α^∗₂(Lβ2)^⊗i)

∼= π^α1∗(L1)⊗π^α2∗(L2)⊗

n

M2

i=1

M2 j=1

π_α^∗(Lβ)^2i+n2j

∼= (π^α1∗(L1)⊗π^α2∗(L2))⊗ Mn i=1

π_α^∗(Lβ)ⁱ (4.4) Putting (4.3) and (4.4) together, the theorem of Krull-Remak-Schmidt gives thatπ^α1∗(L1)⊗π^α2∗(L2)∼=L⊗π_α^∗(Lβ)^⊗ifor somei, and hence:

πα∗(π^α1∗(L1)⊗π^α2∗(L2))∼=πα∗(L)⊗L^⊗i_β ∼=E1⊗E2.

Remark4.25. So far we have silently assumed thatE1⊗E2is semistable and has the right determinant. (Only through the choice of notation, though. -We have not used it otherwise.)

Notice that it now follows from the above lemma in our special case, where Eiis a fixed point under the action ofαias well asβi.

As explained in the proof of proposition 3.20, the action ofζ ∈ µn onΣ^α covers the action ofζ²onΣ^α1 as well as the action ofζⁿ² onΣ^α2,µ2 ⊆µn and µn˜⊆µnbeing the Galois groups for the coveringsπ^α1andπ^α2respectively.

Lemma 4.26. The map:|M(˜n,O)|α1,β1 × |M(2,O)|α2,β2 → |M(n,O)|α,β given by (E1, E2)7→E1⊗E2is bijective.

Proof. To prove surjectivity, assume thatE ∈ |M(n,O)|α,β. By proposition 4.6, we may pick a line bundleLonΣ^αwith πα∗(L)∼= Eandζ^∗(L)∼= π_α^∗(Lβ)for some primitive elementζ∈µn. Thus, for alli∈ {1,2, . . . , n}:

(ζⁱ)^∗(L)∼=π^∗_α(Lβ)^⊗i.

Now letL1= (Nm^α1(L)⊗π^∗_α1(Lα))^⊗n+1˜2 andL2= Nm^α2(L)⁻¹. First observe that:

π^α1∗(L1)⊗π^α2∗(L2) ∼= O2 i=1

(ζ^˜ni)^∗L

!^˜n+1₂

⊗

˜

On i=1

(ζ²ⁱ)^∗L

!−1

∼= O2 i=1

L⊗π^∗_α(Lβ)^ni˜

!^n+1˜₂

⊗

˜

On i=1

L⊗π_α^∗(Lβ)²ⁱ

!−1

∼= L^2·n+1˜2 ⊗L^−˜n⊗ O2

i=1

π^∗_α(Lβ)^ni˜

!^n+1˜₂

⊗

˜

On i=1

π^∗_α(Lβ)²ⁱ

!−1

∼= L⊗π^∗_α(Lβ)^k

∼= (ζ^k)^∗(L) –for somek∈ {1,2, . . . , n}. Consequently,

πα1∗(L1)⊗πα2∗(L2)∼=πα∗(π^α1∗(L1)⊗π^α2∗(L2))∼=πα∗(L)∼=E.

One needs to check thatπα1∗(L1)andπα2∗(L2)have the right determinants.

For this, observe that sinceOΣ ∼= det(πα∗(L))∼= Nmα(L)⊗L^⊗˜_αⁿ(by propo- sition 3.23), we get: (Using proposition 3.23 with the coveringπα1 in the first step)

det(πα1∗(L1)) ∼= Nmα1(L1)

∼= Nmα(L)⊗Lⁿ_α^˜n+1˜₂

∼= L^˜n_α⊗L^˜n_α^n+1˜₂

∼= OΣ

And also: (Using proposition 3.23 with the coveringπα2.) det(πα2∗(L2)) ∼= Nmα2(L2)⊗Lα2

∼= Nmα(L)⁻¹⊗L^˜n_α

∼= OΣ

This proves surjectivity. Finally, by proposition 4.6 the number of elements in|M(n,O)|α,β is:

a(α, β,0)·n^2g−2=φ(n)·n^2g−2.

