Questions tagged [wild-representation-type]

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Tame-Wild dichotomy; why can't tame algebras be wild?

I would like to understand the Tame-Wild dichotomy, and in particular why an algebra cannot be tame and (semi-)wild at the same time. I've looked in the papers by Drozd and Crawley-Boevey [D80, CB88]. ...
Jacob FG's user avatar
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6 votes
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Condition of indecomposability in the definition of wild representation type

In Tame and wild matrix problems, Drozd defines an algebra $A$ to be wild if there is an $A-k\langle x,y \rangle$-bimodule $M$, such that ${M \otimes}-$ reflects isomorphisms. Usually I've seen the ...
Jacob FG's user avatar
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2 votes
1 answer
213 views

Tame/wild classification of *cyclic* quivers?

There is a famous classification of the path algebras of finite acyclic quivers into finite, tame, and wild representation types. For quivers with cycles, it is standard that the 2-loop quiver (with ...
Joshua Grochow's user avatar
3 votes
1 answer
153 views

Explicit proof that algebra is derived wild

Following the terminology of Drozd, Yuriy A., Derived tame and derived wild algebras, Algebra Discrete Math. 2004, No. 1, 57-74 (2004). ZBL1067.16028. let $A$ and $R$ be algebras over a field $k$. A ...
Jacob FG's user avatar
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1 vote
0 answers
358 views

Infinite-dimensional representation theory of $K[x]$

Let $K$ be an algebraically closed field. The finite-dimensional representation theory of the polynomial algebra $K[x]$ is tame and completely understood, which I shall first summarise. It's ...
Iteraf's user avatar
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3 votes
0 answers
97 views

Infinite-dimensional wild commutative algebras with non-trivial units

Let $A$ be an arbitrary (not necessarily finite-dimensional) associative algebra over an algebraically closed field $K$, and let $\mathrm{fin\,}A$ denote the category of finite-dimensional $A$-modules....
Iteraf's user avatar
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1 vote
0 answers
50 views

Regular partial tilting modules in wild hereditary algebras

Let $k$ be an algebraically closed field. Let $Q$ be a connected wild quiver. Let $\mathcal{R}_1$, $\mathcal{R}_2$ be two regular components of the Auslander-Reiten quiver of $kQ$ that contain stones. ...
Ying Zhou's user avatar
  • 417
2 votes
0 answers
37 views

Morphisms from quasi-simple regular rigid $\Lambda$-module $M$ to $\tau^m M$ when $\Lambda$ is wild hereditary

Let $\Lambda$ be a finite dimensional basic wild hereditary algebra and $M$ be an indecomposable regular quasi-simple right-$\Lambda$ that is rigid in a regular component $\mathcal{R_i}$ of the ...
Ying Zhou's user avatar
  • 417
2 votes
0 answers
40 views

Predecessors and Successors of regular silting objects in bounded derived categories of wild hereditary algebras

Let $\Lambda$ be a wild hereditary algebra and let $T$ be one of its regular silting objects (i.e. all indecomposable direct summands of $T$ are shifts of indecomposable regular modules). What do we ...
Ying Zhou's user avatar
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3 votes
0 answers
107 views

Tensor Functors and Representations of Wild Quivers

A finite-dimensional $K$-algebra $A$ is of wild representation type if and only if there exists a $K\left<t_1,t_2\right>$-$A$-bimodule $_{K\left<t_1,t_2\right>}M_A$ such that the left $K\...
Iteraf's user avatar
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3 votes
1 answer
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Is it possible to classify the indecomposable representations of the wild quiver with one vertex and two arrows using infinite sets of parameters?

I have read several other questions about wild representation type but may I ask..what actually can be done and what have been done (such as partial results) about classifying indecomposable ...
Ying Zhou's user avatar
  • 417
13 votes
1 answer
689 views

Why is the A6 preprojective algebra of wild representation type?

As mentioned in the title, I would like to know a proof of the "well known" fact that the A6 preprojective algebra is of wild representation type. Ideally, I would like to see an explicit two-...
Peter McNamara's user avatar