Dimension of preprojective algebra of Dynkin type Fix a field $\Bbbk$. Let $Q$ be a Dynkin quiver and let $\Pi(Q)$ be its preprojective algebra. It is well-known that in this case $\Pi(Q)$ is finite-dimensional, but I've been unable to find a reference for what the dimension actually is for each type. I'm also interested in knowing the dimension of $e_i \Pi(Q)$, where $e_i$ is the idempotent corresponding to vertex $i$ (i.e. the dimension of the space of all paths starting at vertex $i$).
I've already calculated by hand that the dimension of  $\Pi(\mathbb{A}_n)$ is given by ${n+2 \choose 3}$, and that the dimension of $e_i \Pi(\mathbb{A}_n)$ is given by the $i$th entry of the $n$th diagonal in the multiplication table for positive integers. However, the other Dynkin types seem to be more difficult.
 A: It is something well-known but I do not know where it is written. q-deform the Cartan matrix into
$$C(q)=I+qB+q^2I.$$
$B$ has zeroes on the main diagonal; substituting $q=1$ gives you Cartan. Invert the thing:
$$C(q)^{-1}=I+qA_1+\ldots q^nA_n+\ldots.$$
Let $k$ be the smallest integer such that $A_k$ has negative coefficients. Then
$$\sum_{n=0}^{k-1}q^nA_n$$
is the Poincare series of the preprojective algebra, i.e., all dimensions of all graded components of all $e_i\Pi e_j$ are written down.
A: As a module over $kQ$, a finite-type preprojective algebra is a direct sum of each of the indecomposable $kQ$-modules once.  Thus, the total dimension is the sum over all positive roots of the height of the root (where the height is the sum of the coefficients in the simple root expansion).  If we write $\lambda_i$ for the number of roots of height $i$, the partition conjugate to $\lambda$ consists of the exponents of the root system.  (This fact was first observed empirically by Shapiro and Steinberg, then proved by Kostant.)
Thus, we can think of an exponent $e$ as accounting for $e$ roots, one of height 1, one of height 2, etc., up to one of height $i$.  Thus, each exponent $e$ contributes $1+2+\dots+e={e+1\choose 2}$ to the total dimension.  
In type $A_n$, the exponents are 1 to $n$, so the answer is $\sum_{e=1}^n {e+1 \choose 2} = {n+2\choose 3}$, as you found.   
