1. science
2. mathematics
3. geometry

MATH260 TUTORIALS
SABIN CAUTIS
1. Week 1
1.1. Why is Algebraic Geometry Beautiful. We begin with problem number
two on the homework which asks us to find 27 lines on the cubic x3 +y 3 +z 3 +1 = 0.
This leads us to the following beautiful story of algebraic geometry.
Every cubic in P3 (you can think of this as A3 if you want) contains exactly 27
lines. In fact, since the variety of lines in Pn is simple to understand (it is just the
Grassmanian) we know that:
• Every degree 3 polynomial in P3 contains 27 lines.
• Every degree 5 polynomial in P4 contains 2875 lines.
• Every degree 2d-1 polynomial in Pd+1 contains a known finite number of
lines.
Now let’s concentrate on the degree 5 hypersurfaces in P4 and ask how many degree
k curves are contained in these hypersurfaces. We restrict our attention to these
varieties because they are Calabi-Yau which means ”very nice”. In this case the
answer is much harder to figure out:
• If k = 1 there are 2875 curves.
• If k = 2 there are 609, 250 curves (1985).
• If k = 3 there are 317, 206, 375 curves (1991).
The methods used in figuring out these numbers are generalizations of the ones
used for lines. However, they fail beyond very small values of k. In 1991 however,
physicist in the area of mirror symmetry announced answers for all k. This lead
mathematicians to try to translate their work into mathematics and lead such
people as Kontsevich to develop ideas such as stable maps. Thus we see the beautiful
connexions which algebraic geometry has to other areas.
1.2. Affine vs Quasiaffine.
• X = A2 − {0} is an example of a quasiaffine variety which is not isomorphic
to an affine variety. This is because if it were then A(X) = k[x, y] = A(A2 )
which would mean that x 7→ x and y 7→ y induces an isomorphism between
X and A2 which it clearly does not.
• The variety xy = 1 in A2 and A1 − {0} give an example where a quasiaffine
variety is in fact isomorphic to an affine one.
Such is not the case in projective space where all morphisms are closed. This is
one more reason to work in Pn and not An .
1.3. Isomorphisms vs Homeomorphisms. Consider the curve x2 = y 3 and
A1 . Then the map t 7→ (t3 , t2 ) is a homeomorphism between curves but it is
not an (algebraic) isomorphism because the two coordinate rings of the curves are
nonisomorphic.
1
2
SABIN CAUTIS
1.4. Bezout’s Thm. Bezout’s theorem in the plane states that two general curves
of degree d and e intersect in de points. This generalizes. There are two ways to
think of this result. First is to visualize a degree d curve as d lines. Then clearly d
lines meet e lines in de points. This, though naive, can probably be pulled through
for a proof.
The other way to think of this is by looking at resultants. The resultant of two
polynomials of degrees d and e is a polynomial of degree de and hence the two
polynomials have de common solutions. There is an interesting philosophical point
to be made here. That is that for specific polynomials the resultant will not be of
degree de and will not do what you’d expect. If it always did then the study of
algebraic geometry would be reduced to the study of curves which to many extents
is well understood is somewhat trivial.
One can use this idea to show for instance that the curve (t3 , t4 , t5 ) in A3 cannot
be carved out by only two polynomials.
2. Week 2
2.1. Solution to Problem 1.11 from Hartshorne. The question is to show that
the curve Y given (t3 , t4 , t5 ) satisfies:
• I(Y ) (the ideal associated with Y ) is prime and of height 2
• I(Y ) cannot be generated by two elements
To show it is prime we need to show that Y is irreducible. We use the following
trivial but useful result.
Lemma 2.1. If f : X → Y is a surjective morphism and X is irreducible then so
is Y .
In this case A1 → Y and A1 is irreducible so Y is irreducible. Also the height of
I(Y ) is at least two since it is prime and contains the prime ideal (say) (y 2 − xz).
Moreover, the height of I(Y ) plus the dimension of Y (which is at least 1) equals
the dimension of k[x, y, z] which is three. Thus indeed the height of Y must be
precisely 2.
Now suppose I(Y ) can be generated by two elements f and g. The intersection
of Y by a general hyperplane is 5 points (this is commonly known as the degree of
Y ). The intersection of a general plane with f = 0 gives a plane curve of degree
d where d is the degree of f . Similarly the intersection with g = 0 gives a curve
of degree e = deg(g). These two curves meet in general in de = 5 points. Thus
without loss of generality we may assume d = 1. But Y is not a planar curve so we
get a contradiction. (NO! - this only works in P3 and not A3 )
2.2. Solution to Problem 3. Show that the set of invertible n × n matrices is an
affine variety. To do this you simply use the same idea as to show that C − {0} is
affine. You map (xij ) 7→ (xij , 1/ det(xij ) from the set of invertible matrices to the
variety det(xij )t = 1. This is an irreducible affine variety and is isomorphic to the
set of invertible matrices via the inverse map (xij , t) 7→ (xij ).
2.3. Some Examples of Sheaves. One of the most important things to remember
about sheaves is that a surjective sheaf map f : F → G does not mean fU : F (U ) →
G(U ) is onto for every U . The thing to remember is that f is onto iff. fp : Fp → Gp
is onto for every point p. Thus to check that f is onto one needs to check that
MATH260 TUTORIALS
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for every point p and pair (φ, U ) where p ∈ U and φ ∈ G(U ) there exists a subset
V ⊂ U such that φ|V is in the image of fV : F (V ) → G(V ).
In particular, it is enough to show that for any p we can find a small neighbourhood U around p such that fU : F (U ) → G(U ) is onto. This is often the case in
practice as we show next.
Suppose X is a manifold. If we denote by I i (U ) the group of i-forms on the open
set U and by Z i (U ) the group of closed i-forms on U then we obtain the sheaves
I i and Z i .
Proposition 2.2. We have the following exact sequence of sheaves:
0 → R → C∞ = I 0 → Z 1 → 0
where R is the constant sheaf R(U ) = R. More generally we get the long exact
sequence:
0 → R → I 0 → I 1 → · · · → I n → Z n+1 → 0
This is because locally, any closed 1 form is exact. This is actually called the
Poincare lemma. In fact it is not hard to show that for instance in any star shaped
region in R2 every closed form is exact. You just integrate along straight lines
from your centre x0 to and point x and say that the value at x is the value of this
integral.
