I'm working on a paper about representation of quivers and Gabriel's theorems. See this .pdf if you're interested ; but I guess you can answer my question without knowing anything about quivers, or at least I'll try to simplify it so that quivers are not involved.
Let $n \in \mathbb{N}$ and $\alpha = (\alpha_{1}, \dots, \alpha_{n}) \in \mathbb{N}^{n}$. I'm asked to prove that $\prod\limits_{i = 1}^{i= n-1} \text{Hom}(\mathbb{C}^{\alpha_{i}}, \mathbb{C}^{\alpha_{i+1}})$ is an affine variety which is isomorphic to $\prod\limits_{i = 1}^{i= n-1} \mathbb{C}^{\alpha_{i}}\times \mathbb{C}^{\alpha_{i+1}}$ , and thus it is irreducible and its dimension is $\sum\limits_{i=1}^{i=n-1} \alpha_{i}\alpha_{i+1}$.
Problem is I know nothing about algebraic geometry, and so I really find it hard to understand truely the words I put in bold.
Then, assume there is a group action of a given group $G$ on $\prod\limits_{i = 1}^{i= n-1} \text{Hom}(\mathbb{C}^{\alpha_{i}}, \mathbb{C}^{\alpha_{i+1}})$.
I'm told two things which, I guess, are close to each other :
- if the codimension of every orbit is $\geq 1$, then there is an infinity of orbits .
- if there is a finite number of orbits, then one of those orbits must be a dense open set.
How would you justify that ?
Thanks for any help or reference.
Your question seems to indicate that you are working over $\mathbb{C}$, so the answer will be restricted to that setting as well. Note, however, that all the definitions work equally well if you replace $\mathbb{C}$ by an arbitrary (but fixed) field $K$ and that the proof will then still be sound if K is an infinite field.
"Geometric" preliminaries
(Here and sometimes in the sequel I write "geometric" to indicate that, while the concepts used are geometric, this answer is designed to bypass geometric intuition and to avoid using results from algebraic geometry proper. Instead, I will try to frame everything in the language of commutative algebra and will deal with just those special cases that are needed for the proof of Gabriel's theorem. As a result, the following exposition will appear rather crude to readers familiar with algebraic geometry).
Definitions
Let $\mathsf{Rings}_\mathbb{C}$ denote the category of commutative and unitary $\mathbb{C}$-algebras, $\mathsf{Sets}$ the category of sets and $\mathsf{Groups}$ the category of groups.
Note that, using the above notation, the Yoneda Lemma implies that $A$ is uniquely determined by $X$ up to isomorphism. I will write $A = K[X]$ and call $A$ the representing algebra of $X$.
An affine scheme is called integral if and only if its representing algebra is an integral domain. (You will be able to work with this notion instead of that of an irreducible variety.)
The dimension of an integral affine scheme $X$ is $$ \dim X = \operatorname{trdeg}_\mathbb{C} \operatorname{Frac} K[X], $$ i.e. the transcendence degree of the field of fractions of its representing algebra over $\mathbb{C}$. (You will only need the notion of dimension for integral affine schemes.)
An affine group scheme is a representable functor $G\colon \mathsf{Rings}_\mathbb{C} \to \mathsf{Groups}$. Note that an affine group scheme also "is" an affine scheme (by composing it with the forgetful functor $\mathsf{Groups} \to \mathsf{Sets}$) and we can thus speak of its representing algebra, integral affine group schemes and their dimension.
Let $G$ be an affine group scheme and let $X$ be an affine scheme. A group action of $G$ on $X$ is a natural transformation $$ G \times X \to X $$ (where $G \times X$ denotes the usual product of $\mathsf{Sets}$-valued functors) such that for all $\mathbb{C}$-algebras $B$ the map of sets $$ G(B) \times X(B) \to X(B) $$ is an action of $G(B)$ on $X(B)$ in the usual sense.
Basic facts and examples
You will encounter only products of the following integral affine schemes:
The universal properties of polynomial rings, localization and cokernels show that, for a $\mathbb{C}$-algebra $A$, the sets $\operatorname{Mat}_{m \times n}(A)$, $\operatorname{GL}_n(A)$ and $\operatorname{SL}_n(A)$ can be identified with those groups of matrices having entries in $A$ that are usually denoted by this notation (i.e., the set of all $m \times n$-matrices, the general linear group and the special linear group, respectively). In particular, we see that $\operatorname{GL}_n$ and $\operatorname{SL}_n$ are indeed affine group schemes.
