Temperley-Lieb algebras

12 Pages·2009·0.204 MB·English

by  Phillip Linke

Temperley-Lieb algebras
Phillip Linke
Karlstads Universitet
October 2009
1. Introduction
This paper aims to wrap up the information on Temperley-Lieb algebras given in the
presentation. The first part will concern the definition and multiplication of Temperpley-
Lieb diagrams, which form a basis of the Temperley Lieb algebra. The second part will be
about Temperley-Lieb algebras and give some examples for them. The last Part will then
give a brief introduction to the representation theory of Temperley-Lieb algebras. Most
of the information that are presented originate from the paper ”Representations of the
Temperley-Lieb algebra” by A. Moore. See [Moo08] in the bibliography.
2. Temperley-Lieb diagrams
We try to take two aproaches here at the same time. The graphical via Temperley-Lieb
diagrams and the structural via generators and relations.
Definition 2.1. A Temperley-Lieb diagram (TLD) of length n is a graph consisting of
two rows of n dots and edges that connect one dot to exactly one other dot. The dots are
placed in a way such that both rows are directly above each other. An edge can now be
drawn from the top to the bottom row or between dots in one row.
Definition 2.2. The set of all those diagrams with n dots in each row, that fullfil an
additional requirement, is denoted by TL . This requirement is that the resulting graphs
n
have to be planar when no dot is moved and all the lines are drawn between the rows of
points.
• • • • • • • • •
, , ,
• • • • • • • • •
• (cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)•(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)• , • (cid:111)(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)•(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)•
• • • • • •
1
2
Are the diagrams of TL . Counterexamples are:
n
• (cid:64) • • • (cid:64) • •
(cid:64) (cid:64)
(cid:64) (cid:64)
(cid:64) (cid:64)
(cid:64) (cid:64)
(cid:64) (cid:64)
(cid:64) (cid:64)
,
• • • • • •
We also want to donate the diagrams for TL and TL . For the first we only get
1 2
•
•
For the second we have:
• • • •
and
• • • •
2.1. The number of diagrams. We can recursively find all the diagrams in TL . The
n
number of all diagrams is obviously finite, since there are 2n dots and n edges involved.
We enumerate the dots clockwise, starting in the top left corner.
Lemma 2.3. An even-labeled dot can only be connected to an odd-labeled dot and vice
versa.
Proof. If we would have an edge between two odd-labeled dots then we would have an odd
number of dots on every side of the edge. If we now compose pairs from the remaining dots
under the edge we see that there would have to bee a crossing which is not permitted. (cid:3)
We now start to count: We observe, that to keep the planarity requirement the even-
labeled dots can only be connected to odd-labeled dots.
1 2 3 4
• • • •
• • • •
8 7 6 5
So there are n choices for an edge from the first dot to an even-labeled dot. This first
edge already devides the diagram into two parts. On one side of the edge we got k dots
and on the other side 2n−2−k dots. Now we get that the number of diagrams that have
this edge is
|TL |·|TL |.
k/2 (2n−2−k)/2
3
1• (cid:76)(cid:76)(cid:76)(cid:76)T(cid:76)(cid:76)L(cid:76)1(cid:76)2•T(cid:76)L(cid:76)(cid:76)2(cid:76)(cid:76)(cid:76)(cid:76) 3• 4•
• • • •
8 7 6 5
Example 2.4. We want to show the recurrence for TL . To get to |TL | we need to know
4 4
all the values of TL for n < 4. We have five diagrams for TL and two for TL . Threre is
n 3 2
one diagram in TL and we set |TL | to 1 to get the recursion to work. We therefore get:
1 0
|TL | = |TL ||TL |+|TL ||TL |+|TL ||TL |+|TL ||TL |
4 0 3 1 2 2 1 3 0
= 1·5+1·2+2·1+5·1 = 14
Note 2.5. This recurrence relation is well known in combinatorics. The values are the
Catalan numbers C . So we have |TL | = C . The formula for these numbers is
n n n
(2n)!
