This first pdf has stuff about families of group schemes and stuff about representations of groups that look like where .

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The last post left off with the definition of semisimple groups and a basic result about them. Continuing in this direction:

Thm: If is a semisimple group over a field then is the connected component of . Further .

It is classical that any element can be decomposed uniquely as . invertible implies is invertible. Then where is unipotent.

Thm: If is any affine algebraic group then any can be decomposed into . An analogous decomposition applies to the Lie algebra. Further is closed.

The key to proving this is verifying it for and noting that any affine algebraic group can be embedded as a closed subgroup of . For the statement about note that as is closed so is the intersection of two closed subset.

Thm (structure thm): Let is any commutative affine algebraic group. If is connected then so are . Further the multiplication map is an isomorphism and and .

A torus is an algebraic group isomorphic to the diagonal matrices in some . A d-group is an affine algebraic group such that has a basis consisting of characters. E.g. consider . Then . The character group of is isomorphic to ; i.e. is . This also gives all monomials in , i.e. a basis.

Thm: If is a d-group then where is a torus and is a finite group whose order is not divisible by the characteristic of the field.

Analogously to how these notions are defined with Lie algebras there is the upper series and the lower series where is the group generated by all commutators .

is solvable if $D^nG = e$ for some . It is nilpotent if for some .

Thm: If is a positive dimensional nilpotent group then is also positive dimensional. Further if is a proper closed subgroup then .

proof : This is not so hard to prove once its established that

Lemma: If are closed subgroups of with connected then .

this in turn is not hard to establish modulo

Prop. Let be the inverse morphism. If is any family of morphisms such that

1) is also part of the family

2) the are irreducible varieties.

3)

then the group generated by the is closed and connected. Further there is a finite sequence such that .

This is proved in Humphrey’s book, section 7.5. The proof is a little technical but also another good example of what Chevalley’s thm is good for. No proof for now.

proof (of lemma): Simply consider the family of morphisms for given by . Apply the prop. QED.

let be the largest number such that . By the lemma its connected and hence hence positive dimensional.

For the second statement use induction on . Set . Either in which case replace with and use induction hypothesis or showing the dimensions are not equal. QED.

The following is the group theoretic analogue of Lie and Engel’s theorem.

Thm: If is finite dimensional and it is unipotent or solvable then has a common eigenvector.

It implies any connected solvable is a subset of , the upper triangular matrices over a vector space of dimension over a field . There’s a split short exact sequence

where are the unipotent matrices in vector space of dimension over a field . Intersecting with gives

is a closed connected subgroup of hence also a torus. This is some intuition for the following result

Thm: Let be a connected solvable group. Then is closed connected normal and contains . Set . The maximal tori of are all conjugate under and fixing a maximal torus: .

Thm(fixed point): If is connected and solvable and acts on a complete variety then it has a fixed point.

Let be an affine algebraic group. The identity component of the maximal normal solvable subgroup is called the radical of . Then is the unipotent radical. If is connected, then a borel subgroup is a maximal solvable closed subgroup.

Suppose is connected. It is semisimple if . It is reductive if .

Thm: Let be a borel subgroup. Then is projective and all other borel subgroups of are conjugate. Further iff is projective.

I think the proof of this statement is quite enlightening but this post is already too long…

The groups in the thm are called parabolic subgroups. Reductive groups have the following nice property

Thm: Let be reductive; a maximal torus and the set of roots; let . There exists a unique -stable subgroup of having . There is an isomorphism such that . Note .

In the case of matrix lie groups the first claim is easy to see: The group in question is roughly . Verify that’s is -stable.

If then

which is in since .

proof (of the assertion): the group is one dimensional and the only such groups are . So general theory says there is some isomorphism . For and for consider

this is an automorphism . So gives a character of . More explicitly for a character . Or

So there is a commutative diagram formed by and by . Looking at the map on differentials and writing it follows that

.

Let be a reductive group. Fix a maximal torus and a borel subgroup containing it. = the unipotent elements of , and the Weyl group; for I use the convention to say that maps to .

More notation

; a root subgroup.

is independent of choice of lift.

Technical proposition:

One version of Bruhat decomposition says for fixed and and for any there are unique such that . Using and I see also that

a more common version is that .

this actually has a short axiomatic proof, but that’s for another time.

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(vector fields) left invariant vector fields on = point derivations at = .

