Geometry of Hilbert schemes
Professor Kai Behrend
Notes by: David Steinberg, Pooya Ronagh
Department of mathematics, University of British Columbia
Room 121 - 1984 Mathematics Road, BC, Canada V6T 1Z2
E-mail address: pooya@math.ubc.ca
Abstract. The goal of this course is to define/construct the Hilbert scheme. The applications are to Donaldson-Thomas invariants.
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18.
Representable functors
Projective space
Closed immersions/subpresheaves
OX -modules
Hilbert schemes
Flatness
Hilbert scheme of n points
Sheaves/Schemes
Grassmannians
Properties of the Hilbert schemes of n points
̃
Tangent space to H
n
s
Points of Hilb (A ) as locus of n points in As
Euler characteristics
Computing the Euler characteristic of Hilbn (Y (C))
Euler characteristic of schemes over C
Hilbn (A3 ) as a critical locus
Singular loci of hypersurfaces
Behrend function for general Hilbert schemes of n points
Contents
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1. REPRESENTABLE FUNCTORS
3
1. Representable functors
Here we develop an alternative point of view to schemes; that is we view schemes of functors
(functor of points). This leads to construction of moduli spaces.
Convention: All rings are commutative with unit, and morphisms of rings map unit to unit.
An R-algebra is a ring A with a ring map R → A. Consequently, note that a Z-algebra is
just any ring (for any ring R there is a unique morphism of rings Z → R).
ϕ
ϕ∗
Let X be a functor X ∶ R − alg → Set via A ↦ X(A) and on morphism A → B ↦ X(A) →
X(B) we call these functors presheaves.
Example 1. View the affine scheme cut out by y 3 = x3 − x over Z by a functor
X ∶ Z − alg → Set
A ↦ X(A) = {(x, y) ∈ A × A ∶ y 3 = x3 − x}.
One can check that this is actually a functor. More generally let R be a ring and f1 , ⋯, fr ∈
R[x1 , ⋯, xn ]. Then let X be the presheaf
X(A) = {(x1 , ⋯, xn ) ∈ An ∶ f1 (x1 , ⋯, xn ) = f2 (x1 , ⋯, xn ) = ⋯ = fr (x1 , ⋯, xn ) = 0}
this is a presheaf and in fact this functor is the affine scheme defined by f1 , ⋯, fn . This
functor is in fact representable by Z(f1 , ⋯, fr ) ⊂ AnR .
One interesting special case is when r = 0. Then X = AnR :
AnR (A) = {(x1 , ⋯, xn ) ∈ An }
and if r ≠ 0 then X is a subfunctor of AnR .
More exclusively for an R-algebra A we correspond the functor Spec A given by
Spec A ∶ R − alg → Set
B ↦ Spec A(B) = HomR (A, B)
This is a functor representable by A (i.e. it is hSpec A ). Recall that an isomorphism X ≅
hSpec A is given by a universal object x0 ∈ X(A).
Definition 1. An affine scheme over R is a representable presheaf X ∶ R − alg → Set i.e.
there exists an R-algebra A and x0 ∈ X(A) representing X.
Remark. The category of presheaves with morphisms being natural transformations among
them, admits fibered products. Here is a construction: given
Z
X
f
/Y
f
2. PROJECTIVE SPACE
4
define W ∶ R − alg → Set via A ↦ X(A) ×Y (A) Z(A) is the fibered product promised. This
complete the diagram to a square
W
/Z
X
f
f
/Y
which satisfies the universal mapping property in the category of presheaves.
Remark. Our most interesting case if when R = C and then X(C) is the set of C-valued
points of X.
2. Projective space
We know that
Pn ∶ R − alg → Set
is given by
n
Pn (A) = {(L, s0 , ⋯, sn ) ∶ L is an invertible A-module, s0 , ⋯, sn ∈ L such that L = ∑ Asi }/ ∼ .
i=0
and on mappings we have
A → B ↦ (L, s0 , ⋯, sn ) ↦ (L ⊗A B, s0 ⊗A 1, ⋯, sn ⊗A 1).
Definition 2. An A-module L is invertible if for any prime ideal p of A, the localization
Lp over Ap is free of rank one (i.e. there is s ∈ Lp such that Ap → Lp via a ↦ as is an
isomorphism).
 Note 2.1. If L is invertible over A and ϕ ∶ A → B is a morphism, and is a prime
ideal of B, then
(L ⊗A B)q = L ⊗A B ⊗B Bq = L ⊗A Bq = L ⊗A Ap ⊗Ap Bq = Lp ⊗Ap Bq
and Lp ⊗Ap Bq is free of rank 1 on Bq . Thus L ⊗A B is invertible over B if L is invertible
over A.
⌟
Definition 3. The isomorphism of the above data is given by (L, s0 , ⋯, sn ) ≅ (M, t0 , ⋯, tn )
if the is an isomorphism of A-modules ϕ ∶ L → M such that ϕ(si ) = ti for all i.
Example 2. If A = k is a field, an invertible module over it is just a 1-dimensional vector
space. Thus
Pn (k) = {(s0 , ⋯, sn ) ∈ k n−1 ∶ s0 , ⋯, sn ∈ k such that not all are zero }/k ∗
by a canonical isomorphism.
3. CLOSED IMMERSIONS/SUBPRESHEAVES
5
3. Closed immersions/subpresheaves
Let’s first consider the affine case: Let A be an R-algebra and I ⊂ A is an ideal of it.
The projection A → A/I induces a morphism of affine schemes Spec A/I → Spec A and
consequently a morphism HomR (A/I, .) → HomR (A, .) by composition. We can regard
HomR (A/I, .) as a subfunctor of HomR (A, .): any morphism A/I → B is a morphism
A → B that vanishes on I, hence for all B, Spec A/I(B) is a subset of Spec A(B) so it is a
subpresheaf of Spec A.
Now let f ∶ Y → X be a morphism of presheaves/R.
Definition 4. f is a closed immersion if for every R-algebra A and every x ∈ X(A) there
is an ideal I ⊂ A such that
/ Spec A
Spec A/I
x
Y
f
/X
is a fibered product.
Example 3. Let f = zy 2 − x3 + z 2 x ∈ R[x, y, z] be the homogeneous polynomial given.
Define Z(f ) ⊂ P2 by (L, x, y, z) ∈ P2 (A) such that zy 2 − x3 + z 2 x ∈ L⊗3 in zero. Check that
Z(f ) ⊂ P2 is a closed immersion.
 Note 3.1.
Recall that GLm ∶ R − alg → Set via A ↦ Gln (A) is a representable
functor and the special case of m = 1 is denoted by Gm . Besides, the additive group
object Ga is the functor A ↦ A+ . Recall also the notion of action of a group presheaf G on
presheaf X. As an example we have the action of Gm on An+1 − {0} where
An+1 − {0} ∶ A ↦ {(a0 , ⋯, an ) ∈ An+1 ∶ ⟨a0 , ⋯, an ⟩ = 1}
⌟
More generally let I ⊆ A be an ideal, the corresponding closed subscheme is Spec A/I →
Spec A when
Spec A/I(B) = {A → B ∶ IB = 0}
and its open complement is given by
U (B) = {A → B ∶ IB = B}.
Example 4. Gm (A) acts of An+1 − {0}(A) by
λ.(a0 , ⋯, an ) ↦ (λa0 , ⋯, λan )
also we have a morphism of presheaves An+1 − {0} → Pn given by
(a0 , ⋯, an ) ↦ (A, a0 , ⋯, an )
3. CLOSED IMMERSIONS/SUBPRESHEAVES
6
Note that the automorphism group of A and A-module is the same as A∗ i.e. the group of
units of A. Thus for all A, we have an injection
An+1 − {0}(A)/Gm (A) ↪ Pn (A)
which is not always surjective.
In fact, let Y be any affine variety over an algebraically closed field k. Let A = O(Y ) be
the affine coordinate ring which is a finitely generate k-algebra but an infinite dimensional
k-module. Let K be the quotient field of Y . Then all localizations are contained canonically
in K, A ↪ K. If K is an invertible A-module then L → L ⊗A K is an injection, and L ⊗A K
is a 1-dimensional K vector space hence isomorphic to L. So think of L as an A-submodule
of K.
If L = ⟨m1 , ⋯, mn ⟩, mi ∈ K, multiplying by common denominator we get L as a sub-module
of A. So every finitely generated invertible A-module is isomorphic to an ideal of A, which
is locally principal. We conclude that
Pn (A) = {(L, a0 , ⋯, an ) ∶ L ⊂ K as an A-submodule, locally principal }/A∗ .
▸ Exercise 1.
(1) Y = Spec (C[x, y]/(y 2 − x3 + x)) is an affine elliptic curve. Then
(x, y) ⊂ A is a locally principle ideal but not principle.
(2) Let A = C[x, y] and Y = A2 then (x, y) is not locally principle.
By uniqueness of extension Hom(Af , B) → Hom(A, B) is injective. The image is all ϕ ∶ A →
B such that ϕ(f ) ∈ B ∗ . Think of Spec Af ↪ Spec A as a subfunctor: Say F ∶ Sch/k → Set
via X ↦ Hom(X, Spec A) then G ∶ Sch/k → Set via X ↦ Hom(X, Spec Af ) is a subfunctor
of it.
Spec A/(f )  q#
/ Spec A o
O
? _ Spec Af
ϕ#
Spec k
for any field k and any ϕ ∶ A → k. Then ϕ(f ) = 0 if ϕ# factors through Spec A/(f ) and
ϕ(f ) is invertible if ϕ# factors through Spec Af . This justifies to a certain extent calling
Spec Af the open complement of Spec A/(f ). In fact Spec Af ↪ Spec A is an open immersion
and it is affine.
