What classes of polygons are equivalent up to affine transformation? Let's call two planar figures "equivalent" if each of them is an affine transformation of the other. What are the equivalence classes of the convex figures in the plane?
Some classes which I found are:


*

*All ellipses;

*All triangles;

*All parallelograms.


However, not all quadrangles are in the same class. For example, a trapezoid is not equivalent to a parallellogram, since the former has only one pair of parallel sides while the latter has two, and it is known that affine transformations preserve parallelism. Moreover, a convex shape cannot be equivalent to a non-convex shape, since affine transformations preserve convexity.
So my questions are:


*

*What are the equivalence classes of all quadrangles? (how many classes are there? If there are infinitely many, how many parameters are required to characterize them?)

*What are the equivalence classes of all convex polygons with $n$ vertices?

 A: I will be considering labelled convex quadrilaterals $Q$ in the affine plane ${\mathbb A}^2$ with vertices $A, B, C, D$. Two such quadrilaterals $Q=ABCD, Q'=A'B'C'D'$ are affine-equivalent if there is an affine transformation $T$ which sends $A\mapsto A', B\mapsto B', C\mapsto C', D\mapsto D'$. Similarly, regarding ${\mathbb A}^2$ as an affine patch in the real projective plane ${\mathbb P}^2$, we define projectively equivalent quadrilaterals, by allowing projective transformations of 
${\mathbb P}^2$. 
Now, observe that the affine group $Aff({\mathbb A}^2)$ acts simply transitively on the set of non-collinear triples of points in ${\mathbb A}^2$. Therefore, fixing three points in general position $A_0, B_0, D_0\in {\mathbb A}^2$, every convex quadrilateral in ${\mathbb A}^2$ is affine-equivalent to a quadrilateral of the form $A_0B_0CD_0$, where $C$ lies in an open unbounded convex region $R$ in the affine plane, bounded by the lines $A_0B_0, B_0D_0$ and $A_0D_0$. Hence, the space of affine equivalence classes (with your favorite topology) is homeomorphic to the region $R$, which, in turn, is homeomorphic to the affine plane itself. 
On the other hand, if we consider quadrilaterals up to projective equivalence, we can use the fact that the pointwise stabilizer of $\{A_0, B_0, D_0\}$ in $PGL(3, {\mathbb R})$ acts simply transitively on the region $R$. In order to see this, identify the line $B_0D_0$ with the "line at infinity" in the projective plane and identify the complement to this line with the affine plane. Then the stabilizer of
$\{A_0, B_0, D_0\}$ in $PGL(3, {\mathbb R})$ is identified with the group of diagonal matrices (where $A_0$ serves as the origin in the affine plane and the lines $A_0B_0$, $A_0D_0$ serve as the coordinate axes; the region $R$ becomes an open coordinate quadrant). Since the group of diagonal linear transformations acts simply transitively on each open coordinate quadrant, the claim follows. Therefore, up to projective equivalence, there is exactly one convex quadrilateral. 
More generally, the space of convex $n$-gons modulo projective equivalence is homeomorphic to the open $2(n-4)$-dimensional ball. if you only consider them modulo affine equivalence, you get the open $2(n-3)$-dimensional ball.