C = .
n
(n+1)!n!
While for now this is just useful to determine if we found all the diagrams, it will later
be helpful to determine if a representation is irreducible.
2.2. Multiplication of diagrams. Two diagrams of the same length are multiplied by
stacking them togther vertically and connecting the edges from the top row of the bottom
diagramtothebottomrowofthetopdiagram. Wethenuseflexibilitytobringthediagram
again into a normalised form. If a loop with no connection to the top or bottom of the
resulting diagram should occour, then it is omitted.
• • • • (cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)•(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)• • • • • • •
• (cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)•(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)• = • • • , • • • = • • •
• • • • • •
This multiplication is obviously associative, because it can be thought of as following
paths throught a stack of them. What we don’t have in general is commutativity.
We have an identity element for this multiplication which is given by:
• • • ··· •
• • • ··· •
This is easily checked because multiplication with this diagram will just continue all
paths and therefore leave the diagram unchanged.
4
The next question that arises natuarally is if there is an inverse to each or at least some
diagrams. Theanswerisno(exeptfortheidentity). Toprovethiswefirstneedtointroduce
the rank of a diagram, which is the number of lines connecting the top row of dots to the
bottom row. Now similar to matrix multiplication the rank of the product of two diagrams
can only be as high as the minimum of the ranks or smaller. So, since the rank of the
identity is n and the rank of a non identity element x is smaller then n, we cannot find an
element x−1, such that xx−1 = id.
• • • • • • • • •
• • • = • • • = • • •
• • • • • •
2.3. Generator diagrams. AlldiagramsinTL canbemadefromthesocalledgenerator
n
diagrams of TL . These are n−1 generators labeled e ,e ,...,e . The generator e is
n 1 2 n−1 i
the diagram that has an edge from the i-th to the i+1-th dot in the top and bottom row.
Besides from that, every dot is connected to the one right above respectively below it.
• • • • • • • • • • • •
, ,
• • • • • • • • • • • •
2.4. Algortihmic approach to make any diagram from generators. If we want to
make this figure:
• (cid:103)(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)•(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)•(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)(cid:103)•(cid:103)(cid:111)(cid:103)(cid:111)(cid:103)(cid:111)(cid:103)(cid:111)(cid:103)(cid:111)(cid:103)(cid:111)(cid:111)•(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)(cid:111)•
• • • • • •
We start with those edges which stay in the top row. Among them we start with edges
connecting neighbors because the could be nested in bigger edges.
• • • • • •
• • • • • •
• • • • • •
Next we try to connect the dots from the top row to the bottom row. We start with the
connection that connects the left most dot in the bottom row.
5
• • • • • •
• • • • • •
• • • • • •
• • • • • •
The last step would be to work on nested edges in the bottom row. So in conclusion:
The trick is to start with connections at the top and work downward.
• • • • • •
• • • • • •
• • • • • •
• • • • • •
• • • • • •
2.5. Generator relations. We dan deduce three relations between the generators:
(R1(cid:48)) e2 = e by dropping a loop
i i
(R2) e e e = e
i i+1 i i
(R3) e e = e e if |i−j| ≥ 2
i j j i
To prove those, we can restrict ourselves to examples in TL , because the general case
4
would just mean an extension of the diagrams by vertical edges. But these edges do not
affect the outcome.
(R1(cid:48)) e2 = e by removing a loop
i i
• • • • • • • •
• • • • = • • • •
• • • •
6
(R2) e e e = e
i i+1 i i
• • • • • • • •
• • • • = • • • •
• • • •
• • • •
(R3) e e = e e if |i−j| ≥ 2
i j j i
• • • • • • • •
• • • • = • • • •
• • • • • • • •
3. The Temperley-Lieb algebra
For n ∈ Z and x ∈ C we definfe the Temperley-Lieb algebra TL (x) as the vector
>0 n
space with a basis given by the diagrams d in TL .
n
 (cid:12) 
(cid:12)
TLn(x) = (cid:104)d | d ∈ TLn(cid:105)C =  (cid:88) add(cid:12)(cid:12)ad ∈ C
(cid:12)
 
d∈TLn (cid:12)
Thisdefinitiondoesnotgiveanycluewhatweneedthexfor. Whenweweremultiplying
so far, we forgot about loops we removed. But now the x allows us to keep track of them.
