User dinakar muthiah - MathOverflow

Quotients of Schemes by Free Group Actions

2009-10-21T02:51:22Z

I've often seen people in seminars justify the existence of a quotient of a scheme by an algebraic group by remarking that the group action is free. However, I'm pretty sure they are also invoking something else. So my question is: when you can you quotient a scheme by a free action and get a scheme? In particular, when do the coset spaces of a subgroup of an algebraic group exist as a scheme? And in these cases, how do you construct the quotient?

Inverting the Weyl Character Formula

2010-04-02T20:27:25Z

The Weyl Character formula tells us how to write the character of a representation as a linear combination of integral weights. Since characters are invariant under the action of the Weyl group, $W$, we can write a character as a linear combination of $W$-symmetrized dominant integral weights. It is know that the representation ring of a Lie algebra is isomorphic to $\mathbb{C}[P]^W$ as a vector space, where $P$ is the weight lattice of some Cartan subalgebra.

So we have two bases for the same vector space: the $W$-symmetrized dominant integral weights and the the character basis. The Weyl character formula tells us how to go from the former to the latter. My question is: is there much known about the matrix of going from the latter to the former? I've gone through a few low rank examples, and many of the coefficients are coming out to be zero. Does anyone know of a reference for this question in general?

Addendum: Jim makes a good point. The Weyl character formula isn't really needed. Perhaps we should just say that the matrix from the weight basis to the character basis is precisely the matrix of weight multiplicities. From this point of view it is clear that the matrix will be "upper triangular" (since weight multiplicities are zero above the highest weight). Thus the inverse should also be upper triangular. So my modified question is there any way to interpret the coefficients of the inverse matrix as counting anything interesting? As the matrix is upper triangular, we can certainly give recursive formulas for the coefficients. Does anyone have any other insight?

Longest Element of an Affine Weyl Group

2009-10-26T16:20:46Z

I know that the Weyl groups of affine Lie algebras don't have a longest element, but are there any good substitutes for w_0. In particular, is there any good substitute for a reduced decomposition of the longest element?

Comparing colimits in schemes with colimits in sheaves of sets

2011-02-14T01:43:23Z

Suppose I have a diagram of schemes, and I know that the colimit exists in the category of schemes. How does this colimit compare with the colimit of the corresponding sheaves (I'm being nonspecific about the topology on purpose)? We always have a map from the colimit of sheaves to the colimit of schemes. Are then any conditions I can impose on my diagram so that this map is an isomorphism? Is there any reference where this issue is discussed?

Constructing Affine Kac-Moody Groups

2009-10-28T14:55:36Z

Does anyone know a simple construction for Affine Kac-Moody groups? There is a book by Kumar ("Kac-Moody groups, their flag varieties, and representation theory") that does the construction for the general Kac-Moody case, but I find the presentation dense. There is also a section that constructs a one-dimensional extension of the loop group by loop rotation, which is a fairly transparent definition. However, I don't know how to add on the final central extension.

Even if the answer to my question is "There is no simpler construction," could someone also tell me about a fruitful way to get my hands on Affine Kac-Moody groups?

Intersection Cohomology of Coordinate Hyperplanes

2010-11-27T13:06:48Z

I'm trying to learn how to compute stalks of IC sheaves, and I was wondering about the following example:

Fix $n$. Let $X \subset \mathbb{C}^n$ be the variety cut out by the equation $x_1 \cdots x_n =0$, i.e. the coordinate hyperplanes. What are the stalks of $\mathrm{IC}(X)$ at the various points of $X$, in particular at the origin?

This seems like a natural toy example, but if the general answer is difficult, I'd be happy to know how to compute this for small $n$.

Intuition behind moduli space of curves

2009-12-04T17:07:32Z

For a genus g compact smooth surface $M$, an algebraic structure is the same as a complex structure is the same as a conformal structure. So the moduli space of smooth curves should be the same as the moduli space of conformal structures on $M$. A conformal structure is an equivalence class of Riemmanian metrics that give the same angle measurements.

If I embed $M$ in $\mathbb{R}^3$, I get a metric on $M$. This gives me a conformal structure, hence a point in the moduli space. The moduli space is known to have complex dimension 3g-3 (except for g=0, where the moduli space is a point, and g=1, where the moduli space is 1-dimensional).

My question is: can we visualize the 6g-6 real dimensions as deformations of the embedding of $M$ in $\mathbb{R}^3$. In particular, how can we see that small deformations of a 2-sphere are conformally equivalent to the original 2-sphere.

Why is Maps(X,Y) an open subfunctor of Hilb(X x Y)?