Not all open immersions are affine:
3. CLOSED IMMERSIONS/SUBPRESHEAVES
7
Example 5. An = {0} → An is not affine. Because the subfunctor
An − {0} 
? _ ∗ = Spec R
/ An o
O
q#
ϕ#
Spec k
Then for all ϕ ∶ R[x1 , ⋯, xn ] → k, if exists ϕ(xi ) ≠ 0 then factors through An − {0} and if
ϕ(xi ) = 0 for all i then factors through 0. In case R = k, algebraically closed field, if Y is
an affine variety, then the morphism of varieties Y → An factors through An − {0} if and
only if ∀ρ ∈ Y , ai (ρ) ≠ 0 for any i. ⇒ ⟨a1 , ⋯, an ⟩ is contained in no maximal ideal. ⇒
⟨a1 , ⋯, an ⟩ = A motivating the definition of An − {0}.
Definition 5. f ∶ X → Spec A is an open immersion iff
(1) X(B) → Spec A(B) is injective for all B (i.e. f is a mono)
(2) There exists an ideal I ⊂ A such that ϕ ∈ Hom(A, B) is in X(B) iff ϕ(I)B = B.
Example 6. An − {0} ↪ An = Spec [x1 , ⋯, xn ] is an open immersion and I = (x1 , ⋯, xn ).
Example 7. Spec Af → Spec A is an open immersion I = (f ).
Definition 6. f ∶ X → Y is an open immersion iff
(1) X(B) → Y (B) is injective for all B (i.e. f is a mono)
(2) ∀A, y ∈ Y (A) the fibered product U → Spec A is an open immersion from
/ Spec A
U
X
f
/Y
Remark. The definitions (1) and (2) agree if Y is affine: Given ϕ ∶ B → A, U ↪ Spec B
is the complement of Spec B/f ↪ Spec B then V = U ×Spec B Spec A is the complement of
Spec A/IA ↪ Spec A.
Remark. If A is an R-algebra, f1 , ⋯, fn ∈ A such that (f1 , ⋯, fn ) = A then Spec Af ⊂ Spec A
ϕ#
form an open cover in the sense that every Spec k → Spec A for any field k factors through
at least one Spec Afi ⊂ Spec A; i.e. ϕ ∶ A → k can’t map each fi to zero, otherwise ϕ(A) = 0
but we know that ϕ(1) = 1.
Remark.
(1) The previous definition is actually correct only for noetherian schemes.
The correct definition is as follows: Let L be an A-module, is invertible if ∃f1 , ⋯, fn ∈
A such that (f1 , ⋯, fn ) = 1 (or elementary open covering {Spec Afi ⊂ Spec A}i=1,⋯,n )
such that ∀i = 1, ⋯, n Lfi free rank 1 over Afi .
?
4. OX -MODULES
8
(2) If i ∶ Z ↪ X is a closed immersion, define the subfunctor U ⊂ X by U (A) = {x ∈
X(A) ∶ I(x) A = A} (this comes from definition of closed immersion). Then U ↪ X
is an open immersion and ideal I ⊂ A for which "IB = B iff I n B = B" defines
open the open complement. It is not clear that every open immersion has a closed
complement.
Lemma 1 (Baby descent theory). ⟨f1 , ⋯, fn ⟩ = A and M be an A-module. Then
α,β
γ
M ↪ ∏ M fi ⇉ ∏ M fi fj
i
i,j
where Mab = (Ma )b , and α(m)i,j is the image of mi ∈ Mfi fj and β(m)i,j is the image of mj
in Mfi fj . Then
(1) γ is injective
(2) If α(x) = β(x) then x ∈ im(γ).
that is the sequence is exact.
Note that we have the commutative diagram
/ Ma
M
Mb
/ Mab
Theorem 3.1. An → Pn via
An (A) → Pn (A)
(a1 , ⋯, an ) ↦ (A, 1, a1 , ⋯, an )
is an open immersion and is the complement of the closed immersion Z(x0 ) ↪ Pn given by
Z(x0 )(A) = {(L, `0 , ⋯, `n ) ∈ Pn (A) ∶ `0 = 0}.
Proof. TO DO
4. OX -modules
Definition 7. Let X be a presheaf. A presheaf M together with
π
(1) M → X is a morphism.
(2) . ∶ A1 × M → M such that
.
/M
HH
HH
π
H
π○proj HHH #
A1 × M
H
X
4. OX -MODULES
9
(3) + ∶ M ×X M → M such that
+
/M
II
II
II
II π
I$ M ×X M
I
X
(4) + ∶ M (A) ×X(A) M (A) → M (A) such that
+
/ M (A)
PPP
PPP
PPP
π
PPP
(
M (A) ×X(A) M (A)
X(A)
get induced maps A × Mx → Mx and Mx × Mx → Mx making Mx an A-module.
?
The induced map comes from Hom(Spec A, A1 ) ≅ Hom(k[x], A) ≅ A × Mx → Mx
satisfying
(I) For any A and x ∈ X(A) the set
Mx = {m ∈ M (A) ∶ m ↦ x ∈ X(A) under π}
we have
×
/ M (A)
NNN
NNN
NN
π
π○proj NNN'
A1 (A) × M (A)
X(A)
(II) For every ϕ ∶ B → A and equivalently
ϕ#
/ Spec B
JJ
JJ
JJ
y
ϕ∗ y JJJ
J% Spec A
X
get map My → Mϕ∗ y which is B-linear.
Such an object is called an OX -module.
Definition 8. In (II) get induced morphism of A-modules My ⊗B A → Mϕ∗ y . If this is an
isomorphism for all ϕ and y, then M is called quasi-coherent.
Example 8. In the case M = A1 × X, and M → X is the projection on second factor, then
A1 × M → M is the mapping
A×A×X →A×X
(a, b, x) ↦ (ab, x)
4. OX -MODULES
10
and the product is the mapping
(A1 × X) ×X (A1 × X) → A1 × X
((a, x), (b, x)) ↦ (a + b, x)
⊕n
Then Mx = A1 × {x}, more generally OX
is given by
M = An × X → X.
(*** A1 action on An etc.)
⊕n
Remark. For OX
= An × X the structure map is
proj
An × X → X
An ×O X
proj
/X
O
x
Spec A[x1 , ⋯, xn ]
a
/ Spec A
(*** this diagram appears without explanation!)
Remark. The interesting case: affine X and coherent OX -module
Let A be an R-algebra and M is an A-module, then there is an A-algebra S and module
map ψ ∶ M → S such that for any A-algebra B
HomA−alg (S, B) → HomA−mod (M, B)
ϕ↦ϕ○ψ
is bijective. In fact we shall take S to be the symmetric product
n
M = ⊕n≥0 M ⊗n /(x ⊗ y − y ⊗ x).
SymA M = ⊕n≥0 SA
This is because for any α ∶ A → B a map of A-modules, there is a unique extension of α to
̃
α
SymA (M ) → B given by
̃(m1 ⋯mn ) = α(m1 )⋯α(mn ).
α
Example 9. M = An , HomA−alg (A[x1 , ⋯, xn ], B) = B n = HomA−mod (An , B) therefore
SymA (An ) ≅ A[x1 , ⋯, xn ].
Given A, M as above we have a natural morphism Spec SymM → Spec A. Let M =
Spec SymM and X = Spec A. We now show that M is an OX -module. We need a morphism
A1 × Spec SymM → Spec SymM , i.e. a morphism
π
SymM → SymM ⊗ k[x]
which is given naturally. Let y ∶ Spec B → Spec A be an element of X(B). Then
My = {m ∶ Spec B → M ∶ π ○ m = y}.
4. OX -MODULES
11
So if y ∶ ϕ# ∶ Spec B → Spec A then
Mϕ# = HomA−alg (SymA M, B).
And this turns M → Spec A into an OSpec A -module. We also desire that M → X is quasicoherent. So we need for any mapping A → B that
HomA−mod (M, A) ⊗A B → HomA−mod (M, B)
(4.1)
be an isomorphism. BUT this is not true for an arbitrary A-module M.
Definition 9. AN A-module N is projective if for any epimorphism π ∶ N → M of Amodules there exists a section s ∶ M → N .
Example 10. Free modules are projective!
Example 11. Invertible modules are projective.
Proof. TO DO
If M is a projective finitely generated module then the mapping 4.1 is an isomorphism.
Step 1. First we prove this when M is free of finite rank.
A⊕n ⊗A B ≅ HomA (A⊕n , A) ⊗A B → HomA (A⊕n , B) ≅ B ⊕n .
Step 2. Say M is finitely generated and projective, then we have a surjection A⊕n → M → 0
and there is a section of it s ∶ M → A⊕n . But we can also complete the sequence with its
kernel K by
0 → K → A⊕n → M → 0
and the fact that this short exact sequence is split, implies that M ⊕ K ≅ A⊕n . We know
from previous step that HomA (A⊕n , A) ⊗A B ≅ HomA (A⊕ , B) therefore
HomA (M ⊕ K, A) ⊗A B ≅ HomA (M ⊕ K, B)
But both sides split
(HomA (M, A) ⊗ B) ⊕ (HomA (K, A) ⊗ B) ≅ HomA (M, B) ⊕ HomA (K, B)
the result follows because the isomorphism is induces by direct sum of maps as in ??.
Definition 10. The OX -module M → X is a vector bundle if M → X is affine and
∀A, x ∈ X(A) there is a projective A-module P of finite rank making the following diagram
cartesian:
/X
MO
O
x
Spec SymP
/ Spec A
5. HILBERT SCHEMES
12
If M → X is a vector bundle, then for any A, x ∈ X(A),
Mx = HomA−alg (SymP, A)
, i.e. sections Spec A → Spec SymA
HomA−mod (P, A) =∶ P
∨
Example 12 (Eventual example). An+1 − {0} → Pn is a vector bundle.
