Archive for the ‘Examples/exercises’ Category

Another example

December 17, 2008

example-polytopes-1
The purple box is one polytope colored two different ways. In general a polytope is colored according to the following scheme:

  • Light blue dots are from the previous polytope.
  • The red dot is the “moving” dot.
  • Dark Blue dots come from a reflection (connected by a blue arc to its preimage).
  • Green dots are added so the shape will be “Pseudo-Weyl”.

Where “Pseudo-Weyl”, in this case, means that non-vertical edges have integer slope and any point in the closed polytope minus its vertical lines must be included.

The process for constructing a new polytope goes in that order:

  1. Include all previous points
  2. Move the “moving point” in the desired direction
  3. Reflect any corners on the appropriate side of the reflection line and include those, and
  4. Include any additional points necessary to make it Pseudo-Weyl

Where the reflection line is placed so that the first corner before the red dot (in the direction it moved) would be reflected onto the red dot. Only corners on the side opposite the red dot get reflected.

Note: I use “reflect” loosley. In this case all reflection lines will be vertical, but when moving diagonally the reflection will be shear-reflection.

Lsl2 MV-Polytopes: Inductive Approach

December 5, 2008

The following picture has the first few levels of a crystal, like those discussed in the 2008 Kamnitzer paper, that constructs a class of MV Plytopes.

lsl2-mv-polytope-crystal1

The two examples at the top indicate how this method differentiates the elements h_2 and h_1h_1. In a sense its as though the weight in in the middle of that line is sometimes included in the polytope and sometimes not. (Though, how this plays out for more complicated partitions I don’t yet know.)

Let me describe the inductive process. For a polytope P with highest weight mu we define new polytopes F_i(P) with highest weight \mu+\alpha_i (where \alpha_i is a fundamental coroot.) F_i(P) is characterized by the fact that it is the smallest PW-Polytope containing all the weights of P as well as \mu+\alpha_i.

For example the purple box in the picture above outlines the polytopes with highest weight 2\delta. If you test, you will see that only two of them will fit into the parabola for the basic representation.

Non Parabola Verma module

December 4, 2008

Probably nothing important, just a calculation I was doing last night. In the Verma module where h_0 acts by -2 , (c by 1 and d by zero). I calculated that:

e_{-1}^nf_1^n\cdot v =ne_{-1}^{n-1}f_1^{n-1}\cdot v+\sum e_{-1}^{n-1}f_1^ih_0f_1^{n-i-1}\cdot v
=ne_{-1}^{n-1}f_1^{n-1}\cdot v+(\sum2(n-i))e_{-1}^{n-1}f_1^{n-1}\cdot v
=(-n^2)e_{-1}^{n-1}f_1^{n-1}\cdot v

So inductively, beginning with e_{-1}f_1\cdot v=-1\cdot v, none of these are zero.

We do have f_0^3e_0^3\cdot v =0, so the outline looks like \_/, a truncated cone, not a parabola.

I want to know the shapes and weights of various representations so I can determine how paths pair up to become MV-Polytopes — more on this with pictures to come this week.

Some explanation for recent matrix rank calculations

December 1, 2008

Let

L =\mathbb{C}[t,t^{-1}]
L\mathfrak{sl}_2 = L\otimes \mathfrak{sl_2}
\mathfrak{g}=hat L\mathfrak{sl}_2 = \mathbb{C}d\otimes L\mathfrak{sl}_2\otimes\mathbb{C}c
 ,[d,t^k\otimes x]=kt^k\otimes x
 ,[t^{-k_1}\otimes x,t^{k_2}\otimes y]=t^{k_2-k_1}[x,y]+k_1\delta^{k_1}_{k_2}(x,y)c

Where (,) is the killing form:(e,f)=1, (h,h)=2 and (e,h)=(h,f)=(e,e)=(f,f)=0.

To abbreviate take the convention: h_k=t^k\otimes h (similarly for e_k,f_k). (Don’t confuse it with similar notation used for higher rank Kac-Moody Algebras.)

Keeping in mind the previous post about Borels, split \mathfrak{g}=\mathfrak{n}_-\oplus\mathfrak{h}\oplus\mathfrak{n}_+ where \mathfrak{h} (the large green dot) is generated by h_0,c, and d.

Now we are ready, given a character \lambda\in\mathfrak{h}^*, to define a Verma module V^\lambda=\mathcal{U}(\mathfrak{g})\otimes\mathbb{C}_\lambda, where the product is taken over \mathfrak{b}_-=\mathfrak{n}_-\oplus\mathfrak{h} (\mathfrak{n}_- acts trivially on \mathbb{C}_\lambda and \mathfrak{h} action is given by \lambda).