For example, consider the starshaped region given by [0, 1] × [0, 1] and the with
the origin at (0, 0). To show that the closed form ω = xdx+ydy is exact we consider
the function f (x, y) where f (a, b) is the value of the integral of ω from (0, 0) along
a straight line to (a, b). The line is (ta, tb) and so the integral is:
Z 1
Z 1
f (a, b) =
(ta)(dx/dt) + (tb)(dy/dt)dt =
ta2 + tb2 = 1/2(a2 + b2 )
0
0
2
2
Thus f (x, y) = 1/2(x + y ) and indeed df = xdx + ydy = ω.
But clearly, if our space is X then 0 → R → I 0 (X) → Z 1 (X) → 0 is in general
√ 2 2 is
not exact. For instance, if X = R2 − {(0, 0)} then the closed 1-form −ydx+xdy
x +y
not exact. This is because the integral of it around the unit circle is 1 6= 0. The
fact that this sequence is not exact tells us about the topology of the space and
in fact this is how one defines cohomology... (the example above defines the first
1
deRham cohomology group HDR
for those of you who know what that is).
So originally when I read the definition of sheaves being onto I thought lowly of
the situation. My first reaction was to think that perhaps we had chosen the wrong
definition. But in retrospect, it is the perfect definition. This tiny incovenience
for checking onto-ness allows us to develop a cohomology theory which would be
impossible to develop otherwise.
2.4. History.
• Originally people thought of finite groups as subgroups of Sn . Only later
did they realize that they could define a group abstractly and this freed
them from having to always work with Sn and got to the essence of the
group which allowed us to generalize to other groups such as infinite groups.
• Originally people thought of manifolds as a subset of Rn . Only later did
they realize they such think of manifolds as being made up of pieces of Rn .
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SABIN CAUTIS
• Originally people thought of varieties as subsets of An carved out by polynomials. Later they realized it is better to consider a variety as being made
up of pieces of subsets of An carved out by polynomials.
3. Week 3
I wanted an introduction to Grassmannians. I decided to first give the example
below to give them a feel of things in G(1, 3) and later generalize to the more general
grassmannian G(k, n) as opposed to giving them the general definition and present
this as an example.
3.1. How many lines pass through 4 general lines in P3 . We want to count
the number of lines in P3 which intersect 4 general lines. Exercise: why is this
number expected to be finite.
Here’s the plan:
• Find a good description of the space of all lines in P3
• Given a general line l find all lines which intersect l (this will be a subset
of the set of lines in P3 determined above).
• Figure out how 4 of these subsets intersect (ie. in how many points)
3.1.1. 1. A line in P3 is described by two vectors v1 , v2 in A4 . Two sets of vectors
(v1 , w1 ) and (v2 , w2 ) describe the same line iff v1 ∧ w1 = v2 ∧ w2 are linearly
independent. So we have a map
φ : {lines in P3 } → P(∧2 (A4 )) = P5
This map is an embedding. If A4 has basis e1 , e2 , e3 , e4 then a point α ∈
P(∧2 (A4 )) will have basis xij where 1 ≤ i, j ≤ 4. So, for example, the points
(1, 2, 3, 0, 0, 0) corresponds to the vector e1 ∧ e2 + 2e1 ∧ e3 + 3e1 ∧ e3 . So, the space
of lines in P3 is isomorphic to the image of φ in P(∧2 (A4 )) = P5 . What is this
image?
Suppose α ∈ P5 is a point corresponding to line α′ = α12 e1 ∧ e2 + · · · α34 e3 ∧ e4 .
P
Lemma 3.1. If V is an n dimensional vector space then an element α = vi ∧wi ∈
∧2 V can be written in the form v ∧ w iff. α ∧ α = 0.
Proof. Clearly if α = v ∧ w then α ∧ α = 0. Conversely suppose α ∧ α = 0 and that
α = v1 ∧ w1 + · · · vl ∧ wl where l is as small as possible.P
Then {v1 , w1 , . . . , vl , wl } is
a set of independent
vectors.
Otherwise
we
can
write
ai vi + bi wi = 0 and then
P 1
v
∧
(a
v
+
b
w
)
we
can
reduce
l
by
one (contradiction). All
rewriting α as
i
i
i
i
i
ai
this means that α ∧ α cannot be zero since (say) the coefficient of v1 ∧ w1 ∧ v2 ∧ w2
is nonzero. The result follows.
Consequently, α is in the image of φ iff. α′ can be written as v ∧ w which is
true iff. α′ ∧ α′ = 0. Thus the image of φ is a hypersurface S in P5 given by
x12 x34 − x13 x24 + x14 x23 = 0.
3.1.2. 2. Now lines l1 ↔ v1 ∧ w1 and l2 ↔ v2 ∧ w2 intersect iff v1 ∧ w1 ∧ v2 ∧ w2 = 0.
Then given a line l corresponding to a point (a12 , . . . , a34 ) the condition that the line
corresponding to the point (x12 , . . . , x34 ) intersects it is a34 x12 − a24 x13 + · · · = 0.
This is a plane in P5 .
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3.1.3. 3. Thus the number of lines intersection 4 lines is the number of points of
intersection of S with 4 general planes. But 4 general plane carve out a general line
so that the number of points is 2.
3.2. More general Grassmannians. More generally the Grassmannian G(k, n)
is defined to be the variety of all k-planes in Pn . The embedding corresponding to
φ above is given by the Pluker embedding:
(v1 , . . . , vk+1 ) → v1 ∧ · · · ∧ vk+1 ∈ P(k+1 )−1
n+1
Generally it is hard to write down the equations of this variety. Instead people
look at its cohomology which is extremely beautiful and has various relations to
representation theory.
4. Week 4
4.1. More on Grassmannians. Recall, we wanted a parameter space of all k
planes in Pn which we denoted G(k, n). To do this we used the Plucker embedding
n+1
< v1 , . . . , vk+1 >→ v1 ∧ · · · ∧ vk+1 ∈ P(∧k+1 An+1 ) = P(k+1)−1
Some examples:
• G(1, 3) we saw was given by the equation x12 x34 − x13 x24 + x14 x23 in P5 .
• More generally
G(1, n) is given by the condition w ∧ w = 0 which translates
to n+1
quadric
equations. Thus G(1, n) is carved out by n+1
quadric
4
4
n+1
−1
(
)
2
.
relations in P
• There are also the trivial examples G(0, n) = Pn .