Proof. TO DO
5. Hilbert schemes
cl.im.
op.im.
Suppose X is a quasi-projective scheme, i.e. X ↪ Z ↪ Pn .
Hilb(X)(A) = { closed subschemes Z of X × Spec A, flat over Spec A}
in particular if k if a field, then
Hilb(X)(k) = { closed subschemes Z ⊂ X}.
We can complete to a diagram
Z

/ X × Spec A
/ Spec A
cl. imm.
/ X × Hilb(X)
Z
Z
/ Hilb(X)
using Z such that both squares are cartesian.
Z(A) = {(s, Z) ∶ Z ⊂ X × Spec (A) is a closed immersion and
s ∶ Spec A → Z a morphism such that the following is commutative.}
Zo
Spec A
r8
rrr
r
r
r
rrr
s
X × Spec A
We can think of a close subscheme Z ⊂ X × Spec A as a family of subschemes of X
parametrized by Spec A:
Zt
Z
/X
/ Spec k
/ X × Spec A
/ Spec A
cl. imm
6. FLATNESS
13
6. Flatness
Flatness is the rigorous mathematical expression for the notion of a “continuously varying
family”.
Definition 11. A is a ring and M is an A-module. M is said to be flat if the functor
M ⊗A − is exact.
Note that tensor product is right-exact, hence we only need to check left-exactness. In
particular, if I ⊂ A is an ideal, we have an exact sequence
0 → I → A → A/I → 0
and if M is flat then
0 → I ⊗A M → M → M /IM → 0
is exact. Hence I ⊗A M → IM is an isomorphism. Conversely if this is true for all I ⊂ A
then M is flat.
Example 13. Free modules are flat. Projective modules are also flat (since they are direct
sums of free modules). Every localization S −1A is flat.
Proposition 1. M is flat over A iff Mm if flat over Am for any maximal ideal m ⊂ A.
Proof. TO DO
Lemma 2. Let M be a flat A-module. Then every linear relation ∑ni=1 ai xi = 0 in M
(ai ∈ A, xi ∈ M ) can be lifted to some free module of finite rank. i.e. there exists As , ϕ, bij
ϕ
such that As → M via ei ↦ yi , xi = ∑j bij yj for all i and ∑ ai bij = 0.
Proof. Matsumura 2, Ch. 2, Section 3 [ey too rooooohet joghd!]
Corollary 1. If A is an integral domain, then M is torsion-free if it is flat.
Proof. If M has torsion element x, i.e. ax = 0 for some a ≠ 0, then by the lifting
property, there is b1 , ⋯, br ∈ A and y1 , ⋯, yr ∈ M such that ∑ bj yj = x and abj = 0 for all j.
But if a ≠ 0 then bj = 0 for all j, thus x = 0.
From this we learn that:
Corollary 2. If A is a local ring and M is a finitely generated A-module then M is flat
iff M is free. In some sense flatness is a globalization of being free!
For the prove we recall
Lemma 3 (Nakayama’s lemma). Let I be an ideal in R, and M a finitely-generated module
over R. If IM = M , then there exists an r ∈ R with r ≅ 1( mod I), such that rM = 0.
6. FLATNESS
14
Proof of Corollary. (Matsumura (2), Ch. 2, Section 3) We only need to show that
flatness implies being free. In fact it suffices to show that for a minimal set of generators
x1 , ⋯, xn of M , they form a basis as well. First we need to show that x1 , ⋯, xn ∈ M are
such that their images x1 , ⋯, xn ∈ M /m are linearly independent over k then they are
linearly independent over A. For this use the previous lemma together with induction on n.
Secondly, we should show that if x1 , ⋯, xn generate M /m then their lifts x1 , ⋯, xn generate
M . If r ∈ M is not generated by xi ’s then rM is an A module such that m(rM ) = 0 and
thus by Nakayama’s lemma rM = 0, contradicting it’s nontriviality.
Definition 12. An A-algebra is flat if it is flat as an A-module.
Lemma 4 (Support of M is closed!). Let A be a ring, M a finitely generate A-module. Let
p be a prime ideal, such that Mp = 0. Then there exists s ∈/ p such that Ms = 0.
 Note 6.1. We are not going to use noetherian condition!
⌟
Proof. Let m1 , ⋯, mn generate M . Then m11 , ⋯, m1n ∈ Mp are all zero! Thus there are
s1 , ⋯, sn ∈/ p such that si mi = 0 in M . Let s ∶= ∏ si , then smi = 0 for all i. Also note that
s ∈/ p. Hence sm = 0 for all m ∈ M and Ms = 0.
Proposition 2. Let A be a ring and M and A-module. Let n ∈ Z be an integer. Then the
following are equivalent:
(1) M is Zariski-locally free of rank n (i.e. there are f1 , ⋯, fr ∈ A such that ⟨f1 , ⋯, fr ⟩ =
1 with Mfi free of rank n over Afi for all i).
(2) M is finitely generated and at every maximal ideal Mm is free of rank n.
(3) M is finitely generated, flat and for all maximal ideal m, M /mM is n-dimensional
vector space over A/m.
Remark. The tensor product of noetherian rings is not necessarily noetherian. That is
why we insist on not using the noetherian conditions. As an example not that R ⊗Q R is
not noetherian.
If we restrict to noetherian rings, then the fiber product of affine schemes may not be affine.
This is inconvenient: it forces you to restrict to certains kinds of fiber products: at best we
have under Hilbert’s basis theorem that if A, B are noetherian and B is a finitely-generated
A-algebra then B ⊗A C is noetherian.
As a result restriction to noetherians, forces restriction to affine morphisms of finite type
(i.e. morphisms X → Y such that for all cartesian diagrams
Spec B
/ Spec A
/Y
X
7. HILBERT SCHEME OF n POINTS
Proof. TO DO
15
Remark. The scheme Spec Am can be thought of as the germ of Spec A at m and Spec A/m
as the point at m. The above is a passage between information at a point and information
of an infinitesimal neighborhood. We can conclude that the vector bundles over Spec A are
in one-to-one correspondence with the finitely generated flat modules such that all fibers
are vector spaces of some dimension.
7. Hilbert scheme of n points
Hilbn (X)(A) = { closed subscheme Z ⊂ X×Spec A with the following properties holding for it }
Properties: That Z → Spec A is flat and finite. And for every maximal ideal m of A with t
corresponding to the A/m-valued point of Spec A and Zt for the fiber product,
Zt
/ Spec A/m
t
/ Spec A
Z
then Zt = Spec B for an A/m-algebra B and dimA/m B = n as a vector space.
Definition 13. An affine morphism X → Y is flat/finite if for any cartesian diagram
Spec B
/ Spec A
/Y
X
B is flat/finite A-algebra. Note that being a finite A-algebra means being finitely generated
as an A-module.
Example 14. X = A1 and A = k a field. Then any Z = Spec k[t]/I is flat over k obviously,
and is finite since k[t] is a PID and hence I = (f ) for some polynomial f , thus k[t]/(f ) is
finite dimensional.
Hilbn (A1 )(k) = { closed subscheme Z ⊂ A1k eq. I ⊂ k[t] such that Z = Spec (k[t]/I) has our properties}
It remains to note that we want the fibers (here being only Spec k[t]/I = Spec k[t]/(f ) ) be
n-dimensional k-vector space, so we demand f to be a degree n polynomial. So we showed
that
Hilbn (A1 )(k) ≅ k n
in a canonical way.
We will prove now the following
8. SHEAVES/SCHEMES
Proposition 3.
16
Hilbn (A1 ) = An
.
Proof. It suffices to show that Hilbn (A1 )(A) = An (A) in a natural way! We are
looking for ideals I ⊂ A[t] such that B = A[t]/I is flat and finitely generated as A-module
and for every maximal ideal m ⊂ A, we have that B/mB = B ⊗A A/m is a vector space of
dimension n over A/m. By proposition 2 this corresponds precisely to ideals I ⊂ A[t] such
that B = A[t]/I is Zariski locally free of rank n.
Claim. If I is Zariski locally free of rank n over A, then I = (f ) for a unique monic f ∈ A[t].
Consider 1, t, ⋯, tn−1 ∈ B; these give a basis of B/B over A/A for any maximal m ⊂ A as
B/B ≅ A[t]/mA[t] ≅ A/m[t].
 Note 7.1. If k is an algebraically closed field then for such f we have the splitting
f = ∏ni=1 (t−αi ) hence any Z ⊂ A1k corresponds to the finite set of points α1 , ⋯, αn ∈ A1k .
But we always a functor in the other direction:
An →Hilbn (A1 )
n
(α1 , ⋯, αn ) ↦ ∏(t − αi )
i=1
So we can think of Hilb (A ) as the quotient of An by Sn as the permutation of αi ’s is
irrelevant.
⌟
n
1
Remark. Hilbn (P1 ) = Pn .
8. Sheaves/Schemes
Recall that a presheaf is just a functor X ∶ R − alg → Set.
Definition 14.
(1) Let {Ui → X}i∈I be a family of open immersions of presheaves.
The family is a (Zariski) open covering if for any A and for all x ∈ X(A) when we
complete to cartesian diagrams
Vi
/ Spec A
/X
Ui
the fibered products Vi satisfy {Vi → Spec A} is a Zariski open covering of Spec A
in the following sense!
?
8. SHEAVES/SCHEMES
17
(2) {Vi → Spec A} is an open covering if for any field k and all R-algebra morphism
ϕ
A → k there exists i such that ϕ# factors through Vi making the following diagram
commute
Spec (k)
i
Vi
|x
x
x
x
x
ϕ#
/ Spec A
Definition 15. The presheaf X is a sheaf if for all R-algebra A, and for all open coverings
{Ui → Spec A}i∈I , the following sequence is exact.