What we are really interested in is a quotient V^\lambda / Q where
Q={q\otimes v | (\mathcal{U}(\mathfrak{g})\cdot q\otimes\mathbb{C}_\lambda) \cap (1\otimes\mathbb{C}_\lambda) = 0}.
This will be the irreducible representation of \mathfrak{g} (by left multiplication) of lowest weight \lambda.

Of course V^\lambda is itself a \mathfrak{g}-representation and it diagonalizes under the action of \mathfrak{h}, V^\lambda=\bigoplus_\gamma  V^\lambda_\gamma.

Let W^\lambda be the Verma module with a highest weight \lambda (i.e. with tensor taken over \mathfrak{b}^+), with a similar decomposition W^\lambda=\bigoplus_\gamma  W^\lambda_{-\gamma}. With a choice of basis for each W^\lambda_{-\gamma} and V^\lambda_{\gamma} we define a pairing \phi^\lambda_\gamma such that p\cdot q \otimes v =\phi(p\otimes w, q\otimes v) for basis elements p\otimes w\in W^\lambda_{-\gamma} q\otimes v\in V^\lambda_{\gamma}. The rank of \phi^\lambda_\gamma gives the dimension of the \gamma-weight space of the quotient V^\lambda / Q.

Homework

September 13, 2008

To familiarize myself more with the distinction between roots and coroots, I’ve been given a homework assignment.(WARNING: notation in this article differs from other notation I use, particularly the use of checks)

SL_3:

The weigth lattice (denoted, for this post only, as \check\Lambda_{SL_3}) is \text{Hom}(T;\mathbb{C}^\times) and will be generated by the following maps.

\check X_1: \left(\begin{array}{ccc}z_1 & 0 & 0\\ 0 & z_2 & 0 \\ 0 & 0 & z_1^{-1}z_2^{-1} \end{array}\right) \mapsto z_1
\check X_2: \mapsto z_2
\check X_3: \mapsto z_1^{-1}z_2^{-1}

The coweight lattice (denoted, for this post, as \Lambda_{SL_3}) is \text{Hom}(\mathbb{C}^\times;T) and will be generated by

X_1: z\mapsto \left(\begin{array}{ccc} z & 0 & 0 \\ 0 & z^{-1} & 0 \\ 0 & 0 & 1\end{array}\right)
X_2: z\mapsto \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & z & 0 \\ 0 & 0 & z^{-1}\end{array}\right)
X_3: z\mapsto\left(\begin{array}{ccc} z^{-1} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & z\end{array}\right)

The action of T on \mathfrak{sl}_3 breaks down like this:
\left(\begin{array}{ccc}z_1 & 0 & 0\\ 0 & z_2 & 0 \\ 0 & 0 & z_1^{-1}z_2^{-1} \end{array}\right)\left(\begin{array}{ccc}a & b & c\\ d & e & f \\ g & h & -a-e \end{array}\right)\left(\begin{array}{ccc}z_1^{-1} & 0 & 0\\ 0 & z_2^{-1} & 0 \\ 0 & 0 & z_1z_2 \end{array}\right) =\left(\begin{array}{ccc}a & \frac{z_1}{z_2}b & \frac{z_1}{z_3}c\\ \frac{z_2}{z_1}d & e & \frac{z_2}{z_3}f \\ \frac{z_3}{z_1}g & \frac{z_3}{z_2}h & -a-e \end{array}\right)
where z_3=z_1^{-1}z_2^{-1}.

So the roots are \check R_{SL_3}=\{\check\alpha_{i,j}=\check X_i-\check X_j\}_{i\neq j\in\{1,2,3\} }.

sorry for the mixed conventions.

Circled dots are Roots, red lines are Weyl reflections numbered Lambdas are the fundamental weights: sorry for the mixed conventions.

I determine the roots also to be R_{SL_3}=\{\alpha_{i,j}=X_i-X_j\}_{i\neq j\in\{1,2,3\}} by solving the equations:
\langle \check X_i-\check X_j,\alpha_{i,j}\rangle=2 and
\langle \check X_i+\check X_j,\alpha_{i,j}\rangle=0
(notice in the diagram that \check X_1+\check X_2 is orthogonal to \check X_1-\check X_2)

(The Coweight diagram is almost identical to that for the Weight space.)

