Seventh Annual Louise and Richard K. Guy Lecture. The Mathematics of Doodling. Ravi Vakil, Stanford University.

Transcription

1 September 20, 2012 Seventh Annual Louise and Richard K. Guy Lecture The Mathematics of Doodling Ravi Vakil, Stanford University vakil/

2 Thanks. Everyone doodles in their own particular way. I d like to tell you about the way I doodle, and some of the mathematics that I realized later that it connected to. There is a lot of room for exploration, and some of the answers are quite sophisticated. [Props: a couple of boxes. 2 small blue struts and connector. stray transparencies, pens, ball, planet, line, tangrams. toothpick with bump? tangrams, line, boarding pass, pens, blank slide [here]. Oregondoodling12.tex] [This looks like play, what mathematics is really about. Finding patterns in nature, explaining them, and extending them. Much of what mathematics is about is asking the right questions. We ll do that today, finding interesting questions buried in every day doodles. These questions actually lead to deep math, in lots of fields. Not my own! I m not an expert! But we ll get to Hilbert problems, fields medals, current cutting edge research. No answers! Do doodle! Vote. ] 2

3 Drawing circles around something. First question: [from age 5]Are things getting more circular? Why? Prior question: What precisely (mathematically) are we doing? Given a plane set X, if r 0, define the rth neighborhood of X:[step of size r; dust cloud] N r (X) = {y : y x r for some x X}. Reworded question: In some sense does N r (N r ( (N r (N r (X))) )) become more and more circular? 3

4 [Prior question:]how does our choice of r affect the question? How does N 1 (N 1 (X)) compare to N 2 (X)? Answer: N a+b (X) = N a (N b (X)). [Reason soon.] 4

5 [Use:] The triangle inequality: The shortest distance between two points is a straight line. More mathematically, AB + BC AC. 5

6 N 1 ( (N 1 (N 1 (N 1 (X)))) ) = N 1,000,000 (X) Reworded question: get more circular? As r, does N r (X) Answer: Yes! [Next slide. We ll use:] Key idea: If A is contained in B, then N r (A) is contained in N r (B). [Explain why. We ll use this simple principle a lot.] 6

7 Suppose p is a point of X. Let C t be the circle of radius t around p. Pick an r such that X C r. Then {p} X C r, so C R = N R ({p}) N R (X) N R (C r ) = C R+r. Thus as R, the border of N R (X) is squeezed between two circles whose ratio of radii goes to 1. We ve answered the question! Are we done with this doodle? No, we re just getting started! 7

8 For simplicity, let X be a convex polygon. A convex polygon:[no dents] A non-convex polygon: 8

9 Let X be a convex polygon. Let P denote perimeter. What is P (N r (X))? [Ask them. P (N r (X)) = P (X) + 2πr.] Let A denote area. What is A(N r (X))? [A(N r (X)) = A(X) + rp (X) + πr 2.] 9

10 Side question: [(if you know calculus)] d dr A = P. Why?! Subtle observation: A(N r (X)) is a quadratic polynomial in r, and its coefficients have good geometric meaning.[point: A, P, π] 10

11 Let s generalize. Suppose X is convex, but not a polygon. It will be more convenient to define N r in terms of the normal vector. [Picture: sticking out your right hand. [Is this the same?]] Fact: P (N r (X)) = P (X) + 2πr A(N r (X)) = A(X) + rp (X) + πr 2 still hold! (The proof uses Green s Theorem from multivariable calculus.) 11

12 String-around-the-Earth puzzle [This relates to a fun puzzle I ve always liked.] (For the purposes of this puzzle, the Earth is a perfect sphere.) String is wrapped tightly around the equator of the Earth. Someone comes along and adds 1 m more of string, and raises the string above the earth at a constant difference. How high off the ground is the string? A millimeter? A micrometer? A nanometer?[vote! Answer: P (N r (X)) = P (X)+2πr, so P (N r (X)) P (X) = 2πr = 1 m. Also: not sphere.] 12

13 Generalize! Let s give up convexity. What happens? How should we change the statement in order to have a good solution? [Round U-shaped thing] [Cover! Or more likely tell.] New concept: Area with multiplicity. But nothing changes: our formulas for perimeter and area still hold. 13

14 What about a figure eight? (We have to make sense of area!) [Draw it with arrow. Ask them! Negative area!] P (N r (X)) = P (X). A(N r (X)) = A(X) + P (X)r. We ve lost the πr 2!? What happened? [Don t answer yet] 14

