Author: Eiko

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

Time: 2024-10-10 09:43:43 - 2024-10-10 09:52:30 (UTC)

Correspondence

Given a representation $V$ of ${\mathrm{\Pi }}_Q^{\prime }$, we can form a representation of $e_0{\mathrm{\Pi }}_Q^{\prime }{e}_0$ by simply taking

$$V\mapsto e_{0}V.$$

To recover the reverse process, from a representation $V_0$ of $A_0=e_0{\mathrm{\Pi }}_Q^{\prime }{e}_0$ to a representation of ${\mathrm{\Pi }}_Q^{\prime }$, we can use the tensor product (the obvious base change operation)

$$\mathrm{Rep}(A_0)\to \mathrm{Rep}({\mathrm{\Pi }}_Q^{\prime })$$

$$V_0\mapsto {\mathrm{\Pi }}_Q^{\prime }{\otimes }_{A_0}{V}_0$$

Example

Consider $Q=A_2$, Let $V_0$ be a representation of $A_0=k[x_{01}{x}_{10},x_{01}{y}_{10},y_{01}{x}_{10},y_{01}{y}_{10}]$

If $V_0=⟨v_i⟩$ is a basis of $V_0$, then we have that

$$V={\mathrm{\Pi }}_Q^{\prime }{\otimes }_{A_0}{V}_0=e_{1}V\oplus e_{0}V=V_1\oplus V_0$$

where $V_1$ is generated by $x_{10}{V}_0+y_{10}{V}_0$.

i.e. the vectors $x_{10}{v}_i,y_{10}{v}_i$ certainly generates