# The hierarchy of geometric constructibility: can we go back?

I have recently become interested in the hierarchy of relative constructibility in geometry (see the discussion on my Twitter thread). From a given collection of points in the plane, we may proceed to construct other points using the classical tools of straightedge and compass. This induces a hierarchy of relative constructiblity, namely, for any two sets of points $X$ and $Y$ in the plane, we may say that $X$ is constructible from $Y$, written as $$X\trianglelefteq Y$$ if for every point $A\in X$, there is a construction procedure using only straightedge and compass proceeding from points in $Y$ and constructing $A$. And so we are faced with a hierarchy of constructibility.

Perhaps one might ask immediately whether this is genuinely a hierarchy. A central case occurs with pairs of points, since one often proceeds in geometry with the idea of a unit segment having already been fixed.

Question 1. Given two distinct points A and B, suppose that distinct points C and D are constructible by straightedge and compass. Can we go back? That is, must A and B both be constructible from C and D?

The answer, surprisingly, is Yes! This is a sense in which the relative constructibility notion is an equivalence relation rather than a hierarchy, since it shows that if AB can construct CD, than it is also true conversely. Since constructibility is also transitive, this shows that the relative constructibility is an equivalence relation on pairs of points in the plane.

Before addressing the question, let me first dispense with a distracting line of reasoning that suggests a wrong answer to this question. Suppose that we have a segment AB of length $\sqrt[3]2$, a nonconstructible number. Since the constructible numbers are closed under multiplication, this might suggest that from AB we can construct a segment of length 2, and therefore also construct a segment CD of length 1. But from a segment of length 1, we can never construct a segment of length $\sqrt[3]2$, because the duplication of the cube is impossible. So this would suggest a negative answer to the question. Is it right?

But no, this distracting counter-argument is not correct. The reason is that although the constructible numbers are indeed closed under multiplication, the construction methods requires the consultation of a fixed unit segment. That is, to prove that from a segment of length $a$ and a segment of length $b$ one can construct a segment of length $ab$, we require the use of fixed unit segment of length $1$. Indeed, it will follow from our analysis here that one cannot omit this consultation of a unit segment from the construction—we cannot in general construct the cube of a segment just from the segment itself.

So let me prove the answer to question 1 is positive. Suppose that from distinct points AB we can construct distinct points CD. Performing excactly the same construction process again, but proceeding from points CD as input, we construct points EF. From points CD we can see how CD are related to EF, forming a quadrilateral CDFE.

The point is that we may now construct the points A and B from C and D by simply inverting this similarity. (This was also observed by user @ZenoRogue on Twitter.) Namely, the two quadrilaterals ABDC and CDFE are similar, and so have all the same angles. So the line AC is constructible from CD by the angle ACD, which is the same as CEF, and the distance AC is scaled from CE by the same factor that CD is related to EF. So from CD we can construct this similar quadrilateral, and thereby construct both A and B.

Question 2. Given three distinct points forming a triangle ABC, suppose that we can construct triangle DEF by straightedge and compass. Can we go back? That is, must A,B, and C be constructible from D,E,F?

The question is of course similar to question 1, which had a positive answer. Nevertheless, the answer here is negative. The reason is that the original idea we had with question 1 concerning $\sqrt[3]2$ now works here to refute question 2. Namely, let ABC be a right triangle where AB has length 1 but AC has length $\sqrt[3]2$.

From the triangle ABC, we may easily construct the triangle ABD, where AD has length 1, since in fact the point D is constructible from the segment AB alone. But from ABD, which is a right triangle with both legs of length 1, we shall be unable to construct the number $\sqrt[3]2$, and consequently unable to construct the point C. So although ABD is constructible from ABC, the point C is not constructible from points ABD. This is therefore a counterexample to question 2, which therefore must be answered in the negative.

Another argument, offered by Ed Wagstaff, is that from an angle trisection we can construct the original angle, but we can’t necessarily construct the trisection from the original angle.

Finally, I should like to understand the nature of the relative hierarchy of constructibility better. $$X\trianglelefteq Y$$ What is the nature of relative constructibility of sets of points in the plane? For example, one question I didn’t know how to answer at first is the following:

Question 3. Is the hierarchy of relative constructibility a dense order? That is, if the set $X$ is constructible from $Y$ but not conversely, must there be a set of points $Z$ strictly in between? $$X\lhd Y\quad\implies\quad\exists Z\quad X\lhd Z\lhd Y?$$

In an exchange on Twitter with David Madore, I had thought that we answered the question negatively, but now I am not at all sure of that argument (and thanks to Gabin for the comments below!), and so I have deleted the wrong argument here. Question 3 seems currently to be open.

I have now asked question 3 on MathOverflow: Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order?

## 3 thoughts on “The hierarchy of geometric constructibility: can we go back?”

1. Nice!
I didn’t understand why the quadratic closure of $Q(\sqrt[3]{2})$ is simply $K(\sqrt[3]{2}$. For example, do we really have the square root of $\sqrt[3]{2} + 1$?
Some obvious thoughts. If we choose two points a and b, and we look at the preorder of sets containing a and b, we have an isomorphism of partial orders between the equivalence classes a and the quadratically closed fields, and it becomes completely an algebra question. Moreover, the equivalence classes containing a finite set are in correspondance with the finitely generated quadratically closed fields. In caracteristic 0, we know that finite field extensions have a primitive element. Hence, all the finite sets with at least 3 points corresponding to an algebraic extension are equivalent to a set with exactly 3 points. It would be cool if between some set of points and the same set of points + an element which is “algebraic” with respect to the first set of points, there are only finitely many equivalence classes. It is equivalent to the following field conjecture: let K be a quadratically closed field, α an element of degree n on K, and K’ the quadratic closure of K(α), then there are only finitely many quadratically closed fields between K and K’. I don’t know if it’s true or not. Perhaps the fact that I didn’t understand is also true here, and we could conclude.

For any transcendental r on a field K, r is not contained in the quadratic closure of K(r^3), and r^3 is transcendental, so here we would have infinitely many intermediate fields.

• Oh dear, I think you are right. Perhaps we don’t know that $K(\sqrt[3]2)$ is quadractically closed? The new distance will be in the quadractic closure of this field. I have to think about it more carefully, but perhaps this might be good enough, since it can’t be purely degree $2^n$ over $\mathbb{Q}$ and so degree 3 will come into it. I’ll think about it and post more later.

• I think the argument is broken, and I am updating the post with edits to say as much. Thank you very much for your comment!