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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?
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    There are more equivalence classes than possible descriptions. That is not to claim that it would be impossible to describe the classes in a general way. In fact "the equivalence classes of convex figures in the plane" would be one such description, but I think you might get more response if you could further specify what kind of description you are looking for. – 2017-03-01
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    @MartinRattigan I am interested in a description based on geometric featuers, such as "the class of triangles". Perhaps I should start with a narrower question: how many classes of *quadrangles* are there? – 2017-03-02
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    The question is too broad. As for quadrangles, the space of affine classes is 2-parameter, homeomorphic to the open 2-disk. If you mod out by projective transformations, there is only one convex quadrangle. – 2017-03-02
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    @MoisheCohen I edited the question to focus on polygons. What do you mean by "mod out by projective transformations"? – 2017-03-03
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    I meant to consider quadrilaterals up to their projective equivalence: Take the space $Q$ of convex quadrilaterals and then its quotient $Q/PGL(3,R)$. – 2017-03-03

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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.

enter image description here

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    Where is the convexity assumption used in this argument? Apparently, if we allow non-convex quadrangles, the point $C$ can be in a larger region, but this region is still homeomorphic to the entire affine plane, right? – 2017-03-07
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    @ErelSegal-Halevi: If the polygon is not assumed to be convex then C can be in another region of the plane (other than $R$), say, inside the triangle ABD. For affine equivalence this is not a big deal (the topology of the space of quadrilaterals modulo the affine group does not change), but for projective equivalence, it is: Instead of one point, you get several; the quotient is not Hausdorff, which I find unpleasant. – 2017-03-07
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    @ErelSegal-Halevi: Actually, I have to amend my comment: If you allow nonconvex polygons, you get two connected components (up to affine equivalence), both are homeomorphic to $R^2$. (The "exotic" component is where $A$ is inside of the triangle $BCD$.) In my answer I assumed that you are interested in convex polygons. – 2017-03-07
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    I followed the argument for $n=4$, but what's the argument for bigger $n$? – 2017-03-08
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    @BenBlum-Smith: It is a bit more complicated: You identify the space of $n$-gons with a fiber bundle over the space of $n-1$-gons whose fiber is homeomorphic to $R^2$. Then, inductively, the base of the bundle is contractible, so the bundle is trivial. Do these words have any meaning to you? – 2017-03-08
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    @MoisheCohen - yes, and I mostly buy it. There's one detail unclear to me. I see that any one point of a convex $n$-gon can vary in an $R^2$-homeomorphic region if the other points are fixed. But it's not clear to me from this that the space of $n$-gons is an $R^2$-bundle over the space of $n-1$-gons, because the choice of $n$th point isn't canonical. If I just take a convex $n-1$-gon, there will be $n-1$ disjoint $R^2$-homeomorphic regions in which I could add an $n$th point. Some of the choices could be affinely equivalent but it doesn't seem to me they'd all be? – 2017-03-08
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    @BenBlum-Smith: You have to make a choice, say, the vertex $A_n$ of your polygon. Cut off the triangle $A_{n-1}A_nA_1$ from your $n$-gon. This will be your map: $(A_1,....,A_n)\mapsto (A_1,....,A_{n-1})$. One still has to check that this map is a fiber bundle and this is indeed an unpleasant part: There will be two charts that one has to use. – 2017-03-08