Today's News

Today's headlines in the New York Times:

• Lockheed Martin got another government contract.
  
• Bush said something he said last week too.
  
• Folks post stuff online.
  
• Someone in Chicago wants to be mayor.

No news is good news?

Young boys and a man

While looking out the window at a rainy Newark Airport and waiting for a very delayed flight, I found myself standing next to a young boy — perhaps five or six — eating a large roll of bread. I struck up a conversation, and we were soon joined by his older brother — six or seven. I let the conversation go wherever it wandered, and learned quite a lot: that their father is a pilot; that the Yankees are the best baseball team, pitching is the best position, and next year they won't use the tee until you get six strikes; that the bushes below the hotel in Hawaii with the big rooms (three balconies in the suite!) now house a favorite action figure; that the police climbing the stairs into the jet-way were probably entering the airplane, because if there were a bad guy in the terminal, the security would have caught him in the initial screening (in fact, they were there to escort a very drunk passenger, who had repeatedly opened an alarmed door, from the terminal to the hospital).

After a while, their farther joined us at the window. "Tell the man next to you" — me — "what the kind of plane with the bump on top is," he asked his younger son. "I'll give you a hint: it starts Seven...."
"Um, Seven Seven?"
"No, Seven Forty-Seven."
"Seven Forty-Seven."
"And if there are [a particular kind of wing flaps]" — here my memory of the technical terms, which he used, has gone — "then it's a 747-400."


What I found most memorable about this discussion was not the ease with which we changed topics — an ease I normally associate with the uniformly brilliant kids at Mathcamp; an ease often pathologized as ADHD and ruined with drugs such as speed ritalin — nor the freedom with which these kids would talk to a complete stranger. What stuck with me was one particular piece of language: "Tell the man next to you..."

Those who've known me for a while may remember previous discussions I've had (though I think not here) about the different words "boy", "man", "kid", etc., which I find fascinating. I've intentionally used some throughout this entry: Mathcamp students and five-year-olds I've both described as "kids," for instance, whereas my first companion was a "young boy." I generally insist that periodicals refer to high school, and certainly college, students as "men" and "women": my freshman roommate was on the men's swim team, and in my brother's CS class there are only six women, as opposed to "boys'" and "girls." Mathcampers, on the other hand, and even my housemates, I often think of as "boys and girls". Not "children," perhaps, but "kids."

What's hardest, though, is self-identity — I'm good at holding multiple contradictory beliefs about the external realty — I had never before defined myself as someone who could be a "man [standing] next to you." Perhaps, when discussing sexual and gender politics, I've identified myself as a "(suitably adjectived) man," but more often as a "male." Categories like "men who have sex with men" are so entirely foreign and don't seem to apply to me or any of my peers. People in my socioeconomic class don't become "adults" until closer to 26, but I'm definitely no longer a "young adult." I'm a "student" or a "guy," not a "man."

One reason for my sojourn to New York was to attend a ninetieth birthday party and family reunion, where I spent some time chatting with various second cousins whom I haven't seen in ten years. My father, an older brother, is younger than his cousins, so while I played cards and board games with my fourteen-year-old cousin, the majority of "my generation" were three to ten years older than me. One announced the wonderful news of her pregnancy, making the matriarch whose birthday we were celebrating extremely happy. I'm used to my peers consisting of younger siblings and students exactly my age; I'm used to understanding those classmates only a few years older than me as significantly closer to adult, since they tend to be grad students when I'm an undergrad, or undergrads when I'm in high school.

But I'll be graduating in four months, and dreaming of my own apartment, and, eventually, house and family. I watch my fresh-out-of-college friends with their jobs in Silicon Valley, and can't help but think how similar that life is to college — they have roommates, come to campus, go on dates. They're no more "adults" than I am.

I have no trouble being "mature", or "old", or even relatively "grown up". But I'm twenty-one years old, and have a hard time thinking of myself as an "adult". Identifying as a "man" is impossible, and it is my current self-descriptor.

Angst and graduate school

I'm taking the GREs tomorrow (today), so instead of sleeping I'm avoiding looking up the rules and instructions. To do well on tests, it's best to go in knowing the structure of both the individual questions and the test as a whole. I don't yet, because I've been procrastinating with such useful time-sinks as listening to all of these pieces (link from the most excellent TWF234, about math and music). Oh, and actually getting things done --- I've written thank-you cards, answered e-mails --- there's no better way to be truly productive than to avoid something you really, really have to do.

One of my major accomplishments was writing back to a mathematical physicists from whom I had asked advice about grad schools. My e-mail ended up doing a decent job of outlining both some of my angst and some of my intellectual excitement; I thought you might enjoy it, and I'd gladly here your advice as well:


I'm at what I imagine to be the hardest part of grad school applications (and academic life? I'm sure there are harder things that my fantasy of the "easy life after grad school" leaves out) before actually writing them: figuring out what I want to do. This seems to come in two parts: 1. what am I interested in (and how to formulate it, and how much to formulate it or leave interests undecided as yet)? 2. and what people and departments are right for me given my interests?

I know I want to go into a career in mathematics; I want to teach calculus, and, though I enjoy the amorphous lands between math and physics, I've been ultimately happiest in math departments. At the same time, I know that the mathematics I want to study should have obvious connections with the physics: I'm happiest when I can use language and ways of thinking from physical theories, and when it's clear how the mathematical objects I'm playing with are connected to various attempts at fundamental theories. I want an anthropologist to conclude that my epistemology involves a real world that I'm studying (as opposed to those mathematicians who study platonic, nonexistent ideals). I assume that such is "mathematical physics" --- I definitely enjoy the material in John Baez's This Week's Finds. I would not be interested in studying interesting applied math such as fluid mechanics (or cryptography).

