Question regarding proof of convex body theorem

[Peter_Newman]

Hello,

I am currently working on the proof of Minkowski's convex body theorem. The statement of the corollary here is the following:

If $S$ is a convex symmetric body of volume $vol(S) > 2^m det(B)$, then $S$ contains a non-zero lattice point.

Now in the proof the following is done:

We consider the set $S/2 = \{x : 2x \in S\}$. The volume of $S/2$ satisfies $vol(S/2) = 2^{-m} vol(S) > det(B)$

My questions are as follows: First, why does the equality $vol(S/2) = 2^{-m} vol(S)$ hold here and second what allows us to define $S/2$ in that way, what does $S/2$ represent, so how can you imagine that? I can't quite figure out why this is so. I would be glad if someone here could unravel this.

 

[Office_Shredder]

$S/2=\{x : 2x \in S\}$, equivalently, $\{ x/2 : x\in S\}$

You just scale every dimension down by a factor of 2.

 

[Peter_Newman]

@Office_Shredder thank you for your answer. I can understand that with the scaling, related to a convex body we reduce it by a factor of 2, if I understand that correctly?

But why exactly is this relation $vol(S/2) = 2^{-m}vol(S)$ valid? Assuming $m = 1$ then this makes sense, because by scaling my original $S$ by factor of $2$, the total volume of $S$ must halve, so $vol(S)$ must halve, so $2^{-1}vol(S)$. But why exactly do we have $m$ in the exponent, so how do we get $vol(S/2) = 2^{-m}vol(S)$? Does this $m$ consider the dimensions here?

 

[Office_Shredder]

You scale each of the m dimensions by a factor of 2. Let $S$ be a square with side length 2 centered at the origin. It has area 4. $S/2$ is a square with side length 1, so has area 1, not 2. $m$ is the dimension of the space.

 

[Peter_Newman]

@Office_Shredder thanks for the correction, I understand!

But again back to $vol(S/2) = 2^{-m}vol(S)$, why can one say here that this is equivalent? Does $m$ stand for the dimension here (In the 2D case this would make sense e.g. if one views $S$ as a square for example)?

 

[Office_Shredder]

I think you need to build some intuition here. Try the following:
Let $S$ be a circle. How does the volume scale?

Let $S$ be a sphere in 3d. How does the volume scale?

Let $S$ be an octahedron in 3 dimensions. How does the volume scale?

You can prove this in general for any integrable region $S$ (No need for convexity or symmetry or containing the origin)

Write the volume of $S/2$ as an integral over $S/2$, and do a change of variables to turn it into an integral over $S$. What Jacobian pops out?

 

[Peter_Newman]

Let $S$ be a circle. How does the volume scale?

With "volume" of a circle you mean the "area", right? It scales by the radius, same for a sphere in 3D. The octahedron scales by the length of a side (since all sides of the triangle are the same).

Write the volume of S/2 as an integral over S/2, and do a change of variables to turn it into an integral over S. What Jacobian pops out?

Here I can no longer follow you...

 

[Office_Shredder]

With "volume" of a circle you mean the "area", right? It scales by the radius, same for a sphere in 3D. The octahedron scales by the length of a side (since all sides of the triangle are the same).

No, the area of a circle and the volume of a sphere do not scale the same way when the radius doubles.

 

[Peter_Newman]

@Office_Shredder can you demystify this, I don't understand at the moment...

OK the circle scales in one direction depending on the radius. With the sphere it looks differently, since this has a further dimension (But also here there would be a radius dependence, which would not be comparable however with the circle...). What you try to say then with the integral, I do not understand however.

 

[Office_Shredder]

The area of a circle is $\pi r^2$, so when you double the radius, the volume quadrupled. What happens with a sphere?

 

[Peter_Newman]

The area of a circle is πr2, so when you double the radius, the volume quadrupled. What happens with a sphere?

@Office_Shredder oh, that's what you mean, good!

So what you say about the circle fits and is understandable. About the sphere: Its volume is $(4/3) \pi r^3$ i.e. if I double the radius, the volume increases to the $2^3$ third power (I do not know if that is the right word) $(4/3) \pi (2r)^3 = (4/3) \pi 2^3 r^3$.

 

[Office_Shredder]

Yes. Similarly, the volume of an octahedron is multiplied by 8 if you double every dimension.

 

[Peter_Newman]

Yes. Similarly, the volume of an octahedron is multiplied by 8 if you double every dimension.

@Office_Shredder so if I now relate this to $S/2$, then this means that the volume $vol(S/2)$ on the other side is halved by $2^{-m}$, where $m$ is the dimension here, right? Then $vol(S/2) = 2^{-m}vol(S)$ makes sense? So this factor $2^{-m}$ shrinks/scales our volume in case of $S/2$?

 

[Office_Shredder]

Yes, that's right.

If you know multivariate calculus, $vol(S) = \int_S dx$, and $vol(S/2) =\int_{S/2} dx$. Doing the change of variable $u=2x$, $dx=\det(\frac{1}{2} I ) du$ so
$vol(S/2) = (\frac{1}{2})^{m} \int_S dx$If you don't know calculus, basically you can estimate the volume of $S$ by covering the shape with a bunch of little cubes. When you divide all the dimensions by 2, you still have a covering by cubes, but each cube has volume divided by $2^m$, which givesthe result you want. The integrals make it a little more rigorous but you don't need them to understand the result in arbitrary dimensions.

 

[Peter_Newman]

Yes, that's right.