In the sequel, I will use this identification without mention, e.g. by saying that one can multiply an element of $\operatorname{GL}_m(A)$ from the left with one of $\operatorname{Mat}_{m \times n}(A)$ (usual matrix multiplication).
You will also need some facts about products of affine schemes, namely the following:
Application to the proof of Gabriel's theorem
(I will freely use definitions and notation from the PDF linked to in the question. For the benefit of possible other readers, I will include some background on the context that is likely known to you, @krirkrirk. Feel free to just look at the proof of the dimension inequality in the end.)
Fix a quiver $Q$ whose underlying undirected graph is conneceted. Gabriel's theorem says
Outline of how to prove the "only if" part
Proof strategy
An elementary argument shows that the underlying graph of $Q$ is an $ADE$-Dynkin diagram if and only if the Tits form $q$ of $Q$ is positive definite. The "only if" part of Gabriel's theorem can therefore be proved by the following strategy (if I recall it correctly, Gabriel wrote in his original paper on Gabriel's theorem that the proof by this strategy is due to Tits):
For any non-zero vector $\alpha \in (\mathbb{Z}_{\geq 0})^{Q_0}$, find an integral affine group scheme $G_\alpha$ and an integral affine scheme $X_\alpha$ ($G_\alpha$ and $X_\alpha$ will, as most objects we consider, also depend on the quiver) together with a group action of $G_\alpha$ on $X_\alpha$ satisfying the following properties
The second property will yield $\dim G_\alpha \geq \dim X_\alpha$ by a "geometric" proof. Thus, $q(\alpha) \geq 1$ by the first property. One can then make an elementary argument that this implies that $q$ is positive definite (note that one still has to extend the inequality to the case where $\alpha$ is an arbitrary non-zero vector in $\mathbb{R}^{Q_0}$).
Definition of $G_\alpha$ and $X_\alpha$
How to carry out this strategy? The following will do: Fix any vertex $y \in Q_0$ and put
$$ G_\alpha = \operatorname{SL}_{\alpha_y} \times \prod_{x \in Q_0 \setminus \lbrace y \rbrace} \operatorname{GL}_{\alpha_x} , $$
$$ X_\alpha = \prod_{a \in Q_1} \operatorname{Mat}_{\alpha_{ta} \times \alpha_{ha}} . $$
By the dimension formulas given above, the first of the desired two properties holds for this choice.
What about the group action? For a $\mathbb{C}$-algebra $A$ it is defined by
$$G_\alpha(A) \times X_\alpha(A) \to X_\alpha(A), ((\varphi_x)_{x \in Q_0},(f_a)_{a \in Q_1}) \mapsto (\varphi_{ha} \cdot f_a \cdot \varphi_{ta}^{-1})_{a \in Q_1} , $$
where $\cdot$ and $^{-1}$ denote matrix multiplication and the inverse matrix respectively.
To see why this group action has the desired property, note that $X_\alpha(\mathbb{C})$ is the "space of $\alpha$-dimensional $\mathbb{C}$-representations of $Q$" and that two $\alpha$-dimensional $\mathbb{C}$-representations of $Q$ are isomorphic if and only if they lie in the same $G_\alpha(\mathbb{C})$-orbit. Thus, the group action has the desired property if and only if the category of $\mathbb{C}$-representations of $Q$ contains only finitely many isomorphism classes of dimension $\alpha$. This last assertion clearly holds if said category contains only finitely many isomorphism classes of indecomposable objects because any object in it can be written as a direct sum of indecomposable objects.
Proof of the dimension inequality
To sum up, it remains to carry out the "geometric" proof for $\dim G_\alpha \geq \dim X_\alpha$ alluded to above. Since the representing algebra of our $X_\alpha$ is a polynomial ring in finitely many variables, we are done if we can prove the following lemma:
Proof: Working with the definition of dimension used in this answer it is clear that it suffices to show the existence of an injective morphism of $\mathbb{C}$-algebras $\mathbb{C}[X] \hookrightarrow \mathbb{C}[G]$.