2010-09-29T18:51:42Z

Let $X$ and $Y$ be projective schemes. Then we can define the mapping scheme between them, $\rm{Maps}(X,Y)$ as follows:

To any map $f:X\rightarrow Y$ we consider the graph $\Gamma_f$ as a closed subscheme of $X \times Y$. So $\rm{Maps}(X,Y)$ is the set of all subschemes of $X \times Y$ that are graphs of morphisms. (Concretely, a subscheme $Z \subset X \times Y$ is the graph of a morphism iff the projection to $X$ is an isomorphism) Of course this all makes sense in families, so $\rm{Maps}(X,Y)$ is a subfunctor of the Hilbert scheme $\rm{Hilb}(X \times Y)$.

Now at this point, I have seen a number of sources casually claim that $\rm{Maps}(X,Y)$ is actually an $\it{open}$ subfunctor and is hence representable. None of these sources even remark on why this is true? So my question is: why is this true?

Why is homology not (co)representable?

2009-10-20T15:09:52Z

This is in the same vein as my previous question on the representability of the cohomology ring. Why are the homology groups not corepresentable in the homotopy category of spaces?

Answer by Dinakar Muthiah for What are your favorite instructional counterexamples?

2010-03-03T02:54:55Z

The Cantor set is a nice source of counterexamples:

The first measure zero sets you meet are usually countable. However, the Cantor set is uncountable and measure zero.

It is totally disconnected, yet it is not a discrete space. In particular, this shows that connected components of a topological space need not be open sets.

Answer by Dinakar Muthiah for How to Compute the coordinate ring of flag variety?

2010-05-04T15:37:18Z

$G/B$ is most naturally a multi-projective variety, embedding in the product of projectivizations of fundamental representations: $\prod \mathbb{P} (R_{\omega_i})$. So there is a multi-homogeneous coordinated ring on $G/B$. You mentioned that this ring is $\bigoplus_{\lambda\in P_+}$ $R_\lambda$. This is correct, and the grading is also apparent. It's given by the weight lattice. (To be more canonical, you should take the duals of every highest weight representation, but what you've written down is isomorphic to that.)

So all that remains is giving the multiplication law on $\bigoplus_{\lambda\in P_+}$ $R_\lambda$. You need to specify maps $R_\lambda \otimes R_\mu \mapsto R_{\lambda+\mu}$. There's a natural candidate: If you decompose $R_\lambda \otimes R_\mu$ into a direct sum of irreducible representations, $R_{\lambda+\mu}$ will appear exactly once. The multiplication law is simply projection onto this factor.

@Shizuo: For $sl_n$, the situation is more explicit. The flag variety here is the set of flags $0 = V_0 \subset V_1 \cdots V_{n-1} \subset V_n = \mathbb{C}^n$ with $\mathrm{dim} V_i = i$. So the flag variety is a closed subvariety of the product of Grassmannians $Gr(1,n) \times \cdots Gr(n,n) $. Each of these Grassmannians have a explicitly Plucker embedding into the projectivization of the exterior power of $\mathbb{C}^n$. In particular, the homogeneous ideal is explicitly given by the Plucker relations. So the multi-homogeneous coordinate ring of $Gr(1,n) \times \cdots Gr(n,n) $ is just the tensor product of the known homogeneous coordinate rings.

Finally, to get the multi-homogeneous coordinate ring for the flag variety, we need to specify an incidence locus inside $Gr(1,n) \times \cdots Gr(n,n) $. Namely, we need to specify those tuples of subspaces that form a flag. But this is easy: it corresponds to certain wedge products being zero. Just impose those additional relations, i.e. mod out by the corresponding multi-homogenous ideal. Now you should have an explicit, albeit fairly long description of the homogeneous coordinate ring. The above answer for $SL_3$ looks like a special case of this construction.

Pushouts in the Category of Schemes

2009-11-12T03:01:57Z

When does it make sense to glue schemes together along subschemes?

In particular: is there a way to glue two schemes together along a closed point (say we're working over a field)? Can you glue two closed points of the same scheme together? Is it easier to glue in the category of algebraic spaces?

Line Bundles on Torus Quotient

2010-03-26T18:47:30Z

Suppose you have a scheme $X$ that is acted on by a torus $T$. Then the action induces a grading on the functions on $X$ by the character lattice of $T$. So for a fixed character $\lambda$, we can consider $\mathcal{O}_{X,\lambda}$, the $\lambda$ graded part. Assuming the quotient $X/T$ exists, these graded parts should descend to quasicoherent sheaves on the quotient.

My question is, when are these sheaves line bundles?

