# volume growth of tubular neigbhorhood of critical values of an algebraic/differentiable map

Edit: One can also assume $M,N$ compact in the following, and that $M$ is equipped with a Riemannian metric as well. Incorporated Mike's suggestion.

Sard's theorem says that for every smooth map $f: M \to N$ between two manifolds, the set of critical values of $f$ is a set of measure $0$ in $N$. Now assume $N$ is equipped with a Riemannian metric, and let $A$ be the set of critical values of $f$, and $T_\epsilon(A)$ the tubular neighborhood of $A$ in $N$ with radius $\epsilon$, are there any results on the volume growth of $T_\epsilon(A)$ under certain conditions on $f$? For instance, $f$ could be an algebraic map of bounded degree (in the sense of the lowest degree of a representing polynomial of $f$ under suitable coordinates), which is in fact the case I care about.

If $f$ is an algebraic map between two algebraic varieties $M,N$ that are also smooth manifolds, is it true that the set of critical values of $f$ is a union of submanifolds (or perhaps closed submanifolds with boundary) of $N$ with dimension strictly less than $\dim N$? If that's the case, then it appears the volume growth can be estimated once we know the Riemannian induced volume of each of those lower dimensional component. But maybe there is a way to estimate volume growth more intrinsically?

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I came across these notes at Shmuel Weinberger's website a while back when I was looking for something else. See Theorem 1.8 (p 18), for instance, to see if it could be of use to you. I am not sure it they are relevant, but just in case. math.uchicago.edu/~shmuel/QuantCourse%20/LNM1834.pdf –  Keivan Karai Nov 6 '10 at 8:33
In the first question of the last paragraph, were you envisioning that the submanifolds in the union would be closed? If so then the answer is no because the critical set can certainly have significant singularities: for instance the critical values of $(x,y)\mapsto (x^3-3xy,y)$ are the points on the cuspidal cubic $w^2=4z^3$. I suppose you could say that the cuspidal cubic is a union of submanifolds (one consisting just of the singular point and the other of everything else), but it seems like any volume estimates that you'd want to make would degenerate somewhat near the singularity. –  Mike Usher Nov 6 '10 at 20:51
Hi Mike. I guess I am not envisioning them to be close now. But maybe I should include manifolds with boundaries as well, which is harmless for my volume estimate. Basically I would like to say the critical set in $N$ is union of $k$ dimensional submanifolds, $k < n$, such that the sum of the $k$ dimensional volume of these components are bounded as a function of the polynomial degree of the map and dimension of $M,N$, as well as the metric of course. So the cuspidal cubic would probably have a small tubular growth. –  John Jiang Nov 6 '10 at 22:13
Hi Keivan. I think you pointed me to the right literature for this problem. Thanks. –  John Jiang Nov 7 '10 at 0:03

EDIT: This answer is wrong: as pointed out by Alfonz, this map is not algebraic. The similar one with a rational approximation of $\sqrt2$ is, but then the degree is unbounded.

Consider the following map $f$ from $\mathbb R^2$ to the torus $\mathbb T^2=\mathbb R^2/\mathbb Z^2$: $$f(x,y) = (x, \sqrt2 x) \bmod \mathbb Z^2 .$$ Its set of critical values equals its image and it is dense in the torus, so the $\epsilon$-neighborhood has full measure for any $\epsilon>0$.

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Thanks. I guess by degree I really meant the smallest possible degree of the polynomial representation of the algebraic map. Also in my application $M$ can be assumed to be compact and also equipped with a Riemannian metric. –  John Jiang Nov 6 '10 at 4:34
The $y$ is here to make the values non-regular (ok, formally they are not regular for the map of a line either, but it is even more confusing then). With rational coefficients, it can be made a polynomial map of a circle. The length of the image can be large but then the degree is large too, this is what I wanted to say. –  Sergei Ivanov Nov 6 '10 at 15:22

If $f:M\to N$ is an algebraic morphism between algebraic varieties (defined over a field of characteristic $0$) such that $M$ is smooth, then there exists a non-empty Zariski open subset of $N$, $U\subseteq N$ such that $f^{-1}U\to U$ is a smooth morphism in the algebraic sense. If I am not mistaken, this means that it has no critical points/values.

The algebraic statement above is known as Generic Smoothness and it is proved for example in III.10.7 of Hartshorne's Algebraic Geometry.

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It is true that $f$ restricted to $U$ has no critical values in its image. But I am actually interested in the complement of this Zariski open subset $N \setminus U$, and how its tubular neighborhood grows in volume. –  John Jiang Nov 6 '10 at 4:41
I can't comment on the volume growth, but your question in the last paragraph was whether the critical values of $f$ were contained in a union of submanifolds of $N$ with dimension strictly less than $\dim N$. This follows. –  Sándor Kovács Nov 6 '10 at 7:16
@John: $f$ is not restricted to $U$, but $f^{-1}U$; $U$ is a subset of the target. In particular, all critical values of (the original) $f$ are contained in $N\setminus U$. –  Alfonz Nov 6 '10 at 7:43