I am having trouble understanding the proof of lemma 3.2 in Nakajima's paper "Quiver varieties and tensor products," available on arxiv. This is where he shows that the fixed components of a subtorus of the residual action on a quiver variety are themselves quiver varieties.
Fix a decomposition of dimension vector $w = w' + w''$, and the corresponding decomposition of framed spaces $W = W^1 \oplus W^2$. Define a one parameter subgroup $\lambda: \mathbb C^* \to G_W$ by $$ \lambda(t) = \text{id}_{W^1} \oplus t \text{id}_{W^2} \in G_{W^1} \times G_{W^2} \subset G_W $$
Lemma 3.2: The fixed point set of $\lambda(\mathbb C^*)$ in $\mathfrak M(v, w)$ is isomorphic to $\sqcup_{v^1 + v^2 = v}\mathfrak M(v^1,w^1) \times \mathfrak M(v^2,w^2)$.
Proof: $[B,i,j] \in \mathfrak M(v,w)$ is fixed by $\lambda(\mathbb C^*)$ iff there exists a one-parameter subgroup $\rho: \mathbb C^* \to G_V$ such that $$ (\star) \quad \quad \lambda(t) * (B,i,j) = \rho(t)^{-1}(B,i,j) $$ This is just unwinding the definitions of $\mathfrak M(v,w)$, but I am not certain what the action on either side of this equation is. I think Nakajima defines $*$ towards the top of page 13 where he says:
Let $G_W = \Pi_{k \in I}GL(W_k)$. It acts naturally on M, $\mathfrak M$ and $\mathfrak M_0$. We define $\mathbb C^*$-actions on $\mathfrak M$ and $\mathfrak M_0$ by $$ B_h \mapsto t^{m(h) + 1}B_h,\quad i \mapsto ti, \quad j \mapsto tj \quad (t \in \mathbb C^*) $$
For simplicity, let's take $m \equiv 0$.
We denote this action by $(B,i,j) \mapsto g * (B,i,j)$ or $[B,i,j] \mapsto g*[B,i,j]$ for $g \in G_W \times \mathbb C^*$.
- Here I believe B is the collection of morphisms $V \to V$, and $i$ is the collection $V \to W$ and $j$ is the collection $W \to V$. Concretely, what is this action? It seems to me that both $G_W$ and $\mathbb C^*$ want to act on the $i$ and $j$ maps. Nakajima writes "this action" but it looks to me like he just described two different ones.
Let $V^1$ (resp. $V^2)$ be the eigenspace of $V$ with eigenvalue 1 (resp. $t$). Let $V'$ be the sum of other eigenspaces. The above equation implies that $$ 1) B(V^1) \subset V^1, B(V^2) \subset V^2, B(V') \subset V' $$ $$ 2) i(W^1) \subset V^1, i(W^2) \subset V^2 $$ $$ 3) j(V^1) \subset W^1, j(V^2) \subset W^2, j(V') = 0 $$
- The equation above is equation $(\star)$. Am I right that these are eigenspaces with respect to $\rho^{-1}(t)$? And how do I see this conclusion?
Assuming $m(h)=1$ for all $h$, there is a $\mathbb G_m$ action on $\mathfrak M$ given on $$ E(V,W) = E(V,V)\oplus L(W,V)\oplus L(V,W), $$ by $t.(B,i,j) = (tB,ti,tj)$. Since it commutes with the $G_V$ action, it descends to the quotient $\mathfrak M$. Similarly, $G_W$ acts on $E(V,W)$ by $g(B,i,j) = (B,ig^{-1},gj)$, and this also descends to $\mathfrak M$. Since they commute, these define an action of $G_W\times \mathbb G_m$, which is the action that Nakajima denotes by a "$*$". Thus $$ (t,g)\star(B,i,j) = (t.B,tig^{-1},tgj). $$
The action of $\rho(\mathbb C^\times)$ is via $G_V$, whereas the action of $\lambda$ is via $G_W$. As you pointed out, the $G_W$-action leaves $B$ invariant, hence if $\rho(t)(B,i,j) = \lambda(t)*(B,i,j)$ we must have $\rho(t)(B)=B$. But this exactly means that the map $B_h$ for an edge $h=(i\to j)$ must map $V_i^{\alpha}$, the $\alpha$-weight space of $\rho(\mathbb G_m)$ in $V_i$ into the $\alpha$-eigenspace of $\rho(\mathbb C^\times)$ in $V_j$, which yields the inclusions $(1)$ $(2)$ and $(3)$ that you quote.