关于箭图$1\to 2\leftarrow 3$（即$Q = \bullet \to \bullet \leftarrow \bullet$）的表示分解的技术咨询

Hey there! Let's break this down step by step—you're already halfway there with the linear algebra basis argument, so we just need to connect that to the quiver representation direct sum decomposition.

First, let's clarify what a representation of this quiver actually is: for $Q = \bullet_1 \to \bullet_2 \leftarrow \bullet_3$, a representation consists of three vector spaces $(V_1, V_2, V_3)$ plus linear maps $f: V_1 \to V_2$ and $g: V_3 \to V_2$. As you noted, Kirillov reduced this to classifying triples where $V_1, V_3$ are subspaces of $V_2$—but that's specifically when $f$ and $g$ are injective. For general representations, we also need to account for the kernels of $f$ and $g$ (the parts of $V_1$ or $V_3$ that map to zero in $V_2$).

Let's map every piece of your linear algebra decomposition to the indecomposable representations Kirillov listed:

1. The kernel cases (the "zero-middle" representations) #

• Any vector in $\ker(f) \subseteq V_1$ (vectors that map to 0 in $V_2$) corresponds to the representation $\overset{k}{\bullet}\to \overset{0}{\bullet} \leftarrow \overset{0}{\bullet}$. Each such vector is a 1-dimensional copy of this indecomposable, so if $\dim(\ker(f)) = k$, we take $k$ direct sums of this.
• Similarly, vectors in $\ker(g) \subseteq V_3$ correspond to $\overset{0}{\bullet}\to \overset{0}{\bullet} \leftarrow \overset{k}{\bullet}$. The dimension of the kernel tells you how many copies to include in the direct sum.

2. The injective/image part (your basis decomposition) #

Now look at the images of the maps: let $Im(f) = U_1 \subseteq V_2$ and $Im(g) = U_3 \subseteq V_2$. You already know we can decompose $V_2$ into four disjoint subspaces relative to $U_1$ and $U_3$:

• $U = U_1 \cap U_3$ (vectors present in both images)
• $U_1' = U_1 \setminus U$ (vectors only in $Im(f)$)
• $U_3' = U_3 \setminus U$ (vectors only in $Im(g)$)
• $W = V_2 \setminus (U_1 + U_3)$ (vectors in $V_2$ not in either image)

Each of these subspaces maps directly to an indecomposable representation:

• $U$ (dimension $k$): Every vector here comes from both $V_1$ and $V_3$, so we have 1-dimensional spaces in all three vertices, with $f$ and $g$ acting as identity maps. This is exactly $\overset{k}{\bullet}\to \overset{k}{\bullet} \leftarrow \overset{k}{\bullet}$.
• $U_1'$ (dimension $k$): Vectors here only come from $V_1$, so we have a 1-dimensional $V_1$, 1-dimensional $V_2$ (containing the vector), and $V_3 = 0$. The map $f$ is identity, $g$ is zero. This is $\overset{k}{\bullet}\to \overset{k}{\bullet} \leftarrow \overset{0}{\bullet}$.
• $U_3'$ (dimension $k$): The mirror case—1-dimensional $V_3$, 1-dimensional $V_2$, $V_1=0$, map $g$ is identity, $f$ is zero. That's $\overset{0}{\bullet}\to \overset{k}{\bullet} \leftarrow \overset{k}{\bullet}$.
• $W$ (dimension $k$): Vectors here are in $V_2$ but don't come from either $V_1$ or $V_3$, so $V_1=0$, $V_3=0$, $V_2$ is 1-dimensional, both maps are zero. This is $\overset{0}{\bullet}\to \overset{k}{\bullet} \leftarrow \overset{0}{\bullet}$.

The big picture #

When you take the direct sum of all these pieces, you get back your original representation. The basis you chose for $V_0$ (which is $V_2$ in our notation) is exactly picking one basis vector from each of these primitive subspaces—each vector corresponds to one copy of an indecomposable representation, and putting them all together gives the full decomposition.

For example: suppose $V_1 = \mathbb{R}^2$, $V_2 = \mathbb{R}^3$, $V_3 = \mathbb{R}^1$. Let $f$ map $(1,0)\mapsto (1,0,0)$ and $(0,1)\mapsto (0,0,0)$ (so $\ker(f)$ is 1-dimensional), and $g$ map $1\mapsto (1,1,0)$. The decomposition would be:

• 1 copy of $\overset{1}{\bullet}\to \overset{0}{\bullet} \leftarrow \overset{0}{\bullet}$ (from $\ker(f)$)
• 1 copy of $\overset{1}{\bullet}\to \overset{1}{\bullet} \leftarrow \overset{0}{\bullet}$ (from $Im(f)$)
• 1 copy of $\overset{0}{\bullet}\to \overset{1}{\bullet} \leftarrow \overset{1}{\bullet}$ (from $Im(g)$)
• 1 copy of $\overset{0}{\bullet}\to \overset{1}{\bullet} \leftarrow \overset{0}{\bullet}$ (from the leftover part of $V_2$)

Adding these up gives exactly the original representation.

备注：内容来源于stack exchange，提问作者staedtlerr