To be pricise, every time we remove a loop we get an x in return.
In particular the first relation between generators of TL becomes
n
(R1) : e2 = xe
i i
in the algebra.
However, we should avoid certain choices of x in order not to run into zero divisions. For
our treatment of the content it wont be necessary to know what these values are and we
just assume we made a good choice.
Example 3.6. We look at TL (x), the simplest case, in more detail. This algebra has a
2
basis given by
• • • •
and
• • • •
7
So let us take a look at the following two vectors in the resulting vector space:
• • • •
a + b
• • • •
and
• • • •
c + d
• • • •
Multiplication of these two elements gives us:
• • • • • • • •
ab• • + ad• • + bc• • + bd • •
• • • • • • • •
Applying our rules for multiplication, including the one for removal of a loop, gives us:
• • • •
ab +(ad+bc+xbd)
• • • •
Note 3.7. If more that one loop is removed, then this is reflected in the exponent of the x
in the resulting term. It is easy to remove simple loops, that is loops that are just existing
beween adjacent dots as above. It gets much more complicated when we have nested loops
or multiple edge loops. But we do not want to go into too much detail here. What is
important to remember is, that the removal of a loop originates only from the relation
e2 = xe and that the other relations are unchanged.
i i
4. Representations of TL (x)
n
In order to find representations of TL (x) we use a structure called Young tableaux.
n
Definition 4.8. Young diagram A Young diagram of length n is a collection of n boxes,
that represent a partition of an integer n. Ex:
Such a diagram represents a partition of an integer into positive integers. Here 8 = 3+
2+2+1. The partition is also called the shape of the diagram. As an abreviation for the
shape we can just write (3,2,2,1).
8
Definition 4.9. Young tableauAYoungtableauisaYoungdiagramfilledwithintegers.
Ex:
9 8 3
5 5
6 0
1
Definition 4.10. Standard tableau A standard tableau of length n contains every in-
teger from 1 to n exactly once and the values of the numbers are strictly increasing from
top to bottom and from left to right in every row and every column.
1 2 3
4 5
6 7
8
Definition 4.11. The hook length of a box in a Young diagram is the number of boxes
under the given box plus the boxes to the right plus one.
(cid:81)
Definition 4.12. The number of tableaux to a given shape is n!/( h), where the h’s are
the hook lengths for each box.
Example 4.13. We take the diagram of shape (3,2,2,1) from above. The number of all
tableaux is:
8!/(6·4·1·4·2·3·1·1) = 70
4.1. Connection to the Hecke algebra.
Note 4.14. From history: Young tableaux were used by A. Young to find representations
of the symmetric group. The q-generalized Hecke algebra H (q) or Iwahori-Hecke algebra
n
has its representations indexed by Young tableaux. The connection between H (q) and
n
TL (x) was first noted by V. Jones in 1987.
n
ThereisasurjectivealgebrahomomorphismfromH (q)toTL (x), whichwewilluseto
n n
push down representations of H (q) to representations of TL (x). We use the presentation
n n
of Halverson, Mazzoncco and Ram in [HMR09] for the homomorphism from H (q) to
n
TL (x). The relation between the parameter x from the Temperley-Lieb algebra and the
n
parameter q from the Hecke algebra is x = q+q−1.
