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Ben Webster
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I don't know where this is written, but in Crawley-Boevey's notes (http://www1.maths.leeds.ac.uk/~pmtwc/quivlecs.pdf), there's a classification which is uniform for the different cases, and which should be easy to extend. I believe the answer is just that you take one of the "obvious" representations over a finite extension of your field, and restrict scalars. This shows that reps are classified by orbits under the absolute Galois group of representations of the algebraic closure of your field. Note, this is really just taking seriously the idea that simple reps are classified (up to a few weird points) by points in $\mathbb{P}^1_k$. For a non-algebraically closed fields, you have to include the points with residue field given by a finite extension.

EDIT: To put this is more universal terms: following Crawley-Boevey, for a choice of extending vertex, we have modules $P$ and $L$, which are defined over the integers, with $\operatorname{Hom}(P,L)\cong \mathbb{Z}^2$. Let $\alpha,\beta$ be a basis of this space. Then, we have a universal family over $\mathbb{P}^1_{\mathbb{Z}}$, given by the cokernel of the map from the trivial bundle with fiber $P$ to the tensor product of $L$ with $\mathcal{O}(1)$, given by $x\alpha+y\beta$ (for $x,y$ a basis of the sections of $\mathcal{O}(1)$). For any field $k$, we can base-change to $\mathbb{P}^1_{k}$. The residue at a given closed point (with finitely many exceptions) in $\mathbb{P}^1_{k}$ is a simple regular representation and every simple regular representation appears as a composition factor at a unique point (and at the points where the rep is not simple, it is uniserial, with all simples over that point appearing once). This tells you almost everything you need for an arbitrary classification, since you only have extensions between regular simples corresponding to the same point in $\mathbb{P}_k^1$.

It's just important to note that $\mathbb{P}^1_{k}$ doesn't just have points given by lines in $k^2$: there's one point for each irreducible homogeneous polynomial in $k[x,y]$, up to scalar. Such a polynomial factors over the algebraic closure into a product over an orbit of the absolute Galois group (raised to some power if the polynomial isn't separable). Hence, the appearance of those.

I don't know where this is written, but in Crawley-Boevey's notes (http://www1.maths.leeds.ac.uk/~pmtwc/quivlecs.pdf), there's a classification which is uniform for the different cases, and which should be easy to extend. I believe the answer is just that you take one of the "obvious" representations over a finite extension of your field, and restrict scalars. This shows that reps are classified by orbits under the absolute Galois group of representations of the algebraic closure of your field. Note, this is really just taking seriously the idea that simple reps are classified (up to a few weird points) by points in $\mathbb{P}^1_k$. For a non-algebraically closed fields, you have to include the points with residue field given by a finite extension.

I don't know where this is written, but in Crawley-Boevey's notes (http://www1.maths.leeds.ac.uk/~pmtwc/quivlecs.pdf), there's a classification which is uniform for the different cases, and which should be easy to extend. I believe the answer is just that you take one of the "obvious" representations over a finite extension of your field, and restrict scalars. This shows that reps are classified by orbits under the absolute Galois group of representations of the algebraic closure of your field. Note, this is really just taking seriously the idea that simple reps are classified (up to a few weird points) by points in $\mathbb{P}^1_k$. For a non-algebraically closed fields, you have to include the points with residue field given by a finite extension.

EDIT: To put this is more universal terms: following Crawley-Boevey, for a choice of extending vertex, we have modules $P$ and $L$, which are defined over the integers, with $\operatorname{Hom}(P,L)\cong \mathbb{Z}^2$. Let $\alpha,\beta$ be a basis of this space. Then, we have a universal family over $\mathbb{P}^1_{\mathbb{Z}}$, given by the cokernel of the map from the trivial bundle with fiber $P$ to the tensor product of $L$ with $\mathcal{O}(1)$, given by $x\alpha+y\beta$ (for $x,y$ a basis of the sections of $\mathcal{O}(1)$). For any field $k$, we can base-change to $\mathbb{P}^1_{k}$. The residue at a given closed point (with finitely many exceptions) in $\mathbb{P}^1_{k}$ is a simple regular representation and every simple regular representation appears as a composition factor at a unique point (and at the points where the rep is not simple, it is uniserial, with all simples over that point appearing once). This tells you almost everything you need for an arbitrary classification, since you only have extensions between regular simples corresponding to the same point in $\mathbb{P}_k^1$.

It's just important to note that $\mathbb{P}^1_{k}$ doesn't just have points given by lines in $k^2$: there's one point for each irreducible homogeneous polynomial in $k[x,y]$, up to scalar. Such a polynomial factors over the algebraic closure into a product over an orbit of the absolute Galois group (raised to some power if the polynomial isn't separable). Hence, the appearance of those.

Source Link
Ben Webster
  • 44.7k
  • 12
  • 126
  • 260

I don't know where this is written, but in Crawley-Boevey's notes (http://www1.maths.leeds.ac.uk/~pmtwc/quivlecs.pdf), there's a classification which is uniform for the different cases, and which should be easy to extend. I believe the answer is just that you take one of the "obvious" representations over a finite extension of your field, and restrict scalars. This shows that reps are classified by orbits under the absolute Galois group of representations of the algebraic closure of your field. Note, this is really just taking seriously the idea that simple reps are classified (up to a few weird points) by points in $\mathbb{P}^1_k$. For a non-algebraically closed fields, you have to include the points with residue field given by a finite extension.