4.2. What is the dimension of G(k, n)? Answer: (k + 1)(n − k). To see this
notice GLn+1 acts on G(k, n) transitively. The elements of GLn+1 which fix the
k + 1-plane < e1 , . . . , ek+1 > are of the form... and whence are of dimension
(n + 1)2 − (k + 1)(n − k). Thus
dim(G(k, n)) = dim(GLn+1 ) − ((n + 1)2 − (k + 1)(n − k)) = (k + 1)(n − k)
Note: at this point one asked the question of what does it mean to talk of the
dimension of GLn ? I told him you can think of it as the Krull dimension.
4.3. The degree.
Definition 4.1. If X ⊂ Pn variety, dim(X) = m then the degree of X is
• the number of points in X intersect a general plane of dimension n − m
• equivalently the number of points in X intersect m general hyperplanes Hi .
One way you can think about or even define dim(X) is as the integer m such
that the number of intersection points of X with m general hyperplanes is finite
and positive.
Here are some examples...
• If X ⊂ Pn is a hypersurface given by X = V (f ) then the degree of X is the
number of points of X intersect a general line which is mboxdeg(f ) (whence
the definition of degree)
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SABIN CAUTIS
• The veronese surface S given by v : P2 → P5 where (a, b, c) 7→ (a2 , b2 , c2 , 2ab, 2ac, 2bc).
Since dim(S) = 2 we want S ∩ H1 ∩ H2 . Since v is a 1-1 map we have
#{S ∩ H1 ∩ H2 }
= #{v ∗ (S) ∩ v ∗ (H1 ) ∩ v ∗ (H2 )}
= #{P2 ∩ v ∗ (H1 ) ∩ v ∗ (H2 )} = 4
The last line follows since if H = a1 x1 +· · ·+a6 x6 is a general hyperplane
then f ∗ H = a1 (a2 ) + a2 (b2 ) + · · · + a6 (2bc) which is a polynomial of degree
2 in P2 (now use Bezout).
exercise: generalise to other veronese embeddings vd : Pn → PN .
This is a more general way of calculating the degree if the embedding is
given by a morphism.
• Since G(1, 3) is a hypersurface of degree 2 we have deg(G(1, 3)) = 2. Note
that when we talk of degree we are implicitly talking of a variety as well as
a map to projective space.
4.4. Calculating the degree of G(1, 4) by specializing. G(1, 4) ⊂ P9 and
dim(G(1, 4)) = 6. The degree is not so easy to calculate. If you recall, in the
case of G(1, 3) a hyperplane corresponded to all the lines through a given line in
P3 . Similarly here, by a similar calculation, the a hyperplane corresponds to all
lines through a given 2-plane in P4 . Then the degree of G(1, 4) is the number of
lines which intersect 6 general 2-planes in P4 . Notice how before we used the algebra to solve a geometry problem but now we use the geometry to solve an algebra
problem. However, we will have to cheat.
First, here is a way of solving our old problem: finding how many lines pass
through 4 lines in P3 . Suppose we specialize two of the lines so that they lie on a
plane. Then a line passing through all four either lies in the plane in which case
it must go through the two points of intersection of the other two lines with the
plane (1 line) or it passes through the intersection of these two lines in the plane (1
line). In total we get 2 lines. In this case we could have specialized even further by
assuming that the first two lines lie in a plane and that the last two lines also lie
in a plane. Then we also get the answer of 2. But if we specialize too much then
we get nonsense...
In our case we specialize so that our 6 2-planes satisfy:
• H1 ∩ H2 = l1 (a line) and so span(H1 , H2 ) = Γ1 (a 3-plane)
• H3 ∩ H4 = l2 (a line) and so span(H3 , H4 ) = Γ2 (a 3-plane)
• H5 ∩ H6 = l3 (a line) and so span(H5 , H6 ) = Γ3 (a 3-plane)
Then we have 4 cases for a line l passing through all 6 2-planes:
• l passes through l1 , l2 , l3 . In this case there is 1 line since there is a unique
line passing through 2 lines and a point in P3
• l passes through l1 , l2 and lies in Γ3 . There is also a unique line here since
l1 ∩ Γ3 and l2 ∩ Γ3 are points (so it total we get 3 lines here)
• l passes through l1 and lies in Γ2 , Γ3 . There are no lines here since l1 ∩Γ2 ∩Γ3
is empty in general
• l lies in Γ1 , Γ2 , Γ3 . There is a unique line here since Γ1 ∩ Γ2 ∩ Γ3 is a line.
Thus in total we get 1 + 3 + 0 + 1 = 5 lines and so deg(G(1, 4)) = 5.
4.5. The degree of G(k, n). So we have:
• deg(G(1, 2)) = 1
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• deg(G(1, 3)) = 2
• deg(G(1, 4)) = 5
• deg(G(1, 5)) =?
Guess? In general we get deg(G(1, n)) = n1 2n−2
n−1 which is a Catalan number.
Recall that one way to view Catalan numbers is as the number of rectangular
tableaux of shape 2 × (n − 1). This pattern continues and in general deg(G(k, n))
is the number of rectangular tableaux of size (k + 1) × (n − k). Incidentally there
is the hook formula which allows one to write a precise answer.
Note: someone asked if this follows from cohomology theory. I told them that it
does and that this is the first, most natural way of stating what cohomology tells
you (ie. once you compute the cohomology theory there is no more combinatorics
necessary).
5. Week 5
5.1. From last week. I want to finish example from last week. Recall we were
trying to find the degree of G(1, 4). To do this we had to calculate some geometric
intersections. One of these was: the number of lines l in P4 which go through two
lines l1 and l2 and line in a 3-plane Γ.
So, let l ∩ l1 = P1 and l ∩ l2 = P2 . Then P1 , P2 ∈ Γ. What is l1 ∩ Γ? It is a point
so l1 ∩ Γ = P1 . Similarly l2 ∩ Γ = P2 and so there is exactly one such line l (the
line trough P1 and P2 ).
The general idea is to reduce more complicated geometrical problems (which are
usually nonlinear) to simpler (more linear) geometric ones.
5.2. Examples. Suppose we have a variety V ⊂ Pn with coordinate ring M =
k[x0 , . . . , xn ]/I. We want information about M .
idea: M has a natural grading given by degree M = ⊕d≥0 Md so look at φM (d) =
dimk (Md ).
