Suppose that $X$ is an affine real algebraic variety. In particular $X$ is the zero locus of a set of $k$ degree-two polynomials in $\mathbb{R}^N$. (If it makes any difference, we also happen to know that $X$ is compact in the subspace topology.)

Let $F : X \to \mathbb{R}$ be projection on to one of the coordinates. My understanding is that even if $F$ is not flat and proper, the set of $r \in \mathbb{R}$ such that $F$ is not smooth at $r$ is still constructible, and for a one-dimensional variety such as $\mathbb{R}$ the Zariski topology is the cofinite topology, so any constructible set that is not the whole space must be finite.

What I would like is to make this effective.

Can we give an upper bound on the number of critical values of $F$, in terms of $k$ and $N$?

For example, is there some way to apply Milnor-Thom?

Added: I do not need an exact answer, but only asymptotic upper bounds. For example if $k \approx N$ is it true that the number of critical values of $F$ is bounded by $e^{cN}$ for some constant $c$?


2 Answers 2


Hi M@:

I kind of doubt that you can get a bound like that without using more specific information. This is not a proof that you cannot, just some issues to consider:

First of all, it seems to me that it might be more beneficial to consider the question over $\mathbb C$. If you get an estimate there it would work in your case and I don't see why you would get a better estimate working over $\mathbb R$. On the other hand working over $\mathbb C$ you would have more powerful methods available. Then again, I might be wrong and being only motivated by my preference for complexity over reality. :)

Let's work over some unspecified field for now. Take an arbitrary flat projective morphism $f:Z\to \mathbb P^1$ and let $H\subset X$ be a fixed very ample divisor such that it does not contain any of the fibers of $f$. Let $G$ denote the divisor class of a fiber of $f$ (you already took $F$ for something, #@%!). Assume that $H$ was chosen so that $H+G$ is very ample. In other words, there exists an embedding $Z\hookrightarrow \mathbb P^N$ such that $H+G$ is a pencil of hyperplane cuts, so you get a projection to a coordinate axis that realizes the original $f$. Taking $X=Z\setminus H$ gives a map like you're considering.

Mumford proved (actually this is not too hard) in "Varieties defined by quadratic equations", Questions on algebraic varieties, C.I.M.E. Varenna, 1969 , Cremonese (1970) pp. 29–100, that possibly combined with a Veronese embedding every variety (over $\mathbb C$ now) is the intersection of quadrics (i.e., "the zero locus of a set of $k$ degree-two polynomials"). It seems to me that if you combine this with a Veronese you still get a projection to a coordinate axis that realizes the original map.

So, the point is that you can make the original map to have an arbitrary number of critical values, so there seems to be no chance giving a universal upper bound.

Possible way out: The only thing this example does not give you (at least not directly) is that $k\approx N$. Obviously Mumford's theorem does not say that everything is a complete intersection. I am not sure if there are any known estimates on the number of quadrics needed. The obvious upper bound for $k$ that I can see is exponential. So, this gives you a chance, but it means that somehow you need to use the $k\approx N$ condition in an essential way.

I guess another issue is that Mumford's theorem is over $\mathbb C$ and not over $\mathbb R$, but I can't imagine that that would actually make a difference.

  • $\begingroup$ Hi Sándor --- usually things in the real world are quite a bit more complex than over ${\Bbb C}$, but somehow it seems this might be one of those exceptions... $\endgroup$ May 16, 2011 at 21:56
  • $\begingroup$ Dave, I agree with both statements. Really! $\endgroup$ May 16, 2011 at 22:24

Hi Matt,

I'm not sure this helps, but you could write your $X$ as the zero locus of a single degree $4$ polynomial. Of course, if your coordinate projection is sufficiently general, it's a Morse function and the Milnor-Thom bounds on betti numbers should work for you. For a special projection, it seems like critical points would collide (and so decrease in number) but I could be completely wrong.


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