In the basic examples I know, they are always line bundles. For example, if you take $X= \mathbb{A}^{n+1} - 0$ and $T = \mathbb{C}^*$, then on $X$ you get the ordinary grading by homogeneous degree. When you descend to the quotient, you get the line bundles $\mathcal{O}(k)$ on $\mathbb{P}^n$. You can also take $G$ a complex semi-simple group, $B$ a Borel subgroup, $U$ the maximal unipotent. Then $G/U \rightarrow G/B$ is a torus quotient, and the graded pieces descend to line bundles. I think, a similar story is true for all homogeneous spaces, but I'm having a little trouble phrasing it in terms of torus quotients.

In fact, in these situations, these are all the line bundles.

So more generally, my question is, what properties can you require of the general $X$ so it behaves like the two examples above?

When is tensoring with a module representable by a scheme?

2009-11-25T04:53:01Z

Consider the following: Let $A$ be a commutative ring, let $M$ be an $A$-module. When is the functor from $A$-algebras to Sets given by $R \mapsto R \otimes M$ representable by an $A$-scheme?

Unless I've made a mistake, this is always be an fpqc sheaf. When $M$ is a finitely generated free A-module, then $\mathrm{Spec}( \mathrm{Sym}^\bullet M^*)$ does the trick.

How Does a Borel Subgroup Know Which Weights Are Dominant

2009-11-11T03:40:24Z

Let $G$ be a simple group (say $SL_n$) and let $B$ be a Borel subgroup (say upper triangular matrices). Then all irreducible representations of $G$ are induced from one-dimensional representations of $B$, i.e characters of $B$. However, only some of the weights will induce to non-zero representations. Relative to the standard maximal torus of diagonal matrices, these weights appear to be the anti-dominant weights. However, $B$ contains other tori as well, and no one choice is canonical.

My question is: since induction from $B$ didn't require a choice of torus, how are some characters of $B$ already deemed to be anti-dominant? Where is the symmetry broken?

Answer by Dinakar Muthiah for What are some fundamental "sources" for the appearance of pi in mathematics?

2010-03-14T18:01:08Z

As for the normal distribution, you can characterize it as the unique distribution with the following properties:

Let $X_1, X_2, \cdots X_n$ be independent identically distributed normal random variables. Then the joint distribution of the vector $X=(X_1, X_2, \cdots X_n)$ is the same as that of $AX$ where $A$ is any orthogonal matrix. So the normal distribution is intimately related to the geometry of real inner product spaces.

The $\pi$ comes from the fact that you can integrate such a distribution by first integrating over a sphere and then integrating over $[0,\infty]$. Because the distribution is orthogonally invariant, you pick up a constant corresponding to the area of the sphere. For $n=2$ you get the circle, and this is the usual calculation for computing the normalization constant for the normal distribution.

So then the mystery becomes: given that the normal distribution is so closely tied to inner product spaces, why does it show up all the time? The central limit theorem tells us that all that really matters in large scale limits are the first and second moments. The first moment can always be eliminated by re-centering. So all that matters is the second moment. But the second moment comes from the covariance, which is an inner product! (technically, only once you restrict to re-centered random variables, but we are doing that)

I'd venture a guess that most, if not all, appearances of $\pi$ in statistics boil down to this fact that covariance is an inner product, and the fact that spheres, which are the norm-level sets for inner product spaces, have areas related to $\pi$

Answer by Dinakar Muthiah for Expository treatment of Schubert Cells Paper

2010-03-08T05:56:34Z

The Lecture Notes in Mathematics number 1689, "Schubert Varieties and Degeneracy Loci" by Fulton and Pragacz seems to be exactly what you're looking for. I think chapter 6 is particularly relevant.

Answer by Dinakar Muthiah for Quasi-coherent sheaves in the Functor-of-points approach

2010-02-25T19:06:12Z

You can define a quasicoherent sheaf on a functor $X : \mathrm{AffSch^{op}} \rightarrow \mathrm{Set}$ as a choice of a module $R$ module $F_x$ for every $x \in X(\mathrm{Spec}R)$ along with some compatibility isomorphisms. If $X$ is the functor of points of scheme, and $F$ is a an honest quasicoherent sheaf on this scheme, then $F_x$ is just the pullback to $\mathrm{Spec} R$ via the map $x$. The compatibility isomorphisms that we require are the ones that naturally arise from pseudo-functoriality of the pullback. The details are given in the second page of the following notes of a lecture by Jacob Lurie http://www.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/Nov17-19(Crystals).pdf.