5.2.1. eg. V = Pn . Then M = k[x
dim
0 , . . . , xn ]. So n+d
k (M1d ) nis the number of
monomials of degree d which is n+d
.
So
φ
(d)
=
= n! d + · · · .
M
n
n
5.2.2. eg. V curve given by x2 − yz. Then M = k[x, y, z]/(x2 − yz).
•
•
•
•
M0 has basis 1 and so dimk (M0 ) = 1.
M1 has basis x, y, z and so dimk (M0 ) = 3.
M2 has basis xy, xz, y 2 , yz, z 2 and so dimk (M0 ) = 5.
Md has basis xy d−1 , xy d−2 z, . . . , xz d−1 , y d , y d−1 z, . . . , z d and so dimk (M0 ) =
2d + 1.
Hence φM (d) = 2d + 1.
m−1
5.2.3. eg. Hypersurface S = V (f ) in Pn . Suppose f = xm
(. . . ) + · · · . For
0 + x0
m−i
d > m we have basis given by monomials of the form x0 ·(monomial of degree d−
8
SABIN CAUTIS
m + 2 − i in x1 , . . . , xn ). Thus
d+n−m
d+n−m−1
d+n−1
φM (d) =
+
+ ···+
n−1
n−1
n−1
d+n
d+n−m
=
−
n
n
1
n(n + 1)
1 n n(n + 1) n−1
(d +
d
+ · · · ) − (dn + (
− nm)dn−1 + · · · )
=
n!
2
n!
2
m
dn−1 + · · ·
=
(n − 1)!
5.3. Notice. Notice: in these cases φM (d) is a polynomial for d >> 0. Also
φM (d) = ak!k dk + · · · where k = dim(V ) and ak = degree(V ).
5.4. General Setup. Let S = k[x0 , . . . , xn ] viewed as a graded ring (ie. S = ⊕Sj ).
Let M be a graded ring over S. (ie. M = ⊕Mi such that if s ∈ Sj and m ∈ Mi
then sm ∈ Mi+j ).
Definition 5.1. φM is defined for each d ∈ Z by φM (d) = dim(Md ).
Theorem 5.2. If M is a finitely generated S-module then φM (d) is a polynomial
for large d (ie. φM (d) = P (d) for d >> 0 where P is a polynomial).
Proof. We use induction on n. In the base case n = −1 so Mj = 0 for j >> 0 thus
φM (d) = 0 for d >> 0.
Consider the exact sequence:
0 → Kd → Md → Md+1 → Cd+1 → 0
where the middle map is multiplication by xn and were Kd is the kernel of
Md → Md+1 and Cd+1 = Md+1 /xn Md . Now Kd and Cd are graded S modules on
which xn acts by 0. So they are k[x0 , . . . , xn−1 ]-modules.
Now from the exact sequence we get that
dim(Md+1 ) − dim(Md ) = dim(Cd+1 ) − dim(Kd )
which means that
φM (d + 1) − φM (d) = φC (d + 1) − φK (d)
But φC (d + 1) and φK (d) are polynomials by induction. Consequently φM (d) is a
polynomial for d >> 0 (exercise).
Definition 5.3. Given a variety V ⊂ Pn with coordinate ring M , the Hilbert
polynomial PV is defined by PV (d) = φM (d) for d >> 0.
5.5. Results. We know that:
• The degree of PV is equal to dim(V ).
deg(V ) dim(V )
d
. So we get another way
• The leading coefficient of PV (d) is
dim(V )!
of viewing degree.
•
(−1)dim(V ) (PV (0) − 1)
is called the arithmetic genus of V . We will see that it is independent of
the embedding (unlike degree but like dimension).
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For example, if V is a curve in P2 of degree m then the arithmetic genus equals
− 1)(m − 2) (use our prior calculation). This corresponds with the normal
definition of genus if we view this curve as a surface. To see this think of the curve
as d lines. Each line is a sphere and every two spheres are tangent. This gives
1
2 (m − 1)(m − 2) holes... (eg. m = 3)
1
2 (m
6. Week 6
6.1. Rational Maps.
Definition 6.1. X an affine irreducible variety. The function field K(X) contains
equivalence classes of pairs (U, f ) where U is a nonempty open set and F a regular
function on U . Furthermore, (U, f ) ∼ (V, g) iff. f = g on U ∩ V . The elements of
K(X) are called rational functions.
Theorem 6.2. If X ⊂ An then K(X) equals the quotient field of A(X). If X ⊂ Pn
then K(X) contains the elements of degree 0 in the quotient field of S(X).
Aside: More generally, for integral schemes X, K(X) is defined to be the local
ring of the generic point. This means K(X) ∼
= K(U ) where U is any open affine
subset. Even more generally, for arbitrary schemes X we replace K(X) by a sheaf...
(see Hartshorne)
For example, we have for X the zero set of x2 − y in A2 we have that K(X)
consists of all rational functions fg where f, g ∈ k[x1 , x2 ] and x2 − y 6 |g.
Definition 6.3. X, Y varieties. A rational map φU : X → Y is an equivalence
of pairs (U, φU ) where U ⊂ X is open and U 6= ∅ such that (U, φU ) ∼ (V, φV ) iff
φV = φV on U ∩ V .
For example, f : P2 → P1 is a rational morphism since it is a morphism on z 6= 0.
Definition 6.4. X, Y varieties. X and Y are birationally isomorphic (birational
for short) if there exist rational maps and f : X → Y and g : Y → X such that
f ◦ g = idY and g ◦ g = idX .
Just as with morphisms we try to understand varieties up to isomorphism so
with rational maps we want to classify varities up to birational equivalence.
• With morphisms we have X ∼
= A(Y ).
= Y iff. A(X) ∼
• With rational maps X is birational to Y iff K(X) ∼
= K(Y ).
So this gives us an algebraic criterion for birationality.
We can use this idea and the following lemma to give us a neat result about
rational maps.
Lemma 6.5. If L → K is a finite separable extension then L = K(α) (ie. L is
generated by one element).
Proposition 6.6. Any variety X of dimension r is birational to a hypersurface Y
in Pr+1 . (thus in a way it is enough to study hypersurfaces)
Proof. We know k ⊂ K(X) is a finitely generated extension with transcendence
degree r. Hence we can find x1 , . . . , xr such that k(x1 , . . . , xr ) → K(X) is a finite
separable extension (it is separable since for convenience we’re working over a field
of characteristic zero).