0 → X(A) → ∏ X(Ui ) ⇉ ∏ X(Uij )
i,j
here we have more generally than ever defined X(U ) = Morpresh (U, X). The last maps
are easy to describe on the image objects, the first one X(A) → ∏ X(Ui ) is given vie
x ∶ X → Spec A ↦ x∣Ui
; XO
xx
xx
x
x
xx
xx
/ Spec A
Ui
x∣Ui
Remark.
(1) These are Zariski sheaves; we can use different notions of coverings and
get different notions of sheaves.
(2) Let X be a Zariski sheaf. Then we get an induced sheaf X∣Spec A on the topological
space Spec A, for every A.
X∣Spec (A) (U ) = Morpresh (U, X)
where U ⊂ Spec A is Zariski open subset.
(3) X is a “big sheaf” and X∣Spec A is a “small sheaf”.
Lemma 5. For X to be a sheaf it suffices to check that for any A and a1 , ⋯, an ∈ A such
that ⟨a1 , ⋯, an ⟩ = A then following is exact
n
0 → X(A) → ∏ X(Spec Aai ) ⇉ ∏ X(Spec Aai ∩ Spec Aaj )
i=1
i,j
Proof. This follows from sheaf theory on any space and Spec Af ⊂ Spec A form a basis
of topology.
Lemma 6. Let B be an R-algebra, then Spec B is a sheaf.
Proof. From the previous lemma we need only to take some R-algebra A, where
⟨a1 , ⋯, an ⟩ = A and show exactness of
Hom(B, A) → ∏ Hom(B, Aai ) ⇉ Hom(B, Aai aj )
8. SHEAVES/SCHEMES
18
ϕ,ψ
For the first arrow suppose ϕ, ψ ∶ B → A and the compositions ϕi , ψi ∶ B Ð→ A ↪ Aai are
identical for all i. Thus ψi (b) − ϕi (b) = 0 ∈ Aai for all b ∈ B and for any i. So ψ(b) − ϕ(b) ↦ 0
in ∏ Aai . But A → ∏i Aai is injective hence ϕ = ψ. In fact in the same paradigm, the
exactness follows from the exactness of the following diagram
0 → A → ∏ Aai ⇉ ∏ Aai aj .
i
i,j
Definition 16. A presheaf X is a scheme if
(1) X is a sheaf, and
(2) There exists an open covering {Ui → X} of X such that each Ui is an affine scheme.
Remark. Let (S, OS ) be a scheme in the usual sense. This induces a functor X ∶ R − alg →
Set via A ↦ Morsch (Spec A, S). We claim that X satisfies the new definition of scheme.
Example 15. Pn is a scheme. For any R-algebra, A, such that ⟨a1 , ⋯, ak ⟩ = A,
0 → Pn (A) → ∏ Pn (Aai ) ⇉ ∏ Pn (Aai aj )
should be shown to be exact. Given (Li , `0i , ⋯, `ni ) a collection of descent data for i = 1, ⋯, n,
such that we have isomorphisms
ϕij ∶ (Li )ai aj → (Lj )aj ai
sending `ki ↦ `kj for all ki ’s, we need to construct an invertible A-module L with sections
`0 , ⋯, `n an isomorphisms ϕ ∶ Lai → Li mapping `j ↦ `ji . Thus we have
∏ Li ⇉ ∏(Li )ai aj
i
(xi )i ⇉
i,j
(xi )i,j , ϕ−1
ij ((xj )j,i )
Define
L = {x = (xi )i ∈ ∏ Li ∶ α(x) = β(x)}
This has the structure of an A-module because of A-linearity of α and β. Now we want to
check that L is invertible: suffices to prove
Lai ≅ Li
since invertibility is a local property. This follows from the isomorphism ϕi ∶ Lai → Li
which we will define promptly. Let’s first define the sections of L. Let `k = (`ki )i which is a
well-defined element of L because of the definition of ϕij . We also demand that (`0 , ⋯, `n )
generate L. Construction:
ϕi
Lai _ _ _/ Li
=
O
L
||
||
|
|
|| proj
?
8. SHEAVES/SCHEMES
19
Since Li is an Aai -module get a unique map: L ⊗A Aai → Li via ` ⊗ a ↦ a.proj. Verify
locally that ϕi is an isomorphism. Without loss of generality now suppose Li is free of rank
1. The remaining claims are local in Spec A and can be checked in a Zariski neighborhood
of any maximal ideal m of A. Thus without loss of generality let Li = A, then
α−β
L = ker(∏ Aai Ð→ ∏ Aai aj ) = A✓
So Pn is a sheaf!
ϕi
Next we show Pn is locally affine. Let Ui ∶ An Ð→ Pn via (a1 , ⋯, an ) ↦ (A, a1 , ⋯, 1, ⋯, an )
defines an open immersion and {Ui → Pn }i=0,⋯,n is an affine Zariski open cover. It suffices
to show the following: Given a field k and element ⟨a0 , ⋯, an ⟩ ∈ Pn (k) ≅ k n+1 − {0}/k ∗ , we
can find ai non-zero such that
a0
an
⟨a0 , ⋯, an ⟩ = ⟨ , ⋯, 1, ⋯, ⟩.
ai
ai
So Spec (k) factors through Ui and this completes the proof.
Definition 17. A scheme X is quasi-projective if there exists an open and closed subscheme
U, Z ⊂ Pn such that X = U ∩ Z.
Proposition 4. If X is quasi-projective then Hilbn (X) is a (quasi-projective) scheme.
Proof. First we show that Hilbn (X) is a sheaf. We know that Hilbn (X)(A) is the
set of closed immersions Z ↪ X × Spec A such that Z → Spec A is finite, affine, flat and the
fibers are all of length n. (Note: We say an affine scheme Spec B over field k has length n if
dimk B = n.) We want to consider algebras A = ⟨a1 , ⋯, an ⟩. Suppose that we have the local
data that
/ X × Spec Aai
Zi K
KKK
KKK
KKK
K%
Spec Aai
ϕij
such that for any i, j we have an isomorphism Zi ∣Spec Aai aj → Zj ∣Spec Aaj aj and want to
construct a global affine Z = Spec B.
/ X × Spec A
II
II
II
II
$ Z II
Spec A
The idea is to show that Z = Spec B where B is constructed from the local data, as a locally
free A-module. To construct it locally we use sheaf properties, then we check that it is a
closed immersion locally!
?
9. GRASSMANNIANS
20
Define Z ↪ P1 × Pn as the zero-locus of t0 xn + t1 xn−1 y + ⋯ + tn y n , where ti ’s are coordinates
on Pn and x, y are coordinates on P1 . This polynomial is homogeneous of degree 1 in t and
of degree n in x and y. The claim is that Z ↪ P1 × Pn → Pn is affine, finite, flat and the
fibers are of length n. So Z defines a morphism Pn → Hilbn (P1 ). We now want to check
that this is an isomorphism. We only need to check this locally as both are sheaves. This
completes the proof that
Proposition 5. Hilbn (P1 ) = Pn .
The last thing to prove in this section is that
Proposition 6. Hilb` (An ) is a quasi-projective scheme.
9. Grassmannians
The presheaf G(m, n) is defined by
G(m, n)(A) = { short exact sequences of A-modules 0 → U → An → Q → 0
such that U and Q are finite, flat of degree m and n − m respectively }/ ≅
The equivalence relation on the short exact sequences is existence of the following commutative diagrams:
/Q
/0
/ An
/U
0
/ U′
0
/ An
/ Q′
/0
Note that a quotient Q as above (finite flat), determines the finite, flat, degree n − m kernel
U in other words:
G(m, n)(A) = { submodules of An such that the quotient is finite flat degree n − m }
= { s.e.s 0 → U → OSnpec A → Q → 0 where U, Q
are locally free coherent of ranks m and n − m }
Proposition 7. G(m, n) is a projective scheme.
Proof.
ρ
n
G(m, n) → P(n−m)−1
n−m
n−m
A
A
(An ↠ Q) ↦ ( ⋀ An ↠ ⋀ Q)
The canonical basis of ⋀
P
n
(n−m
)−1
n−m
n
A gives
n
(n−m
)
. Note that ⋀n−m An is again free.
elements `i ∈ ⋀n−m
Q so (⋀n−m
Q, `0 , ⋯, `(
A
A
n
)−1
n−m
)∈
10. PROPERTIES OF THE HILBERT SCHEMES OF N POINTS
21
Claim: ρ is a closed immersion. For this it suffices to show that ρ0 is a closed immersion
n
for all standard open affines of P(m)−1 .
Am(n−m)
G(m, n)
ρ0
/
n
A(m)−1
ρ
i0
n
/ (m
P )−1
Tracing on an R-algebra A we have
Hom(Am , An−m )
G(m, n)(A)
ρ0
ρ
/
n
A(m)−1
n
)−1
/ P (m
(A)
The left downwards arrow is ϕ ∈ Mn−m×m (A) ↦ (In−m×n−m ∣ϕ) ∶ An ↠ An−m . The upper
right arrow is
ϕ ↦ all determinants of n − m × n − m minors of (I∣ϕ) .
It is easy to check that the diagram is cartesian. It remains to prove that ρ0 is a closed
immersion. This is equivalent to having that the corresponding maps of rings is a surjection.
This follows of existence of π such that π ○ ρ0 = id. The determinant of minors of (I∣ϕ)
gives ϕ, and this is the mapping π.
Now: Hilb` (An ) is a locally closed subscheme of some G(M, N ). For the general case of
the proof that Hilbp (X) is a quasi-projective scheme look at [Bertram]. Here we give a
proof of this specific case along the same ideas.
TO DO
10. Properties of the Hilbert schemes of n points
Up to here we have shown that if X is a quasi-projective scheme over R then X [n] ∶=
Hilbn X is a quasi-projective scheme by showing that it is a locally closed subscheme of a
Grassmannian. We have seen that if k is a field, then X [1] = X, Hilbn (A1 ) = An . Now we
will show that Hilbn (A2 ) is a smooth variety if R is a field k.