Fix one member of each of the finitely many $G(\mathbb{C})$-orbits of $X(\mathbb{C})$, say $x_1,\dotsc,x_r$. If $A$ is any $\mathbb{C}$-algebra, then applying $X$ to the structure morphism $\mathbb{C} \to A$ yields a map $X(\mathbb{C}) \to X(A)$; I will denote the image of $x_i$ under this map again by $x_i$. This explains how to understand the definition
$$ \xi_{i,A}\colon G(A) \to X(A), g \mapsto g . x_i $$
(where $.$ denotes the group action). As the action of $G$ on $X$ is required to be a natural transformation it is easy to see that the collection of those $\xi_{i,A}$ is as natural transformation as well; it will be denoted by $\xi_i$. By the Yoneda Lemma, the natural transformations $\xi_i\colon G \to X$ correspond to morphisms of $\mathbb{C}$-algebras $\zeta_i\colon \mathbb{C}[X] \to \mathbb{C}[G]$ that are characterized by the formula $$ \xi_{i,A}(g) = g \circ \zeta_i , $$ where $g \in G(A)$.
Remember that we are done if one of the $\zeta_i$ is injective. So let us assume that this is false, i.e. no $\zeta_i$ is injective. Fix an element $f_i \in \ker \zeta_i \setminus \lbrace 0 \rbrace$, $i=1,\dotsc,r$ and put $f = \prod_{i=1}^r f_i$. Let $x \in X(\mathbb{C})$; $x$ lies in one of the finitely many $G(\mathbb{C})$-orbits, i.e. there is a $j$ and an $g \in G(\mathbb{C})$ such that $\xi_{j,\mathbb{C}}(g) = x$. Thus:
$$ x(f) = \prod_{i=1}^r x(f_i) = \prod_{i=1}^r (\xi_{j,\mathbb{C}}(g))(f_i) = \prod_{i=1}^r (g \circ \zeta_j)(f_i) = 0 $$
Now what does that mean? Remember that $f$ is just a polynomial in $\mathbb{C}[T_1,\dotsc,T_n]$ and $x$ is just a morphism of $\mathbb{C}$-algebras $\mathbb{C}[T_1,\dotsc,T_n] \to \mathbb{C}$. That is, $x$ is given by inserting a complex number for any of the $T_i$, i.e. by evaluating a polynomial at some fixed point. Conversely, every such evaluation gives rise to a morphism of $\mathbb{C}$-algebras $\mathbb{C}[T_1,\dotsc,T_n] \to \mathbb{C}$, i.e. to an element of $X(\mathbb{C})$. (This is just the universal property of the ring of polynomials.) We have thus seen that evaluating the polynomial $f$ at any point yields zero. Therefore, $f$ is the zero polynomial. But this is a contradiction, because $f = \prod_{i=1}^r f_i$ in an integral domain with all $f_i$ being non-zero.
(Caution: The argument that $f$ is the zero polynomial because it has value zero everywhere is clearly sound for $f \in \mathbb{C}[T_1,\dotsc,T_n]$. In fact, an easy proof by induction on the number of indeterminates shows that this argument is sound for any infinite field $K$ instead of $\mathbb{C}$. However, over finite fields this argument - and thus this proof of one direction of Gabriel's theorem - breaks down: e.g. the polynomial $T^2 - T$ is not the zero polynomial but has constant value zero over the field with two elements.)
Further notes
Gabriel's theorem holds for any field. The original paper by Gabriel (1972) (correcting an earlier erroneous statement and proof by Yoshii) contains a proof which is, for both directions of the theorem essentially a clever but tedious case by case analysis, and mentions briefly the proof outlined above (which, as indicated, only works for infinite fields). However, Bernstein, Gel'fand and Ponomarev (English translation 1973) soon published a new, beautiful proof for the "if" part of Gabriel's theorem (the one this answer doesn't deal with at all) using certain "reflection functors" which act on the dimension vectors of indecomposable representations just like the Weyl group of the Dynkin diagram's root system. For the "only if" part, they explain essentially the proof outlined in this answer (in slightly fancier terms). They are aware that it only works for infinite fields but, in a footnote, present an alternative that is supposed to deal with the finite field case. Essentially, the overall strategy is the same, but the geometric lemma used in the end is supposed to be replaced by a combinatoric argument counting the numbers of isomorphism classes of representations of a fixed dimension and the number of elements of matrix groups over finite fields, respectively. However, I think that I can prove that their suggested modification fails because it can only show the Tits form to be positive semidefinite which is of no use. The only sound proof of the "only if" part for arbitrary fields I am aware of is contained in Gabriel's original paper. If someone happens to know another one (maybe one can fix Bernstein et. al.'s suggestion somehow?), I'd love to hear it.