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SABIN CAUTIS
Then by our lemma K(X) ∼
= k(x1 , . . . , xr , y) where f (x1 , . . . , xr , y) = 0 (after
we clear fractions this is a polynomial with coefficients polynomials in xi ’s). Let
S be the hypersurface defined by f = 0 in Ar+1 . Then K(S) ∼
= K(X) which
means that S is birational to X. Whence the closure of S in Pr+1 is the required
hypersurface.
The problem of classifying varities up to birational equivalence is difficult. However, for curves it has been “solved”. Every birational equivalence class of curves
contains exactly one nonsingular curve. Thus classifying curves up to birational
equivalence is the same as classifying nonsingular curves up to isomorphism.
By the way, a representative of a birational equivalence class is called a model.
6.2. Singularities.
Definition 6.7. (naive) Suppose X ⊂ An is defined by (f1 , . . . , ft ) and has dimen∂fi
sion r. Then P ∈ X is a nonsingular point if the rank of || ∂x
j (P )|| is n − r. X is
nonsingular if it is nonsingular at every point.
∂f
For example, f (x, y) in A2 is nonsingular at P if ( ∂f
∂x (P ), ∂y (P )) 6= (0, 0). Thus
xy = 0 is singular at (0, 0). and in general f g = 0 is singular at all points P ∈
V (f ) ∩ V (g). Thus any nonsingular curve in P2 is irreducible.
Notice that nonsingularity is a local condition. Also, nonsingular varieties are
locally homeomorphic to Cn so they are manifolds. Whence a nonsingular projective
variety is a compact manifold (a closed subset of a compact space is compact) and
hence is a good candidate for cohomology.
Incidentally, this definition of nonsingular points is naive because we make use
of the embedding. We don’t want to do this so the fancier definition is to first
introduce the cotangent bundle and take the tangent bundle to be its dual. We will
try to do this next time. However, this way of thinking about the tangent bundle
has its advantages because it is more straightforward.
7. Week 7
7.1. A couple of words about genus. There are two definitions of genus:
• geometric genus: pg (X) = Γ(X, ωX ) where ωX is the canonical sheaf.
• arithmetic genus: pa (X) = (−1)r (χ(OX ) − 1) where r is the dimension of
X and χ(OX ) = dimk (H 0 (X, OX )) − dimk (H 1 (X, OX )) + · · · .
At this point someone asked me if the χ is the same as the Euler function defined
in diffgeo. The answer is that it is not. The genus of a curve (Riemann surface) is
the same but the Euler characteristic in diffgeo of such a surface is 2−2g = 1−2g+1
whereas χ in this case is 1 − g.
What can be said:
• pa and pg agree for nonsingular curves where they are birational invariants.
• for nonsingular curves which are surfaces if we are working over C they
agree with the usual definition of a genus for surfaces.
• they are defined algebraically so that they work over any field (this is why
we bother with such technicalities)
If we always worked over C they things might be a bit easier. For example,
one can show by triangularization that a curve of degree n in P2 has genus n−1
2
and thus any 2 such curves of degree n 6= m are not homeomorphic and hence not
isomorphic. However, one has to be more clever to extend this proof to other fields.
MATH260 TUTORIALS
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7.2. The Tangent Space.
7.2.1. The Intuition. We first consider lines tangent to an affine variety.
Definition 7.1. P ∈ X variety. Suppose line L is given by P + tQ and IX =
(f1 , . . . , fn ). Then the multiplicity of L with X at P is the multiplicity of t as a
root of hcf(f1 (P + tQ), . . . , fn (P + tQ)).
For example, for y − x2 = 0 and line L = (t, 0) we get t2 and hence a multiplicity
of 2. For line L = (0, t) we get t and hence multiplicity 1.
Definition 7.2. A line L is tangent to X at P if it has multiplicity ≥ 2 with X at
P . The locus of all points on lines tangent to X at P is the tangent space to X at
P denoted (T X)P .
Now suppose X is given by f = 0. If P = (x1 , . . . , xN ) ∈ X then (α1 , . . . , αN )
is a point in the tangent space if (x1 , . . . , xN ) + t(α1 − x1 , . . . , αN − xN ) is a line
tangent to X at P . This is true if
satisfies
g(t) = f (x1 + t(α1 − x1 ), . . . , xN + t(αN − xN ))
N
X ∂f
d
(P )(αi − xi ) = 0
g(t)|t=0 = 0 ⇔
dt
∂ti
i=1
So if X is defined by IX = (f1 , . . . , fn ) then the tangent space is composed of the
intersections of the tangent spaces described above - ie. of all points (α1 , . . . , αN )
such that for k = 1, . . . , n we have
N
X
∂fk
i=1
∂ti
(P )(αi − xi ) = 0
For example, the tangent space of y − x2 = 0 at (a, b) is given by all points
(α1 , α2 ) such that (−2a)(α1 − a) + (1)(α2 − b) = 0. So at (0, 0) the tangent is
α2 = 0. Similarly for y(y − x2 ) = 0 the tangent space at (0, 0) is A2 despite the
fact that the tangent space of both y and y − x2 at (0, 0) are y = 0.
7.2.2. Intrinsic Description.
Definition 7.3. The differential of a polynomial f (t1 , . . . , fN ) at a point x =
(x1 , . . . , xN ) is
N
X
∂f
(x)(ti − xi )
dx f =
∂t
i
i=1
which is a first order approximation of f .
Notice that dx : k[X] → (T X)∗x . Moreover dx (c) = 0 for c a constant. So we get
a map dx : Mx → (T X)∗x where Mx = {f ∈ k[X] : f (x) = 0}.
Now if f ∈ M2x then f = f1 f2 where f1 (x) = f2 (x) = 0. Then
dx f =
N
X
i=1
So dx :
Mx /M2x
f1 (x)
∂f2
∂f1
(x)(ti − xi ) + f2 (x)
(x)(ti − xi ) = 0
∂ti
∂ti
→ (T X)∗x .
Proposition 7.4. dx : Mx /M2x → (T X)∗x is an isomorphism.
12
SABIN CAUTIS
PN
PN
Proof. ONTO: Suppose
hi (ti − xi ) ∈ (T X)∗x . Then f =
i=1
i=1 hi (ti − xi )
PN
satisfies dx f = i=1 hi (ti − xi ) which proves dx is onto.