̃ n / Gln where H
̃ n is a scheme and Gln acts freely on
We would like to write Hilbn (Ar ) = H
̃ n is smooth.
it. We will then show that in the case when r = 2, H
Proposition 8.
Hilbn (A2 ) = {(V, B1 , B2 , v0 ) ∶ V is an n-dimensional vector space, Bi ∈ End(V )
and [B1 , B2 ] = 0 and v0 ∈ V such that there is no W ⊊ V containing v0 and B1 , B2 are stable}/ ≅
10. PROPERTIES OF THE HILBERT SCHEMES OF N POINTS
22
The rest of this section is the proof of this theorem. More precisely we will show
(10.1)
Hilbn (A2 )(A) = {(V, B1 , B2 , v0 ) ∶ V is a finite flat A-module of rank n, Bi ∈ EndA (V )
(10.2)
and [B1 , B2 ] = 0 and v0 ∈ V such that A[x1 , x2 ] → V via xi ↦ Bi (v0 )}/ ≅
By definition Hilbn (A2 )(A) is the set of commutative diagrams
ϕ
A[x1 , x2 ]
B dIo o I
O
II
II
II
II
I
A
where B is a finite flat A-algebra of rank n. Associat to this the following data:
V = B,
Bi = multiplication by ϕ(xi ),
v0 = ϕ(1) = 1B
Note that under this, [B1 , B2 ] = 0 and under ϕ ∶ A[x1 , x2 ] → B we have xi ↦ ϕ(xi ).1 =
B(v0 ). Moreover,
f (x1 , x2 ) ↦ f (B1 , B2 )(v0 )
and is surjective. Thus the properties in 10.1 holds here.
Conversely given data as in 10.1,
I = {f (x1 , x2 ) ∈ A[x1 , x2 ] ∶ f (B1 , B2 ) = 0 ∈ End(V )}.
To show that 0 → I → A[x1 , x2 ] → V → 0 is exact where the last mapping is given by
f (x1 , x2 ) ↦ f (B1 , B2 )(v0 ), beside trivial things, we have to show that f (B1 , B2 )(v0 ) = 0
implies f (B1 , B2 ) = 0. Let v ∈ V be arbitrary vector, given by v = g(B1 , B2 )(v0 ), then
f (B1 , B2 )v = f (B1 , B2 ).g(B1 , B2 ).v0 = 0.
This completes the proof that V ≅ A[x1 , x2 ]/I as A-modules. Now we want to endow V
with a ring structure to get that ϕ ∶ A[x1 , x2 ] → V is an A-algebra morphism. We define
the multiplication by
(f (B1 , B2 )v0 ).(g(B1 , B2 )v0 ) = f (B1 , B2 )g(B1 , B2 )v0 .
Now define
̃ n (A) = {(B1 , B2 , v0 ) ∈ Mn×n (A)×Mn×n (A)×An , such that [B1 , B2 ] = 0, A[x1 , x2 ] ↠ An }
H
2 +n
which is a subfunctor of Mn×n × Mn×n × An ≅ A2n
.
2
The condition [B1 , B2 ] = 0 defines a closed subscheme of A2n +n . The surjectivity of
A[x1 , x2 ] ↠ An translates to the fact that if we associate to A[x1 , x2 ] the corresponding
quasi-coherent sheaf on the affine space, and to An the coherent sheaf. The cokernel which
is hence coherent is supported away from our desired points! The support of a coherent
̃ n is an open subscheme of a closed
sheaf is closed hence this is an open condition. So H
2
subscheme of A2n +n .
10. PROPERTIES OF THE HILBERT SCHEMES OF N POINTS
23
π
̃n →
Define the morphism of schemes H
Hilbn (An ) via
(B1 , B2 , v0 ) ↦ (An , B1 , B2 , v0 ).
Next step is to involve the action of Gln . Define an action
̃ n sigma
̃n
Gln ×H
→ H
̃ n (A) → H
̃ n (A)
Gln (A) × H
g.(B1 , B2 , v0 ) ↦ (gB1 g −1, gB2 g −1, gv0 )
Lemma 7. We have a cartesian diagram:
̃n
Gln ×H
proj
σ
̃n
H
/ ̃n
H
π
π
/ Hilbn (A2 )
Proof. The proof of this is easy; remember that one step of it is to show that given
̃ n (A) there is an isomorphism g ∶ An → An in g ∈ Gln (A) sych
(B1 , B2 , v0 ), (B1′ , B2′ , v0′ ) ∈ H
′
that g.(B1 , B2 , v0 ) = (B1 , B2′ , v0′ ).
Remark. This g is unique; we know g on B1 v0 and B2 v0 and this determines g on all of
An .
̃ n → Hilbn (A2 ) is sheaf surjective.
Lemma 8. π ∶ H
Proof. This means that for all R-algebra A and all mappings Spec A → Hilbn (A2 )
there is an open covering of Spec A, ⟨f1 , ⋯, fn ⟩ = 1 such that for all i ∈ {1, ⋯, s} there is a
dotted map making the following diagram commutes:
̃n
i i4 H
i i
i
i i
i i
i
i
/ Hilbn (A2 )
/ Spec A
Spec (Afi ) Take a covering over which V becomes free. Then over Spec Afi choose isomorphism Anfi ≅
Vfi , via this isomorphism B1 , B2 map to matrices in Mn×n (Afi ) which we take as B1′
and B2′ . And v0 maps to an element of Afi which we take as v0′ . This defines a lift
̃ n.
Spec (Afi ) → H
̃ n → Hilbn (A2 ) is a principle bundle with structure
Remark. The above lemmas say that H
group GLn ; i.e. a GLn -bundle. This definition makes sense for X is a scheme and Y a
π
GL-scheme and Y Ð→ X a GLn -invariant maps. For other groups the Zariski topology on
X may not be fine enough, and we must use the etale topology instead. Zariski topology
however works for GLn , SLn and Sp2n .
?
10. PROPERTIES OF THE HILBERT SCHEMES OF N POINTS
24
We conclude that
Corollary 3. There exists a Zariski open covering {Ui } of Hilbn (A2 ) such that the following is cartesian:
Gln ×Ui
open
/ ̃n
H
proj
Ui
π
/ Hilbn (A2 )
open
Proof. For this note that Hilbn (A2 ) is a scheme. So it has an affine covering. Refine
this covering to get another one by the previous lemma which has lifts
̃
9H
ss
si sss
sss
s ss
/ Hilbn (A2 )
n
Ui
We get cartesian squares:
GLn ×Ui
/ GL ×H
̃n
n
/ ̃n
H
Ui
si
/ ̃n
H
π
σ
/ Hilbn (A2 )
Since π ○ si is the inclusion it is an open immersion, implying that the top composition is
also an open immersion.
The trivial principal GL-bundle over X is GLn ×X → X with GLn action by left multiplication. Every GLn -bundle is locally trivial by definition.
Remark. All the above works with Ar not just A2 , get principal GLn -bundle
̃ n → Hilbn (Ar )
H
where
̃ n = {(B1 , ⋯, Bn , v0 ) ∈ M r × An ∶ ∀i, j[Bi , Bj ] = 0, An generated by Bi (v0 )s}.
H
n×n
However the following depends heavily on r = 2:
Theorem 10.1. Let R = k an algebraic closed field, char(k) = 0 and consider Hilbn (A2 ).
̃ n is a smooth subvariety of M 2 × An .
Then H
n×n
k
Here on, we assume R = k is algebraically closed.
Definition 18. A small extension is a pair I ≤ A1 where A1 is a finite dimensional local
k-algebra and I is an ideal of A1 with dimension one over k.
10. PROPERTIES OF THE HILBERT SCHEMES OF N POINTS
25
Let A = A1 /I. Then from the exact sequence 0 → I → A1 → A → 0 we have that A1 is an
extension of A by I. Let m be the maximal ideal of A and m1 be the maximal ideal of A1 .
Then m1 is a flat A1 module and hence m1 ⊗A1 I ≅ Im1 = 0 hence I 2 = 0. However I is an
A1 -module and since I.I = 0 we get a well-defined action of A on I. Moreover this action
factors to give an action of A/m on I. So I ≅ A/m as a vector space over A/m.
Example 16. Small curvilinear extension: A1 = k[t]/(t`+1 ), A = k[t]/t` and I = ⟨t` ⟩.
Definition 19. A k-presheaf is smooth if for every small extension I → A′ → A, X(A′ ) →
X(A) is surjective.
If p ∈ X(k) is a point, X is smooth at p if considering X(A) → X(A/m) = X(k) for any
x ∈ X(A) that maps to p ∈ X(k) under X(A) → X(A/m) then there is x′ ∈ X(A′ ) such
that x′ ↦ x.
Spec A′
O
?
G
G x′
G
G
G#
/X
w;
w
w
ww
wwp
w
ww
Spec A
O
?
x
Spec A/m
Example 17. Take A′ = k[t]/(t3 ) and A = k[t]/(t2 ) and I = ⟨t2 ⟩. Take X = Spec k[x, y]/(y 2 −
x3 ). THen X(A′ ) → X(A) is not surjective; take (t, t) ∈ X(k[t]/(t2 )). Then (t, t) is living
above the point p ∶ Spec Am → X given by the origin, and hence X is not smooth at the
origin.
Suppose chark ≠ 2.
̃ is smooth.
Proposition 9. H
Lemma 9. Consider the trace form for a small extension I → A′ → A given by
Mn×n (A′ ) × Mn×n (I) → I
(ξ, X) ↦ tr(ξX).
If W ⊂ Mn×n (I) is an A-submodule, then W → W ⊥ ⊥ is sujrective.
