# Problem 7 (Catalan Algebras)

I’m working on problem 7, and I must be doing something wrong, because when I try to work out some low dimensional examples, I don’t get catalan numbers.  Takning $n=1$, we get the empty set, (a zero dimensional basis, while $C_1=1$,) and for $n=2$ we get the basis $\{e_1\}$.  But as far as I reason, $n=3$ gives $\{e_1, e_2, e_1e_2\}$, $(e_2e_1=e_1e_2)$.  For $n=4$ we get $\{e_1, e_2, e_3, e_1e_2, e_1e_3, e_2e_1, e_2e_3, e_3e_2, e_1e_2e_3, e_1e_3e_2, e_2e_1e_3, e_3e_2e_1, e_3e_2e_3, \ldots\}$.  This may not be an exhaustive list, but it I don’t think it contains any equivalent elements, and it certainly has more than 5 elements.  If anyone has looked at this problem and wants to give me some advice, I’d appreciate it.  Thanks,

-Zach

1. Michael J

Be careful about the relations. The relation \$e_i e_j = e_j e_i\$ only applies if \$i\$ and \$j\$ differ by more than one place. If they are one place apart, then you have to use the third line. So for \$n=2\$, you have \$e_1^2 = e_1, e_2^2 = 2, e_1 e_2 e_1 = q e_1, e_2 e_1 e_2 = q e_2\$. You should now be able to find a basis here with \$C_3 = 5\$ elements. Hope that gets you back on track. 🙂

2. I think I was pretty careful about this, but let’s double check for $n=3$. The third Catalan number is 5. Using the notation 1, 2, for our basis elements, and just listing all possible combinations of one or two elements, we have, before relations are applied,

1, 2, 11, 12, 21, 22

Our second relation does not apply, since all elements have adjacent indices, so using rule 1, we eliminate 11 and 22, so our list becomes

1, 2, 12, 21 (so yes, I made an error earlier by eliminating 21.)

Now, as for any possibility of a three element string, the only cases are of the form 1xx and 2xx. If I start a string 1xx, then if I extend to 11x, rule 1 reduces me to the two element string 1x, so I must extend as 12x. Now, if I complete the string 122, rule 1 again reduces this to a two element string, so I complete as 121, which becomes q1 by rule 3. Since q is non-zero in the field, the span of q1 is not distinct from the span of 1. The string beginning 2xx is symmetric. So our basis in the $n=3$ case is

1, 2, 12, 21

Still, we don’t have 5 basis elements.

3. Karen Walters

Don’t we have to have a “1” Isn’t that what A1 and A2 are?

4. Brian Cruz

Ditto what karen says. The case of $A_4$ which has three generators has the following 14 basis vectors:

one
1, 2, 3, 21, 32, 321
12, 23, 13, 132, 213, 2132
123

5. @Karen
When you say “1”, do you mean $e_1$ or a multiplicative unit? If we include the unit as a basis element, doesn’t that throw off Brian’s count for $A_4$?

6. Duquec

The object you are dealing with is called a Temperley-Lieb algebra.

Since I was also confused as to how to get the right number of basis elements, I decided to look up this algebra. I haven’t found anything yet.

7. So how do you guys solved it? Here in Bogotá we managed to write a bijection with dyck paths. What did you try?