Answer by Dinakar Muthiah for Various concepts of "closure" or "completion" in mathematics

2010-02-07T17:40:33Z

Projection operators on a linear space are precisely the idempotents. All these other examples are somewhat like linear projections in that they are projecting from a category to a subcategory.

Reference for Tate vector spaces

2010-01-29T01:35:27Z

... aka locally linear compact vector spaces. The one reference I know is http://www.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/Nov3-10(CentExt).pdf. Does anyone know another good reference?

How to do Computations Using the Decomposition Theorem for Perverse Sheaves

2009-10-21T15:10:33Z

This is a follow-up to this post on the Decomposition Theorem. Hopefully, this will also invite some discussion about the theorem and perverse sheaves in general.

My question is how does one use the Decomposition Theorem in practice? Is there any way to pin down the subvarieties and local systems that appear in the decomposition. For example, how do you compute intesection homology complexes using this theorem? Does anyone have a link to a source with worked out examples?

Another related question: What is the deep part of the theorem? Is it the fact that the pushforward of a perverse sheaf is isomorphic to its perverse hypercohomology? Is it the fact that these pieces are semisimple? Or are these both hard statements? And what is so special about algebraic varieties?

When is a commutative ring the limit of its local rings?

2009-12-02T05:32:02Z

Let $A$ be a commutative ring. Then we get local rings $A_p$ by localizing at each prime ideal $p$. Moreover, we get $A_p \rightarrow A_q$ when $p$ contains $q$. So we get a big diagram indexed by the inclusion poset of prime ideals. When is $A$ the limit of this diagram?

When $A$ is a local ring or an integral domain it's true. I don't see any reason why it should be true for arbitrary rings. What's going on here?

Answer by Dinakar Muthiah for Heuristic explanation of why we lose projectives in sheaves.

2009-12-01T18:52:58Z

One reason is that surjectivity of a map of sheaves is a weaker condition than surjectivity of a map of presheaves. For a map of sheaves to be surjective, it need only be surjective on stalks.

Recall the definition of a projective sheaf $\mathcal{P}$: Suppose $\mathcal{N} \rightarrow \mathcal{M}$ is a surjective map of sheaves and $\mathcal{P} \rightarrow \mathcal{M}$ is a sheaf map. Then we require that there exists a lifting $\mathcal{P} \rightarrow \mathcal{N}$ making the obvious diagram commute. Because of the definition of surjectivity for sheaves, there's probably an open set $U$ for which the map $\mathcal{N}(U) \mapsto \mathcal{M}(U)$ isn't surjective. So if $\mathcal{P}(U)$ doesn't map into the image, then there is no hope for a lifting. In all but the trivial cases (like discrete spaces), it will be easy to cook up a map $\mathcal{N} \rightarrow \mathcal{M}$ to do this.

For presheaves, surjectivity means surjectivity on each open set, so this problem doesn't happen. But presheaves as an abelian category aren't very interesting. For example, the strictness of surjectivity means there is no cohomology.

Answer by Dinakar Muthiah for Thorough Introduction to Singular Value Decomposition

2009-11-27T22:46:06Z

This course at Stanford covers singular value decomposition in lectures 15-17. The notes are very good, and the lectures are online too.

Total Spaces of Quasicoherent Sheaves

2009-10-29T18:57:19Z

You can construct a total space of a quasicoherent sheaf on an scheme by taking relative spec of the symmetric algebra of the dual sheaf. For locally free sheaves, you get vector bundles, and every vector bundle arises this way. What about sheaves that are not locally free? Are there any other sheaves for which the total space is a useful construction?

Answer by Dinakar Muthiah for Theorems for nothing (and the proofs for free)

2009-11-13T16:16:41Z

Faithfully-flat descent:

It tells you that you can construct quasicoherent sheaves locally on a faithfully-flat cover. This is pretty amazing, because quasicoherent sheaves are, a priori, only Zariski local. So to specify a sheaf it requires a lot less data than it initially appears.

Answer by Dinakar Muthiah for Notions of Matrix Differentiation

2009-11-12T19:01:10Z

Since you say you are doing an undergraduate research project, I think the following document, The Matrix Cookbook, might be useful for you. There is a whole section devoted to computations of matrix derivatives. There is no deep mathematics going on, but it's a great reference.

Littlewood-Richardson-Type Rule for Cohomology Ring of Grassmannians

2009-11-09T17:15:49Z

The ordinary Grassmannian of k-planes in n-space is a coset space for $GL_n$. It is $GL_n$ mod a maximal parabolic. Here there is a nice basis given by Schubert varieties, which can be indexed by Young diagrams that fit in an (k)x(n-k) box. The structure constants for the cup product are then given by Littlewood-Richardson numbers.