–Whereas|M(˜n,O)|α1,β1hasφ(˜n)·n˜^2g−2elements, and|M(2,O)|α2,β2has2^2g−2 elements. Hence, the number of elements in|M(˜n,O)|α1,β1× |M(2,O)|α2,β2is

φ(˜n)·˜n^2g−2·2^2g−2=φ(n)·n^2g−2.

Next, we ask how the map in lemma 4.26 relates the components of the fixed point varieties.

Lemma 4.27. Supposea ∈ ^α₂. IfE1 ∈ |M(˜n,O)|^ζ_α¹₁ for someζ1 ∈ µn˜ andE2 ∈

|M(2,O)|^ζ_na˜² for someζ2∈µ2, thenE1⊗E2∈ |M(n,O)|^ζ_a^1·ζ2. Proof. Assume

E1∼=πα1∗(L1)∈ |M(˜n,O)|^ζ_α¹₁ E2∼=πα2∗(L2)∈ |M(2,O)|^ζ_na˜². This is to say thatL1∈P_α^ζ1₁ andL2∈π_α^∗₂(L˜na)⊗P_α^ζ2₂.

For allζ∈µn, we have:

ζ_α^∗◦π^α1∗=π^α1∗◦(ζ²)^∗_α1 ζ_α^∗◦π^α2∗ =π^α2∗◦(ζ^n˜)^∗_α2

Since pull-back takesPick(Σ^α1)to Pic2k(Σ^α) and Pick(Σ^α2) to Pic˜nk(Σ^α), proposition 3.31 shows that:

π^α1∗(P_α^ζ1₁)⊆P_α^ζ1 π^α2∗(P_α^ζ2₂)⊆P_α^ζ2

Consequently,π^α1∗(L1)∈P_α^ζ1andπ^α2∗(L2)∈π_α^∗(L˜na)⊗P_α^ζ2. But π_α^∗(L˜na)∼=π_α^∗(La)^⊗˜n ∼=π^∗_α(La)⊗π_α^∗(Lα)^⊗n−1˜2 ∼=π_α^∗(La), soπ^α1∗(L1)⊗π^α2∗(L2)∈π_α^∗(La)⊗P_α^ζ1·ζ2, and hence:

E1⊗E2∈ |M(n,O)|^ζ_a^1·ζ2.

Theorem 4.28. Assumen= 2˜n, wheren˜is odd. Supposeα, β ∈J⁽ⁿ⁾are primitive elements withhαi∩hβi= 0andλn(α, β) = 1. Leta∈ ^α₂,b∈ ^β₂, and defineγ=α+β andc=a+b. For each tripleζ, ζ^′, ζ^′′∈µn, we have:

|M(n,O)|^ζ,ζ_a,b,c^′,ζ′′6=∅ ⇔ ζ^′′

ζζ^′

=λ2n(a, b)ⁿ².

Proof. SupposeE∈ |M(n,O)|^ζ,ζ_a,b,c^′,ζ′′for someζ, ζ^′ζ^′′∈µn.

Denoteα1, α2, β1, β2, γ1, γ2as in the preceding lemmas. Leta2= ˜na,b2= ˜nb andc2= ˜nc.

By lemma 4.26,E ∼= E1⊗E2 for some bundles, E1 ∈ |M(˜n,O)|α1,β1 and E2∈ |M(2,O)|α2,β2.

Suppose that

E1∈ |M(˜n,O)|^ζ_α¹₁^,ζ_,β^1′₁^,ζ_,γ^′′1₁ E2∈ |M(2,O)|^ζ_a²₂^,ζ_,b^2′₂^,ζ_,c^2′′₂

– for certainζ1, ζ₁^′, ζ₁^′′ ∈ µn˜ andζ2, ζ₂^′, ζ₂^′′ ∈µ2. Then by lemma 4.27,ζ1ζ2 =ζ, ζ₁^′ζ₂^′ =ζ^′andζ₁^′′ζ₂^′′=ζ^′′. But by theorem 4.17,

ζ₁^′′

ζ1ζ₁^′ = 1

and by corollary 4.22 (or simply theorem 2.3 in [1]), ζ₂^′′

ζ2ζ₂^′ =λ4(a2, b2) =λ4˜n(a, b)^n˜ =λ2n(a, b)ⁿ².

