Archive for the ‘Uncategorized’ Category

A remark on a problem from my undergrad

November 6, 2009

Sometime, I think early, in my years at University of Michigan. I discovered somehow the following phenomenon.

Begin with any positive integer a_1=a_2 then define for n\ge 2 let a_{n+1}=a_n+c_n where a_n=b_n\times n+c_n, 0\le c_n  <n.  If ever b_n=c_n then b_N=c_N=b_n, \forall n>N In fact, it would seem that, no matter which number you start with this will always happen.

For example if you begin with a_1=1 it takes 397 steps to stabalize at a_397=97\times 397 + 97. On the other hand if you begin with a_1=3 then a_2=1\times 2+1 so a_3=1\times 3 + 1 etc. In otherwords, sometimes it may happen soon, sometimes it may take rather a long time.

Only recently I realized that the possibility that this always stabilizes should not be very surprising. Suppose that prior to stability the value \frac{c_n}{n} behaved randomly. If b_n<n then there would be a \frac1n chance that c_n=b_n. Since a_n grows by no more than n, at some point we must have a_N<N^2 and so b_n<n, \forall n>N. At that point, if we haven’t already stabilized the probability that we eventually stabilize is \sum_{n=N}^\infty \frac1n\prod_{m=N}^{n-1}\frac{m-1}{m}. Noting that the partial sums of this series are \frac {M}{N+M-1} we deduce that the limit is equal to 1. In otherwords the probability of never stabilizing is zero.

Second Calculation level=rank=2

August 8, 2009

This time we will try to find the quiver variety components corresponding to the four figures:
4Lsl2basiselements
Found in the level two \mathfrak{sl}_2 representation weight diagram:
Lsl2example

If my level-rank duality calculations are correct (very similar to those from last post) the dimension vectors will be \mathbf{w}=(1,1),\mathbf{v}=(2,2).

As before we have a_1\neq 0 \neq b_2 and a_2=0=b_1 by the “limit” condition. Thus we can again deduce that x\overline x=\overline y y and \overline x x=y\overline y by the \mu^{-1}(0) condition. The limit condition additionally tells us that each of these compositions, and the composition \overline x\cdot \overline y, must be nilpotent. This might be sufficient to satisfy the limit condition — technically we should require that any loop involving at least one bar must be nilpotent.

So the only thing left to answer is the “stability” condition. I’m still working on this but I believe there are two basic ways the stability could be satisfied:(1) there is a path beginning with a_1 and ending at b_2 such that the partial paths give a basis for the whole space (see below for an example) or (2) there are two paths from a_1 to b_2 who together span the whole space.

Example: one way to satisfy the stability condition is if a_1(1),xa_1(1),yxa_1(1) and xyxa_1(1) form a basis for V and if b_2xyxa_1(1)\neq 0. In that case I have used the sage notebook (sagenb.org) to apply the composition and nilpotency conditions to get four subcases:

Subcase 1:

x=\left(\begin{matrix}0 & 1\\1 & 0\end{matrix}\right), \overline{y}=\left(\begin{matrix}\overline y_{11} & 0\\-\overline y_{11}y_{21} & 0\end{matrix}\right)
y=\left(\begin{matrix}0 & 0\\y_{21} & 1\end{matrix}\right), \overline{x}=\left(\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right)

Subcase 2:

x=\left(\begin{matrix}0 & 1\\1 & 0\end{matrix}\right), \overline{y}=\left(\begin{matrix}\overline y_{11} & \overline y_{12}\\ \overline y_{12} & 0\end{matrix}\right)
y=\left(\begin{matrix}0 & 0\\ 0 & 1\end{matrix}\right), \overline{x}=\left(\begin{matrix}0 & 0\\ 0 & \overline y_{12}\end{matrix}\right)

Subcase 3:

x=\left(\begin{matrix}0 & 1\\ 1 & 0\end{matrix}\right), \overline{y}=\left(\begin{matrix}\overline y_{11} & \frac{1}{2}\overline y_{11}y_{21}\\ -\frac{1}{2}\overline y_{11}y_{21} & -\frac{1}{4}\overline y_{11}y_{21}^{2}\end{matrix}\right)
y=\left(\begin{matrix}-\frac{1}{4}y_{21}^{2} & 0\\ y_{21} & 1\end{matrix}\right), \overline{x}=\left(\begin{matrix}-\frac{1}{8}y_{21}^{3}\overline y_{11} & -\frac{1}{4}\overline y_{11}y_{21}^{2}\\ \frac{1}{4}\overline y_{11}y_{21}^{2} & \frac{1}{2}\overline y_{11}y_{21}\end{matrix}\right)

Subcase 4:

x=\left(\begin{matrix}0 & 1\\ 1 & 0\end{matrix}\right), \overline{y}=\left(\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right)
y=\left(\begin{matrix}y_{11} & 0\\ y_{21} & 1\end{matrix}\right), \overline{x}=\left(\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right)

Where the basis for writing the matrices is chosen to be the basis mentioned above.