15 To answer this question, consider a more complicated shape. [cover] P (N r (X)) = P (X) + 4πr, A(N r (X)) = A(X) + P (X)r + 2πr 2. [cover]this has winding number 2. (The winding number is the number of times you turn around when you walk around a path.) This is the winding number! [arrow up.] 15

16 What happens with many holes? Many pieces? Answer: A(N r (X)) = A(X) + rp (X) + χ(x)πr 2 where χ(x) is the Euler characteristic:[topology] the number of pieces minus the number of holes. [topology] More generalizations: What happens in three dimensions?[prop] [first:]what happens in one dimension? [then:]what happens in n dimensions? 16

17 Let s do this with a box, of height h, length l, and width w. Now let V denote volume and A denote surface area. [Go back to old definition. Ask them; cover] V (N r (X)) = V (X) + A(X)r + (h + l + w)πr πr3 This works for a convex body in general. We get a magic polynomial. We know the meaning of three of the coefficients.[cover] We have a fourth! (Exercise: Calculate it for a sphere.[cover]) 17

18 A beautiful Russian problem A Russian train company has a rule: you are not allowed packages (in a box) whose sum of dimensions (length plus width plus height) is more than 1 m. Is it possible to cheat by taking an illegal box, putting it in a larger box that is legal?[boarding pass: 62 linear inches] [Cover! Love it. Pf very surprising.] Answer: No! Why?! Suppose X and Y are boxes, and X Y. Then N r (X) N r (Y ). 18

19 V (X) + A(X)r + (h X + l X + w X )πr πr3 V (Y ) + A(Y )r + (h Y + l Y + w Y )πr πr3 [Cancel last term]now let r. π(h X + l X + w X ) π(h Y + l Y + w Y ) from which h X + l X + w X h Y + l Y + w Y as desired. Problem solved! 19

20 In dimension n, V (N r (X)) = V (X) + A(X)r + (???)r 2 + These higher invariants are very interesting. [Set of Riemann surfaces. Space (Deligne- Mumford). How big? Shape (topology)? Witten. Kontsevich. Okounkov. Mirzakhani: add holes. Polynomial in l. Cutand-paste. Leading coefficients.] [Let s see some now.] 20

21 [[Move to 2nd last slide, on Hilbert.]] [How to interpret them all in terms of typical shadows. First we reinterpret:] n = 2, onedimensional invariant (aka perimeter) Consider the average length of the shadow of a convex body. Multiply it by π. You get the perimeter![draw pentagon.] Example: A circle of radius R always has a shadow of area 2R, so its perimeter is 2πR.[draw. Then ask: Pf?] 21

22 Another example: What is the average length of the shadow of a line segment of length 1? [ask: perimeter of segment?] [cover]if the answer is x, then πx is the perimeter, i.e. 2, so x = 2/π. (So the average length of the shadow of a 1 inch toothpick is 2/π!)[calculus: hard!] 22

23 Buffon s needle problem This connects to the famous ancient experimental method of determining the value of π by dropping needles of length 1 on a table marked with parallel lines 1 unit apart, and counting how often a needle meets a line. [draw] The odds of crossing a line are 2/π. So take 1000 trials and see what happens! You ll get around

24 n = 3, two-dimensional invariant (aka surface area) Consider the average area of the shadow of a convex body. Multiply it by 4. You get the surface area![draw cube] Example: a sphere of radius R always has shadow of area πr 2, so its surface area is 4πR 2! Hard question: prove this for a box. Astronomy: Finding the surface area of a planet. 24

25 n = 3, one-dimensional invariant (π(h + w + l) for a box): Take the average length of its shadow on a 1- dimensional screen. Multiply by a certain constant. [prop] 25

26 A related Hilbert problem [23 problems, start of 20th c]can you dissect a cube and rearrange the pieces to obtain a regular pyramid? [tangrams] (In dimension two, the answer is yes: you can dissect any polygon and rearrange the pieces to make any desired other polygon with the same area. In other words, the only invariant up to dissection is area.) Answer: No. The proof uses a pumped-up version of this one-dimensional invariant. In dimension three, there are two invariants for dissection: volume, and this new one-dimensional invariant![demaine] 26

27 Summing up Lurking behind even the most trivial-looking doodles can be mathematics of surprising beauty and power. Thank you! 27