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From Frank Morgan: Geometric Measure Theory, Fourth Edition: A Beginner's Guide, page 13,the $2$-dimensional density of the cone $x^2+y^2=z^2$ at $0$ is $\sqrt{2}$. I feel strange of that,roughly speaking,the density is

$$\lim \frac{m(E \cap B)}{m(B)}$$

how is it larger than $1$?

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Hi Strongart, I think you should work on reformulating your question a bit! I'm not sure what you mean by "How does it large than 1". –  Bruno Joyal Jul 28 '11 at 6:54
@Bruno, I think OP is asking, how can anything have density greater than one. And I suspect the answer lies in a close reading of the source, paying attention to how exactly the density is defined and how it's calculated for that example. –  Gerry Myerson Jul 28 '11 at 6:56
Removing the density tag since there are several different interpretations of that word in mathematics. There's the density as you wrote above, which agrees with the colloquial usage of the word, but there are also densities in the sense of weighted tensor distributions in differential geometry. –  Willie Wong Jul 28 '11 at 11:28

3 Answers 3

up vote 3 down vote accepted

Roughly speaking, the answer is "because you have a (bi)cone".

Any smooth embedded manifold (in fact any $C^2$ manifold) will have the property that the appropriately defined density is precisely equal to $1$. But if you have something that is not smooth, you can have other values.

If you have just a single cone $z = \sqrt{x^2 + y^2}$, near the origin you see that locally there is a constant angle defect: a circle of radius $r$ from the origin on the cone has circumference $\sqrt{2}\pi r < 2 \pi r$. For smooth manifolds the angle defect is related to the integral of the curvature inside the circle, so if the curvature is a continuous function (which can be guaranteed by the manifold being $C^2$), the angle defect must become zero as the circle shrinks. The single cone, on the other hand, has a curvature singularity at the origin, which allows it to have an angle defect even as the circle shrinks to a point.

Notice that a computation will show you that the single cone has density $1/\sqrt{2}$ at the origin. (The double cone, then, will have twice that.)

Remark 1 It is not necessary to have a multisheeted surfaces to get density $>1$. Consider $\mathbb{R}^3$ in spherical coordinates $(r,\phi,\theta)$. Define a surface by $\theta = f(\phi)$ with $|f(\phi)| < \pi/2$, this will by definition be a graph over $\mathbb{R}^2$ and thus single sheeted. Choose $f(\phi) = \sin (k\phi)$ for $k\in\mathbb{Z}$. The circumference of the unit circle then is given by $$ \int_0^{2\pi}\sqrt{1 + k^2\cos^2(k\phi)}\cos\circ\sin(k\phi)d\phi > 2\pi $$ if $|k| > 1$. So one gets that the graph defines a $C^0$ embedded surface with density $> 1$.

Remark 2 Very roughly, the density is a statement of how "crunched up" our set is in an infinitesimal setting. You should compare the notion, and the statements given in Florian's answer, to the notion of fractal dimension.

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Good discussing!Maybe we can call the reason is "branches" instead of "bicone",but Remark 1 is another situation which is rather interesting. –  Strongart Jul 30 '11 at 10:46
The thing is, even with branches you don't need to have density bigger than 1, and remark 1 shows that branches are not necessary for having density bigger than 1. (Imagine a bicone that is very narrow: something like $z^2 = 99 (x^2 + y^2)$. Then the density is something like $2/10 < 1$.) So while having multiple sheets is a mechanism that can lead to large densities, it is neither necessary nor sufficient. –  Willie Wong Jul 30 '11 at 12:37
Yes,you are right.I find a new example:the sine curve of topologist y=sin(1/x)…… which is a basic example to distinglish the connection and path-connection(From Munkres:Elements of Algebraic Topology),the density can go to infinite and not just at one point. –  Strongart Jul 31 '11 at 6:25

I just checked the book and I found out that the author defines the $m$-dimensional density of a set $E\subset \mathbb{R}^n$ near a point $x$ for $m\le n$ by $$\lim_{r\to 0} \frac{\mathcal{H}^m(E\cap B^n(x,r))}{|B^m(0,r)|}$$ Note that in the numerator, the $n$-dimensional ball appears, whereas in the denominator it's the $m$-dimensional ball! For $m=n$ the density becomes $$\lim_{r\to 0} \frac{|E\cap B(x,r)|}{|B(x,r)|}$$ and this is indeed $\le 1$, but for $m<n$ it may be greater than 1. This occurs for example if several surfaces intersect at some point, and in this case the density counts the number of such surfaces. The situation for the vertex of the cone is similar.

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Thank you,the dimension is different.I am too careless. –  Strongart Jul 30 '11 at 10:35

Probably what is going on (I do not have access to the book) is the author is using a different definition of density. In geometric measure theory, we care about sets $E \subset \mathbb{R}^d$ which have, for example, positive (upper) Lebesgue density:

$$ \overline{D}(E) = \limsup_{R \to \infty} \frac{m(E \cap B_R)}{m(B_R)}. $$

However, some authors (and I suspect Morgan is among them) define positive upper Lebesgue density as

$$ \overline{d(E)} = \limsup_{R \to \infty} \frac{m(E \cap B_R)}{R^d}. $$

Notice that $\overline{D}(E) > 0$ if and only if $\overline{d}(E) > 0$. Also, $\overline{D}(E) \in [0,1]$, whereas $\overline{d}(E)$ can in fact be larger than $1$.

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I recall that the limits were approaching $0$ when I took the course in measure theory, is my memory faulty or do you have a typo? (or the third option, your answer and my understanding are currently incompatible :-)) –  Asaf Karagila Jul 28 '11 at 8:08
@Asaf: I supposed it depends on the definition of density. The only notion of density which I have come across in my research looks at the size of a set intersected with a suitable rectangle or ball as the size of the ball (or rectangle) goes to infinity. –  JavaMan Jul 28 '11 at 8:10
I suddenly notice that you define it for sets, not for points. I guess approaching infinity makes more sense :-) –  Asaf Karagila Jul 28 '11 at 8:12
@Asaf: On the other hand, when one defines the Hausdorff or Minkowski dimension, you look at how the size of the $\delta$-neighborhood of a set behaves as $\delta \to 0$. Perhaps this is what you mean? –  JavaMan Jul 28 '11 at 8:13
@Asaf: Sounds like your memory is right on. –  JavaMan Jul 28 '11 at 14:44

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