Mnev's Universality Type Theorem

Question

In order to state properly Mnev's universality type theorems, one has to understand the definition of stable equivalence. I have some questions to the definition. Here is the definition as in Oriented Matroids from Björner et.al.

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Let $V \subseteq \mathbb{R}^n$ and $W \subseteq \mathbb{R}^{n+d}$ be semi-algebraic sets with $\pi(W) = V$, where $\pi : \mathbb{R}^{n+d} \rightarrow \mathbb{R}^n$ is the canonical projection that deletes the last $d$ coordinates. $V$ is a stable projection of $W$ if $W$ has the form $$W = \{(v,v') \in \mathbb{R}^{n+d} : v \in V,\ \phi_i(v) \cdot v' > 0; \ \psi_j(v) \cdot v' = 0 \textrm{ for } i \in X; j \in Y \} .$$ Here $X$ and $Y$ denote finite (possibly empty) index sets. For $i \in X$ and $j \in Y$ the functions $\phi_i$ and $\psi_j$ have to be polynomial functions $$ \phi_i= ( \phi_i^1 , . . . , \phi_i^d ) : \mathbb{R}^n \rightarrow (\mathbb{R}^d)^* \text{ with } \phi_i^k \in \mathbb{Z}[x_i, \ldots , x_n] \quad \mbox{and}$$

$$\psi_j = (\psi_j^1,\ldots,\psi_j^d) :\mathbb{R}^n \rightarrow (\mathbb{R}^d)^* \text{ with } \psi_i^k \in \mathbb{Z}[x_i, \ldots , x_n],$$ that associate to every element of $\mathbb{R}^n$ a linear functional on $\mathbb{R}^d$. Two semialgebraic sets $V$ and $W$ are rationally equivalent if there exists a homeomorphism $f : V \rightarrow W$ such that both $f$ and $f^{-1}$ are given by rational functions. Two semialgebraic sets $V$ and $W$ are stably equivalent if they are in the same equivalence class with respect to the equivalence relation generated by stable projections and rational equivalence.

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Question 1:

What is exactly meant by homeomorphism in this case?

What is exactly meant by rational function in this case?

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Question 2:

OK, to unwrap and really understand the concept, how can I show that the following two sets are stably-equivalent?

$S = \{(x,y,z)\in \mathbb{R}^3 : x y = z\}$

$T = \{(x,y,z,a)\in \mathbb{R}^4 : x y = z ; a = (x+y)^2\}$.

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Question 3:

Intuitively the following two sets should be stably equivalent: $S\subset \mathbb{R}^n$ and $S'= \{(x,1)\in \mathbb{R}^{n+1} : x\in S\}.$

I don't see how to show this.

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Question 4:

Are the following two sets stably-equivalent?

$S = \{x\in \mathbb{R} : x>0\}$

$T = \{(x,y)\in \mathbb{R}^2 : x y^2 - 1 = 0\}.$

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thanks Till

Answer 1

Ok, it seems I can answer the first question. In this context, it seems the following is meant:

A homeomorphism is a function that is continuous and is invertible and has a continuous inverse.

A rational function is a function that can be written as ratio of polynomials, i.e. f(x_1,...,x_n) = p(x_1,...,x_n)/q(x_1,....,x_n) where p and q are polynomials.

Answer 2

I found an answer to the third question.

Consider the sets $S\subset \mathbb{R}^n,$ $$S'= \{(x,0)\in \mathbb{R}^{n+1} : x\in S\}\quad \text{and}$$ $$S''= \{(x,1)\in \mathbb{R}^{n+1} : x\in S\}.$$

In order to show that $S''$ and $S$ are stably equivalent, we will show that

(1) $S'$ and $S''$ are rationally equivalent. Furthermore, we will show that

(2) there is a stable projection from $S'$ to $S$.

We start with (1). The homeomorphism $\phi : S'\rightarrow S''$ maps $(x,\ldots,x_n,t) \mapsto (x,\ldots,x_n,t+1)$. It is easy to see that $\phi$ and its inverse are a rational functions. This shows (1).

For (2), we define $\psi_1$ as the polynomial that is everywhere constant to $1$. It holds that $$S' = \{(x_1,\ldots,x_n,t) \in \mathbb{R}^n : (x_1,\ldots,x_n) \in S \text{ and } \psi(x_1,\ldots,x_n) t = 0\} =$$ $$ = \{(x_1,\ldots,x_n,t) \in \mathbb{R}^n : (x_1,\ldots,x_n) \in S \text{ and } 1\cdot t = 0\} =$$ $$ = \{(x_1,\ldots,x_n,0) \in \mathbb{R}^n : (x_1,\ldots,x_n) \in S \}.$$

This finishes the proof and answers my third question.

Answer 3

To show that

$$S = \{(x,y,z)\in \mathbb{R}^3 : x y = z\}$$

and

$$T = \{(x,y,z,a)\in \mathbb{R}^4 : x y = z ; a = (x+y)^2\}$$

are stably equivalent, we show that they are in fact rationally equivalent. We define $f : S\rightarrow T$ as follows $$f(x,y,z) = (x,y,z,(x+y)^2).$$

Note that its inverse is $$g(x,y,z,a) = (x,y,z).$$

Both functions are bijective, continuous and rational. Thanks Michael Dobbins for clarifying this for me.

Answer 4

OK, Let me also answer Question 4. It may be a little silly, but the two sets are not stably-equivalent as they have a different number of connected components as can be seen in the picture below.