Let $V$ be a real affine variety in $\mathbb R^n$, i.e. the zero set of a real polynomial $p(x_1,\dots,x_n)$. Consider the following three definitions of the dimension of $V$, $dim(V)$.
Definition 1: if $I$ is the ideal of
polynomials vanishing on $V$, then
$dim(V)$ is the maximum dimension of a
coordinate subspace in $V(\langle
> LT(I)\rangle)$ with a given graded
order $>$ on the monomials
($\sum\alpha_i>\sum\beta_i$ implies
$x^\alpha>x^\beta$) (see the book
"Ideals, Varieties, and Algorithms"
chapter 9)
Definition 2: $dim(V)$ is the largest $d$ such that there exists an injective semi-algebraic map from $(0, 1)^d$ to $V$ (see "Algorithms in Real Algebraic Geometry" chapter 5)
Definition 3: $dim(V)$ is the largest $d$ such that there exists a
subset of $V$ homeomorphic to $(0,
> 1)^d$
Are these three definitions, for the case of real affine varieties, pairwise equivalent, or pairwise different, or something else?
Best Answer
They are all equivalent, including definition 1, to the Krull dimension of $S/I$, where $S=\mathbb{R}[x_1,\ldots,x_n]$ is the polynomial ring $I$ lives in. This is very good news for people like me who want to apply algebraic geometry to statistics, where numbers are mostly real.
Here's how it goes:
Definition 1 is a way of computing the Krull dimension of $S/I$ via Groebner bases. See, e.g., p. 250 of Computing in algebraic geometry: a quick start using SINGULAR by Wolfram Decker and Christoph Lossen
Definition 2 is shown equivalent to Krull dimension in Corollary 2.8.9 of Real Algebraic Geometry by Bochnak, Coste and Roy. Note that they define the "dimension" of a real variety to be Krull dimension of its coordinate ring. I recommend reading the whole chapter.
Definition 3 is equivalent to Defintion 2 because any semialgebraic set admits a decomposition into finitely many pieces homeomorphic to $(0,1)^d$ (see Theorem 2.3.6 in RAG), and a finite union of semialgebraic sets of dimension less than $d$ cannot contain a set of dimension $d$. I.e., the only way a real variety can contain an open set homeomorphic to $(0,1)^d$ is by containing a semialgebraic set homeomorphic to $(0,1)^d$ in its semialgebraic cell decomposition.