On the other hand if we consider modules for which the path x\overline x x generates a basis we only get two subcases:

Subcase 1:

x=\left(\begin{matrix}0 & 1\\ 1 & 0\end{matrix}\right), \overline{y}=\left(\begin{matrix}\overline y_{11}\neq0 & 0\\ 0 & 0\end{matrix}\right)
y=\left(\begin{matrix}0 & \frac 1 {\overline y_{11}}\\ \frac 1 {\overline y_{11}} & y_{22}\end{matrix}\right), \overline{x}=\left(\begin{matrix}0 & 0\\ 0 & 1\end{matrix}\right)

Subcase 2:

x=\left(\begin{matrix}0 & 1\\ 1 & 0\end{matrix}\right), \overline{y}=\left(\begin{matrix}\overline y_{11} & \overline y_{12}\neq0\\ \overline y_{12} & 0\end{matrix}\right)
y=\left(\begin{matrix}0 & 0\\ 0 & \frac 1 {\overline y_{12}}\end{matrix}\right), \overline{x}=\left(\begin{matrix}0 & 0\\ 0 & 1\end{matrix}\right)

(Notice that this subcase and subcase 2 above intersect when \overline y_12=1.)

But for the path \overline yyx there will be only one case:

x=\left(\begin{matrix}0 & -\frac{y_{11}}{\overline{x}_{12}^{2}}\\ 1 & \frac{y_{11}}{\overline{x}_{12}}\end{matrix}\right), \overline{y}=\left(\begin{matrix}\frac{\overline{x}_{12}}{y_{11}} & 1\\ 0 & 0\end{matrix}\right)
y=\left(\begin{matrix}y_{11}\neq0 & 0\\ -\overline{x}_{12} & 1\end{matrix}\right), \overline{x}=\left(\begin{matrix}0 & \overline{x}_{12}\neq0\\ 0 & -\frac{\overline{x}_{12}^{2}}{y_{11}}\end{matrix}\right)

I was on going to do all possible paths — but I begin to realize this isn’t a good way to approach things. For starters I don’t have a good idea when these subcases lie on the same component of the quiver variety.

Calculation when level=rank=2

July 27, 2009

I will use the formula on the top of page 35 from Nakajima’s “Quiver Varieties and Branching”:
\mathbf{w}=\sum w_i\Lambda_i = \sum \Lambda_{\mu_p}; \mathbf{w}-\mathbf{v}=\sum w_i\Lambda_i - v_i\alpha_i =\overline{\sp{t}\lambda}+t\delta^Y.
Where the first “t” in the second equation indicates transposing (according to the process described on page 33) the generalized young diagram \lambda and the second “t” is given by the formula (on page 34):
t=\langle d^X,\bar\mu\rangle-\langle d,M(\mu)\rangle.
d^Y, \delta^Y, d^X, \delta^X are suitable choices for “d” and “\delta” in affine \tilde{L\mathfrak{sl}_2} (which in this case is on “both sides” of the level-rank duality).
And \langle d, M(\mu)\rangle is (I believe) the coefficient of M in the formula for d(M) found near the bottom of page 32.

My goal is to find the cycles corresponding to the first two triangles in my list of polytopes so I use:
\mu=\delta^X+\Lambda_0+\Lambda_1
\overline\lambda=\overline{\sp{t}\lambda}=\Lambda_0+\Lambda_1
\mu_1=0, \mu_1=-1
Some of the notation is very confusing, I understand. I’m sorry, I don’t know what to do with it: at the end of the day the dimension vectors we are concerned about are \mathbf{w}=(1,1) and \mathbf{v}=(1,1)

\mathbb{C} \leftarrow \mathbb{C}
\rightarrow
\rightarrow
\leftarrow
\downarrow\uparrow \downarrow\uparrow
\mathbb{C} \mathbb{C}

We want to replace the arrows in this diagram with maps (in order of appearance top to bottom left to right) y,\bar y,x,\bar x,b_1,a_1,b_2,a_2 in such a way that it satisfies three conditions which I abbreviate as the “\mu^{-1}(0)” (this is not the same \mu… sorry), “stability” and “limit” conditions.

