A finite-dimensional associative $\mathbf{k}$-algebra $\mathbf{k}Q/I$ is of tame representation type if for each dimension vector $d\geq 0$, with the exception of maybe finitely many dimension vectors $d$  representations**\***, the indecomposable representations of $\mathbf{k}Q/I$ with that dimension vector can be described up to isomorphism as finitely many one-parameter families, the parameter coming from $\mathbf{k}$. 


What is an illustrative example of a tame algebra? Specifically, what's an example of a quiver $Q$ and admissible ideal $I$ such that (1) for some dimension vectors $d$ the indecomposables can't be described as finitely many one-parameter families**\***, and (2) for some dimension vectors there is more than just one one-parameter family.


I ask because my go-to example of a tame algebra now is the path algebra of the Jordan quiver, the quiver having one vertex and one loop, over an algebraically closed field. But this example doesn't utilize all the wiggle-room that the definition of a tame algebra allows. So I'm hoping there is a better quintessential example to keep in mind.




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**\*** Note that, as it was originally written, this was not the correct definition of an algebra having tame representation type, and actually condition (1) is not possible. See [the comments below](https://mathoverflow.net/questions/352687/whats-an-illustrative-example-of-a-tame-algebra#comment884944_352687) for the correct definition.



What's an illustrative example of a tame algebra?
I think the following is an example of a tame algebra where there is more than one component to a moduli space of fixed dimension. I don't know any examples where there are dimension vectors with moduli of dimension $>1$.


Take a quiver with two vertices $1$ and $2$, two arrows $x\_1$ and $x\_2$ from $1$ to $2$ and two arrows $y\_1$ and $y\_2$ from $2$ to $1$. Impose the relations $x\_i y\_j = 0$ and $y\_j x\_i = 0$ for $1 \leq i,j \leq 2$. I believe every indecomposable representation either satisfies $x\_1=x\_2=0$ or $y\_1=y\_2=0$. Thus, every indecomposable representation is a representation of either the Kronecker quiver $1 \rightrightarrows 2$ or else $1 \leftleftarrows 2$, both of which are tame, so this is tame. 


For each dimension vector of the form $(n,n)$, we get two families of indecomposable representations, coming from choosing whether to make $x\_1=x\_2=0$ or else $y\_1=y\_2=0$.