1-1: Suppose dx f = 0 and f ∈ Mx . Assume that x = (0, . . . , 0) for simplicity.
Then f = F1 + F2 + · · · where Fi are terms of degree i in t1 , . . . , tN . Then dx f = 0
implies F1 = 0 which means f ∈ (t1 , . . . , tN )2 and so f ∼
= 0 in Mx /M2x .
This description of the tangent space (T X)x = (Mx /Mx2 )∗ still depends on the
embedding (as an interesting note, this is the usual way of defining the cotangent
space in differential geometry). Fortunately we have:
Theorem 7.5. (T X)∗x ∼
= mx /m2x where mx is the maximal ideal in Ox .
Proof. We can write any function in Ox as F/G with F, G ∈ k[t1 , . . . , tN ]. Then the
differential is dx f = dx (F/G)|(T X)x and the same argument as before works.
Hence we have the equivalent definitions:
Definition 7.6. mx /m2x is the cotangent space at x and (mx /m2x )∗ is the tangent
space.
For example, take y − x2 = 0 again. At (0, 0), O(0,0) =
Then m = m(0,0) =
2
f
g where f (0, 0) = 0
2
ax+by
= ax+bx
= ax
g
g
g
2
f
g
where g(0, 0) 6= 0.
and g(0, 0) 6= 0. Then any element of
m/m looks like f =
for some g = a0 + a1 x + . . . . But then
a
a
ax
=
x
since
g
=
ax.
Thus
m/m
is
one
dimensional with basis x. This is as
g
a0
a0
expected since a natural linear functional on y = 0 in A2 is x.
Now suppose f : X → Y is a morphism with f (x) = y. Then f ∗ : Oy → Oy with
means that f ∗ : my /m2y → mx /m2x (check this). Hence we get a map (mx /m2x )∗ →
(my /m2y )∗ of tangent spaces (ie a map (T X)x → (T Y )y ).
We finish off with an interesting curve C. It is defined as the image of A1 in An
by x1 = tn , x2 = tn+1 , . . . , xn = t2n−1 .
Proposition 7.7. Tangent space to C at (0, . . . , 0) is An .
Proof. At this point I ran out of time and couldn’t fit in the proof...
It is enough
Pn terms. SupPn to show that if F ∈ IX then F contains no linear
pose F = i=1 ai ti + higher terms Then letting ti = ti we get i=1 ai ti+n−1 +
terms in t of degree ≥ 2n which must equal zero. Thus ai = 0 for all i.
So C is a curve which sits in An and cannot be embedded in Am for any m <
n. Notice how it is much easier to figure out the tangent space from elementary
definitions...
All we did today defines the tangent space. To define the tangent bundle we
need to use differential forms which are essentially dx g where x is allowed to vary.
This way we get the cotangent bundle whose dual is the tangent bundle.
8. Week 8
8.1. Blowups. Start with an example. Consider xy = 0 in A2 . This has a singularity at (0, 0). Why?
• Reason 1:
∂f
∂x (0, 0)
= 0 and
∂f
∂y (0, 0)
= 0 where f = xy.
MATH260 TUTORIALS
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• Reason 2: The tangent space at (0, 0) is too large. That is
2 = dimk m(0,0) /m2(0,0) > dimO(0,0) = 1
(recall that a local ring O is regular if dimO/m m/m2 = dimO. so a point P
is nonsingular iff. OP is regular)
Now let’s consider the surface M in [x, y; z, w] ∈ A2 × P1 given by xw = yz. We
have a map p : M → A2 where (x, y; z, w) 7→ (x, y). Now p−1 (x, y) = (x, y; x, y)
unless x = y = 0 in which case p−1 (0, 0) = (0, 0; r, t) ∼
= P1 .
What is the preimage of y = mx?
If x 6= 0 then p−1 (x, mx) = (x, mx; x, mx) = (x, mx; 1, m). As x → 0 we get
−1
p (x, mx) → (0, 0; 1, m). Thus
p−1 (y = mx) = (x, mx; 1, m) ∪ (0, 0; r, t)
where the component (0, 0; r, t) is called the exceptional divisor. Thus in our case
of xy = 0 we get:
p−1 (xy = 0) = (x, 0; 1, 0) ∪ (0, y; 0, 1) ∪ (0, 0; r, t)
which is the union of two disjoint lines and the exceptional divisor. If we now throw
away the exceptional divisor we have resolved our singularity at (0, 0) obtaining two
copies of our lines (picture).
Generally, if one blows something we hope to obtain something which is less
singular.
8.2. example. Take y 2 = x3 in A2 which is given by (t2 , t3 ) and has a singularity
at (0, 0). Then preimage of in A2 × P1 is (t2 , t3 ; t2 , t3 ) = (t2 , t3 ; 1, t) union the
exceptional divisor. Now C : (t2 , t3 ; 1, t) lies in A3 and is carved out by x = w2 and
y = w3 . This variety is no longer singular.
8.3. example. Take y 2 = x5 in A2 which is given by (t2 , t3 ). Blowing up we get
as before (t2 , t5 ; 1, t3 ) which sits in A3 as the image of A1 via (t2 , t5 , t3 ). But this
is still singular at (0, 0, 0) (see lemma below).
So we blow up again at (0, 0, 0). The blowup of A3 at (0, 0, 0) is the subvariety of
[x, y, z; u, v, w] ∈ A3 × P2 given by xv = yu, xw = zu, yw = zv. Then the preimage
of point (t2 , t5 , t3 ) is (t2 , t5 , t3 ; u, t3 u, tu) = (t2 , t5 , t3 ; 1, t3 , t). Thus the preimage
of our curve (t2 , t5 , t3 ) ⊂ A3 is (t2 , t5 , t3 , t3 , t) ⊂ A5 which is no longer singular at
(0, 0, 0, 0, 0).
Hence in the case of y 2 = x5 it took 2 blowups to get rid of the singularity at
(0, 0).
8.4. lemma. The following is a useful little lemma whose proof provides a great
exercise for working with the definitions of singularities from last class.
Lemma 8.1. If (te1 , . . . , ten ) is a parametrized curve in An with gcd(e1 , . . . , en ) = 1
and e1 ≤ e2 ≤ · · · ≤ en then it is nonsingular at the origin iff. e1 = 1.