Calculations: Extended Loop Lie Algebra

June 19, 2008

Let G be a Lie group. We are interested in the infinite dimensional Lie group LG=\text{Map}(S^1;G) where composition is done pointwise. One way to understand a group is by understanding its representations. In this particular case our interest is quickly narrowed to smooth, projective, positive energy representations. It turns out that a better way object of study is the semidirect product \mathbb{T}\tilde\times\tilde{LG} where \tilde{LG} is a particular one dimensional central extension and \mathbb{T} acts on LG by rotating loops (that is precomposing \gamma\in\text{Map}(S^1;G) with a rotation).

This thing’s Lie Algebra will be (as a vectors space) \mathbb{C}_\text{rot}\oplus L\mathfrak{g}\oplus\mathbb{C}_\text{cent}. My charge, by this Sunday, is to calculate the Lie bracket. Suffice we will consider the (dense?) subalgebra \mathbb{C}_\text{rot} \oplus \mathfrak{g}[t^{-1},t] \oplus \mathbb{C}_\text{cent}

To begin with:
\left[(z_1,0,0),(z_2,0,0)\right]=(0,0,0)
because rotation is commutative. The centeral extension is, well, central so we have:
\left[(z,t^k\alpha,w_1),(0,0,w_2)\right]=(0,0,0)
What’s left are
\left[(z,0,0),(0,t^l\beta,0)\right]
\left[(0,t^k\alpha,0),(0,t^l\beta,0)\right]

Lets start with the first. We’ll take what I’ll call the “scenic route” doing as much explicit calculation as possible.

\left[(z,0),(0,t^l\beta)\right] =ad_{(z,0)}(0,t^l\beta)
=\frac {d}{d\rho}|_{\rho=0}Ad_{\gamma(\rho)}(0,t^l\beta)
where \gamma(0)=(1,id)
and \gamma'(0)=(z,0)
e.g. \gamma(\rho)=(e^{\rho z},id)
=\frac {d}{d\rho}|_{\rho=0}\frac {d}{d\xi}|_{\xi=0} (e^{z\rho},id)(1,exp(\xi t^l\beta)(e^{-z\rho},id)
=\frac {d}{d\rho}|_{\rho=0}\frac {d}{d\xi}|_{\xi=0}(1,exp(\xi(e^{z\rho}t)^l\beta)
=\frac {d}{d\rho}|_{\rho=0}(0,(e^{z\rho}t)^l\beta)
=(0,k(e^{z\rho}t)^{k-1}\alpha\cdot te^{z\rho}z)|_{\rho=0}
=(0,kzt^k\alpha,0)

The second calculation is actually fixed for us by the particular type of central extension we’re using. I’ll explain tomorrow.

Intersection (Co)homology

April 10, 2008

Part 1: About this post

This post to be updated throughout the day today, and finished by this evening. UPDATE: Finished with pictures by this weekend.

Based on a conversation I had a few weeks ago, I thought it worthwhile to give an outline the inductive method for calculating of intersection homology I was using last year.

Briefly, we allow closed chains living in the smooth part of the stratified space, and need only conisder whether they should be allowed to “cap off” to the lower strata, which is determined inductively and based on dimension: an already allowable chain, living in the cone over a lower strata is allowed to cap down to the strata if it is the product of an allowable lower strata chain, and the cone of a link of dimension better than half the dimension of the link.

More elaboration on what that means, and some examples later today.

Part 2: Stratified spaces

We consider a topological space X=X_n\supset X_{n-1}\supset X_{n-2}\supset\dots such that

  1. each X_k is closed,
  2. X_k\setminus X_{k-1} is a manifold of dimension k, and
  3. X_{n-1}=latex X_{n-2}$.

We also may write the space in terms of open pieces U_n\subset U_{n-1}\subset\dots\subset U_0=X where U_k=X_{n-k+1}^C.

We also require that each strata M_k=X_k\setminus X_{k-1} is covered by open sets in X_{k+1} such that each open set V is of the form V\cong (V\cap M_k)\times C^o(L_k) where L (called the “link”) is a stratified space depending only on the strata (or possibly on the component of the strata) and C^o indicates the open cone ( (0,1]\times L) / (1\times L).

Part 3: Admissible (co)chains

First, any closed chain that lives entirely in the “smooth” part of our stratified space, U_n is called admissible. A chain, \eta, that intersects X_{n-2} will be called admissible if it can be written as the product \gamma\times C^o(\lambda) where \gamma=\eta\cap X_{n-2} is an admissible chain (defined inductively) for the space X_{n-2}, and \lambda is a chain in L with sufficiently large dimension (small co-dimension). Sufficiently large dimension isn’t mysterious; for most cases (the standard case I think) we require it to have half the dimension of the link.