We repeat the generators and relations of the Hecke algebra very brifly. The generators
are 1,T ,...,T and the relations, for fixed q, are:
1 n−1
T2 = (q−q−1)T +1
i i
T T = T T for |i−j| ≥ 2
i j j i
T T T = T T T
i i+1 i i+1 i i+1
Lemma 4.15. The homomorphism from H (q) to TL (x), found for example in [HMR09],
n n
sends T to q−e . We check if the relations in TL (x) hold for this map.
i i n
9
Sketch of Proof. We just do the calculation for the relation e2 = xe .
i i
e2 = (q−T )2 = q2−2qT +T = q2−2qT +(q−q−1)T +
i i i i i i
= q2−2q(q−e )+(q−q−1)(q−e )+1 = q2−2q2+2qe +q2−qe −1+q−1e +1
i i i i i
= qe +q−1e = (q+q−1)e
i i i
So we must have x = q +q−1, since we know, we have e2 = xe , to make this map work.
i i
We will use this identity to find representations of the Temperley-Lieb algebra. (cid:3)
The representations of H (q) are known to be indexed by partitions of n. When we now
n
pass to TL (x) the representations on partitions of more than two rows are in the kernel
n
of the map H (q) (cid:16) TL (x). The representations connected to the diagrams with two
n n
rowsarecalledtheseminormalrepresentations. Thesearedescribedin[HMR09]. Fromthe
actionofH (q)onstandardtableaxwegetanactionofthee onthestandardtableauxwith
n i
at most two rows. We can now directly prove that this gives a representation of TL (x)
n
without having to look back at H (q). We can also show that these representations form
n
a complete set of pairwise non-isomorphic irreducible representations of TL (x).
n
4.2. The vector space of Young tableaux. We try to find an appropriate description
of a vector space for TL to act on. When we defined the action of the diagrams, we only
n
need to check if all the relations in TL (x) are preserved.
n
We now form a vector space with a basis indexed by the tableaux of a given shape. For
that vector space we will be able to show that it is indeed a TL (x) module.
n
Example 4.16. We look at the shape (2,1). From the formula above we get that there
are 2 tableaux of that shape. So every vector in the vector space with the basis given by
these tableaux will be of the form:
1 2 1 3
v = a 3 +b 2 , a,b ∈ C
Let now d be an element of TL (x). Then we got the action:
n
 1 2   1 3 
mv = am 3 +bm 2 
Therefore we need to know how en element d acts on the basis vectors of that vector space.
AndsinceeveryelementofTL isgeneratedbytheelementse , weneedtoknowhowthese
n i
act on the Young tableaux of a given shape. Therefore we got different representations of
every TL (x) for different shapes.
n
10
4.3. Defining the action on the basis. Our first aproach is via the action of the sym-
metric group on the tableaux. We denote by s the simple transposition, that interchanges
i
i and i+1. Therefore s interchanges the numbers i and i+1 in a given tableau t.
i
Example 4.17.
 1 2  2 1
s1 3  = 3
 1 3  1 2
s2 2  = 3
As we can see s (t) can sometimes be no longer standard.
i
The action of e now sends t to a linear combination of t and s (t) if s (t) is standard
i i i
and to a multiple of t else.
(cid:26)
Ct+Ds (t), if s (t) is standard
e ·t = i i
i Ct, else
4.3.1. The polynomial [d]. To find the values of C and D we need the polynomial [d] of x.
To find the value of [d] we need the variable q from the Hecke algebra. We have:
qd−q−d
x = q+q−1 and [d] = = qd−1+qd−3+···+q−(d−1)
q−q−1
The first values of [d] are:
q0−q−0
[0] = = 0
q−q−1
q1−q−1
[1] = = 1
q−q−1
q2−q−2
[2] = = q1+q−1 = x
q−q−1
These are the only values we need. Further results are can be found in the article [Moo08].
The constants C and D:
We denote by d the number of steps it takes (boxes to cross) to get from i to i+1.
(cid:40) [d−1]
if i+1 is d steps N/E from i
C = [d]
[d+1]
if i+1 is d steps S/W from i
[d]
(cid:112)
[d−1][d+1]
D =
[d]