8.5. Summary. So what we get in the end is a map f : X ′ → X which is a
birational morphism. Here are some reasons for studying blowups:
• to resolve singularities
• to “resolve” rational maps
• as a construction (eg. every cubic 2-fold in P3 is obtained from blowing up
P2 at 6 general points)
14
SABIN CAUTIS
8.6. Blowing up along subvarities. Now we can also blow up An (or rather Pn
for technical reasons) along closed subvarities Y (a point being the case above).
Suppose Y ⊂ Pn with IY = (r1 , . . . , rm ). Then the blowup of Pn along Y is the
subvariety of Pn × Pm−1 carved out by the homogeneous polynomials generating
the kernel of φ where
φ : k[x1 , . . . , xn ; y1 , . . . , ym ] → k[x1 , . . . , xn ]
with xi 7→ xi and yi 7→ ri . So in the case of blowing up along Y = (0, . . . , 0)
we have ri = xi and the homogeneous polynomials are xi yj = xj yi . (notice that
yi = xi is not homogeneous in y)
For example, let’s blow up A3 along line x = y = 0. It is the image in A3 × P1
carved out by xv = yu (ie the kernel of [x, y, z; u, v] 7→ [x, y, z; x, y]). As before
we have p : A3 × P1 → A3 with p−1 (x, y, z) a point unless x = y = 0 when it is
P1 . So consider y 2 − x3 in A3 . This is singular when x = y = 0 (on a line). If we
blow up along this line then the preimage of this surface, which is parametrized by
(t2 , t3 , r), is
(t2 , t3 , r; t2 , t3 ) ∪ P1
Ignoring the P1 component as before we get (t2 , t3 , r; t, 1) which is the same as
(t2 , t3 , t, r) in A3 . This is given by x = z 2 and y = z 3 which means it is nonsingular.
At this point I ran out of time so I postponed the statement of two theorems for
next time. I promissed them that next time I’ll talk about these theorems, blowing
down, the connexion of this to the definition of blowing up from class and about
DuVal singularities and their relation to representation theory.
9. Week 9
9.1. A couple of Theorems from last week.
Theorem 9.1 (Resolution of Singularities (Hironaka)). Given variety X there exist
varieties Xi and maps Πi : Xi+1 → Xi
Xn → · · · → X2 → X1 = X
such that Πi is the blowup of Xi along some subvariety Zi ⊂ Xi and Xn is smooth.
(ie. there exist smooth variety Xn Y and a regular birational map Π : Xn → X).
Theorem 9.2 (“Resolving” rational maps). Suppose φ : X → Y is rational. Then
there exist a sequence of blowups Πi : Xi+1 → Xi such that the composition
Xn → · · · → X2 → X1 = X → Y
from Xn to Y extends to a regular map.
For an example of the 2nd theorem consider φ : A2 → P1 given by (x, y) 7→ [x, y].
Then the blowup at (0, 0) given us a map V ⊂ A2 × P1 → A2 → P1 by [x, y; u, v] 7→
(x, y) 7→ [x, y] where V is given by xv = yu. This map can be extended to a regular
map by mapping [x, y; u, v] 7→ [u, v].
9.2. Connection To Definition of Blowup From Class. Recall that we defined
the blowup of Pn along Z where IZ = (r1 , . . . , rm ) as the subvariety of Pn × Pm−1
carved out by the homogeneous generators of the kernal of k[x0 , . . . , xn ][y1 , . . . , ym ] →
k[x0 , . . . , xn ] where xi 7→ xi and yi 7→ ri .
MATH260 TUTORIALS
15
In class we defined the blowup as P roj(S) where S = k[x0 , . . . , xn ]⊕IZ ⊕IZ2 ⊕. . . .
Now, we have the map
φ : k[x0 , . . . , xn ][y1 , . . . , ym ] → S
given by xi 7→ xi and yi 7→ ri ∈ S1 . This map is onto. Whence P roj(S) is the
subvariety in P roj(k[x0 , . . . , xn ][y1 , . . . , ym ] (where deg(xi ) = 0 and deg(yi ) = 1)
with ideal the kernel of φ (the same as before).
So the only reason we introduced S is so that we can write down the blowup
explicitly as P roj(S) without having to use an embedding.
The result we are using here is that if φ (onto) φ : A[x0 , . . . , xn ] → S is a
graded morphism (preserves degree) of graded rings then P roj(S) is the subvariety
of P roj(A[x0 , . . . , xn ]) = PnA carved out by the homogeneous generators of ker(φ).
9.3. Blowdown.
˜ is the blowup of X then π : X
˜ → X is the blowdown of X
˜ to
Definition 9.3. If X
X.
Theorem 9.4 (Castelnuovo’s Contractibility Criterion). If a nonsingular projective
surface X contains a −1 curve C (a curve whose self intersection is −1) then there
exists a blowdown f : X → Y such that Y is nonsingular and f (C) is a point.
9.4. Blowup More generally. Suppose we want to blow up X about a subvariety
Y (before we took X to be An or Pn ). Then if X ⊂ Pn with IX = (f1 , . . . , fl ) and
IY = (r1 , . . . , rm ) then the blowup is the subvariety of X × Pm−1 carved out by the
homogeneous polynomials in the kernel of
φ : k[x0 , . . . , xn ][y1 , . . . , ym ]/(f1 , . . . , fl ) → k[x0 , . . . , xn ]/(f1 , . . . , fl )
where xi 7→ xi and yi 7→ ri .
9.5. Connexions to Representation Theory. Let X be a surface with singularity x ∈ X. Then one can blow up X to resolve this singularity: f : Y → X with
f birational, regular which contracts curves C1 , . . . , Cr ⊂ Y to x ∈ X.
Using Castelnuovo’s criterion one can show that we can get rid of the −1 curves
among the Ci . Doing this gives us thet minimal resolution of X.
Definition 9.5. A point x ∈ X of a normal surface is a Duval singularity if
there exists a minimal resolution f : Y → X contracting curves Ci to x such that
KY (Ci ) = 0. (think of it as a special type of singularity).
How do we get a graph from a resolution?
• vertices vi ↔ curves Ci
• two vertices (vi , vj ) i 6= j are connected if Ci and Cj intersect (ie. Ci Cj ≥
1). (Ci2 is the self intersection and has special meaning).
Theorem 9.6. The scalar product Ci Cj is negative definite.