Part 4: Eg. Banana Space

Rotating Banana Space

The banana space is the torus with one of its belts pinched to a point. So called because one way of drawing it looks like a banana bending around so its tips meet. Also you may call it a circle with two of its antipodes identified.

It has two stratum. One the singular point, and the other of dimension 2 (everything else).

The link, L, over the singular point consists of two circles (one on each side of the banana). No 1-chains can hit the singularity. Only two chains can meet the singularity.

Part 5: Eg. Three Complex 2-Planes

Next we consider the case of three complex hyperplanes complex 3-space. Or rather the one-point compactification (for technical reasons we like working on compact spaces only).

Here the “smooth part” consists of three copies of (\mathbb{C}^x)^2, the next strata consists of three copies of \mathbb{C}^x at their intersections, and the final strata consists of two points, the origin and point of compactification.

Over any point in the M_2 the link is two circles, one for each hyperplane. Once again, 1-chains cannot cross the singular stratum. Also no 2-chain can touch unless it wraps around the singular part.

[I need to dig up in my notes, I don't remember what happens near the origin.]

Part 6: Eg. Suspended 3-Torus

This example is the simplest where the link is more than one dimensional. In this case the smooth part is just a thickened three torus. The singular stratum consists of two points, one at each end of the suspension. The link around either of these points is simply a 3-torus.

We allow ourselves to cap off a chain in the 3-torus only if its dimension is 2 or 3. In other words a 1-chain (which is the cone of a 0-chain) is not allowed, only certain 3-chains (the cones of 2-chains) and 4-chains are allowed.

When we take the (co)homology of the resulting intersection (co)chain complex we will get:

0 0-cycle \mathbb Z
1 \mathbb Z^3
2 0
3 \mathbb Z^3
4 \mathbb Z

Some calculations

March 19, 2008

Sorry if this is boring to those who hate calculations, or a spoiler for those who need to do them (like me). But I want to post something so here’s one calculation I’ve finished and another that I’m still puzzling through. These are both from Loop groups chapter 4.

Proposition 4.3.2 says:

The adjoint action of L\mathfrak{g} on \tilde{L\mathfrak{g}} comes from an action of LG given by

\gamma . (\xi,\lambda)=(\gamma . \xi,\lambda - \langle\gamma^{-1}\gamma',\xi\rangle ).

Here $latex\gamma . \xi$ denotes the adjoint action of \gamma\in LG on \xi \in L\mathfrak{g}.

The book says verifying this is a group action is straightforward. I agree, having done it, that it is straightforward, but it is not obvious – and the formula anyway seems rather mysterious. But here we go.

Let me recall:

  • G is a Lie group, L\mathfrak{g} is the Lie algebra of its loop group.
  • We are constructing a Lie algebra \tilde{L\mathfrak{g}}=L\mathfrak{g}\oplus\mathbb{R} with the bracket: \left[(\xi,\lambda),(\eta,\mu)\right] =([\xi,\eta ],\omega (\xi,\eta )) (notice that \mathbb{R} is central in the sense of not contributing to the bracket.)
  • \omega:L\mathfrak{g}\times L\mathfrak{g}\rightarrow\mathbb{R} is given by (WLOG in some sense):\omega(\xi,\eta)=\frac{1}{2\pi}\int^{2\pi}_{0}\langle\xi(\theta),\eta'(\theta)\rangle d\theta
    derived from a symmetric invariant form \langle\cdot,\cdot\rangle on \mathfrak{g}. (Invariant means that  \langle g.\xi,g.\eta\rangle=\langle\xi,\eta\rangle.)
  • (We also use \langle\cdot,\cdot\rangle to denote a form on L\mathfrak{g} which is the average over loops of the form on \mathfrak{g}.)

OK, so to my calculation: I want to verify that

(\alpha\beta).(\xi,\lambda)=\alpha.(\beta.(\xi,\lambda))

On the first component this is obvious, the second component is the one where anything interesting happens.

Begining by writing out complicated bit from the left hand side we have:

  \lambda - \frac{1}{2\pi}\int_0^{2\pi}\langle \beta(\theta)^{-1}\alpha(\theta)^{-1} (\alpha(\theta)\beta'(\theta) +\alpha'(\theta)\beta(\theta)) ,\xi(\theta)\rangle d\theta
= \lambda - \frac{1}{2\pi}\int \langle \beta(\theta)^{-1}\alpha(\theta)^{-1}\alpha(\theta)\beta'(\theta),\xi(\theta)\rangle d\theta
    - \frac{1}{2\pi}\int\langle \beta(\theta)^{-1}\alpha(\theta)^{-1}\alpha'(\theta)\beta(\theta),\xi(\theta)\rangle d\theta
= \lambda - \frac{1}{2\pi}\int \langle \beta(\theta)^{-1}\beta'(\theta),\xi(\theta)\rangle d\theta
    - \frac{1}{2\pi}\int\langle \alpha(\theta)^{-1}\alpha'(\theta),\beta.\xi(\theta)\rangle d\theta

which is exactly the complicated bit on the right hand side.