The whole point of saying ‘left invariance’ is that such vector fields are determined by their value at the identity, so it seems superfluous to think of it as vector fields, but then again it connects well with the next definitions)

(derivations) left invariant derivations on =

The notation is to more explicitly differentiate between the definitions. The correspondence between them is given by

where is the evaluation at the identity function and the derivation sends a function to the function

.

This is plausible because and ; but it still an exercise to show that is actually a derivation.

To get a sense of the yoga that goes on in keeping these notions straight consider the adjoint action of on its lie algebra. In normal matrix land this described as where everything is a matrix the juxtaposition is just matrix multiplication. Compare in contrast

Prop: If then where .

proof: Set i.e. . Then . Let be the backwards map on on coordinate rings, then the map is . Note . So . By left invariance

QED.

1. ; the (unsurprising) claim is that

note also can write . From it follows that .

Note in general that :

Consider the point derivation on .

latex \delta(f\otimes 1 \cdot 1 \otimes g)$

so the map gives the desired isomorphism.

…

So for and

2. is the inverse map. Using that the composition is trivial (hence also on tangent spaces) its easily deduced that .

3. Fix . Set by . Set . Then .

From it follows that

using the prop and and 1,2 and that . Let be the constant map with image . Then from which the claim follows.

Thm: is a closed, normal subgroup then is an ideal.

proof: normal stabilizes stabilizes . So if is a basis of and extends it to a basis of then in this basis

Then the differential also has this form. One way to see this is that the image of a 1-parameter subgroup will look like

so differentiating and evaluating at shows the claim. Now apply this to then we get for arbitrary . QED.

Thm: if in addition char and are both connected then is an ideal iff is normal.

p oof: It remains to show is normal when is an ideal. Consider the subgroup .

this result depends on the following results

a) The association gives a 1-1 inclusions preserving map

b) Another powerful result implies that

Now b) . Now consider ; I have for every that the lie algebra of is . So by a) .QED.

Thm and algebraic group and then the lie algebra of is contained in

Thm. If char and connected then the lie algebra of is exactly . Further and the lie algebra of is .

This last result is just for characteristic zero. Let be a connected an algebraic group; it’s semisimple if it has no nontrivial closed connected normal commutative subgroups.

Thm: is semisimple iff is semisimple.

I’m not giving a proof but the intuition is to prove not one implies not the other. For if is not semismiple then it has a closed connected normal commutative subgroup and then is a commutative ideal in so it’s not semisimple. Going the other way, from a commutative ideal consider .

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I’m not being careful or detailed, I’m not even going define algebraic groups; that’s done in many other places.

is an algebraic group. If is a decomposition into irreducible components, then there is a unique one containing the identity:

say then is irreducible and the product must map into one of the irreducible components, but the image contains so they must be equal.

The identity component is denoted . It is normal: is irreducible and contains the identity.

Prop 1. If are dense and open subsets then .

Proof: have dense open or .QED

Prop 2. If is any subgroup then is a subgroup.

Proof: recall a function is continuous if . Apply this to inversion morphism: . Next, then repeating with the translation by morphism get . Thus is , thinking of then . QED.

Prop 3. If is constructible then

Proof: contains dense open in . By prop 1. .QED.

Thm (Chevalley for varieties) If is a dominant morphism of varieties then contains an open set of

Rmk: Compare with ex II.3.7 in Hartshorne. I’m not proving it but it can be proved using Noether Normalization; ultimately you reduce to affine and show for distinguished affines there is a surjection .

Thm (Chevalley for Noetherian Schemes) If is a finite type morphism of Noetherian schemes then the image of a constructible set is constructible.

Again not going to prove it but some consequences for algebraic groups:

Thm Let be a morphism of algebraic groups. Then

a) is a closed subgroup.

b) is a closed subgroup.

c) (here is the connected component containing identity.)

d)

Proof: a) . b) Note is constructible, so simply apply Chevalley’s thm. c) is closed connected and finite index in . Its cosets partition and its the complement of the union of all the cosets non containing , so its open. Since the cosets are disjoint, open and cover any irreducible space must lie inside one i.e. and equality follows. d) I’m not proving, it depends on a nontrivial dimension theorem about varieties. Maybe I’ll come back to it later.QED.

In passing note that if is a closed subgroup then it can be characterized algebraically: Let be the ideal of , then

Here is defined by and . This can be proved by following your nose.

