5
$\begingroup$

Let $Q$ be a quiver without loops (cycles of length 1). Kac proved that if $K$ is algebraically closed, the dimension vectors of indecomposable representations of $Q$ over $K$ are exactly the positive real roots and the imaginary roots; further, there is a single isomorphism class of representations if the root is real, and infinitely many if the root is imaginary.

My question is: what is known if we drop the assumption that $K$ is algebraically closed? For real roots, this was answered by Schofield (The field of definition of a real representation of a quiver $Q$. Proc. AMS 116 (1992), no. 2, 293--295) --- he showed one can drop the assumption and nothing changes.

My go-to textbook for quiver representations over non-algebraically closed ground fields (Deng, Du, Parshall, Wang, "Finite dimensional algebras and quantum groups") in this case only gives the result for algebraically closed fields.

Edited to add: I am writing an expository note which mostly works over an arbitrary ground field, but I would also like to mention Kac's theorem, so I feel like it's incumbent on me to say something about what the state of our knowledge is about Kac's theorem for arbitrary ground fields. I guess mainly my focus is on the question of what vectors can appear as dimension vectors of indecomposable representations --- is even that much known generally?

Also: there is a remark by Jan Schröer here from 2016 to the effect that in full generality, we do not have an analogue of Kac's theorem.

$\endgroup$
2
  • 2
    $\begingroup$ This is probably not the case you are interested in, but for $Q$ of (extended) Dynkin type, the result should follow from the explicit classification of modules, see e.g. [Dlab, Ringel: Indecomposable representations of graphs and algebras] or [Ringel: Tame algebras and integral quadratic forms, Section 3.6]. $\endgroup$ Commented Feb 27, 2018 at 16:29
  • 2
    $\begingroup$ Do you want to reduce to infinite fields (or what level of generality)? For finite fields, the number is of course finite for trivial reasons. In that case, another result by Kac shows that the number is given by a polynomial in $q$, the cardinality of the ground field. $\endgroup$ Commented Feb 27, 2018 at 16:51

0

You must log in to answer this question.

Browse other questions tagged .