# Type A quivers and flag varieties from Kirilllov

I'm trying to understand section 10.7 of Kirillov's "Quiver representations and quiver varieties", which shares its title with this question. We let Q be a type A quiver of length $$\ell$$. We consider a framing $$W = (0,\dots,0,r)$$ and the framed then doubled quiver $$Q^\sharp$$. We want to show $$\mathcal M(\vec v, \vec w) \cong T^*\mathcal F(\vec v_1,\dots,\vec v_\ell,r)$$, the cotangent bundle to the partial flag variety with same dimension vector in $$\mathbb C^n$$. I understand the semistability conditions, but I don't understand the moment map condition. In Kirillov, the moment map in this setting is written as, for $$(z,i,j)$$ a representation of $$Q^\sharp$$, $$\mu(z,i,j) = \sum_{h \in H} \epsilon(h)z_hz_{\bar h} - \sum_k i_kj_k$$ where $$H$$ is the set of edges of $$Q$$ doubled, $$\epsilon$$ is the orientation function (+1 for $$x$$'s (arrows going right) and $$-1$$ for $$y$$'s (arrows going left)), and $$j_k$$ are the maps to the framings and $$i_k$$ are the maps out of them. This notation is very confusing to me. First of all, I believe the moment map should be a map into $$\mathfrak{gl}(\vec V)$$, i.e. a direct sum of $$\ell$$ components, since our Lie group is the product $$GL(\vec V)$$. I believe this should be more accurately written as $$\bigoplus_{\alpha=1}^\ell \sum_{t(h) =\alpha} \Big(\epsilon(h) z_hz_{\bar h} -i_\alpha j_\alpha\Big)$$ where $$i_\alpha,j_\alpha$$ are the framing morphisms at the vertex $$\alpha$$. Is this the correct reformulation? For example at the first vertex, i.e. the first component, I think $$\mu=0$$ implies $$y_1x_1=0$$, and at the final vertex, I have $$x_{\ell-1}y_{\ell-1} -ij =0$$.

Then Kirillov states the equation $$\mu=0$$ gives $$y_i|_{V_i} = y_{i-1}$$ and $$i|_{V_i} = y_{\ell-1}$$, which is equivalent to $$y_k = i|_{V_{k+1}}$$. I don't see how to obtain these conditions from the moment map condition. I'm not even sure how to make sense of these equations. For example, $$y_i$$ is a linear map $$y_i: V_{i+1}\to V_i$$, so what does $$y_i|_{V_i}$$ mean?

My specific case of the moment map in this scenario is also correct, that is it has the form $$\Big(-y_1x_1,-y_2x_2+x_1y_1,\dots,x_{\ell-1}y_{\ell-1}-ij \Big)$$ so setting $$\mu=0$$ yields $$y_1|_{im\ x_1} = 0, \quad y_2|_{im\ x_2} = x_1y_1,\quad\quad i|_{im\ j} = x_{\ell-1}y_{\ell-1}$$ From the semi-stability condition, one can prove that each $$x_i$$ and the map $$j$$ are all injective. Thus $$im\ x_i \cong V_i$$, so the above conditions can be written $$y_1|_{V_1} = 0, \quad y_2|_{V_2} = x_1y_1,\quad\quad i|_{V_\ell} = x_{\ell-1}y_{\ell-1}$$ Now we can jump straight to proving the final result Kirillov states in this section: $$i|_{V_{k+1}} = y_k$$. To understand this, first we recognize $$i: \mathbb C^r \to V_\ell$$. But by injectivity of $$j$$, we can always embed the image of $$i$$ back into $$\mathbb C^r$$, $$ji: \mathbb C^r\to \mathbb C^r$$. By an abuse of notation, we also call this map $$i$$. Because each $$x_i$$ and $$j$$ is injective, we can consider each $$V_i$$ as a subspace of $$\mathbb C^r$$. Thus we could ask what $$i|_{V_k+1}$$ is. Since each $$x_i$$ is an injective map, it follows that $$V_i$$ is isomorphic to a subspace of $$V_{i+1}$$. Therefore to restrict $$i$$ to the subspace $$V_{k+1}$$, it suffices to first restrict $$i$$ to $$V_\ell$$, then restrict to $$V_{\ell-1}$$, and so on until arriving at $$V_{k+1}$$. Doing this and using a relation from $$(*)$$ at each step, we conclude that $$i|_{V_{k+1}} = x_{\ell-1}x_{\ell-2}\dots x_k y_k$$ which is almost what Kirillov was asking us to show. It is exactly what he is asking us to show when you realize we are using the same abuse of notation that was employed with $$i$$: The maps $$x_i$$ serve to embed the image of $$y_k$$ back into $$\mathbb C^r$$. Because these are all isomorphisms onto their images, we can omit them from the notation. That is the conclusion of this part of the proof.