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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.

user551642
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2 Answers2

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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.

user551642
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  • Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $\bigoplus_{\alpha:j\to i} X_{j} \to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $\dim\bigoplus_{\alpha:j\to i} X_{j} - \dim X_i$, but I can't see much more. – Mike Pierce Dec 18 '18 at 03:56
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    Oh yeah, and $\ker\left(\bigoplus_{\alpha:j\to i} f_{\alpha}\right)\neq\bigoplus_{\alpha:j\to i} \ker(f_{\alpha})$. Consider an $A_3$ quiver $1 \rightarrow 2 \leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1\oplus X_3 \to X_2$ certainly will. – Mike Pierce Dec 18 '18 at 04:03
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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 $i$.

Stephen
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