Proof. The trace form is a non-degenrate symmetric bilinear pairing and we may think
2
of W ⊂ Mn×n (k) ≅ k n in lieu of the identification I ≅ k.
̃ n (A).
Choose (B1 , B2 , v0 ) ∈ H
Lemma 10. Let W = {[B1 , X2 ] − [B2 , X1 ] ∈ Mn×n (I) ∶ X1 , X2 ∈ Mn×n (I)}. Then
W ⊥ = {ξ ∈ Mn×n (A′ ) ∶ ∃f ∈ A[z1 , z2 ], ξ = f (B1 , B2 ) mod m′ }.
?
̃
11. TANGENT SPACE TO H
26
TO DO
̃1 , B
̃2 be arbitrary lefts of B1 , B2 to Mn×n (A′ ). Then [B
̃1 , B
̃2 ] ∈ Mn×n (I)
Lemma 11. Let B
⊥ ⊥
and is in (W ) .
TO DO
̃1 , B
̃2 , ̃
̃1 , B
̃2 ] ∈ W . So there are X1 , X2 ∈
Let (B
v0 ) be any lift of (B1 , B2 , v0 ) to A′ . Then [B
Mn×n (I) such that
̃1 , B
̃2 ] = [B1 , X2 ] − [B2 , X1 ]
[B
̃1 − X1 , B
̃2 − X2 ] = 0 (since [X1 , X2 ] ∈ Mn×n (I 2 ) and I 2 = 0). On the other
Thus [B
hand,
/ A′n
A′ [z1 , z2 ]
A[z1 , z2 ]
/ / An
The lower arrow is surjective and by Nakayama’s lemma for the A′ -module A′n we have
that the above arrow is also surjective (lift generators). This completes the proof of the
fact that
̃1 − X1 , B
̃2 − X2 , ̃
̃ ′)
(B
v0 ) ∈ H(A
̃ n is smooth.
and that consquently H
̃
11. Tangent space to H
̃
̃ at p is given by
Let p = (B1 , B2 , v0 ) ∈ H(k).
The tangent space of H
̃ = preimage of p under H(k[ε])
̃
̃
Tp H
→ H(k).
This preimage consists of all (B1 + εX1 , B2 + εX2 , v0 + εx0 ) such that X1 , X2 ∈ Mn×n (I) and
x0 ∈ k n satisfying [B1 + εX1 , B2 + εX2 ] = 0, i.e.
̃ = {(X1 , X2 , x0 ) ∶ [B1 , X2 ] = [B2 , X1 ]}.
Tp H
For A′ = k[ε], A = k, I = ⟨ε⟩, we replace the two-form tr ∶ Mn×n (k[ε]) × Mn×n (ε) → ⟨ε⟩
by
tr(., .) ∶ Mn×n (k) × Mn×n (k) → k
Then we have W ⊂ Mn×n (k) with the notation as above, k[z1 , z2 ] → W ⊥ via f (z1 , z2 ) ↦
f (B1 , B2 ) is surjective with kernel
K = {f ∈ k[z1 , z2 ] ∶ f (B1 , B2 )v0 = 0}.
but under the mapping f (z1 , z2 ) ↦ f (B1 , B2 )v0 we have the short exact sequence
0 → K → k[z1 , z2 ] → k n → 0
hence W ⊥ ≅ k n , so dim(W ⊥ ) = n and hence dim(W ) = n2 − n.
13. EULER CHARACTERISTICS
27
On the other hand W fits into the short exact sequence
̃ → Mn×n (k)2 × k n → W → 0
0 → Tp H
where the projection is given by (X1 , X2 , X0 ) ↦ [B1 , X2 ] − [B2 , X1 ]. Hence
̃ = 2n2 + n − n2 + n = n2 + 2n
dim Tp H
̃ = n2 + 2n and dim(H/
̃ GLn ) = 2n and finally
and thus dim H
Corollary 4. Hilbn (A2 ) is a smooth scheme of dimension 2n.
12. Points of Hilbn (As ) as locus of n points in As
Definition 20. The point [B1 , ⋯, Bs , v0 ] ∈ Hilbn As is semi-simple if all Bi ’s are diagonalizable. Since they commute, they are simultaneously diagonalizable, so we have that such
a point is equivalent to [D1 , ⋯, Ds , v0′ ].
Note that all entries of
v0′
⎛λ1(i)
0 ⎞
⎟,
⋱
are non-zero from the spanning criterion. Let Di = ⎜
⎜
⎟
(i)
⎝ 0
λn ⎠
and define xj = (λj , ⋯, λj ) ∈ As . Then x1 , ⋯, xn ∈ As are n distinct points (again by
spanning criterion). The converse procedure from n points x1 , ⋯, x0 ∈ As to getting points
on Hilbert scheme, is obvious. So
(1)
(n)
Hilbnss (As ) = (As )n0 .
For simplicity work with s = 2. We wil work on an algebraically closed field k. If
[B1 , B2 , v0 ] ∈ Hilbn (A2 )(k) we get data (Vi , λi , µi )i∈{1,⋯,r} such that V = ⊕ri=1 Vi into generalized eigenspaces. Assume that dim Vi ≥ dim Vi+1 so that we get a partition of n given
by (α1 , ⋯, αr ). This partition corepsonds to [B1 , B2 , v] ∈ Hilbn (A2 ). The subscheme of A2
correpsonding to [B1 , B2 , v] is the ideal
I = {f ∈ k[z1 , z2 ] ∶ f (B1 , B2 ) = 0 ∈ End(k n )}
The polynomials ∏ri=1 (z1 − λi )N and ∏ri=1 (z1 − µi )N are in the ideal and so is ∏I (z1 −
λi )N ∏J (z2 −µj )N for any index sets I ∐ J = {1, ⋯, r}. Thus (a, b) ∈ Z(I) iff (a, b) = (λi , µi )
for some i ∈ {1, ⋯, r}.
13. Euler characteristics
Let X be a scheme over C. Then X(C) is a topological space. We want to compute
χ(X(C)).
Definition 21. The analytic topology:
?
?
13. EULER CHARACTERISTICS
28
(1) X(C) has the Zariski topology: U ⊂ X(C) is open iff there is an open subscheme
X ′ ⊂ X such that U = X ′ (C) ↪ X(C).
(2) X = ∪i Xi open subschemes, Xi ↪ Ani closed subscheme X(C) = ∪i Xi (C). And
Xi (C) ↪ Ani (C) = Cni
the analytic topology on Cni induces a topology on Xi (C). Put the finest topology
on X(C) such that Xi (C) ↪ X(C) are all open subsets.
One would like to prove that the topology is independent of choices: the main point is that
if Y ↪ An is a closed subscheme and likewise is a closed subscheme Y ↪ Am , then Y (C)
gets the same topology from Cn and Cm .
Y
w GGG
GGg
ww
w
w
(f,g) GG
GG
ww
w
w{
#
n o
/ Am
A
An+m
f
We may extend Y → Am to G ∶ An → Am . Use continuity of G in the analytic topology.
Let Y /C be a smooth variety, p ∈ Y (C), then there is an (analytic) open neighborhood such
that p ∈ U ⊂ Y (C), and U → Cm is an analytic isomorphism onto open subset. Algebraically
the best we can do is to take Y smooth and p ∈ Y then there is a Zariski open neighborhood
p ∈ U ⊂ Y and U → Ar is etale. Consider the example of elliptic curves for instance.
Definition 22. If X, Y are smooth then f ∶ X → Y is etale if f is bijective on Zariski
tangent spaces.
/
Spec k
u: X
∃! u
u
u
u
Spec (k[ε])
f
/Y
If X, Y are of finite type schemes over the algebraically closed field k, we will have then
that for all small extensions A′ → A the following diagram completes:
Spec A 
∃! y y
y
y
′
Spec A
Remark.
/X
y<
/Y
(1) If we only get the existence of the diagonal arrow, then f is smooth.
(2) If we only have the uniqueness of the diagonal arrow, then f is called unramified.
(3) If we put Y = Spec k the we get back the notion of smoothness (of X).
14. COMPUTING THE EULER CHARACTERISTIC OF Hilbn (Y (C))
29
Let Y be a smooth variety over C and let U ⊂ Y (C) be an open. We get open subsets
of Hilbn Y (C) by considering all [Z] ∈ Hilbn Y (C) such that Z ⊂ U ⊂ Y . Denote this by
Hilbn (U ) ⊂ Hilbn (Y ).
Remark. The map Hilbn (Y (C)) → Y (C)n /Sn is continuous. For Y = A2 we have that
Hilbn A2 (C) → (C2 )n /Sn
via [B1 , B2 , v0 ] ↦ ∑ri=1 αi (λi , µi ).
Then if U → As is an isomorphism onto an open ste we get an induce map
Hilbn (U ) → Hilbn (As )
from an open subset.1 It is however not true that if Ui is a cover of Y then Hilbn (Ui ) is a
cover of Hilbn (Y ).
Let Z ⊂ Y be a subscheme of length n. Suppose Z = Z1 ∪ ⋯ ∪ Zr were these components are
disjoint, and suppose there exists Ui ⊃ Zi with Ui ↪ As admitting holomorphic coordinates
and for simplicity assume Ui ∩ Uj = ∅. Let length Zi be αi . The partition of n being
α = (α1 , ⋯, αr ), then the morphism
r
ϕ
α
n
∏ Hilb i Ui → Hilb Y
i=1
via (Z1′ , ⋯, Zr′ ) ↦ ∪i Zi′ is an open neighborhood of [Z]. And via coordinates
r
r
r=1
i=1
α
s
α
∏ Hilb i (Ui ) ↪ ∏ Hilb i (A )
is an open embedding. This is how we resolve the problem of finding an open cover of the
Hilbert schemes of n points.
