Question List
Last Updated: 20010426
Maintainer: Nick Pongratz
njpongratz@students.wisc.edu
/** Question 1
***************************************************************/
Submitted by: Michael Lang, 20010424
Setup:
-----
Order n TOADs are in bijection with compatible pairs of ASMs (order
n and n+1). One could also think of the TOAD giving rise to an order
2n+1 ASM by interleaving the compatible pair. Of course, this map is
1-1 but not onto.
Example:
-------
nn
nnwe
wsswee
wwenne
wess
ss
corresponds to the pair
0 + 0 0
0 0 0 + 0 + 0
+ 0 0 0 + - +
0 0 + 0 , 0 + 0
or the single ASM
0 + 0 0 0 0 + 0 0 0 0
0 + 0 0 0 0 + 0 0 0
0 0 0 + 0 0 0 0 0 0 +
+ - + = 0 + 0 - 0 + 0 .
+ 0 0 0 + 0 0 0 0 0 0
0 + 0 0 0 0 + 0 0 0
0 0 + 0 0 0 0 0 + 0 0
Question:
--------
What sort of patterns can be found in this construction?
What can be said about the ASMs that do arise in such a manner?
In particular, what do their DPFLs look like? What do domino
moves and shuffling do to the DPFL? What is the effect of
gyration?
Progress:
--------
None that I know of.
/** Question 2
***************************************************************/
Submitted by: Hal Canary, 20010425
Conjecture:
-----------
Take any two (square) FPLs that have the same endpoint matchings,
x and y.
Is it true that there is a set of basic local moves that will allow us
to morph x into y?
Are these the two basic moves?
1)
0--0 0 0 0--0
| | <==> | | |
0--0 0 0 0--0
2)
0--0 0 0
| |
0--0 <==> 0--0
| |
0 0 0--0
3)
0--0--0 0--0 0
| <==> | | |
0--0--0 0--0 0
4)
0--0 0--0
| | | |
0--0 <==> 0 0
| |
0--0 0 0
/** Question 3
***************************************************************/
Submitted by Michael Lang, 20010426
How many regular hexagonal FPLs are there of order n which pair
terminals 1:2 3:4 ... 6n-1:6n?
Progress:
n #
---------------------------
1 1
2 1
3 9 =3*3
4 602 =2*7*43
5 139118 =2*7*19*523
/** Question 4
***************************************************************/
Submitted by Michael Lang,
20010504
Okay, here's the sort of thing I was talking about yesterday.
=====
Setup:
-----
See the article by Razumov and Stroganov for background.
They assert that the vector Psi is the eigenvector of H
associated with the eigenvalue 2n.
The rows and columns of H (and the entries of Psi) are
indexed by the pairings. Notice that the number of these
is the nth Catalan number.
By Wieland's dihedral symmetry result, the entries of Psi
are constant on symmetry classes of pairings. In the
paper, n=4. So although H is 14x14, there are only three
different entries in Psi.
Let us construct another matrix, which I will write as
HR (Hamiltonian-Reduced). The rows and columns of HR are
indexed by the symmetry classes of pairings, so HR is
smaller than H. The x,y entry of HR is the number of i
such that h_i takes a particular representative of y to a
particular representative of x.
Let N(x) represent the number of pairings in symmetry class
x. Then the sum over y of the products HR[x,y]*N(y) is
2n*N(x).
If you have H, here's how to construct HR: Order the rows
and columns of H so that pairings in the same class are
together. Then divide H into blocks according to these
classes. In each block, the rows should have the same sum.
Replace each block by its row sum and you have HR.
Example:
-------
The reduced version of the Hamiltonian in the paper is this:
4 8 4
1 5 2
0 2 2
Assertion:
---------
The eigenvectors of HR correspond to the eigenvectors of H
that are constant on symmetry classes. Note that this means
the collapsed version of Psi (7 3 1 for n=4) is an eigenvector
of HR. Psi (conjecturally) enumerates the ASMs according to
pairing symmetry.
Question:
--------
What, if anything, do the other eigenvalues and eigenvectors
of HR mean? What is the meaning of the eigenvectors of the
transpose of HR? What, if anything, do the eigenvectors of
H that aren't constant on symmetry classes mean? I've listed
these in decreasing order of likelihood to be interesting.
Progress:
--------
Not much. Here are some data, though.
n=1 and n=2 are boring, of course. HR is 1x1 in each.
n=3:
HR = 3 6
2 2
Besides the eigenvector (2 1) corresponding to eigenvalue 6,
HR has an eigenvector (3 -2) corresponding to eigenvalue -1.
For n=4,5,6 I'll list HR and its eigenvalues/eigenvectors.
Sometimes my entry order becomes a bit odd. I'll list 2n/Psi
to establish it.
n=4:
4 8 4
HR = 1 5 2
0 2 2
eigenvals/vecs = 8 (7 3 1), 2 (-2 0 1), 1 (0 1 -2)
Note: The non-Psi eigenvectors had the same entries. Coincidence?
n=5:
HR:=matrix(6,6,[
[5,10,0,0,10,0],
[1,4,5,2,0,2],
[0,2,3,0,2,0],
[0,4,0,4,4,0],
[0,0,2,1,3,2],
[0,0,0,0,2,2]
])
eigenvals/vecs = 10 (42 17 6 14 4 1), 4 (0 -1 0 -1 1 1)
For n=5, HR has four other eigenvalues, but they're not nice. The roots
of z^4-7z^3+3z^2+47z-52, I think.
n=6:
HR =
6 12 0 0 0 0 12 6 0 0 0 0
1 5 4 2 0 2 0 0 2 0 2 4
0 2 4 0 2 0 0 0 0 0 2 2
0 2 0 4 0 0 0 0 0 2 4 0
0 0 3 0 3 0 0 0 0 0 0 0
0 2 1 0 0 7 2 0 2 0 4 0
0 0 0 0 1 1 4 0 0 2 2 4
0 0 0 0 0 0 0 4 4 0 4 0
0 0 0 0 0 0 0 1 3 2 0 2
0 0 0 0 0 0 0 0 2 2 0 0 *
0 0 1 1 0 1 1 1 0 0 4 2
0 0 0 0 0 0 1 0 1 0 1 4
* This is the row for the "distal" pairing
evals/vecs = 12 (429 169 60 63 20 128 34 23 5 1 41 10),
6 (2, 0, -1, 0, -1, 1, 0, 0, 0, 0, 0, 0),
4 (-3 1 0 1 0 0 0 -1 1 1 -1 0),
-1 (156 -182 160 168 -120 155 -35 252 -147 98 -168 70),
3+-sqrt3 (icky)
4+-2sqrt3 (icky)
roots of z^4-15z^3+69z^2-81z-62 (icky)