# What's an illustrative example of a tame algebra?

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.

* 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 for the correct definition.

• The part 3 of the book by Skowronski and Simson has a chapter on this topic including smoe other tame examples. – Mare Feb 14 at 12:59
• I think this definition of tame is not quite standard (although it is perhaps equivalent to the standard definition). There usually is not an exception for any dimension vectors $d$. Instead it is required that for every dimension $d$, the indecomposable reps of dimension $d$, with finitely many exceptions, are described by a finite number of one-parameter families. If one allows constant one-parameter families, then the phrase "with finitely many exceptions" can be removed. – Alex Dugas Feb 14 at 22:30
• @AlexDugas Yeah, the definition is non-standard, because it's not correct. ;) Reading the definition here I misunderstood; there are not finitely many dimension vectors excepted, but finitely many indecomposables that live outside of a one-parameter family per dimension vector, just as you say. And according to what Hugh Thomas is saying below here and here, this is not equivalent to the correct definition. – Mike Pierce Feb 17 at 16:55

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 $$y_1=y_2=0$$.

• Very nice! And you can't do better: as soon as there is a 2-dimensional family, the representation theory must be wild. – Hugh Thomas supports Monica Feb 14 at 17:23

Any quiver whose graph is affine Dynkin graph $$\tilde{D}_4$$, $$I=0$$. If all arrows look at the center, this is related to the 4-subspace problem, which is tame.

• This has the same features the OP didn't like for the Jordan quiver over an algebraically closed field. For every dimension vector, the family of indecomposable representations is either $0$ or $1$ dimensional. The latter case only occurs for dimension vectors of the form $(n,n,n,n,2n)$; the $1$-dimensional family of modules represented by the following four maps $k^{2n} \to k^n$: $(\mathrm{Id}_n, 0)$, $(0, \mathrm{Id}_n)$, $(\mathrm{Id}_n, \mathrm{Id}_n)$, $(\mathrm{Id}_n, J_n(\lambda))$ where $J_n(\lambda)$ is the $n \times n$ Jordan block with $\lambda$ on the diagonal. – David E Speyer Feb 14 at 15:02
• @DavidESpeyer: I think perhaps you didn't write what you meant. That the families of indecomposable representations be zero or one-dimensional is equivalent to being tame. It is true that the example Bugs gives, and the more general example I gave, can be described as a single one-dimensional family plus finitely many extra points (which you disregard in your comment). I guess it should be possible to find algebras with honestly more than one one-dimensional family of indecomposables of some dimension, but we would have to leave hereditary algebras. – Hugh Thomas supports Monica Feb 14 at 15:18
• @HughThomassupportsMonica It sounds like you are saying that the OP's request (1) is impossible: To get a tame algebra where some of the moduli spaces have dimension $>1$? I thought that too, but I wasn't confident enough to say so. It looks like Bugs' example doesn't achieve the OP's (1) or (2). – David E Speyer Feb 14 at 15:23

For the Kronecker quiver (two vertices, two arrows in the same direction) and dimension vector (1,1), over an algebraically closed ground field, the indecomposables are naturally parameterized by points in $$\mathbb P^1(k)$$. (The representation with the two maps given by $$a$$ and $$b$$ is sent to $$[a:b]$$.

For other tame quivers with no relations over an algebraically closed ground field, the situation is slightly worse: the natural indexing set for the representations whose dimension vector is the null root is $$\mathbb P^1(k)$$ with some points (up to three of them) counted more than once (but finitely many times). This happens in the example Bugs gave: there are three inhomogeneous tubes, each of width two, each containing two representations of dimension vector the null root, whereas the other points of $$\mathbb P^1(k)$$ each correspond to one representation. (With the all-inward orientation, the reason for the indexing by $$\mathbb P^1(k)$$ is that the moduli space of 4 points on $$\mathbb P^1$$—equivalent to representations with dimension vector $$(1,1,1,1,2)$$, i.e., the null root—is again $$\mathbb P^1$$.)

I am not quite sure what you mean by the extra wiggle room of type (1). Are these supposed to be dimension vectors that have only finitely many indecomposables? I would usually think that in that case, they can also be described by one-parameter families: just make the families constant.