My question: is there a similarly nice picture for Grassmannians of arbitrary simple groups. Here the ordinary Grassmannian is replaced by $G/P$ where $G$ is a simple group and $P$ is a maximal parabolic. There are still Schubert varieties in this case, but I don't know how to say anything about the cup product.

How to distinguish between natural and unnatural equivalences of categories

2009-11-05T00:46:18Z

Some equivalences of categories are constructed by explicitly giving a pair of functors that are inverses up to isomorphism. For example, the equivalence between CRing^op and affine schemes is given by the pair (Spec, GlobalSections). I'd say these are "natural", since no choices are made.

Another equivalence of categories is between finite dimensional vector spaces and the category consisting of one vector space of each dimension. The functor in one direction is just the inclusion, but the inverse requires making a bunch of choices. I'd say this is "unnatural".

But my definitions of "natural" and "unnatural" aren't precise. I suppose one of the triumphs of category theory has been the ability to make precise the definition of natural in some contexts. So my question is: how can I make this precise?

Answer by Dinakar Muthiah for equivalence of Grothendieck-style versus Cech-style sheaf cohomology

2009-11-05T05:14:08Z

The problem with Cech cohomology is that even if things are acyclic on open sets of your Cech cover, they may not be when you restrict to intersections of those open sets. The usual fix is to make the cover finer so you don't have that problem. Unfortunately there are topological spaces where no cover will be good enough. That's the bad news.

The good news is that for a lot of spaces and categories of sheaves you're interested in, there will be such a cover. My favorite example is the category of quasi-coherent sheaves on a separated scheme. Then Cech cohomology computed on any affine cover will compute the derived functor cohomology.

The even better news is that there is a way to fix Cech cohomology so that it will work for all situations. This is Verdier's theory of hypercovers, and it computes derived functor cohomology for any category with a Grothendieck topology. I must admit I have not played around much with this, but here is a link to a paper that talks about this circle of ideas.

Comment by Dinakar Muthiah

2010-11-27T18:35:26Z

Can you elaborate on the argument in part 1? I'm sorry to belabor it, but I think I need it spelled out to me.

Comment by Dinakar Muthiah

2010-04-16T19:53:58Z

Great !

Comment by Dinakar Muthiah

2010-03-18T19:33:23Z

This is homework

Comment by Dinakar Muthiah

2010-03-10T06:59:31Z

@Brian: How about this construction of C: Let V be a 2-dim real vector space with an inner product. Pick a pair of orthogonal lines. Then there are exactly two operators on V that preserve the inner product, have positive determinant, and swap the two lines. The algebra generated by these two operators and the identity operator is an algebraic closure of R, but neither of the square roots is special.

Comment by Dinakar Muthiah

2010-03-10T05:56:41Z

I'm confused. How can you orient C without choosing a square root of -1?

Comment by Dinakar Muthiah

2010-03-10T00:56:16Z

You can always get the counterexample from your old question by setting Y=Z, setting f to be the identity map, and X an open subset of Y that isn't closed.

Comment by Dinakar Muthiah

2010-03-07T05:51:19Z

From Ravi Vakil's lecture notes: "Spectral sequences are a powerful book-keeping tool for proving things involving complicated commutative diagrams. They were introduced by Leray in the 1940's at the same time as he introduced sheaves. They have a reputation for being abstruse and difficult. It has been suggested that the name `spectral' was given because, like spectres, spectral sequences are terrifying, evil, and dangerous. I have heard no one disagree with this interpretation, which is perhaps not surprising since I just made it up." ;)

Comment by Dinakar Muthiah

2010-02-26T22:44:46Z

Do you have a reference for this?

Comment by Dinakar Muthiah

2010-02-14T04:08:15Z

Hey, whatever works. It is certainly elegant in its simplicity.

Comment by Dinakar Muthiah

2010-02-12T08:35:18Z

Could you tell me a reference for the Nakajima construction?

Comment by Dinakar Muthiah

2010-01-29T16:16:28Z

Thanks. !

Comment by Dinakar Muthiah

2010-01-29T01:42:24Z

Oops. I didn't realize this version was submitted. Can someone with enough rep delete this when they get a chance.

Comment by Dinakar Muthiah

2009-12-28T08:35:01Z

Wow, that is really beautiful.

Comment by Dinakar Muthiah

2009-12-26T17:27:43Z

This is a great question, but you should make it community wiki because there is no right answer.

Comment by Dinakar Muthiah

2009-12-19T22:56:20Z

Wow !