5
$\begingroup$

I am having trouble proving the following statement, which I think is true (and possibly very basic). Let $M$ be a real differentiable manifold of dimension $(n-1)$ sitting inside $\mathbb{R}^n$. Let $W $ be an algebraic set defined by homogeneous forms in $\mathbb{R}[x_1, ..., x_n]$ for which $W \subseteq \mathbb{A}_{\mathbb{\mathbb{C}}}^n$ has dimension $n-2$ (as an algebraic set in $\mathbb{C}^n$). I want to prove that $$ M \not \subseteq W \cap \mathbb{R}^n. $$

I think one reason why I am having trouble proving this (even though I suspect it might be quite basic) may be that I don't have a good understanding on how the dimension of the manifold and dimension of algebraic sets relate to one another. I would greatly appreciate any comments or references. Thank you.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ If I haven't got my inequalities wrong, then the algebraic tangent space at any point of $W$ has dimension $\ge n - 2$, whereas the geometric tangent space at any point of $M$ has dimension $n - 1$; so you could just show that, if your containment held, then the two notions of tangent space would have to agree. (I'm not sure if this is true, or easy, in general. $p$-adic groups, my speciality, are much easier to work with in this way, since there is no geometric tangent space. :-) ) $\endgroup$
    – LSpice
    Commented Apr 2, 2018 at 15:51
  • 3
    $\begingroup$ Err, sorry, those bounds aren't contradictory. I guess that, even if you showed the tangent spaces agreed, then you would still have to show that $W$ had a smooth real point, where the dimension of its algebraic tangent space was precisely $n - 2$. $\endgroup$
    – LSpice
    Commented Apr 2, 2018 at 16:56
  • $\begingroup$ @JohnnyT.: Since $W$ is made of points in $\mathbb C^n$, its dimension as an algebraic set is a complex dimension, so its real dimension must be $2n-4$; are you sure that the words that you have chosen correctly convey what you mean? (Of course, I might be misunderstanding your usage of the term "dimension of an algebraic set".) $\endgroup$
    – Alex M.
    Commented Apr 2, 2018 at 17:41
  • $\begingroup$ @AlexM At the moment I believe the words have been chosen correctly; this is what I mean en.wikipedia.org/wiki/Dimension_of_an_algebraic_variety $\endgroup$
    – Johnny T.
    Commented Apr 2, 2018 at 17:48
  • 1
    $\begingroup$ @AlexM., note that we are considering not the subset $W(\mathbb C)$ of $\mathbb C^n$, which one would expect to have real dimension $2n - 4$ if it were a smooth manifold, but rather the intersection $W(\mathbb C) \cap \mathbb R^n$, about whose dimension it is not so clear (to me) what our expectations should be. $\endgroup$
    – LSpice
    Commented Apr 2, 2018 at 22:24

1 Answer 1

Reset to default
4
$\begingroup$

You seem to be asking whether it is possible that the topological dimension of the set of real points of an algebraic variety is greater than its algebraic dimension (in your case, the former is $n-1$ and the latter is $n-2.$) It seems to be a standard fact that this is impossible (inequality in the opposite dimension is quite possible, as in the standard example of $x^2+y^2 = -1.$) The standard reference seems to be:

Basu, Saugata; Pollack, Richard; Roy, Marie-Françoise, Algorithms in real algebraic geometry, Algorithms and Computation in Mathematics 10. Berlin: Springer (ISBN 3-540-33098-4/hbk). x, 662 p. (2006). ZBL1102.14041.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .