Author: Eiko

Tags: quiver variety, algebraic geometry, representation theory, Dynkin, extended Dynkin

Time: 2024-12-08 23:28:08 - 2024-12-09 18:28:44 (UTC)

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 $\iff$ Dynkin)

  • Indecomposable Representations $\iff$ 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 $\bar{Q}$ and associated to doubled quiver we have a moment map ${\mu }_{\bar{Q},\alpha }:\mathrm{Rep}(Q,\alpha )\to \mathrm{Lie}(G_{\alpha }{)}^{\ast }$.

• Quiver varieties

  $${\mathfrak{M}}_{\theta }(Q,\alpha )={\mu }_{\bar{Q},\alpha }^{-1}(0)/{/}_{\theta }{G}_{\alpha },\mathfrak{M}(Q,\alpha )={\mu }_{\bar{Q},\alpha }^{-1}(0)//G_{\alpha }.$$

• Parameter function

  $$p_Q(\alpha )=1-⟨\alpha ,\alpha {⟩}_Q=1+\sum _{s\to t\in Q}{\alpha }_s{\alpha }_t-\sum _i{\alpha }_i^2.$$

  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 $\widehat{A_i}$ to denote the extended Dynkin quiver with $i+1$ vertices,

• and $F(\widehat{A_i})$ to denote the framed extended Dynkin quiver with $i+2$ vertices.

$\widehat{A_0}$

The $A_0$ quiver might sounds a bit too funny because it is empty. But we think of the extended $A_0$ quiver $\widehat{A_0}$ 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

$$⟨\alpha ,\alpha ⟩={\alpha }_0{\alpha }_0-{\alpha }_0^2=n^2-n^2=0$$

$$p(\alpha )=1-⟨\alpha ,\alpha ⟩=1.$$

So $\alpha$ corresponds to an isotropic imaginary root, and it has a single parameter.

$F(\widehat{A_0})$

$$⟨\alpha ,\alpha ⟩=1+a^2-a-a^2=1-a.$$ $$p_Q(\alpha )=1-(1-a)=a.$$

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 $\alpha$.

$$\alpha =(1,0)+a_1(0,1)+\cdots +a_k(0,1)$$ $$\begin{array}{rl}{\mathrm{Sym}}^a({\mathbb{C}}^2)& =\underset{a_1+\cdots +a_k=a}{⨆}{S}_{a_1,\dots ,a_k}^n{\mathbb{C}}^2\\ & =\underset{a_1+\cdots +a_k=a}{⨆}\{\sum _{i\le k}{a}_i[x_i]:x_i\text{ distinct}\}\end{array}$$

$F(\widehat{A_1})$

$$⟨\alpha ,\alpha ⟩=1+a^2+b^2-a-2ab=1+(a-b{)}^2-a$$

$$p_Q(\alpha )=a-(a-b{)}^2.$$

$F(\widehat{A_2})$

$$⟨\alpha ,\alpha ⟩=1+a^2+b^2+c^2-a-ab-bc-ac=1+\frac{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2}{2}-a.$$

$$p_Q(\alpha )=a-\frac{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2}{2}.$$

$F(\widehat{A_3})$

$$\begin{array}{rl}⟨\alpha ,\alpha ⟩& =1+a^2+b^2+c^2+d^2-a-ab-bc-cd-da\\ & =1+\frac{(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2}{2}-a.\end{array}$$

$$p_Q(\alpha )=a-\frac{(a-b{)}^2+(b-c{)}^2+(c-d{)}^2+(d-a{)}^2}{2}.$$

Examples Of The Semistable Set ${\mathrm{\Sigma }}_{\lambda ,\theta }$

Computing ${\mathrm{\Sigma }}_{\lambda ,\theta }(F(\widehat{A_1}))$

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

$${\mathrm{\Sigma }}_{\lambda ,\theta }(F\widehat{A_1})=\{(1,n,n-m):m^2\le n\}\cup \{(0,0,k):k\ge 0\}.$$

And the canonical decomposition of $\alpha$ is just $\alpha$.

Stratums corresponds to the decomposition of vectors

$$(1,n,n)=(1,n,n-m)+m(0,0,1),0\le m\le \lfloor \sqrt{n}\rfloor .$$ $$p(1,n,n-m)=n-m^2,p(0,0,1)=0.$$

therefore there will be $\lfloor \sqrt{n}\rfloor +1$ strata.