The “\mu^{-1}(0)” condition in our case (for and for other choices of \mathbf w \mathbf v but still r=l=2) implies:
a_1b_1+\bar x x=y\bar y
a_2b_2+\bar y y=x\bar x

The “Stability” condition has two halves. First that every vector v in the top half has “Origins” i.e. v=\sum f_k(\eta_k) where \eta_k live in the bottom half and f_k are chosen from appropriate paths in the quiver. And second that every non-zero v in the top half has “Futures” i.e. g(v)\neq 0 for some path g terminating in the bottom half.

The “Limit” condition requires that the action of t (not the same as either previous t… sorry) given on page 30 can be “controled” as t\rightarrow\infty by action of G_V described on page 5 (just before eq. 2.1 )

Something to notice right away because it will be useful in future calculations: taking the two halves of “Stability” together gives paths g\circ f:W^j\rightarrow W on which G_V has no control! The “Limit” condition requires that the image of such a composition must lie in W^{<j}. (Superscript refers to the decomposition into 1-dimensional subspaces given by \mu — use of the \le is short hand for the corresponding filtration.)

In our case we have b_2=0,a_1=0. For example if the image of a_1 is non-zero the "Futures" condition says it must escape but, according to the "Limit" condition, in doing so it can only afford to accumulate m_1 orders of t and every possible exiting accumulates at least m_1+1. The argument for b_2 is quite similar replacing "image" with "kernel", "Futures" with "Origins" etc..

Furthermore we get:
a_1\neq0\neq b_2,
\bar y=0\Rightarrow x\neq 0 and
\bar y\neq 0\Rightarrow y=0.
(Moding out by the G_V action in this case we can assume a_1=1=b_2:\mathbb{C}\rightarrow\mathbb{C}

The two components, then, correspond to diagrams:

\mathbb{C} \bar y\rightarrow \mathbb{C}
x\dashrightarrow
\downarrow \uparrow
\mathbb{C} \mathbb{C}

and

\mathbb{C} \dashleftarrow y \mathbb{C}
x\rightarrow
\downarrow \uparrow
\mathbb{C} \mathbb{C}

Where the solid line indicates the map is non-zero and dashed line indicates the map may be zero.

GRTEALA 1: Review of the situation

July 8, 2009

Update: Notes and video from the summer school is available here.

I should have been posting while I was at the conference. But anyway I’ll try to post as much as I can remember, before I forget it.

Geometric Satake gives a correspondence from representation theory to subvarieties of an affine Grassmanian. MV-cycles help give a better handle on them but are still rather abstract. MV-polytopes , introduced by Jared Anderson, are more “hands on” in terms of being able to do direct computations. But you need to know what they are first, and their original definition as moment map images of MV-cycles doesn’t really help. At least if you know they are a convex hill of the torus fixed-points appear in the *closure* of an MV-cycle then you’re done — but this still requires more-or-less calculating the MV-cycles and taking their closure.

Kamnitzer’s thesis provided a few ways to get direct handle on MV-polytopes avoiding MV-cycles entirely.

  • Implicit description via Plucker relations
  • Inductive description (for \mathfrak{sl}_n later extended to types B and C by **FIXME**)
  • Reduction to dim-2: higher dimensional MV-polytopes are all polytopes whose 2-faces are MV-polytopes.
  • Construction from primitives: MV-polytopes are Minkowski sums of “primitives” and sums of primitives from the same “cluster” are MV-polytopes
  • In dimension 2, clusters can be described by networks of non-overlapping cords each parallel to a sides of the Weyl polytope

A conjecture on the Uniqueness of MV-Polytopes

July 6, 2009

Fix a lattice L with \langle , \rangle and a partial order \le. Define
Q^+=\{x\in L|x\ge 0\} = \text{``positive root cone''}
L^+=\{x\in L|\langle x,y\rangle \ge 0 \forall y\in Q^+\} = \text{``dominant weights''}
(Alternatively we could fix Q^+ and define x\ge y \iff x-y\in Q^+. Similarly we might fix L^+ without reference to any inner product.)

