## Gysin morphism

May 24, 2008

When we have a vector bundle ( of finite rank ) over a scheme, that induces a flat pull back map of cycle classes, which can be shown to be isomorphism.

The surjectivity holds for affine bundle as well, this can be reduced to the case where the bundle is trivial using Noetherian induction : the cycle classes of a scheme can be “roughly” divided into the classes in an open subset and those in the complement ( this can be described by an exact sequence, which I would name the structural exact sequence ).

The injectivity is much harder to show, we need extra structures. First we need to complete the bundle projectively, with its projectivization added as the hyperplane at infinity. Then using Noetherian induction, we can show that there is an “O(1) decomposition” of each cycle class in the projective bundle as a ordered sum of classes : each of which is represented by the iterated action of the canonical line bundle on a pull back class. This fits nicely with the description of a projective space recursively as piecing an affine space with a projective space of 1 dimension less at infinity. Then the injectivity comes as a corollary of these added structure, using the “structural exact sequence” on the bundles, which are all compatible with the projection onto the base scheme.

There is a more descriptive description of the Gysin homomorphism using the the universal quotient bundle on the projective completion $\epsilon$. To get the class on the base scheme via the Gysin morphism, first we take the closure of the class inside the projective completion of the bundle, intersect it with the highest Chern class of the universal quotient bundle, and the project it onto the base scheme.

## Construction of Hilbert Schemes

May 16, 2008

We want to parametrize subschemes of projective space $\mathbb{P}^n$. Naturally, those with same Hilbert polynomial will cluster together, by the result which states that any continuous (flat ) family of closed projective subschemes have constant Hilbert polynomial. Now we will construct the moduli space of all subschemes $X$ with a fixed Hilbert polynomial $P$. Here is the idea :
The value of the Hilbert polynomial $P$ at $d$ will coincide with the dimension of the $d-th$ graded piece of the coordinate ring of $X$ when $d$ is large. Another description of the Hilbert polynomial is the Euler characteristic of the structure sheaf of the scheme $X$ twisted by $d$, by Serre vanishing theorem ( why?).
The idea is, if we choose $d$ large enough, then we hope to identify $X$ with the $d-th$ graded piece of its defining ideal $I$. So each scheme $X$ can therefore be identified with a subspace of fixed dimension of the space of all monomials of degree $d$. Which means that we have a map from our “moduli space” to the Grassmanian $G(m,n)$, and then hope that the image is actually a closed subscheme of the Grassmanian. Fortunately, we can do that, but need a lot of results.
First in order to cut out precisely those $X$ with Hilbert polynomial $P$, we require that the image of the map $V\otimes R_{e-d}$ in $R_d$ has dimension $Q(e) = \dim R_e - P(e)$ for all $e \ge d$ ( the dimension of the $e-th$ piece of the defining ideal ). That could be consider a linear map of two vector bundles over the Grassmanian. Unfortunately , this is the same as to require the map of vector bundles has constant rank, which is not in general a closed condition. The best that we can do is to require that the rank is alway not mroe than $Q(e)$. There are three problems we need to solve :
1) As mentioned, this is not a closed condition.
2) We have to do this for infinitely many $e$.
3) It is not clear how we can pick $d$ uniformly, so that the dimension of the graded pieces after $d$ behave as we expect.
There are two important results that have us resolve this :
Thm 1: We can in fact choose such a $d$ uniformly.
Thm 2 : (i) We can choose such a $d$ that for all $V \in Gr(Q(d),R_d)$, the multiplication by $R_1$ map increase the dimension at least 1 ( first test )
(ii) Once any $V$ has passed the first test, it will pass all the required tests:
$V \otimes R_{e-d} \to R_d$ has image of dimension at least $Q(e)$.
Now, with all of the above, to cut out the moduli space, we can just put on the closed condition: the rank of the multiplication maps must not be more than “expected”, i.e. $Q(e)$ and that would cut out precisely the subschemes $X$ having Hilbert polynomial $P$.

## Segre classes and Chern classes ( Overview )

May 8, 2008

For a vector bundle $E$ over a scheme $X$, consider the projective bundle $p : P = P(E) \to X$. Then we can use the canonical line bundle on $P$ to define the $i-th$ Segre class of $E$ as follows : for a cycle $\alpha$, we pull it back, and we let the first Chern class of the canonical line bundle act on the pullback cycle $i + \dim (P/X)$ times ( so that we get the dimensions as expected), and then pushforward the result. Here we utilize the property that the projection from the projective bundle is both flat and proper.
The Segre classes have the following properties inherited from that of Chern class of line bundle :
– Projection formula
– Flat pull back
– Commutativity of classes
It also has vanishing property : that is, the Segre classes whose orders are negative or bigger than the dimension of $X$ are all trivial

Now we can define formally the Segre power series, and then take the inverse to get the Chern polynomials, which then gives us the Chern classes. The Chern classes also have the properties like projection formula, flat pulback, commutativity, vanishing, and one other important property is the Whitney sum formula, which states that the Chern polynomial of the “sum” of two vector bundles ( the middle term is the short exact sequences ) is the product of those of other two.