I think the following is one good reason why people often restrict to talking about connected groups

Prop. Let act on a variety .

a) If is closed and is any subset then is closed.

b) The subgroup is closed.

c) The fixed locus is closed.

d) If is connected then stabilized all the irreducible components of .

Proof: a) the set in question is equal to the closed set where is action by . a) immediately implies b). c) allows follows from a): is closed and is the intersection of these over all .

d) acts on the set of irreducible components. Let be the stabilizer of one of them. It is a closed subgroup. The orbit of an irreducible components is finite, the orbit also is equal to the index of . The same argument as in proof of the prev thm c) shows .

Prop: Let be an affine variety with a action . Let be any finite dim vector space. Then stable under all the translations for . Recall .

proof: Reduce to the 1-dim case . Write

with

by the group action axioms . So . So . Further , so is stable under all translations. QED.

Given a closed a subgroup the goal is to give the coset space an algebraic structure. In the case is a normal subgroup it would be nice if the algebraic structure on also made it into an algebraic group. The key ingredient is

Thm (also Chevalley’s) For as above there exists a representation and an one dimensional subspace such that

Rmk: I’m not proving this (now) because it involves lie algebra stuff that I’ll talk about later. But here’s the idea: Let be the ideal of ; it is finitely generated and the generators all like in a finite dimensional subspace stable under all translations. Set . It is not hard to see that (using the first remark at the start of this subsection) that . A similar statement can be made for the lie algebra. The last step is to make one dimensional by replacing by where .

With this in hand I can take the orbit of as a model of ; its quasiprojective.

In the case is normal note that it acts on by scalar multiplication; this gives a character of . Now replace by subspace of subspaces on which acts by characters (this doesn’t mess up anything).

To recap we have . Consider the adjoint action . Then clearly is in the kernel of the composition ; take the image of as a model for the group .

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(Nullstellensatz) If and is a proper ideal then

a) and

b) .

The key to proving this result is

(Noether Normalization) Let be a finitely generated algebra. Then there are algebraically independent elements such that if a finitely generated module over .

The proof given here rests on the following

Claim: If is nonzero then there are such that is integral over . [1]

PROOF: let be unspecified (for the moment) constant and set . Then

(1)

note a term is plugged into every variable so if has degree then there is necessarily a term . Write w/ homogeneous, then (1) becomes

lower order terms in $latex x_n$. (2)

Assuming is infinite [2], which is fine for me because algebraically closed fields are always infinite, the constants can be chosen so

e.g. use induction on the number of variables, then writing by induction hypothesis there are such that some , so plugging in I’m left with a nonzero polynomial in one variable etc.

Note (2) gives a relation of integral dependence QED.

PROOF: use induction on where for some ideal; the case being clear. If there is nothing to prove. Otherwise there is a nonzero and the claim applies. So is integral over and .

Applying the inductive hypothesis if finite over and is finite over hence also over QED.

Now back to the Nullstellensatz

PROOF: a) follows easily from

Prop. The maximal ideals of are all of the form .

PROOF:Let be a max ideal. Noether Normalization is finite over but its also a field so . So is finite over , in particular algebraic so . Let be the image of in ; then QED.

From the prop is contained in a maximal ideal and then .

b) Suppose is such that . Consider . . so

with . If is the highest power of appearing in the then I can write

plugging in shows , the other inclusion is obvious QED.

[1]Alternatively, if I let represent the image of in then and I can restate the claim: if is nonzero and then there are such that is integral over .

[2] This claim holds for arbitrary fields but the proof is more difficult; see Reid’s book.

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(Orlov’s Result) Any functor which is full, faithful and exact is represented by an object on the product.

The proof is long and complicated so I’ll only attempt to give a flavor of the ideas used in the proof. One central part of the proof is the Beilinson resolution of the diagonal of projective space:

(0)

where is the structure sheaf on and .

Start with what is sometimes called the Euler sequence on $\mathbb{P}^N$ (here is the tangent bundle):

locally:

(1)

Note for and from the les in cohomology. Now from the Kunneth formula in algebraic geometry:

I’ll make use of the global section corresponding to the identity. Now . Locally the identity corresponds to a map

so at this is

and in view of (1) its locally the inclusion followed by the projection:

the only point in this local description is that iff , i.e. , in other words vanishes exactly on the diagonal.

