For the basic notions and results appearing in this example sheet, you should recall here.
A basic summary is put here for quick reminder.
There is an associated root system to a quiver.
For finite type quivers (i.e. for any \(\alpha\), only have finite many indecomposable reps. Finite type \(\Leftrightarrow\) Dynkin)
Indecomposable Representations \(\Leftrightarrow\) Positive roots
Using reflection functors one can obtain all indecomposable from simples, which is the same as obtaining all positive roots from simple roots \(e_i\).
Associated to a quiver \(Q\) we have a doubled quiver \(\overline{Q}\) and associated to doubled quiver we have a moment map \(\mu_{\overline{Q},\alpha}: \mathrm{Rep}(Q,\alpha)\to \mathrm{Lie}(G_\alpha)^*\).
Quiver varieties
\[\mathfrak{M}_\theta(Q,\alpha) = \mu^{-1}_{\overline{Q},\alpha}(0)/\!\!/_\theta G_\alpha, \quad \mathfrak{M}(Q,\alpha) = \mu^{-1}_{\overline{Q},\alpha}(0)/\!\!/G_\alpha.\]
Parameter function
\[p_Q(\alpha) = 1 - \langle \alpha, \alpha\rangle_Q = 1 + \sum_{s\to t\in Q} \alpha_s\alpha_t - \sum_i \alpha_i^2.\]
When \(\alpha\) is a root, \(2p_Q(\alpha)\) is the dimension of \(\mathfrak{M}_\theta(Q,\alpha)\).
Examples Of Computations Of \(p(\alpha)\)
The parameter \(p(\alpha)\) conveys the meaning of number of parameters of representations of dimension \(\alpha\) and is of fundamental importance. Let’s see some examples,
we will use \(A_i\) to denote the Dynkin quiver with \(i\) vertices,
use \(\hat{A_i}\) to denote the extended Dynkin quiver with \(i+1\) vertices,
and \(F(\hat{A_i})\) to denote the framed extended Dynkin quiver with \(i+2\) vertices.
\(\hat{A_0}\)
The \(A_0\) quiver might sounds a bit too funny because it is empty. But we think of the extended \(A_0\) quiver \(\hat{A_0}\) to be the Jordan quiver which is a single vertex with exactly one loop. Whose representations are just assigning a vector space at that node.
Here \(n\) is the dimension of that representation space.
If we compute the inner product we have
\[\langle \alpha, \alpha\rangle = \alpha_0\alpha_0 - \alpha_0^2 = n^2 - n^2 = 0\]
\[p(\alpha) = 1 - \langle \alpha, \alpha\rangle = 1.\]
So \(\alpha\) corresponds to an isotropic imaginary root, and it has a single parameter.
\(F(\hat{A_0})\)
\[\langle \alpha, \alpha\rangle = 1 + a^2 - a - a^2 = 1 - a.\] \[p_Q(\alpha) = 1 - (1-a) = a.\]
For \(\theta=(a,-1)\), this \(\mathfrak{M}_\theta(Q,\alpha)\) is our classical Hilbert scheme \((\mathbb{C}^2)^{[a]}\).
For \(\theta=0\), it reduces to \(\mathfrak{M}(Q,\alpha)\) which is the symmetric product \(\mathrm{Sym}^a(\mathbb{C}^2)\). You can see that the stratifications of \(\mathrm{Sym}^a(\mathbb{C}^2)\) corresponds to the partitions of \(a\), which perfectly matches the decomposition of the dimension vector \(\alpha\).
\[\alpha = (1,0) + a_1(0,1) + \dots + a_k(0,1)\] \[\begin{align*} \mathrm{Sym}^a(\mathbb{C}^2) &= \bigsqcup_{a_1+\dots+a_k=a} S^n_{a_1,\dots,a_k}\mathbb{C}^2\\ &= \bigsqcup_{a_1+\dots+a_k=a} \left\{ \sum_{i\le k} a_i[x_i] : x_i \text{ distinct} \right\} \end{align*}\]
\(F(\hat{A_1})\)
\[\langle \alpha, \alpha\rangle = 1 + a^2 + b^2 - a - 2 ab = 1 + (a-b)^2 - a\]
\[p_Q(\alpha) = a - (a-b)^2.\]
\(F(\hat{A_2})\)
\[\langle \alpha, \alpha\rangle = 1 + a^2 + b^2 + c^2 - a - ab - bc - ac = 1 + \frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2} - a.\]
\[p_Q(\alpha) = a - \frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2}.\]
\(F(\hat{A_3})\)
\[\begin{align*} \langle \alpha, \alpha\rangle &= 1 + a^2 + b^2 + c^2 + d^2 - a - ab - bc - cd - da \\ &= 1 + \frac{(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2}{2} - a. \end{align*}\]
\[p_Q(\alpha) = a - \frac{(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2}{2}.\]
Examples Of The Semistable Set \(\Sigma_{\lambda,\theta}\)
Computing \(\Sigma_{\lambda,\theta}(F(\hat{A_1}))\)
The situation we are considering is \(\lambda = 0\), and \(\theta = (n,-1)\). The case \(\theta=0\) might also be interesting, we might consider it later.
\[ p = n \]
\[ p = n - 1 \]
\[ p = n - 4 \]
\[ p = n - m^2 \]
Since all possible decompositions that can happen in \(R_\theta^+\) look like
\[(1,n,a+b) = (1,n,a) + (0,0,b)\]
We conclude that
\[\Sigma_{\lambda,\theta}(F\hat{A_1}) = \{(1,n,n-m) : m^2\le n\} \cup \{(0,0,k):k\ge 0\}.\]
And the canonical decomposition of \(\alpha\) is just \(\alpha\).
Stratums corresponds to the decomposition of vectors
\[(1,n,n) = (1,n,n-m) + m(0,0,1), \quad 0\le m \le \lfloor \sqrt{n} \rfloor.\] \[ p(1,n,n-m) = n - m^2, \quad p(0,0,1) = 0.\]
therefore there will be \(\lfloor \sqrt{n} \rfloor+1\) strata.