resolving the isospectrality of the dihedral graphs ram band, uzy smilansky a graph g is made of v...
Post on 22-Dec-2015
226 views
TRANSCRIPT
Resolving the isospectrality of the dihedral graphsRam Band, Uzy Smilansky
• A graph G is made of V vertices and B bonds and a connectivity matrix Cij
• Cij≡ # of bonds connecting i and j.
• Bond notation is (i,j)• Valency of the vertex i:
• Boundary vertex is a vertex with vi=1.
• Interior vertex is a vertex with vi>1.2
3
4
5
1
1
3 4
5
6
2
Introduction to graphs
V
jiji Cv
1
• On the bond (i,j) use the coordinate xij:
• The wave function ψij
on each bond obeys
On the interior vertices (vi>1) we demand
• Continuity
• Current conservation
On the boundary vertices (vi=1) we demand
Dirichlet → or Neumann →
Introduction to quantum graphs
ijij
ij
kdx
d 22
2
ijjijiij Lxx ,0
0..)0( ijiji Ctsj
0:0
0;
ij
ij xCj ij
ij
dx
di
0i 00
ijx
ijijdx
d
Examples of several wave functions of kk88 kk1313 kk1616
Resolving isospectrality by counting nodal domains
• Conjecture (by U. Smilansky et. al): The nodal count sequence resolves isospectrality.
• Two types of nodal domains:
1) Metric domains – The connected domains where the wave function is of constant sign.
2) Discrete domains – A discrete domainconsists of a maximal set of connectedinterior vertices where the vertex wave function has the same sign.
• Discrete nodal count resolves isospectrality for some isospectral pairs. Numerical evidence was found [2].
• We look for rigorous proof of resolving isospectrality by counting nodal domains.
Search for new simpler isospectral graphs
a
2b2c
• Constructing isospectral pair using the Dihedral D4 symmetry
• The work was done Following [3] and the notes of Martin Sieber
• We obtain the Dihedral graphs
Theorem 1 – Resolution of the isospectrality by the discrete count
• Theorem 1 – Let GI and GII the graphs below. Denote with {in} the sequence of
discrete nodal count of the graph Gi.
Then {In} is different from {II
n} for half of the spectrum.
Following is a sketch of the proof of theorem 1:
GII
a2ba
2c
D N
GI
2abb
cc
D
D
N
N
The transplantation that takes an eigenfunction of GI with eigenvalue k and transforms it to eigenfunction of GII with eigenvalue k is:
• Cut the graph GI with its eigenfunction along the dashed line. Let , be the two functions defined on the subgraphs.
• Obtain two new functions , by: (appropriate reflections are be needed).
• Glue , together to obtain an eigenfunction on GII
minus
2
1
2
1
11
11~
~
1
• The action of the transplantation on the vertex wave function is
• The transplantation implies that the vector is obtained by rotating counterclockwise by (this is true since the eigenfunction is defined up to a multiplicative factor).
• The number of nodal domains is
• Therefore if the transplantation rotates the vector across the quadrant borders. In other words:
2
1
4
))(tan1(2
11))(1(
2
11
2
1 signsign
III
),1()0,1()tan( III The vector belongs to the colored domains
I=1
I=1
I=2
I=2
2
1
)()( xhxh • Calculation of h(x) the distribution function of yields tan
1
2
{In} is different from {II
n} for half of the spectrum.
• Theorem 2 – Let GI and GII the graphs below. Denote with {in} the sequence of
metric nodal count of the graph Gi.
Then { In} is different from { II
n} for half of the spectrum.
References
Metric count → 8
Wave functionWave function
++
++--
--
++
--Discrete count → 3
Vertex wave Vertex wave functionfunction
Isospectral graphs are graphs that have the same spectrum in spite of being different.
• Below is an example of such isospectral pair of graphs which is constructed out of isospectral domains [1]
• These graphs are not isometricAnd indeed the isospectrality of the dihedral graphs is
resolved by counting nodal domains
kk11
44
kk11
44
kk11
22
kk11
22
Discrete → 1
Metric → 12
Discrete → 2
Metric → 12
++++
++ --
++ ++
++
++
Discrete → 1
Metric → 14
Discrete → 1
Metric → 13
Theorem 2 – Resolution of the isospectrality by the metric count
1 2
1
1
212211~,~
• How can we distinguish between isospectral pairs ?- How can isospectrality be resolved ?
GII
D N
GI
D
D
N
N
N
N
2
1
22
D
D
1~
1
~D
N
N
11
22
2
~2~
D
D
plus
N
1~ 2
~1
~ 2
~
2
1~ 2~
1~ 2~
2
1~
~
2
1
2
1
[1] P. Buser, J. Conway, P. Doyle and K-D Semmler, Int. Math. Res. Notices 9 (1994), 391-400.
[2] T. Shapira and U. Smilansky, Proceedings of the NATO advanced research workshop, Tashkent, Uzbekistan, 2004, In Press.
[3] D. Jakobson, M. Levitin, N. Nadirashvili and I. Polterovich, J. Comp. and Appl. Math. 194 (2006), 141-155.and M. Levitin, L. Parnovski and I. Polterovich, j. Phys. A: Math. Gen., 39 (2006), 2073-2082.