The converse implication follows in precisely the same manner as in corol- lary 4.22.

The action of J ^{( n )} on the Hecke correspondence

So far, I have only treated one value ofdat a time. There is an important in- terplay, known as the Hecke correspondence, between the moduli spaces corre- sponding to different values ofd. I will only treat the case of degrees zero and one. This is done in order to avoid the topic of parabolic semistability. It will be sufficient as a tool in chapter 6 in most cases, including the ones wherenis prime.

5.1 Elementary modification

Recall thatp∈Σwas chosen back in section 3.1. SupposeEis a vector bundle onΣandF ⊆ Epa codimension one subspace. Elementary modification ofE atpin the direction ofFyields a bundleE^′onΣ. It is constructed as follows:

Definition 5.1. Let C_p denote the skyscraper sheaf on Σ with support at p.

Choose an isomorphism: Ep/F ∼= C. Denote byΓ(E) the sheaf of holomor- phic sections inE and define the sheaf morphism: λ: Γ(E) → C_p as follows:

GivenU ⊆^◦ Σand a sectionsonU: λU(s) =

0 , p /∈U

[s(p)]∈Ep/F ∼=C , p∈U 83

The kernel ofλis clearly independent of the chosen isomorphism. According to the following lemma, it is a locally free sheaf of rankn, and hence that is is equal toΓ(E^′)for some bundleE^′onΣof rankn. This bundle is the elementary modification ofEin the direction ofF.

Lemma 5.2. The kernel ofλin the definition above is a locally free sheaf of rankn.

Proof. On a small neighbourhoodUofp, we may choose a coordinatezcentred atpand a holomorphic framee1, e2, . . . , enofEsuch thate1(p)∈ F/ andej(p)∈ Fforj 6= 1. This induces identifications:Γ(E)(U) =O(U)^⊕n andKer(λ)(U) = {(f1, . . . , fn) ∈ E|U | f1(p) = 0}. These are isomorphic (asO(U)-modules) under multiplication in the first term byz. This shows thatKer(λ)is locally free.

Lemma 5.3. We have:

det(E^′)∼= det(E)⊗[−p].

Proof. The exact sequences below show thatdet(E^′) ∼= det(E)⊗det(C_p)⁻¹ ∼= det(E^′)⊗ O_[−p].

0 //Γ(E^′) //Γ(E) ^λ //C_p //0

0 //O[−p] //O ^Evp //C_p //0

(5.1)

Remark5.4. Semistability is not preserved under elementary modification. If for exampleE =L⊕L^′whereLandL^′are line bundles of degree 1 andF =Lp, then E^′ = L⊕(L^′⊗[−p]), which is not semistable. This is where the notion of parabolic (semi-)stability would normally enter naturally. However, ifEis assumed to be of degree 1, then (semi-)stability ofEensures stability ofE^′. Lemma 5.5. In the above construction, ifdeg(E) = 1andEis stable, thenE^′is stable as well.

Proof. Any sub-bundleF^′ofE^′induces (via the sheaf inclusionE^′→E) a sub- bundleFofE.

IfFpis not contained inF, thenF^′is given by elementary modification ofF alongF ∩Fp. In particular,rk(F) = rk(F^′)anddeg(F) = deg(F^′) + 1.Ebeing semistable implies:

µ(F^′) =deg(F)−1

rk(F) ≤deg(E) rk(E) − 1

rk(F)< deg(E)−1

rk(E) =µ(E^′).

IfFpis contained inFthen the sheaf inclusionE^′→Erestricts to an isomor- phism betweenF^′andF. In this case (this is where we needEto be of degree 1):

deg(F^′)rk(E^′) = deg(F)rk(E)<deg(E)rk(F) = 1·rk(F) = rk(F^′).

This can only be the case ifdeg(F^′)rk(E^′)≤0 = deg(E^′)rk(F^′).

No documento Moduli Spaces of Vector Bundles (páginas 76-93)