14. Computing the Euler characteristic of Hilbn (Y (C))
We do this when Y is a smooth algebraic variety over C.
If X is a scheme over C then
χ(X(C)) = ∑(−1)i dimQ Hci (X(C), Q)).
i≥0
Recall that we know the
Theorem 14.1. If U ⊂ X is an open subscheme, Z ⊂ X is a closed subscheme such that
the complete of Z is U then
χ(X(C)) = χ(U (C)) + χ(Z(C)).
Proof. Use long exact sequence of cohomology (possibly with compact support).
1Authors: Hilbn (U ) is the pre-image of U n /S along the Hilbert-Chow map
n
14. COMPUTING THE EULER CHARACTERISTIC OF Hilbn (Y (C))
30
Definition 23. A stratification of a scheme X is
(1) A finite index set I, partially ordered
(2) ∀i ∈ I a locally closed subscheme Zi ⊂ X such that X = ∐i Zi and ∀i ∈ I
⋃ Zj ↪ X is closed and Zi ↪ ⋃ Zj is open.
j≥i
j≥i
Given a stratification (Zi ) of a scheme X, then
χ(X(C)) = ∑ χ(Zi (C)).
i∈I
Remark. Any scheme of finite type over C admits a stratification (Zi )i∈I where each Zi is
smooth: Consider Xred ⊂ X which is a closed subscheme. Then there is an open subset of
Xred which is smooth. Let it be Z0 and Z0 ↪ X is locally closed.
For Hilbn (Y ) we stratify by partition: Let α ⊢ n be a partition of n, α = (α1 , ⋯, αn ) where
αi ≥ αi+1 and αr > 0. Let [Z] ∈ Hilbn (Y )(k) where k is any field. Then we know that
Z = Spec (B) for a finite k-algebra B.
Proposition 10. Every finite dimensional k-algebra B is (up to order) uniquely isomorphic
to a product B ≅ B1 × ⋯ × Br where Bi is a local finite dimensional k-algebra.
Proof sketch. TO DO
So every k-valued point Spec k → Hilbn Y has an associated partition given by (dimk (B1 ), ⋯, dimk (Bn )).
Let Hilbnα Y be the set of points with their corresponding partition being α. We put a partial order on the set of all paritions of n by defining α ≥ β if and only if α is a refinment of
β. The question is whether (Hilbnα Y )α⊢n is a stratification of Hilbnα Y .
It suffices to prove the assertion locally; i.e to show that for any ring A, and any closed
subscheme Z ↪ YA the diagram
/ YA
Z EE
EE
EE
EE
E" Spec A
gives a stratification of Spec A.
Recall that we had defined Hilbnα (As ) is another particular way too. If [B1 , ⋯, Bs , v] ∈
Hilbnα (As )(k) then V decomposes to generalized simultaneous eigenspaces ⊕ri=1 Vi and α =
(dim V1 , ⋯, dim Vr ).
Lemma 12. If I ⊂ k[z1 , ⋯, zr ] is the deal corresponding to [B1 , ⋯, Bs , v0 ] (meaning I = {f ∶
f (B) = 0}) then k[z]/I ≅ ⊕ri=1 Vi as k-algebras, and each Vi is local.
This lemma shows that the two partitions are in fact the same!
14. COMPUTING THE EULER CHARACTERISTIC OF Hilbn (Y (C))
31
Lemma 13. If Z → Spec A is a finite, flat, rank n morphism, there is an affine covering
Spec Afi of Spec A where ⟨f1 , ⋯, fr ⟩ = 1 such that over Spec Afi if Zi fits into the cartesian
diagram
/ Spec Afi
Zi
/ Spec A
Z
then there exists an embedding

Zi F
/ An
Afi
FF
FF
FF
FF
" .
Spec Afi
Proof. TO DO
This last lemma also implies that the verification of out claimed stratification reduces to
the proof that it is so for Y = An . We proceed with the proof of the latter.
TO DO
We restate this result in the following
Theorem 14.2. (Hilbnα (Y ))α⊢n is a stratification of Hilbn Y . Moreover each Hilbnα Y has
a natural scheme structure induced by the flattening stratification.
Corollary 5. χ(Hilbn (Y )(C)) = ∑α⊢n χ(Hilbnα (Y )(C)).
Now let α ⊢ n be any partition. We don’t in general have a morphism
r
α
n
∏ Hilb i Y → Hilb Y
i=1
vai (Z1 , ⋯, Zr ) ↦ Z1 ∪⋯∪Zr . The problem is that it is not in general the case that Z1 ∪⋯∪Zr
is a subscheme of Y . This however is the case when Zi ’s are disjoint! So we pass to

r
αi
Y
V LL / ∏i=1 Hilb
LL
LL
LL
LL
L% Hilbn Y
Let V ⊂ ∏ri=1 Hilbαi Y be the set of all Z1 , ⋯, Zr such that i ≠ j implies Zi ∩ Zj = ∅.
Lemma 14. V is an open subscheme.
Proof. TO DO
15. EULER CHARACTERISTIC OF SCHEMES OVER C
32
So for Spec A → V given by (Z1 , ⋯, Zr ) the scheme Z1 ∐ ⋯ ∐ Zr defines a map Spec A →
Hilbn Y . Let Zα denote the fiber product in the following cartesian diagram
Zα
/V
p
xp p
/ ∏ Hilbαi Y
p
pp
/ Hilbn Y.
Hilbnα Y
Here Zα → Hilbnα Y is a Galois cover with Galois group Aut(α). Then
1
χ(Hilbn Y ) = ∑ χ(Hilbnα Y ) = ∑
χ(Zα ).
#Aut(α)
α⊢n
α⊢n
To compute χ(Zα ) we observe that it fits into another cartesian diagram:
Zα
/ ∏i Hilbαi Y
(αi )
/ Yr
Y0r
Hilbm
(m) Y → Y is a Zariskli locally trivial fibration with fibers Fn being the punctual Hilbert
scheme which is a subscheme of Hilbn (As ) consisting of all subschemes of As supported at
the origin, this corresponds to {(B1 , ⋯, Bs , v0 )} such that all Bi ’s are nilpotent, s = dim Y .
Hence χ(Zα ) = χ(Y0r ) ∏ χ(Fαi ).
15. Euler characteristic of schemes over C
Theorem 15.1. If X is of finite type over C then χ(X(C)) = χc (X(C)).
Remark. For any locally compact space X, z ⊂ X closd, and U = X − Z, then χc (X) =
χc (U )+χc (Z). This is not true for χ without the compact support restriction. For example
X = R, and Z = {0}, then χ(R) = 0, χ(Z) = 1, χ(R − {0}) = 2, while χc (R) = −1, χc (Z) = 1,
χc (R − {0}) = −2.
Proof. TO DO
Proposition 11. If Gm over C act on the finite type C-scheme X without fixed points then
χ(X(C)) = 0.
Proof. TO DO
Remark. The finite stabilizers were used in the proof to pretend Z = X/Gm is a “space” so
that χ(Z) is fintie. The proof works for any connected algebraic group G if all stabilizers
are finite. This will imply that there are no fixed points and hence the quotient stack [X/G]
is a Deligne-Mumford stack. Then [X/G](C) is an orbifold and so χ([X/G](C)) is finite.
We may then conclude that χ(X(C)) = 0 due to the fact that χ(G) = 0.
15. EULER CHARACTERISTIC OF SCHEMES OVER C
33
Remark. Given a complex manifold U ↪ Y being an analytic open immersion and U ↪ As
again analytically, we get and induced diagram
open
/ Hilbn Y
NNN
NNN
N
open NNNN
'
Hilbn (U )
Hilbn (As )
We have to rely on the exitence of the analytic category which X(C) in an object of for
all finite-type C-schemes X. We use this to prove that for an open subscheme U ⊂ Y ,
Hilbn (U ) is an analytic subspace of Hilbn (Y ) and is hence an “analytic space”.
TO DO
The above argumetn shows that if U ⊂ Y is open, there is an analytic open of Hilbn (Y )
which represents Hilbn (U ). And thus:
Corollary 6. All analytic Hilbert schemes we need are representable.
Let’s calculate χ(Hilbn (As )) now. The tool is the natural T = (Gm )s action on As via
weight (1, ⋯, 1):
(λ1 , ⋯, λs )(a1 , ⋯, as ) ↦ (λ1 a1 , ⋯, λs as ).
The T also acts on Hilbn (As )
T (A) × Hilbn As (A) → Hilbn (As )(A) = {(V, B1 , ⋯, Bs , v0 ) ∶ with conditions! }/ ≅
(λ1 , ⋯, λs )(V, B1 , ⋯, Bs , v0 ) ↦ (V, λ1 B1 , ⋯, λs Bs , v0 ).
Recall that an action of Gm on Spec A if an only if A is graded by Z. So the exitence of an
action of (Gm )s on As is equivalent to the existence of the s-fold grading on C[x0 , ⋯, xs ]
where the i-th degree of a polynomial is the degree in xi , (1 ≤ i ≤ s). An element is
homogeneous if and only if each variable occurs with the same power in each monomial, iff
it is a monomial. Each graded piece is a 1-dimensional module and had a canonical basis
(the monic).
The fixed points of T (C) = (C∗ )s on Hilbn (As )(C) = {I ≤ C[x1 , ⋯, xs ] ∶ corankI = n} are
given by :
I is fixed ⇔ I is homogeneous
⇔ I is generated by homogeneous elements
⇔ I is generated by monomials
in the latter case we say I is a monomial ideal. These are in one-to-one correspondence
with the s − D partitions of n (when s = 2 these would be the usual partitions of n, when
s = 3 they are the 3D partitions of it, etc.). Thus the number of T (C)-fixed points of
(n)
Hilbn (As )(C) is equal to Ps , denoting the number of s-partitions of n. We conclude
that
16. Hilbn (A3 ) AS A CRITICAL LOCUS
34
Corollary 7. χ(Hilbn (As (C))) = χ(fixedpointset) + χ(complement) = Pn .
