Classify all isomorphism classes of representations of dimension vector 1 and 2 of the following quiver
The professor briefly did the solution, but I could not understand what was going on. What he did was that he assumed $A$ and $B$ be the associated matrices associated with the linear maps $ K \rightarrow K^2$ and $ K^2 \rightarrow K^2$. Then he assumed $A \neq 0$ WLOG. Finally, he computed all possible Jordan forms of $B$ to identify $A$. Could you guys help me to understand the solution.
Thanks,
So your goal is to classify all isomorphism classes of representations of dimension vector $(1,2)$, that is to write down a representative for each isomorphism class of representations with this dimension vector.
What is a representation of this quiver with dimension vector $(1,2)$?
A representation of this quiver with dimension vector $(1,2)$ is a $1$-dimensional vector space $V$ together with a $2$-dimensional vector space $W$ and linear maps $\alpha:V\to W$ and $\beta:W\to W$.
What does it mean for two such representations to be isomorphic?
Suppose you are given two such representations $(V,W,\alpha,\beta)$ and $(V',W',\alpha',\beta')$, then an isomorphism from $(V,W,\alpha,\beta)\to (V',W',\alpha',\beta')$ is a pair of bijective linear maps $\omega_1:V\to V'$ and $\omega_2:W\to W'$ such that $\alpha'\omega^1=\omega^2\alpha$ and $\beta'\omega_2=\omega^2\beta$.
Why do we just work with $K^n$ and matrices?
I claim that every such representation $(V,W,\alpha,\beta)$ is isomorphic to one of the form $(K,K^2,A,B)$, where $A$ is a $2\times 1$-matrix and $B$ is a $2\times 2$-matrix. To see that, let $\omega_1:V\to K$ be an arbitrary isomorphism (possible since $V$ is one-dimensional) and $\omega_2:W\to K^2$ be an arbitrary isomorphism (possible since $W$ is two-dimensional). Then, it is easy to see that $(K,K^2,\omega_2\alpha\omega_1^{-1}, \omega_2\beta\omega_2^{-1})$ is a representation isomorphic to $(V,W,\alpha,\beta)$. (Check this!). And any linear map $K^n\to K^m$ can be identified with a matrix.
Why did the professor assume $A\neq 0$ wlog?
Suppose $A=0$, then your representation $(K,K^2,A=0,B)$ is the direct sum of two representations, namely $(K,K^2,A=0,B)=(K,0,0,0)\oplus (0,K^2,0,B)$ and the representations $(0,K^2,0,B)$ are representations of the $1$-loop quiver, which are classified by Jordan normal form. So, up to isomorphism, $B$ can be chosen as $B=J_2(\alpha)$ or $B=\begin{pmatrix}\alpha&0\\0&\beta\end{pmatrix}$ (Check this!). So from now on, we can assume $A\neq 0$, because the representations with $A=0$ are already done (and none of them is indecomposable, they are just given by direct sums of representations with "smaller" dimension vector)
Why $B$ in Jordan normal form?
Now we are dealing with the problem of classifying representations of the form $(K,K^2,A\neq 0, B)$. As a first step, there is a matrix $S$, such that $S^{-1}BS$ is in Jordan normal form, i.e. $(K,K^2,A,B)\cong (K,K^2,S^{-1}A,S^{-1}BS)$, where the isomorphism is given by $(1,S^{-1})$. Thus, we can assume that $B$ is in Jordan normal form.
How to finish the classification?
Now, we don't want to change $B$ anymore, since we cannot expect anything easier than Jordan normal form (since this is the classification of just the $1$-loop quiver). What isomorphisms we are allowed to take depends on the shape of $B$. There are three different possibilities, either $B=\alpha\cdot \operatorname{Id}$, $B=\begin{pmatrix}\alpha&0\\0&\beta\end{pmatrix}$, or $B=\begin{pmatrix}\alpha&1\\0&\alpha\end{pmatrix}$. In each of these cases we are allowed to take different isomorphisms $\omega_2$ without changing $B$.
The easiest case is $B=\alpha\cdot \operatorname{Id}$. Since in this case, $B$ is in the centre of the matrix ring, we are allowed to take any matrix $\omega_2$ we want and we will not change the matrix $B$ in the isomorphic representation. So, we choose a matrix, that will sent $A1$ to $\begin{pmatrix}1\\0\end{pmatrix}$ and are left with a representation of the form $(K,K^2,\begin{pmatrix}1\\0\end{pmatrix},\alpha\cdot \operatorname{Id})$
The next case is $B=\begin{pmatrix}\alpha&0\\0&\beta\end{pmatrix}$ with $\alpha\neq \beta$. The matrices which won't change $B$ are of the matrices of the form $\begin{pmatrix}a&0\\0&d\end{pmatrix}$. Hence, by base change in the codomain, we can only achieve that the matrix $A$ is of the form $\begin{pmatrix}1\\0\end{pmatrix}$, $\begin{pmatrix}0\\1\end{pmatrix}$, or $\begin{pmatrix}1\\1\end{pmatrix}$. Some of these representations are isomorphic by changing the position of $\alpha$ and $\beta$. (Figure out which ones exactly!)
The third case is the case $B=J_2(\alpha)$, this is not changed by matrices of the form $\begin{pmatrix}a&c\\0&a\end{pmatrix}$. Hence, by base change with these matrices in the codomain, we can only achieve that $A$ is of the form $\begin{pmatrix}1\\0\end{pmatrix}$ or $\begin{pmatrix}0\\1\end{pmatrix}$. Here are $2$ possible representations for each $\alpha$.
Further remarks
It is also a good exercise to check, which of the above representations are in fact indecomposable.
Somewhat surprisingly the classification of all indecomposable representations (of arbitrary dimension vector) for this quiver turns out to be impossible in some sense (the quiver is wild).