Connected Projections of Configuration Spaces of Linkages

TL;DR

This paper demonstrates that any compact connected semialgebraic set can be represented as a projection of a connected component of a linkage's configuration space, linking geometric structures with mechanical linkages.

Contribution

It establishes a universal representation of semialgebraic sets via linkage configuration spaces, bridging geometry and mechanical linkages.

Findings

Any compact connected semialgebraic set is realizable as a projection of a linkage configuration space.

The result connects geometric set theory with the topology of linkage configurations.

Provides a method to construct linkages for arbitrary semialgebraic sets.

Abstract

We show that any compact connected semialgebraic set is the projection of a connected component of the configuration space of a linkage.

Equations

$$Y=\{(u,v,t)\mid u=x,v_1^2+t^2=v_1,v_i=0i>1\}.$$

$$X\times 0\cup \{(x,0,z)\in {\mathbb{R}}^m\times {\mathbb{R}}^k\times {\mathbb{R}}^{\ell }\mid q(x,z)=0\}$$

$$\{(x,y,z)\in {\mathbb{R}}^m\times {\mathbb{R}}^k\times {\mathbb{R}}^{\ell }\mid q(x,z)=0,p(x,y)=0\}$$

$$\{(x,y,z,t)\in {\mathbb{R}}^m\times {\mathbb{R}}^k\times {\mathbb{R}}^{\ell }\times \mathbb{R}\mid tq(x,z)=1,p(x,y)=0\}$$

Full text

1. Connected images of linkage configurations

Recently Michael Kovalev asked me if any compact connected semialgebraic set is the projection of a connected component of the configuration space of a linkage, i.e., if we can choose some points on the linkage which trace out all of $K$ and the linkage can go continuously from any point of $K$ to any other.

This short note answers this in the affirmative, and also ties up a loose end in [1] for which I did not provide a reference (although a reference of which I am unaware undoubtedly exists). See [1] or [2] for definitions of terms.

Theorem 1**.**

Suppose $K\subset{\mathbb{R}}^{m}$ is a compact connected semialgebraic set. Then for any $n\geq 2$ there is a classical linkage in ${\mathbb{R}}^{n}$ with compact configuration space $C\subset{\mathbb{R}}^{n\ell}$ and a choice of $m$ coordinates of ${\mathbb{R}}^{n\ell}$ so that if $\pi\colon{\mathbb{R}}^{n\ell}\to{\mathbb{R}}^{m}$ is projection to these coordinates then for any connected component $C_{i}$ of $C$, $\pi(C_{i})=K$.

Proof.

We know from Lemma 3.1 of [1] that there is a compact algebraic set $X\subset{\mathbb{R}}^{m}\times{\mathbb{R}}^{k}$ whose projection to ${\mathbb{R}}^{m}$ is exactly $K$, but $X$ may have several connected components, each of which might only project to part of $K$. Choose such an $X$ with the least number of connected components. If $X$ is not connected, there are two connected components $X_{1}$ and $X_{2}$ of $X$ whose projections to ${\mathbb{R}}^{m}$ intersect, so let $(x,y_{i})\in X_{i}$ be points with the same projection $x\in K$. For convenience we may as well assume after an affine transformation that $y_{1}=0$ and $y_{2}=(1,0,\ldots,0)$. Consider the algebraic set $Y\subset{\mathbb{R}}^{m}\times{\mathbb{R}}^{k}\times{\mathbb{R}}$ given by

[TABLE]

Then $Y$ is a circle which intersects both $X_{1}\times 0$ and $X_{2}\times 0$ and projects to $x\in K$. Hence $Y\cup X\times 0$ is a compact algebraic set whose projection to ${\mathbb{R}}^{m}$ is $K$, but it has fewer connected components than $X$, a contradiction. So $X$ was connected. Now by Theorem 1.1 of [2] there is a linkage whose configuration space $C$ projects as a finite trivial analytic cover of $X$. So in particular any connected component of $C$ projects isomorphically to $X$ and thus projects further onto $K$ as desired. ∎

I would imagine that the linkage in the proof above would move around in $X_{1}$, then seem to mostly stop as it traverses $Y$, changing some internal gears so to speak, and then continue moving in $X_{2}$.

2. Unsemiing a semialgebraic set

We’ll say a subset $Y\subset{\mathbb{R}}^{m}$ has property P if there is a real algebraic set $X\subset{\mathbb{R}}^{m}\times{\mathbb{R}}^{k}$ whose projection to ${\mathbb{R}}^{m}$ is exactly $Y$. By the Tarski-Seidenberg theorem we know a subset with property P is semialgebraic. In Lemma 3.1 of [1] we asserted the converse and hinted at a proof, but failed to give a reference which undoubtedly exists. I’ll prove this here, it’s easier than finding a reference. The proof follows from the following two Lemmas.

Lemma 1**.**

Suppose $Y,Z\subset{\mathbb{R}}^{m}$ have property P. Then $Y\cup Z$, $Y\cap Z$, and $Y-Z$ each have property P.

Proof.

Let $X\subset{\mathbb{R}}^{m}\times{\mathbb{R}}^{k}$ and $W\subset{\mathbb{R}}^{m}\times{\mathbb{R}}^{\ell}$ be real algebraic sets which project to $Y$ and $Z$ respectively. Let $p:{\mathbb{R}}^{m}\times{\mathbb{R}}^{k}\to{\mathbb{R}}$ and $q\colon{\mathbb{R}}^{m}\times{\mathbb{R}}^{\ell}\to{\mathbb{R}}$ be polynomials so that $X=p^{-1}(0)$ and $W=q^{-1}(0)$. Then the algebraic set

[TABLE]

projects to $Y\cup Z$, the algebraic set

[TABLE]

projects to $Y\cap Z$, and the algebraic set

[TABLE]

projects to $Y-Z$. ∎

Lemma 2**.**

Let $p\colon{\mathbb{R}}^{m}\to{\mathbb{R}}$ be a polynomial. Then $p^{-1}(0)$ and $p^{-1}([0,\infty))$ each have property P.

Proof.

Note $p^{-1}(0)$ has property P since it is the projection of itself. Also $p^{-1}([0,\infty))$ is the projection of the algebraic set $\{(x,t)\in{\mathbb{R}}^{m}\times{\mathbb{R}}\mid t^{2}=p(x)\}$. ∎