(s)
There is also a generating function that these numbers fit in:
1
⎧
⎪
⎪
1−t
⎪
⎪
∞
F (s) (g) = ∑ χ(Hilbn (As (C)))tn = ∑ Pn(s) tn = ⎨ ∏k=1
⎪
⎪
n=0
n=0
⎪
∏∞
⎪
⎩ k=1
∞
∞
1
1−tk
1
(1−tk )k
s=1
s=2
s=3
16. Hilbn (A3 ) as a critical locus
Proposition 12. M is a smooth scheme.
̃/ Gln where
Proof. We know that M = M
̃ = {(B1 , B2 , B3 , v0 ) ∶ stability condition } ⊂ M 3 × An
M
n×n
is an open subscheme in some basis chosen and has dimension 3n2 + n, hence in particular
smooth.
̃
Gln ×M
act
/ ̃
M
pr
̃
M
π
π
/M
We know that this is cartesian, that π is sheaf surjective, and M is a sheaf. From these
facts it follows that M is a smooth scheme
̃ → M is a principal Gln -bundle. Hilbn (A3 ) ↪ M is a closed subscheme given by
And M
[Bi , Bj ] = 0 and is in general very singular.
f
The goal is now to show that there exists a regular function f ∈ O(M ) (i.e. a map M → A1 )
such that Hilbn (A3 ) = Z(df ). Moreover Hilbn (A3 ) ⊆ Z(f ) = f −1(0). So Hilbn (A3 ) is the
singular locus of the hypersurface Z(f ) ⊂ M .
f ((V, B1 , B2 , B3 , v0 )) ∶= tr(B1 [B2 , B3 ]).
where V is a finite flat module over a C-algebra A.
17. SINGULAR LOCI OF HYPERSURFACES
̃:
M
35
̃ ⊆ M 3 × An . Consider a C[ε]-valued point of
16.1. Computing df . Work on M
n×n
b + εx = (B1 + εX1 , B2 + εX2 , B3 + εX3 , v0 + εx0 )
Then we have,
f (b + εx) = tr((B1 + εX1 )[B2 + εX2 , B3 + εX3 ])
= tr(B1 [B2 , B3 ]) + ε {tr(X1 [B1 , B3 ]) + tr(B1 [X2 , B3 ]) + tr(B1 [B2 , X3 ])}
3
Thus dfb ∶ Mn×n
× Cn → C is the C-linear mapping
(x1 , x2 , x3 , v0 ) ↦ tr(X1 [B1 , B3 ]) + tr(B1 [X2 , B3 ]) + tr(B1 [B2 , X3 ]).
The partial derivatives with respect to the standard coordinates are obtained by setting
⎧
(Eij , 0, 0, 0)
⎪
⎪
⎪
⎪
⎪ (0, Eij , 0, 0)
(X1 , X2 , X3 , x0 ) = ⎨
(0, 0, Eij , 0)
⎪
⎪
⎪
⎪
⎪
(0,
0, 0, ei )
⎩
We know that
tr(Eij [B1 , B3 ]) = (j, i) entry of [B1 , B3 ]
tr(B1 [Eij , B3 ]) = − (j, i) entry of [B1 , B3 ]
tr(B1 , [B2 , Eij ]) = (j, i) entry of [B1 , B2 ]
So the Jacobian is given by
df = ([B1 , B3 ]t , [B3 , B1 ]t , [B1 , B2 ]t , 0).
̃ have f̃ ∶ M
̃ → A1 with Z(df ) = H.
̃ Then follows that for f ∶ M → A1 we have
On M
Z(df ) = Hilbn (A3 ) = M.
Remark. Z(df ) ⊂ Z(f ) by choice of f . So Hilbn (A3 ) is the singular locus of the hyperplane
f −1(0) ⊂ M . This is NOT true in higher dimensions!
17. Singular loci of hypersurfaces
Let f ∶ Cm → C be a holomorphic function. V = f −1(0) and let H = Z(df ). We assume
∂f
m
and
f ∈ ⟨ ∂x
i ⟩. Thus H will be the singular locus of V . Assume p is the origin of C
let
νH (p) = (−1)m (1 − χ(Fp )).
If p is an isolated singularity of V (isolated point of H), then Fp is homotopy equivalent to
a bonquet of (m − 1)-spheres, and
νH (p) = (−1)m (1 − χ( bonquet)) = (−1)m (1 − (1 − (−1)m−1 .# spheres )) = #spheres
18. BEHREND FUNCTION FOR GENERAL HILBERT SCHEMES OF n POINTS
36
and this is known as the Milnor number of the singularity.
Remark. The Milnor number is equal to dimC O(H) so νH (p) is a natural generalization
of Milnor number to non-isolated singularities.
We now define the weighted Euler characteristic χB of H with weight νH to be
χ(H, νH ) = ∑ νH ∣Hα χ(Hα )
Hα
where {Hα } is an stratification of H and ν is constant on each stratum. This is also the
Donaldson-Thomas invariant of H.
2 +n
For p ∈ M n choose an analytic coordinate chart around p ∈ U ⊂ M n and U ↪ C2n
.
Theorem 17.1. Milnor fiber is invariant under biholmorphic maps:
p ∈ CnE
/ Cn ∋ q
y
yy
yy
y
y f
|yy
biholo.
EE
EE
E
g EE
E"
C
(g)
Then Fp
≅ Fq
(f )
is a homeomorphism.
So by this fact every point p ∈ M has a well-defined Milnor fiber Fp , hence we get ν ∶ M → Z
2
via p ↦ (−1)2n +n (1−χ(Fp )) = (−1)n (1−χ(Fp )) as a Z-valued function on M which vanished
outside of Hilbn (A3 ).
18. Behrend function for general Hilbert schemes of n points
The goal is to define and compute ν ∶ Hilbn Y → Z for any smooth C-scheme Y of dimension
3. Let p = [Z] ∈ Hilbn Y (C) for some subscheme Z ↪ Y . Choose holomorphic coordinate
charts for Y , {Ui }. If Z = Z1 ∐ ⋯ ∐ Zr where length of Zi is αi and α ⊢ n is a partition
of n. Assume Zj ⊂ Uj for all j and for simplicity suppose that U1 , ⋯, Ur are pairwise
disjoint.
p = ([Z1 ], ⋯, [Zr ]) ∈ ∏ri=1 Hilbαi Ui open / Hilbn Y ∋ [Z]
∏ri=1 Hilbαi A3
proji
/ Hilbαi A3
ν
/ A1
This shows that νHilbn Y (p) = ∏ri=1 νHilbαi A3 ([Zi ]).
Is this well-defined? For this we need independence of choice of charts and independence
of partition Z1 ∐ ⋯ ∐ Zr .
TO DO
18. BEHREND FUNCTION FOR GENERAL HILBERT SCHEMES OF n POINTS
37
Remark. νHilbn Y is constructible, i.e. there is a stratification of Hilbn Y such that ν is
constant on each stratum.
This follows from the corresponding fact for Milnor fibers! Now χ(Hilbn Y, ν) can be defined
analogous to the above and is well-defined.
Theorem 18.1. χ(Hilbn Y, ν) = (−1)n χ(Hilbn Y ).
So for any smooth scheme Y of dimension 3 over C:
1
)
∑ χ(Hilb Y, ν)t = ∑ χ(Hilb Y )(−t) = ( ∏
k k
n=0
n=0
k=1 (1 − (−t) )
∞
n
n
∞
n
n
∞
χ(Y )
.
χ(Hilbn Y, ν) = ∑ χ(Hilbnα Y, νHilbn Y )
α⊢n
1
χ(Zα , νHilbn Y )
α⊢n #Aut(α)
1
= ∑
χ(Zα , ∏ νHilbαi A3 )
α⊢n #Aut(α)
r
1
χ(Y0r ) ∏ χ(Fαi , νHilbαi A3 )
= ∑
α⊢n #Aut(α)
i=1
= ∑
So until now we have reduced this computation to Hilbn (A3 ).
Recall that G3m acts on M n . Let T ⊂ G3m be defined by λ1 λ2 λ3 = 1 and this is isomorphic
to G2m .
f is constant on T -orbits. As a result T acts on Hilbn (A3 ) and hence as before
χ(Hilbn A3 , ν) = ∑ ν(p)
p
where p ranges on the fixed points of the action of T .
Suppose p ∈ Hilbn (A3 ) is a fixed point of the action of T . Since the T -action on A3 is
compatible with the action of it on points of Hilbn (A3 ), the fact that p is fixed implies that
the support of the corresponding closed subscheme of A3 is also fixed. But the action of T
on A3 has only the origin as fixed point. Thus the closed subscheme corresponding to p is
supported only at the origin. Suppose that the corresponding ideal is I ⊂ C[x, y, z]. Then
by general theory on graded rings, I has to be generated by eigenvectors of the action of
T . But these are all of the form g(xyz)m(x, y, z) where m is a monomial. We can assume
that g(0) ≠ 0. Thus g ∈/ I. Hence V (⟨g⟩ + I) = ∅. Hence αg + β = 1, and mαg + mβ = m. So
the left hand side is also in I, i.e. m ∈ I. Thus I is generated by its monomials. Hence is
any monomial ideal.
18. BEHREND FUNCTION FOR GENERAL HILBERT SCHEMES OF n POINTS
38
Now from the following more general example we have that χ(Fp ) = 0. Thus ν(p) =
n
(−1)dim M (1 − 0) = (−1)n ,
χ(Hilbn A3 , ν) =
n (3)
∑ ν(p) = (−1) Pn .
p fixed