Affine variety and dimension 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. 
 A: 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.


*

*An affine scheme over $\mathbb{C}$ (as everything will be "over
$\mathbb{C}$" I will omit this phrase in the sequel) is an
representable functor $X\colon \mathsf{Rings}_\mathbb{C} \to
\mathsf{Sets}$, i.e. a functor of the form
$$ X = \hom_\mathbb{C}(A,-)\colon \mathsf{Rings}_\mathbb{C} \to
\mathsf{Sets} , $$
where $A$ is a $\mathbb{C}$-algebra. That is, for any
$\mathbb{C}$-algebra $B$ we have 
$$ X(B) = \hom_\mathbb{C}(A,B) $$
and for any morphism of $\mathbb{C}$-algebras $f\colon B \to C$ we
have
$$ X(f) = f_*\colon \hom_\mathbb{C}(A,B) \to \hom_\mathbb{C}(A,C), g
\mapsto f \circ g .$$


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 affine scheme $\operatorname{Mat}_{m \times n}$ (where $m$ and
$n$ are positive integers defined by the representing algebra
$\mathbb{C}[T_{ij} \mid 1 \leq i \leq m, 1 \leq j \leq n]$, i.e. the
ring of polynomials with $mn$ indeterminates. Clearly,
$\operatorname{Mat}_{m \times n}$ is integral and $\dim
  \operatorname{Mat}_{m \times n} = mn$.

*The affine group scheme $\operatorname{GL}_n$ (where $n$ is a
positive integer) defined by the representing algebra
$\mathbb{C}[T_{ij}, \det^{-1} \mid 1 \leq i \leq n, 1 \leq j \leq
  n]$, i.e. the localization of the representig algebra of
$\operatorname{Mat}_{m \times n}$ by the multiplicative set $\lbrace
  \det^k \mid k \in \mathbb{Z}_{\geq 0} \rbrace$. Here, $\det$ denotes
the determinant of the square matrix with entries $T_{ij}$; indeed,
$\det$ is a polynomial in the $T_{ij}$ by the Leibniz formula. It is
easy to see that $\dim \operatorname{GL}_n = n^2$ (use that the
quotient field of a localization agrees with the quotient field of
the original domain).

*The affine group scheme $\operatorname{SL}_n$ (where $n$ is a
positive integer) defined by the representing algebra
$\mathbb{C}[T_{ij} \mid 1 \leq i \leq n, 1 \leq j \leq n] / (\det -
  1)$. A little commutative algebra shows that $\dim
  \operatorname{SL}_n = n^2 - 1$ (try to determine what, more
generally, the dimension of the quotient of any polynomial ring by the ideal
generated by an irreducible polynomial is).


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:


*

*If $X$ and $Y$ are two affine schemes, the representing algebra of
their product $X \times Y$ is $\mathbb{C}[X] \otimes_{\mathbb{C}} \mathbb{C}[Y]$,
i.e. the tensor product of their representing algebras. This follows
easily from the universal property of the tensor product.

*If $X$ and $Y$ are two integral affine schemes such that their
product $X$ and $Y$ are integral and such that both $\mathbb{C}[X]$
and $\mathbb{C}[Y]$ are $\mathbb{C}$-algebras of finite type, we
have $\dim X \times Y = \dim X + \dim Y$. This follows from the
first fact on products and the Noether normalization theorem.


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 

The category of
  $\mathbb{C}$-representations of $Q$ contains only finitely many
  isomorphism classes of indecomposable objects if and only if the
  underlying undirected graph of $Q$ is one of the $ADE$-Dynkin diagrams.

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


*

*We have $q(\alpha) = \dim G_\alpha + 1 - \dim X_\alpha$.

*If the category of $\mathbb{C}$-representations of $Q$ contains
only finitely many isomorphism classes of indecomposable objects, then
the action of $G_\alpha(\mathbb{C})$ on $X_\alpha(\mathbb{C})$ has only
finitely many orbits.


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:

Let $G$ be an integral affine group scheme acting on an integral
    affine scheme $X$. Assume that the representing algebra of $X$ is
    the ring of polynomials $\mathbb{C}[T_1,\dotsc,T_n]$. If the group
    action
$$ G(\mathbb{C}) \times X(\mathbb{C}) \to X(\mathbb{C}) $$
has only finitely many orbits, then $\dim G \geq \dim X$.

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.