## Residue field does not change under restriction to closed subschemes or fibres

May 8, 2008

It does not change when restricting to closed subschemes. It is because the residue ring does not change : $\frac{O}{p} = \frac{O/I}{p/I}$.
When taking fibre of a morphism : $f : X \to Y$, $f(x) = y$
Consider a scheme Spec $T = k(x)$. Then by the universal property of fibre product, there are ( (natural) morphisms
$T \to X_y \to$latex $and since the composition induces isomorphism on residue fields, so does the first map. ## Flat map – fibre dimension May 7, 2008 Flat map has a nice property : the fibre dimension is very well behaved. Assume we have a flat map $f : X \to Y$ . If we take a point $x$ on $X$, $y$ its image, then the dimension of the fibre over $y$ ( going through $x$ ) is exactly as we expect : the difference between the local dimension at $x$ and that at $y$ : $\dim _xX_y = \dim_xX - \dim_yY$ Idea of proof : -The proof use a lot of steps involving simplifying the situation. First we can base change to make $y$ a closed point, $Y$ is a local affine scheme ( Spectrum of a local ring ) , by base change to $O_{y,Y}$. This is in effect focus all attention to the local property at $y$. Note that the local information at $x$ is preserved because we infact take the direct limit of the inverse of a local basis at $y$, which must as not refined as a local basis at $x$. This can be check using using affine open sets at $x$ and $y$ . – Next we can, by base change ( cut out the sheaf of nilradical), get rid of all the nilpotents in $Y$, make it into a reduced scheme. . This does not change dimensions at all. Now we can proceed by induction : we can choose an element $t$ in $O_{y,Y}$ that is not a zero divisor. Then the subscheme defined by $t$ is one dimension less. The pull back of the subscheme ( corresponds to the base change by the map that injects the subscheme into $Y$ ) is defined on $X$ by the inverse image ideal of $(t)$, which is non-zero-divisor at $x$ by flatness, hence define a subschem passing through $x$ which is 1 dimension less, and this agrees with the hypothesis. Since we need the dimension to decrease by exactly one after being cut out by a non zero divisor, we need the schemes to be of finite type over an a.c. field $k$. The above result is a local property. If we need to make claims about the relative dimension between the components of $X$ with their images are all the same, we need to make an assumption that all the fibres have the same dimensions. The two things are infact equivalent. ## Pushforward and flat pull-back of intersections May 6, 2008 These are two quite useful properties of intersecting with divisors. 1- Projection formula : Let $\pi : X' \to X$ be proper. Then if $D$ is a divisor on $X$, we want to pull back $D$, intersect it with a cycle in $X'$, and then push it back. The projection formula gives a shortcut to that : we can just pushforward the cycle, and then intersect it with $D$. $\pi^0_*(\pi^*D \cap \alpha ) = \pi_* \cap \alpha$. ( Where $\pi^0$ is the induced map from the relevant subschemes : from $\pi^{-1}(|D|) \cap |\alpha|$ to $\pi(|\alpha|) \cap |D|$. This is important, because we have to adhere to where we define our intersection classes!) 2- Flat pull back Let $\pi : X' \to X$ be flat. Then if$latex\$ F is a divisor on $X$, we want to pull back the intersection of a cycle $\alpha$ with $D$. So this formula tells us that we can just pull back $D$ and $\alpha$ and do the intersection on $X'$. Again, we need to be a bit careful to restrict maps to relevant subscheme.
$\pi_0^*(D \cap \alpha) = \pi^*(D) \cap \pi^*(\alpha)$

The idea of proof :
1- We can assume $\pi$ is a surjective proper morphism of varieties, and the cycle is just $X'$ itself.

2-

The nice thing about projection formula and flat pull back is that it allows us to define Chern classes of vector bundle : we pushforward the Chern classes (intersection classes) defined by the canonical line bundle on the corresponding projective bundle. The projection map from the projective bundle is both proper and flat.

## Intersecting with pseudo-divisors

May 6, 2008

Suppose we have a pseudo-divisor $D$ on a scheme $X$. Then for each subvariety $V$ of $X$, we can define the intersection $D \cap [V]$ by pulling back ( restricting ) the corresponding line bundle with section of $D$ on to $V$, which will in turn determine a divisor supported in $|D| \cap V$. In case the support of $|D|$ contains $V$, then we can only define that intersection divisor up to rational equivalence ( at level of homology class ). Otherwise, it is uniquely determined at the level of cycle. Note that we want to define a class in $|D| \cap V$, and it is different from defining a class in $V$. A cycle can in fact represent the 0 class in the latter but not in the former. Plus, later we shall need commutativity of intersection, thus we want to define a class in $|D| \cap V$ so that no matter in what order we intersect, the classes always be defined in the same space.

