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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.

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    $\begingroup$ 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
    Apr 2 '18 at 15:51
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    $\begingroup$ 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
    Apr 2 '18 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.
    Apr 2 '18 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.
    Apr 2 '18 at 17:48
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    $\begingroup$ @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
    Apr 2 '18 at 22:24
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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.

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