We say that a set of “characters” \{\chi_\lambda:L\rightarrow\mathbb{N}\}_{\lambda\in L^+} and a set of subsets \{M^\lambda_\mu \subset L\}_{\lambda,\mu\in L} satisfy the “tensor property” if:

  1. \chi_\lambda(\mu)=\#\{m\in M^\lambda_\mu | m\subset \text{supp} \chi_\lambda\}
  2. \chi_{\lambda_1} * \chi_{\lambda_2}=\sum c^{\lambda_1,\lambda_2}_\mu \chi_\mu where
    c^{\lambda_1,\lambda_2}_\mu= \#\{ m \in  M^{\lambda_1+\lambda_2}_\mu  | m\subset ((\text{supp}\chi_{\lambda_1})+\lambda_2)\cap(\mu+\lambda_2-\text{supp}\chi_{\lambda_2}))\}

(Here * indicates convolution and \mathbb{N} includes zero.)

Suppose we also require the following about \{M^\lambda_\mu\}

  1. M^\lambda_\mu\neq\o \iff \lambda\ge\mu
  2. m\in M^\lambda_\mu\Rightarrow \mu\le x\le\lambda \forall x\in m \text{ and } \mu,\lambda\in m
  3. m\in M^\lambda_\mu\Rightarrow m+a\in M^{\lambda+a}_{\mu+a} \forall a\in L \text{ and } (-m)\in M^{-\mu}_{-\lambda}
  4. (*)\sigma a\in M^{\sigma\lambda}_{\sigma\mu} for any isotopy, \sigma of Q^+.

(I’m not sure if the last condition is worded correctly, but I’d like the collection to be invariant under permutations of, for example, minuscule weights.)

And, fixing a group W acting on L for which L^+ is a fundamental domain, suppose we require that:

  1. \chi_\lambda(\mu)=\chi_\lambda(w\mu) \forall w\in W

Conjecture For L,Q^+,L^+,W there is only one \{M^\lambda_\mu\}_{\lambda,\mu\in L} (and corresponding \{ \chi_\lambda\}_{\lambda\in L^+}) satisfying the tensor property and all the above requirements.

For L,Q^+,L^+,W coming from \mathfrak{sl_3} or \mathfrak{sp_4} it seems to be true.

Zero-stability: Examples

May 4, 2009

Example zero

Suppose I consists of a single element. Then a zero stable module is pair of maps a:W\rightarrow V and b:V\rightarrow W such that a is surjective, b is injective and ab=0.

The existence of zero-stable modules is only possible then if the dimension of W is at least twice the dimension of V. In the case of the smallest possible example of this (when W is 2-dimensionl) we are essentially picking two non-zero orthogonal vectors in W.

Example 1

Let (I,\Omega) be a directed graph and V be chosen so that Hom(V_{s(h)},V_{e(h)})=0 for all h. Then zero-stable modules are possible only when \dim W_i \ge 2 \dim V_i for all i\in I, and will correspond to a choice of zero-stable module of the type in example zero for each non-trivial V_i.

As a more specific example consider the directed graph consisting of four vertices and four edges formed into a circle. Let V_1=V_3=\mathbb{C} and V_2=V_0=0 and W_i=\mathbb{C}^2. Then you can think of a zero-stable module as a pair of bases (each an orthogonal basis) for \mathbb{C}^2.

0-stable Modules Part 2

April 25, 2009

Fix a directed graph (I,\Omega) and its double (I,H).  Fix I-graded vector spaces V,W.

Let h_\bullet=(h_n,... h_1) be a path in (I,H).  That is, h_i\in H and s(h_{k+1})=e(h_k) for 1\le k\le n-1.  The length of h_\bullet is n.  Define s(h_\bullet)=s(h_1) and e(h_\bullet)=e(h_n).

P(H) will denote the set of all paths,  P(i,H) will denote the set of paths begining at i, and P(H,j) will denote the paths ending at j.  In practice we will actually these with finite sets depending on the dimension of V — I’ll address this in a moment.

Let (B,a,b) be a module as defined in the previous post.  B_h:V_{s(h)}\rightarrow V_{e(h)} refers to the component of B associated to a particular edge h and B_{h_\bullet}:V_{s(h_\bullet)}\rightarrow V_{e(h_\bullet)} refers a composition of such maps.

Then we may alternatively define zero-stable as follows: (B,a,b) is zero-stable if for all i\in I and every y \in V_i

  1. there exists a path h_\bullet\in P(i,H) such that b_{e(h_\bullet)}B_{h_\bullet}(y)\neq 0 and
  2. there exists a vector x\in\bigoplus_{h_\bullet\in P(H,i)} W_{s(h_\bullet)} such that \sum B_{h_\bullet}a_{s(h_\bullet)}(x_{s(h_\bullet)})=y.

The proof that this is equivelant to the previous definition is straightforward and also demonstrates that we need only consider paths that visit each vertex i no more than \dim(V_i) times.

0-stable Modules

April 24, 2009

For Nakajima’s analogue of MV-cycles we need to understand what it takes for a “module” to be 0-stable. Let me review first what that means.

Fix a finite set I={1 .. r} and a directed graph (I,\Omega). Let H=\Omega\sqcup\bar\Omega. For two I-graded vector spaces V and W we define
E(V,W)=\bigoplus_{h\in H} Hom(V_{s(h)},W_{e(h)}
L(V,W)=\bigoplus_{i\in I} Hom(V_i,W_i)
M(V,W)=E(V,V)\oplus L(W,V) \oplus L(V,W).
(where s and e denote the start and end vertices of an edge.)
We have a function
\epsilon:E(V,V)\rightarrow E(V,V)
which multiplies by -1 components corresponding to h\in\bar\Omega and fixes components corresponding to h\in\Omega. We have a composition
L(V,W)\times L(W,X) \rightarrow L(V,X)
which is straightforward. We have a composition
E(V,W)\times E(W,X) \rightarrow L(V,X)
where (CB)_i=\sum_{e(h)=i} C_hB_{\bar h}. And we have a “moment map vanishing at the origin”
\mu:M(V,W)\rightarrow L(V,V)
where \mu(B,a,b)=(\epsilon B)B+ab.

From now on we fix V and W.

A point (B,a,b)\in\mu^{-1}(0) is called a “module”.

A “sub-module” of (B,a,b) is a B-invariant I-graded vector space V'\subset V which either contains Im(a) or is contained in Ker(b).

A module (B,a,b) is said to be “0-stable” if the only sub-modules are 0 and V.

**EDIT: I should say (B,a,b) is zero-stable if the only sub-module containing Im(a) is V and the only submodule contained in Ker(b) is 0.

Primative Polytopes and Tropical Relations

February 14, 2009

My attention was drawn this week to another part of Kamnitzer’s paper on MV-polytopes were he discusses clusters of primitive polytopes (observed by Anderson).  Primitive polytopes are a finitie set (for the finite dimensional groups anyway) of polytopes that generate all MV-polytopes under Minkowski sum.  MV-polytopes aren’t closed under Minkowski sum, but the primitive polytopes are grouped into clusters such that taking Minkowski sum within a cluster guarantees an MV-polytope.  Prehaps I should emphasize, every MV-polytope can be written as the Minkowski sum of a set of primitive polytopes found in the same cluster.

The clusters correspond to tropical choices.  When we have a tropical formula A=min\{B,C,D\} this can be rephrased as:

A=B,A\le C,A\le D OR
A=C,A\le D,A\le B OR
A=D,A\le B,A\le C

So the solution set to a tropcal formula is a union of cones.  Each of these cones (generally overlapping) correspond to a tropical choice.   If an MV polytope satisfies the tropical Plucker relations with a particular tropical choice — then it can be generated via Minkowski sum by polytopes in the cluster corresponding to that tropical choice.

I’m very interested in this because for $latex  L\mathfrak{sl}_2$ I don’t have a satisfactory sense of tropical Plucker relations, but I do have a notion of “MV-polytopes” (not yet realized as moment map images of cycles, but functioning combinatorially in the same way) and of course Minkowski sum.   So if I can observe primitives and  clusters, I may gain some insight into what sort of tropical relations to expect.

Tensoring Polytopes, Minkowski Sum Method (pictures)

February 6, 2009

Here’s a case in \mathfrak{sl}_3 we should all be familiar with.

If we tensor the standard representation with the standard dual we get a nine-dimensional representation:

sttimesstd

Say the red polytopes come from Standard and the blue come from Standard dual (black is the overlap).  In the Minkowski sum method the MV-polytopes are “added” head-to-tail style.

In this method the action on a basis vector is given by g(x\otimes y)=(gx)\otimes y + x\otimes (gy) so the adjoint representation inside (recall St\otimes St^*=Ad\oplus Tr) looks like this:

adinsttimesstd1

And the trivial representation inside looks like:

trinsttimesstd

The signs will be explained in the next post when I go into what I call “Anderson method”.  For now, notice that this makes Ad\oplus Tr and orthogonal sum with respect to the tensor basis.