If is a vector then there is a contraction map via

.

Now contraction with gives a map and these are exactly the maps that appear in the Beilinson resolution.

Now to Orlov’s result. From the data I need to produce an object .

The first step is to use that is projective to get an embedding and consider the functor

Now that has entered the picture we can utilize the Beilinson resolution (0), and using obtain a complex

(2)

where . I’m brushing a lot under the rug, it takes a bit of work to actually come up with this complex.

Now I need a powerful tool which I don’t have a firm grasp of and I can’t really explain here.

Let be a bounded complex. A left Postnikov system of is a diagram:

where the stared triangles are distinguished and the triangles with circles are commutative. An object is a left convolution of if there is a left Postnikov system such that . Denote by the class of all convolutions of .

Prop. If for and then has a convolution . Further, if and then all convolutions are canonically isomorphic.

Remark: I don’t have a great way of motivating this convolution business but its not a very geometric tool and in this post I’m trying to focus on the geometry that goes into this proof.

Continuing, the idea is now to the use the proposition with the complex (2) to obtain an object . Next one can show that basically by showing that both are convolutions of the same complex. With this result one can ultimately show . I’m not including the details because I want to focus on the rough idea and I want to avoid making an overly long post.

The functor has been represented by an object on the product. Using general Fourier-Mukai properties see e.g. this post, it remains to find an object such that .

The object is produced in much the same way as is produced. Using and ample line bundle on obtain a resolution for the diagonal on and using obtain a complex on much like (2), then use the proposition to get a convolution . The details are a little different and more complicated because is understandably more explicit than a general smooth projective variety .

The object does not represent . Instead, using cohomological properties, one decomposes and and then a another argument is needed go show .

I’m tempted to say that this was much less then even a rough outline of the proof. But I really only wanted to discuss the Beilinson resolution and even though I was brief with Orlov’s result I think its clear that the Beilinson resolution is one of the key ideas behind it.

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In my early readings these were the most prominent results I came across:

- If is an Abelian variety then the Poincare bundle gives an isomorphism .
- (Orlov’s Result) Any functor which is full, faithful and exact is represented by an object on the product.
- If or its inverse is ample then determines .
- For Abelian varieties ,

The proof I plan to outline depends on the following results

1. If and then . See cohom of pic zero in this post.

2. ( Adjunction )For and set

.

Then represent left and right adjoints to respectively.

pf:

this isomorphism comes from the fact that and Grothendieck-Verdier duality:

Let be a morphism of smooth schemes over a field (lets say algebraically closed). There is an isomorphism

.

continuing the hom isomorphisms…

. QED.

3. if fully faithful iff

=

4. (purely category theory) If is a fully faithful, exact functor between triangulated categories and contains objects not isomorphic to and is indecomposable

roughly a triangulated category is decomposable if there subcategories s.t. $\forall O \in ob(C) \exists$ a distinguished triangle with and ; also both subcateogories have to contain objects not isomorphic to 0.

Then is an equivalence iff has left and right adjoints $G,H$ and for , .

A proof of this can be found in Huybrechts book on the Fourier-Mukai transform. This result gives away the idea of the proof: use algebraic geometry results to check the hypothesis of this assertion in the case at hand.

Putting it all together.

Using 2 it is easily checked that hence is exact with left and right adjoints.

Showing that is indecomposable is more category theoretic and I’m skipping that for now.

Unwrapping the definitions,

=

, =

so is fully faithful using 3 and 1.

I’m putting some details for Orlov’s result in a whole other post.

I don’t know how to prove result 3, but I know a few important points regarding its proof. The triangulated consists of all the categorical data plus the data of a Serre functor given by .

Central to the proof are the notions of point like objects and invertible objects. is point like if

- for
- is a field.

An object is invertible if point like $\exists n(L,P) \in \mathbb{Z}$ such that .

My understanding is roughly is picking out this information allows you to pick out and consequently the canonical ring which determines .

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But I wanted to add some notes here. At some point I reference this post about chapter 17 in Polishchuk.

Extensively in the proof base change is used. This says if there is a diagram

where and if flat and is proper then there is an isomorphism

as are flat they don’t need to be derived (Huybrechts pg 85). As an application consider three schemes and a map . Note is flat. Let and consider .

This gives . Also . So

(proj. form.)

This is the result for

There are similar stories for

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