I. What is the question? One can describe the measure on Graff(3,2) as follows: 1) The orientations of the planes (viewed as elements of the projective plane) are uniformly distributed relative to the measure on the projective plane that you get by taking the uniform measure on the 2-sphere and modding it out by the antipodal identification. 2) The restriction of the measure to an orientation-equivalence class of planes looks just like Lebesgue measure on the real line. This is an infinite measure, but if we restrict this measure to the subset of Graff(3,2) consisting of those planes that intersect some ball centered at the origin, we get a finite measure, which can be rescaled to yield a probability measure. Here's a concrete way to sample from that probability measure: 1) Choose a point P uniformly at random on surface of the ball and draw the line segment connecting it to its antipode P'. 2) Choose a random point Q uniformly at random from segment PP', and draw the plane through Q perpendicular to PP'. More generally, if we want to sample from those planes that intersect some compact convex body K, we can find a ball B containing K and use rejection sampling: repeatedly choose a random plane intersecting B until you find one that intersects K. There are other ways of describing the measure on Graff(3,2), and other ways to sample from it, but I like it because the rotational invariance of the measure is obvious. The same cannot be said for schemes that sample a plane ax+by+cz=d by choosing a, b, and c independently and then rejecting them if some inequality fails to hold. Anyway, this well-defined notion of a random plane that cuts a cube allows us to ask "What is the expected perimeter of a random cross-section of the unit cube?" II. Random cross-sections of a square It turns out that answering the preceding question in R^3 requires us to answer a similar question in R^2: "What is the expected length of a random cross-section of the unit square?" Here we are taking the natural measure on Graff(2,1) (the set of lines in the plane), restricting it to those lines that intersect a fixed unit square, and turning it into a probability measure. Here I will cite without proof a couple of results relating to mean width in two dimensions (I won't recite the definition of mean width since you can get that from Wikipedia). First, the mean width of a convex compact polygon A in the plane is its perimeter divided by pi; second, the average length of a random nonempty cross-section of A equals the area of A divided by the mean width of A. (For the proof of the analogous statement in three dimensions, see https://mathoverflow.net/questions/335293/mean-cross-sectional-area .) Since the unit square has perimeter 4, its mean width is 4/pi; and since its area is 1, the average length of a cross-section of the square is pi/4. Now imagine that the Euclidean 2-space E^2 that contains our unit square is actually sitting in a Euclidean 3-space E^3, and consider the subset of Graff(3,2) consisting of all planes that intersect that square. What is the expected length of the intersection of such a plane with the square? It takes a little thought to see that the answer must be the same as it is one dimension down, but it's true because of the symmetry properties of the measure on Graff(3,2). Specifically, if you look at any line in E^2 that intersects the unit square in a particular line segment, then spinning that line around the line segment gives you all the planes in E^3 that intersect the unit square in that particular segment, and the measure is rotationally uniform. (This is the only part of the argument I might have some trouble writing down formally, but I'm sure it's right.) The upshot is, if you look at a random plane that intersects a square sitting in 3-space, the expected length of the intersection is pi/4. III. Vital digression: the Wall of Fire Theorem The Wall of Fire Theorem says that the expected number of sides in a random cross-section of a cube is 4. This tells us that the probability that a random plane that intersects the cube intersects a particular face of the cube is 4/6. That's because we can write the number of sides of the cross-section as a sum of six indicator variables, one for each face of the cube, equalling 1 when random plane intersects the face in question and equalling 0 otherwise. Linearity of expectation, combined with symmetry, tells us that the expected number of sides of the intersection (which we know equals 4) must be equal to 6 times the probability that the random plane intersects a particular face of the cube. So that probability must equal 4/6, aka 2/3. IV. Random cross-sections of a hollow cube Consider a random plane that intersects a unit cube, and consider a specific face of the cube. With probability 2/3, the plane intersects the face, and conditional upon that event, the expected length of the intersection is pi/4. On the other hand, with probability 1/3, the plane doesn't intersect the face, and conditional upon that event, the expected length of the intersection is 0. So the unconditioned expectation of the length of that intersection is equal to (2/3)(pi/4) + (1/3)(0) = pi/6. Lastly, if we sum over all the faces of the cube, we find that the expected length of the intersection between the random plane and the surface of the cube is (6)(pi/6) = pi. So this is indeed the expected perimeter of a random cross-section of the unit cube. V. Random cross-sectional area To prove that a random cross-section of the unit cube has area 2/3, one can appeal to the MathOverflow post I cited before, showing that the average cross-sectional area of a convex compact body in 3-space equals the volume divided by the mean width, and the fact that a cube of side-length s has mean width 3/2. I know of a nice way to prove the latter fact by way of Archimedes' hat-box theorem, but this email has gone for way too long already.) To recap: a random cross-section of the unit cube has 4 sides, has area 2/3, and has perimeter pi. I think that's swell. Thanks for helping me figure this out! Jim Propp On Thu, Aug 8, 2019 at 6:27 PM James Propp <jamespropp@gmail.com> wrote:
Cool!
Better still, I think I know how to prove this. That is, I think I can prove that on average a random cross-section of a unit cube has perimeter pi.
More soon,
Jim
On Thu, Aug 8, 2019 at 10:21 AM Tomas Rokicki <rokicki@gmail.com> wrote:
And here are the final lines after an overnight run. Note that accumulating billions of doubles in this fashion loses precision; I should have done my summing more carefully. But at least five of the digits are probably good.
At 8589934592 pts 4.0000236480264 area 2.66669753208071 perim 6.2832343008038
At 17179869184 pts 4.00001316709677 area 2.66668286448053 perim 6.28322083261956
At 34359738368 pts 4.00000967484084 area 2.66667855572835 perim 6.28320124178968