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Below I describe an infinite (but locally finite) quiver with relations. My question is whether anyone recognizes it and can provide appropriate pointers to the literature. I'm mainly interested in computing its Hochschild homology. The path category of the quiver (with relations) is related to the Temperley-Lieb category at non-generic (non-semisimple) values of the parameter. (At least I think it is -- I haven't double-checked the calculations.)

(This is not my area of expertise, so I apologize in advance if I'm not using standard terminology.)

The quiver

  • A vertex for each natural number $i=0, 1, 2, \ldots$.
  • An arrow $u_i : i \to i+1$ for each $i\ge 0$.
  • An arrow $d_i : i \to i-1$ for each $i \ge 1$.

The relations

  • $u_i u_{i+1} = 0$ for all $i\ge 0$.
  • $d_i d_{i-1} = 0$ for all $i \ge 2$.
  • $u_i d_{i+1} = d_i u_{i-1}$ for all $i\ge 1$.
  • $u_0 d_1 = 0$.

Here's a more geometric description of the quiver. Start with a 2-dimensional mesh: A vertex for each pair of integers $(j,i)$ such that $j+i$ is even; "up" arrows connecting $(j, i)$ to $(j+1, i+1)$; "down" arrows connecting $(j, i)$ to $(j+1, i-1)$. Impose a commutativity relation for the boundary of each square of the mesh. Impose another relation that the composition of two "up" arrows is zero, and similarly for two "down" arrows. Declare that objects $(j, i)$ with $j<0$ are zero (i.e. any path which factors through such an object (vertex) is zero. Finally, mod out by the horizontal translation $(j, i) \mapsto (j+2, i)$.

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Without the $u^2=0=d^2$ relations, it would be the preprojective algebra of $A_{\infty}$. Or, without $u_0d_1=0$ you have the quadratic dual of this preprojective algebra. –  Alex Dugas Feb 10 '12 at 18:31
(The indices in the third set of relations are probably wrong, no?) Without the $u_0d_1=0$ relation, a module is the same as a positively graded mixed complex in the sense of Kassel; see Loday's book on Cyclic Homology, for example. BTW, the geometric description you give in your last paragraph gives something different: you get a vertex for each integer, and then modules are precisely mixed complexes. –  Mariano Suárez-Alvarez Feb 10 '12 at 18:45
@Alex: Thanks for the remarks. –  Kevin Walker Feb 10 '12 at 18:52
@Mariano: I've stared at the third bullet in the relations and I don't see anything wrong. Are you saying that the arrows are not composable? In the geometric description, by "declare that objects...are zero" I mean remove those vertices from the quiver; sorry if that was not clear. I'll have a look at Loday's book, as you suggest. –  Kevin Walker Feb 10 '12 at 18:55
@Kevin: ohh, you are writing arrows in paths in the opposite direction as I do :) –  Mariano Suárez-Alvarez Feb 10 '12 at 18:58
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2 Answers

up vote 3 down vote accepted

The Hochschild homology (for the finite truncations) is computed in:

MR2248284 (2007j:16014) de la Peña, José A. ; Xi, Changchang .

Hochschild cohomology of algebras with homological ideals.

Tsukuba J. Math. 30 (2006), no. 1, 61--79.

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Thanks Bruce, I'll have a look there. –  Kevin Walker Feb 23 '12 at 15:05
Hochschild homology commutes with directed colimits, so if you can compute the induced maps, that should do it. –  Mariano Suárez-Alvarez Feb 23 '12 at 17:23
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This is the category of Schubert smooth perverse sheaves on $\mathbb{CP}^\infty$; the $i$th node corresponds to the projective cover of the constant sheaf of $\mathbb{CP}^i$. This is actually an easy topological calculation; the quadratic dual of your quiver is the same thing with the squared relations removed, which is the quiver of maps between $\mathbb{C}[t]/(t^i)$ for all $i$. This is the same as the Ext spaces between the constant sheaves on the $\mathbb{CP}^i$'s.

Not sure if that will help a lot with calculating Hochschild homology...

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Thanks Ben, that's interesting. I haven't been able to find a definition of the quadratic dual of a quiver, but I assume it is not to take the path algebra (as opposed to path category) of the quiver and then apply the definition of quadratic dual of an algebra. The path algebra has a lot of boring quadratic relations which say that the product of two noncomposable arrows is zero, and we don't want to take orthogonal complements of those, right? (Or am I making a silly mistake here?) Do you have any suggestions for where I can read more about the topics in your answer? –  Kevin Walker Feb 11 '12 at 14:37
When I say quadratic dual, I do mean quadratic dual of the algebra, but it's important that you interpret that correctly. "Quadratic" mean that an algebra $A$ is a quotient of the tensor algebra of $A_1$ as a $A_0$-bimodule by elements of degree 2 in the tensor algebra. The path algebra of a quiver is the tensor algebra of its degree 1 part over its degree 0 part, so it is stupidly quadratic, and it's quadratic dual is the path algebra of the opposite quiver, with all length 2 paths set to 0. The quadratic dual comes from starting there, and as you add relations in length 2 paths, –  Ben Webster Feb 12 '12 at 4:14
you throw out all relations on the dual side that you had non-zero inner product with the one you added on the primal. The dual of your quiver is essentially the same one without the d^2 relations (the length two paths from $i-1$ to $i+1$ is one dimensional, so if you have that as a relation in the primal, there will be no relations there in the dual) and the commutation relations the same after futzing with signs (since the degree 2 space from $i$ to $i$ is two dimensional, and you had 1 relation there in the primal. –  Ben Webster Feb 12 '12 at 4:18
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