In the article "On Being Thurstonized" by Benson Farb (located here), a particular result of Thurston is mentioned.

Namely, suppose a "tinker toy" $T$ is a contraption consisting of a multitude of rods. They can either be bolted to a table, or attached to each other along hinges. The configuration space $C(T)$ consists of all possible configurations of $T$. For example, if $T$ is a single segment, then $C(T)$ is a circle.

The claim is then that any compact, smooth manifold can be obtained as a component of some $C(T)$.

I've thought about it a bit (though perhaps not enough), and this theorem escapes me completely. Are there any notes or explanations out there that describe the proof of this theorem?

  • $\begingroup$ I am not sure about this problem. I heard that for any manifold M (probably one needs compact and smooth) and any positive real epsilon, there was a T (called a linkage) such that C(T) was epsilon close to M, either by embedding both in R^n or by assigning some metric and finding a near isometry. But this is not my field, and I am recalling something from another millennium. Gerhard "Gee, I Feel Really Creaky" Paseman, 2018.03.21. $\endgroup$ Commented Mar 21, 2018 at 21:37
  • 1
    $\begingroup$ See: Geometric Folding Algorithms: Linkages, Origami, Polyhedra by Erik Domaine and Joseph O'Rourke amazon.com/Geometric-Folding-Algorithms-Linkages-Polyhedra/dp/… $\endgroup$ Commented Mar 21, 2018 at 21:37
  • $\begingroup$ Take a look here: arxiv.org/pdf/math/9807023.pdf $\endgroup$
    – Lee Mosher
    Commented Mar 21, 2018 at 21:39
  • $\begingroup$ Thank you for the link to Benson's essay! The first two paragraphs are an absolute gem. $\endgroup$
    – Nik Weaver
    Commented Mar 21, 2018 at 22:37
  • $\begingroup$ Perhaps not "any compact smooth manifold", but "any compact smooth algebraic variety"? For algebraic curves, this is classical. $\endgroup$ Commented Mar 21, 2018 at 23:18

4 Answers 4


I was Thurston's (undergraduate) student in the mid 1980s, when he was thinking about linkages. Here's how Thurston explained the proof to me.

By a theorem of Nash (see https://en.wikipedia.org/wiki/Nash_functions), any smooth manifold is diffeomorphic to a solution space of a set of real polynomial equations. So now all one needs to do is devise planar linkages which implement addition and multiplication of real numbers, and hook them together in a way that mirrors the algebraic equations from Nash's theorem.

I never worked through the details of the above sketch myself.

[Added: Igor Rivin's answer overlaps with mine, and has additional details.]


It was published here:

M. Kapovich, J. Millson, Universality theorems for configuration spaces of planar linkages, Topology 41 (2002), no. 6, 1051–1107.


The result comes by way of Nash's theorem which states that every smooth manifold is a component of a real algebraic variety.

Nash, John, Real algebraic manifolds, Ann. Math. (2) 56, 405-421 (1952). ZBL0048.38501.

The game is now to represent a real algebraic variety as a configuration space of a linkage. This is a very old game, going back to Kempe 1876 - the so called Kempe Universality Theorem which states that any real plane algebraic curve can be thus represented. The result of Kapovich and Millson in Andy Putman's answer is a direct descendant of Kempe's result. As far as I know, Thurston did not prove this result, but did give it to Bill Goldman as a Senior Thesis problem. I was under the impression that Goldman solved the problem, but maybe it was a restricted version.


Sometimes the Kempe Universality Theorem (see Igor Rivin's answer) is expressed as: There is a linkage that signs your name.

          enter image description here

As an indication of how difficult it is to achieve this in practice, here is a little linkage that when driven by $b$ rotating on the pinned circle center $a$, draws a rough approximation to a handwritten letter J (purple), drawn out by joint $z$.

          Fig. 2.16 in How To Fold It. Linkage designed by Don Shimamoto. Computations in Cinderella.


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