In the case of Du Val singularities:
• Ci ∼
= P1
2
• Ci = −2 and Ci Cj ≥ 0 for i 6= j
16
SABIN CAUTIS
Consider Z[C1 , . . . , Cr ] as a Z-module with inner product. The algebraic problem
of understanding Z-modules with negative definite inner product and Ci2 = −2 and
Ci Cj ≥ 0 for i 6= j leads us to root systems and is well understood (comes in in Lie
algebras representation).
Such a module decomposes into irreducible ones whose graphs are connected
and fall into 5 categories (simple Dynkin diagrams): An , Dn , E6 , E7 , E8 where the
subscripts count the number of vertices:
100
0
11 0
1 0
100
11
0 0
1
1 A_n
0000
111
1
000
111
000
111
011
1
0011
0011
0011
00
0
1
D_n
000
011
1
0011
0011
0011
00 000
0
1
111111111
000000000
111111
0000
111
1
0
1
00
11
00
11
00
11
00
11
0
1
0
1
0
1
0
0
1
00
011
1
0011
0011
0011
00 1
0
01
1
0 1
1
0000
1
011
1
00
11
111111111
000000000
11111111
00000000
111
0
0
1
E_7
E_6
00
1
11
00
1
0
1
0
0
0
1
0
0
1
0
0 1
1
0 1
1
0
1
0 1
1
0
1
0 1
1
0
1
0
1
111111111
000000000
1111
0000
0
1
0
1
E_8
0
1
0
1
Figure 1. Simple Dynkin diagrams
The key is:
• DuVal singularities correspond only to connected Dynking diagrams
• Duval singularities are uniquely determined by their graphs
So there are 5 types of Duval singularities which can be given by equations:
•
•
•
•
•
An : x2 + y 2 + z n+1 = 0 for n ≥ 1
Dn : x2 + y 2 z + z n−1 = 0 for n ≥ 4
E6 : x2 + y 3 + yz 3 = 0
E7 : x2 + y 3 + yz 3 = 0
E8 : x2 + y 3 + z 5 = 0
Next time: McKay correspondence.
10. Week 10
Today: McKay correpondence.
Last time we saw: DuVal singularities ↔ Dynkin diagrams
today: DuVal singularities ↔ finite subgroups of SO(3)
conclusion: finite subgroups of SO(3) ↔ Dynkin diagrams
10.1. Quotient Varities. Suppose X is an affine variety and G a finite group of
automorphisms. How do we define X/G? (ie. we know what it looks like as a set
by we need to give it the algebraic structure of a variety)
Let A = k[X]. For every g ∈ G we have g : X → X and hence g ∗ : A → A. So
we define AG = {f ∈ A : g ∗ (f ) = f }. Then we define X/G = spec(AG ) = Y to be
the quotient variety and we have a map AG → A or φ : X → Y which is onto.
MATH260 TUTORIALS
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Proposition 10.1. P2 = g(P1 ) for some g ∈ G iff. φ(P1 ) = φ(P2 ). (hence Y
parametrizes the orbits of X under the action of G as desired)
Proof. Since φ(P1 ) = φ(P2 ) ⇔ IP1 ∩AG = IP2 ∩AG for points P1 = (a1 , . . . , an ) and
AG = IP2 ∩ AG .
P2 = (b1 , . . . , bn ) in X we want to show that g(P1 ) = P2 ⇔ IP1 ∩P
(⇒) Suppose g(P1 ) = P2 for g ∈ G. If f ∈ IP2 ∩ AG then f = (xi − bi )hi and
g ∗ f = f . So
X
X
f=
(g ∗ (xi ) − bi )g ∗ (hi ) ⇒ f (a1 , . . . , an ) =
(g ∗ (ai ) − bi )g ∗ (hi ) = 0
which means that f ∈ IP1 ∩ AG . Hence IP2 ∩ AG ⊂ IP1 ∩ AG . Since P1 = g −1 (P2 )
we get IP1 ∩ AG ⊂ IP2 ∩ AG and we’re done.
(⇐) Suppose g(P1 ) 6= P2 for any g ∈ G. Let {Qi } and {Ri } be the orbits of P1
and P2 . Then {Qi } ∩P
{Ri } = ∅. So one point
P a function
P at a time we can construct
1
f ∈ k[X] such that
f (Qi ) = 0 and
f (Ri ) = 1. Let F = |G|
g∈G f (g(x)).
Then F ∈ AG and F (P1 ) = 0 and F (P2 ) = 1. So IP1 ∩ AG 6= IP2 ∩ AG .
A couple of fact:
• If A is finitely generated over k then AG is also finitely generated.
• If X is nonsingular and G is fixed point free (ie. g(P ) = P → g = 1) then
X/G is nonsingular.
But what happends if G has fixed points?
10.2. Back To DuVal Singularities. Take as an example X = A2 over C and
G cyclic generated by g(x, y) = (ǫx, ǫq y) where ǫ is a primitive nth root of 1 and
gcd(n, q) = 1. Then (0, 0) is a fixed point and we should get a singularity there.
Indeed, A = k[A2 ] = k[x, y]. Suppose q = −1 then AG is generated by u = xn ,
v = y n and w = xy. Then AG = k[u, v, w]/(uv − wn ). This is a surface in A3 with
type (n, −1) singularity (in general we get an (n, q) singularity).
Now recall that Duval singularities fell into 5 categories enumerated by simple
Dynkin diagrams. This was done by a classification using algebra. Here is another
way to view these 5 categories.
Theorem 10.2. G finite group of linear transformations acting on A2 with det(g) =
1 for all g ∈ G. Then the image of 0 is a Duval singularity.
So let’s look at finite subgroups of SL2 (C). But SL2 (C) is homotopy equivalent
to SU2 (C) (Gram-Schmidt process) and SU2 (C) is a double cover of SO3 (R). So
we look at finite subgroups of SO3 (R). These are:
• cyclic order n ←→ An−1
• dihedral group order n ←→ Dn+2
• tetrahedral (symmetry group of tetrahedron) ←→ E6
• octahedral (symmetry group of octahedron or cube) ←→ E7
• icosahedral (symmetry group of icosahedron or dodecahedron) ←→ E8
So we find that:
• Simple Dynkin diagrams correspond to
• DuVal singularities which correspond to
• finite rotation groups of R3 which are really
• the Platonic solids in disguise.