Reflection Functor of a Quiver

Question

Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $j\in Q_{0}$ define $\sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.

Suppose $i\in Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)\to Rep(\sigma_{i}Q).$$

The functor is defined as follows; given a representation $X\in Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $j\in Q_{0}$ and a linear map $f_{\alpha}:X_{j}\to X_{k}$ for each $\alpha: j\to k\in Q_{1}$, define $$(F_{i}^{+}X)_{j}=\begin{cases} X_{j} &j\neq i,\\ \ker\left(\bigoplus_{\alpha:j\to i} f_{\alpha}\right) & i=j. \end{cases}$$

Alternatively one can think of the sequence $$ 0\longrightarrow (F_{i}^{+}X)_{i} \longrightarrow \bigoplus_{\substack{\alpha\in Q_{1}\\ \alpha:j\to i}} X_{j}\longrightarrow X_{i}$$ in which the first map is the trivial map and the second is inclusion.

My question is what is the third map and how do I think about $\ker\left(\bigoplus_{\alpha:j\to i} f_{\alpha}\right)$?

My guess is that you take the direct sum of all the vector spaces $X_{\alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},\ldots,x_{n})\in\bigoplus_{\substack{\alpha\in Q_{1}\\ \alpha:j\to i}} X_{j}$ is mapped to $(f_{1}x_{1},\ldots,f_{n}x_{n})$ where $f_{j}:X_{j}\to X_{i}$ is the map in the representation of $Q$.

Is this correct?

If so does this mean that $\ker\left(\bigoplus_{\alpha:j\to i} f_{\alpha}\right)=\bigoplus_{\alpha:j\to i} \ker(f_{\alpha})$?

Finally, do we have any information about the dimension of $\ker\left(\bigoplus_{\alpha:j\to i} f_{\alpha}\right)$ in general?

As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.

Answer 1

I think I have figured it out;

let $h:\bigoplus_{\substack{\alpha\in Q_{1}\\ \alpha:j\to i}} X_{j} \to X_{i}$ be the third map in the sequence then we define $h(x_{1},\ldots,x_{n}):=f_{1}x_{1}+\cdots+f_{n}x_{n}$, where $f_{j}:X_{j}\to X_{i}$ is the map associated to the edge $\alpha:j\to i\in Q_{1}$. Then we have $$\ker(h)=\{(x_{1},\ldots,x_{n})\in\bigoplus_{\substack{\alpha\in Q_{1}\\ \alpha:j\to i}} X_{j}: f_{1}x_{1}+\cdots+f_{n}x_{n}=0\}.$$

Although I still don't know anything about the dimension in general.

Answer 2

If the representation is indecomposable, the map $\oplus_{\alpha:j \to i} f_\alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.