If you know multivariate calculus, $vol(S) = \int_S dx$, and $vol(S/2) =\int_{S/2} dx$. Doing the change of variable $u=2x$, $dx=\det(\frac{1}{2} I ) du$ so
$vol(S/2) = (\frac{1}{2})^{m} \int_S dx$If you don't know calculus, basically you can estimate the volume of $S$ by covering the shape with a bunch of little cubes. When you divide all the dimensions by 2, you still have a covering by cubes, but each cube has volume divided by $2^m$, which givesthe result you want. The integrals make it a little more rigorous but you don't need them to understand the result in arbitrary dimensions.

@Office_Shredder that is a very very good answer! I understand now what you meant exactly with the integrals. Now that this is here so understandable, before I always wondered how you could apply the integral, but so generally about $S$ is also possible.

But I still have a question, how do you come to this $dx=\det(\frac{1}{2} I ) du$. So where does the determinant come from (as a measure for the area?) and the $I$ (indicator function?).

 

[Office_Shredder]

I is the identity matrix.

This is just how change of variables works in many dimensions. If you're not familiar with it it's probably not worth dwelling on. But it's worth noting given a linear transformation, the determinant is a measure of how much it distorts volume.

 

[Peter_Newman]

@Office_Shredder so completely I have not yet understood your integral conversion, what you have done explicitly. But I understand the sense of it. Because here in the integrand $vol(S) = \int_S dx$ nothing else is defined and we integrate very generally over $S$, the change of the variable as you have done, I do not see here so clearly... Then the determinant of $I$ comes into play, where I also do not see exactly where it comes from.

But:
If we assume that $vol(S) = |det(A)|$, the volume results/can be expressed in terms of the determinant, then we could imagine that a halving of $S$ to $S/2$ causes $vol(S/2) = |det(A/2)| = (\frac{1}{2})^m |det(A)| = (\frac{1}{2})^m vol(S)$ (*But this would be valid only for "angular" figures, so an integral approach would be better.)

 

[Office_Shredder]

@Office_Shredder so completely I have not yet understood your integral conversion, what you have done explicitly. But I understand the sense of it. Because here in the integrand $vol(S) = \int_S dx$ nothing else is defined and we integrate very generally over $S$, the change of the variable as you have done, I do not see here so clearly... Then the determinant of $I$ comes into play, where I also do not see exactly where it comes from.

But:
If we assume that $vol(S) = |det(A)|$, the volume results/can be expressed in terms of the determinant, then we could imagine that a halving of $S$ to $S/2$ causes $vol(S/2) = |det(A/2)| = (\frac{1}{2})^m |det(A)| = (\frac{1}{2})^m vol(S)$ (*But this would be valid only for "angular" figures, so an integral approach would be better.)

For a general shape this is just the formula for a change of variable in the integral. Remember the integral is just summing the volume of a bunch of little cubes, then taking the limit when the cube size goes zero. You can map the cubes covering S and S/2 to each other, and the determinant says how much smaller the cubes in S/2 are compared to S. That's all that's happening in the integral.

 

[Peter_Newman]

@Office_Shredder I tried to write down the thought process about the transition from $S$ to $S/2$ via the Riemann sum (We can approximate the areas/volumes by rectangles/cubes/hypercubes.). What do you say to it:

For the $2$-dimensional case:

Let $\Delta x = x_{i+1} - x_i$ and $\Delta y = y_{i+1} - y_i$ by the Riemann sum:

$$\iint_S(x,y)\,dA=\lim_{m\to\infty}\lim_{n\to\infty} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y\text{,}$$

Now let $S = S/2$, so in this case $\frac{x_{i+1} -x_i}{2}$ and $\frac{y_{i+1} -y_i}{2}$

$$\iint_{S/2}(x,y)\,dA
=\lim_{m\to\infty}\lim_{n\to\infty} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\frac{x_{i+1} -x_i}{2}\frac{y_{i+1} -y_i}{2}\text{,}
=\frac{1}{2^2}\lim_{m\to\infty}\lim_{n\to\infty} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y\text{,}
=\frac{1}{2^2}\iint_S(x,y)\,dA$$

In the $n$-dimensional case:

$$\int...\int_{S/2}(x_1,...,x_n)\,dA
=\lim_{m\to\infty}...\lim_{n\to\infty} \sum_{i=0}^{n-1} ...\sum_{j=0}^{m-1}f(x_1,...,x_n)\frac{x_{1_{i+1}}-x_{1_{i}}}{2} ... \frac{x_{n_{i+1}}-x_{n_{i}}}{2}\text{,}
=\frac{1}{2^n}\lim_{m\to\infty}...\lim_{n\to\infty} \sum_{i=0}^{n-1}...\sum_{j=0}^{m-1}f(x_1,...,x_n)\Delta x_1 ... \Delta x_n
=\frac{1}{2^n}\int ... \int_S(x_1,...,x_n)\, dA$$

 

[Office_Shredder]

##f# is just 1 here!

 

[Peter_Newman]

That $f(x_1,...,x_n)$ is 1 here is because we approximate the whole thing as a rectangle/hypercube? But the rest fits so far, right?

Edit:
Ok, I got it now, the keyword is simply integration over non-constant boundaries. Then one makes almost always this approach $f(x_1,...,x_n) = 1$ and integrates over variable borders... And then it is also clear how to understand S --> S/2, here then each side of a hyper cube is halved...

 

[PeroK]

The area of a circle is $\pi r^2$

Ii seems to me that it's a long way from this to a proof of the convex body theorem!