The calculation is straightforward, boring even — but not obvious to me until I actually write it out. This book is full of things like that. If I write it out just a little but more, that is something I can follow.

Here is another such case I found today but have not managed to untangle yet:
We are considering the map H^3(G)\rightarrow H^3(S^1\times \Omega G) given by pulling back the evaluation map S^1\times\Omega G\rightarrow G. (\Omega G is based loops.) We define \sigma to be a left invariant 3-form on G whose value at the identity is given by \sigma_{id}(\xi,\eta,\zeta)=\langle[\xi,\eta],\zeta\rangle. When we pullback this form and evaluate it at (\theta,\gamma) on the vector (\delta\theta,\delta_1\gamma,\delta_2\gamma) we get, according to the book,
\frac{1}{4\pi}\langle\gamma(\theta)^{-1}\gamma'(\theta),[\gamma(\theta)^{-1}\delta_1\gamma(\theta),\gamma(\theta)^{-1}\delta_2\gamma(\theta)]\rangle\delta\theta
This calculation mystifies me more than the first.

D_\gamma for LSL_3 Part one: Lifting the Weyl group

February 17, 2008

Let me recap D_\gamma a bit. Let \gamma=w\cdot\Lambda_i where w is an element of the Weyl group and \Lambda_i is a fundamental weight. Before calculating D_\gamma we’ll need to choose weight vectors v_\gamma\in V_\gamma such that v_{w\cdot\gamma}=\bar w\cdot v_\gamma where \bar w indicates the lift of w.

For SL_3 W is basically the set of permutation matrices only I feel like there is a trouble with signs. Ignoring signs for now think of it as generated by

\bar{s_1}=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right)

and

\bar{s_2}=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right).

In that case we take

v_{\Lambda_1}=\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right) v_{\Lambda_2}=\left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right)\wedge\left(\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right).

(Note/check that \bar {s_i} v_{\Lambda_j}=\pm v_{\Lambda_j} when j\neq i.)

This information is more completely presented in a set of diagram I have in my notes — The fastest way to get it up will probably be to scan it on Monday.

I’ve made a similary diagram for W_{\text{aff}}. The lifts are much the same as for the finite case only there will be t’s in places. What I don’t have nailed down yet is the v_{w\cdot \Lambda_3} vectors: I don’t even know where they live.

UPDATE: Two of the Scans I promised — (a) the Weyl diagram for SL_3 and (b) the Diagram for the affine Weyl group with my guess at appropriate matrix representations (once again ignoring sign issues).

(a)sl3w.png

(b)lsl3wm.png

D_γ for SL_4

January 7, 2008

The D_\gamma data for SL_3 come in two sets of three, one set for each fundamental weight. For a fixed set of values for D_\gamma the elements of the affine grassmanian corresponding to that data will be the “balance towers” that lie between the “pure towers” described by those two sets.

For SL_2 there’s only one set of two. We can still get two towers, but these will both be described by the same set which is self-dual.

When we move to SL_4 we start getting more intermediate data. We still have the “level 1″ set in the data that describes an outer tower and a a level 3=n-1 set in the data that describes an inner tower. But now we have additional data. I’d like to understand the additional restrictions this set (and further middle sets for higher n) will put on towers.

So far the one thing I’ve noticed is that ignoring a column of the tower (and alowing any parts leaning into that portion to “stand up”) We get a tower like those for SL_3 but not necessarily balanced. There’s a subset of the D_\gamma that can be translated into data about this SL_3\text{-tower}.

Let me take some notation. Let r_i denote row vectors and c_i denote column vectors of a representative in SL_4(\mathscr{K}) of an element in \mathscr{G}r. The r_i are the generators of the subspace represented by our tower. valuations of the c_i and their exterior products for our D_\gamma data. What is not an official part of the data is the valuation of the determinant or the exterior product of all columns.

When we eliminate one of the columns as suggested, we will have 4 rows to generate a tower only three wide, so one of the rows will become superfluous. I argue that the valuations of the wedge products of pairs of 3\text{-vectors} will be unchanged despite the elimination of this row. Its because of this that I say the middle data arising in SL_4 describe these related unbalanced towers’ inner parts. Clarifying exactly how that describes the original tower is one of my current goals.