Now we can extend the definition by linearity to all the cycles. There is a special case where we can definite intersections at cycle level : when the line bundle of $D$ is trivial in a neighborhood of $|D|$. In that case, in the only situation where the previous failed to determine a cycle i.e. when the support of a subvariety is contained $|D|$, we can just define the intersection to be the 0 cycle.

As we should expect, if the line bundle that comes with $D$ is trivial, then intersecting $\alpha$ with $D$ always yield 0 on the groups $A(|\alpha|)$. In fact, if $V$ is any variety, then the restriction of $O_X(D)$ on $V$ is trivial, hence it defines the 0 homology class. Later, we will know that intersecting with divisors yield a homomorphism of Chow groups ( i.e., intersecting with 0 homology class yield a homology class, as a corollary of commutativity of intersection ).

There are two properties of intersection with divisors which are useful but somewhat not quite trivial to prove :
1- Projection formula
2- Flat pullback

In some situation, when we are sure that Cartier divisors can be pulled backed ( i.e., the image scheme is not contained in the support of the divisor, or when the map is flat ( pull back of nonzero divisors are nonzero divisors , hence we can pull back the ration functions that define the Cartier divisor under flat map ). In general, since the support of a Cartier divisor is contained in that of the pseudo-divisor it represents, the equality involving two intersection classes (with Cartier divisor ) is better, since the classes are defined in smaller spaces.

## Blow-up of morphisms

May 6, 2008

Suppose we have a morphism $f :X \to Y$. Let’s say we are going to blow up $Y$ along a coherent sheaf $I$ of ideals, $\pi_Y : Y' \to Y$. We can take the inverse image ideal sheaf $J$ of $I$ in $X$. Blow up $X$ along $J$ : $\pi_X : X' \to X$. Whence $f$ lifts to a morphism $f' : X \to Y$ in the obvious sense.
Furthermore, as a result, the ideal sheaves defining the exceptional divisors are both inverse image ideal sheaves of the $I$, hence they correspond as well according to the lift $f'$, i.e, $f'^*$ pull back the exceptional divisor on $Y'$ to the exceptional divisor on $X'$.

Such a lift exists uniquely, according to the universal property of the blowing up projection : the inverse image ideal sheaf is invertible.

## Inverse image ideal sheaf, blow-up, and exceptional divisor

May 6, 2008

If we have morphism $f : X \to Y$, and a sheaf of ideal $I$ on $Y$, then the inverse image sheaf $f^{-1} (I)$ is therefore a subsheaf of $f^{-1}O_Y$, whose action on $O_X$ makes it an $f^{-1}O_Y$ algebra. Thus $f^{-1}I$ also acts on $X$, and thus we can define the inverse image ideal sheaf of $O_X$ by $f^{-1}I \cdot O_X$.
This sheaf of ideals is exactly the image of the pull back $f^*I$ under the natural morphism :
$f^{-1}I \otimes_{f^{-1}O_Y} O_X \to f^{-1}O_Y \otimes_{f^{-1}O_Y} O_X \cong O_X$

Now let’s look at the situation where we blow up a scheme $X$ along a coherent sheaf of ideals $I$. By definition $\tilde{X} = Spec(\oplus_{n=0}^{\infty} I^n)$

We shall see that the inverse image ideal sheaf of $I$ is exactly the canonical sheaf $O_{\tilde{X}}(1)$, hence is invertible, therefore determines a Cartier divisor on $\tilde{X}$, which we would call the exceptional divisor.

Now we can look at the local picture. Locally over $Spec A$, the blow up along an ideal $I$ is $Proj( A \oplus_{n=1}^{\infty}I^n = S)$, and the canonical invertible sheaf corresponds to the homogenous ideal ( graded $S-$module ) $\oplus_{n=1}^{\infty}$, which is nothing but the homogenous ideal generated by $I$ in $S$. Thus the canonical sheaf coincides with the inverse image ideal sheaf of $I$ in $\tilde{X}$.
This ideal sheaf defines a Cartier divisor, which is called exceptional divisor. This is the same as the inverse image scheme $\pi^{-1} (Y)$ where $Y$ is the closed subscheme of $X$ defined by $I$.

## Normal Cone

May 6, 2008

If $X$ is a closed subscheme of $Y$, with $I$ the defining sheaf of ideals, then the normal cone to $X$ in $Y$ is defined by the spectrum of the sheaf of $O_X$ algebra $S$ defined by :
$S = \oplus_0^{\infty} I^n/I^{n+1}$.

The question is why it is defined this way? First of all, note that this is infact a cone over $X$, as the zero-th grade of $S$ is just $O_X$. Now let’s look a particularly nice case, when $X$ is a regular imbedding in $Y$ of codim $d$, i.e., locally $X$ is cut out by a regular sequences with $d$ elements. Then in this case, we have a normal bundle $N_XY$, the DUAL of the sheaf of sections of which is the locally free sheaf of dim $d I/I^2$. Thus $N_XY$ is the spectrum of the $O_X$ algebra which is the symmetric algebra of $I/I^2$. A well known result in algebra says that this algebra is just $S$.