More precisely? I've been devouring This Week's Finds recently, so have been enjoying Baez's fascination with n-categories. I could happily study those for a while. Mostly as a way for me to record and inspire my own thoughts, I've been working on defining linear algebra entirely in terms of Penrose's tangle notation for tensors. I generally feel like algebraic notations, in which ideas are strung in lines, is restrictive and doesn't take advantage of the page.

My favorite toy is the hyperreal numbers, invented more or less by Abraham Robinson. These beasties have the power to do all of calculus, and provide actual interpretations for divergent sums, concepts like "much smaller than", and other important tools that are generally treated with intuition rather than rigor in most of math and physics. I would love to work on various projects to interpret and rigorize the mathematical footing of modern physics with this type of under-used tool. I hold that Robinson's calculus is more powerful than Cauchy's --- it's a conservative extension, so it can't prove anything that Cauchy can't, but it provides much more elementary meanings to a lot of the intuition. Vector fields really are infinitesimal, etc. The problem is that Robinson didn't get very far in constructing a user interface for his operating system. He can do Leibniz calculus, but no better than Cauchy can, and he didn't go farther. Cauchy is Windows to Robinson's Unix; I want to write Macintosh, incorporating QFT and the like.

What interests me most, beyond the actual mathematics, is the methods and institutions of mathematics and physics, and bettering those. I'm fascinated by the ways people think about math and physics, and the language they use, in a normative way: I want to find ways of understanding objects that get at their meanings, and specifically by combining math and physics intuition. It seems that the physicists are much more willing to take a cavalier attitude towards rigor, instead inventing formalisms that _might_ work in order to answer hard, hands-on questions. Whereas mathematicians may be better able to think extremely abstractly and provide the rigor, thereby arriving at a deeper meaning for the physicists' doodles. I want to help the mathematicians think in terms of particles, local processes, and effective theories (not to mention in terms of two-dimensional diagrams rather than "linear" equations). So what I would actually like to do is act as a translator.

So I've gotten some of the way towards an answer to my first question. Of course my interests will change as I continue to learn more math and physics. But my second question? I need all the advice I can get.

I think that I'm a strong applicant. I've taken the undergrad Intro to String Theory, and I'll be taking QFT this year. In math, I've taken a fair amount of algebra and analysis; most of my math knowledge comes from reading (including lots of math and physics blogs) and attending (now as a counselor) Canada/USA Mathcamp, which tries to expose its students to a wide variety of graduate-level math. So I'm looking at the top: strong departments, with people doing what I'm interest in.

But which are those? And who are the people?

Are there mathematical physics journals I should be reading or glancing at, because they're interesting or because they will suggest people and places I should pursue?

If you have any other advise for an aspiring (and presumably as-yet naive) mathematician with physics envy, please do share. Thank you so much.

A conventional question

I'm in the progress of writing up what I understand about tensors, defining them from scratch, using only intuition and Penrose's graphical notation. Eventually, perhaps I will write a version of my notes for Wikipedia, since their current article on the subject is laughably bad. I first read about them in this post by jao at physics musings; I had started reading Penrose's most recent book, The Road to Reality: A Complete Guide to the Laws of the Universe, which explores them in some depth. I am going to shamelessly reproduce jao's picture of such diagrams, so that you have some idea what I'm talking about:



Incidentally, I wonder what the history of such doodles really is. I hear talk of "einbeins", "zweibeins" and "dreibeins", lit. one-leg, two-leg, and three-leg, and if I knew more German, multi-legs ("mehrbeins"?), which sound like these tensorial pictures. Based on skimming the discussion here, it looks like einbeins are related, but not fully formulated. I wonder why someone would refer to an objects legs, though, unless it had legs.

Anyway, the notational question I wanted to ask was this:

We generally write "vectors" (as opposed to covectors) with raised indices, and covectors with lowered indices. This has physical significance: the Poincare group acts differently depending on whether the index is raised or lowered: on lowered indices, symmetries act by the adjoint, and so it's really a "dual" action, in the sense that it happens in the backwards order. So, although in some sense vectors and covectors are interchangeable, interpretations of diagrams are not.

Since vectors' indices are raised, Penrose proposes that a vector ought to have one "arm" (an edge coming out of the top), whereas a covector ought to have one leg. This makes sense, and closely matches how he thinks of contractions: contracting indices corresponds to drawing curves from the tops of the vectors to the bottoms of the covectors.

On the other hand, as soon as you start playing around with Penrose's diagrams — well, as soon as Josh H. started playing with them, when I introduced them to him over IM — you notice the connection between these diagrams and various ideas from quantum topology. In particular, diagrams like this look an awful lot like tangles.

This is actually no surprise. A n,m-tensor (one with n arms and m legs, so e.g. a vector is a 1,0-tensor), by definition, is a map from V tensored with itself m times to V tensored with itself n times. (By convention, V tensored with itself 0 times is the ground field — no, not a generic one-dimensional vector space, because I do in fact need the special number "1". This is so that there is a natural isomorphism between "V^0 tensor W" and W.)

But this, then, is a problem, because this commits me to reading my morphisms as going up. But my friends who study TQFTs think of their cobordisms as going down (see, for example, the many This Week's Finds starting at Week 73, in which Baez gives a mini course on n-categories).

So clearly one of us is right, and the other is wrong. Either we should think of vectors as having a head and a leg (more than half the time I catch myself drawing them this way anyway), or we should think of cobordisms, tangles, and their cousins as transforming the bottom of the page into the top of the page.

I'm leaning towards the latter, but only because there's one more, very established, case in which this matters. Diagrams of Minkowski space, and more generally of, for example, light cones in curved space, the positive time dimension is always drawn going up the page. And if our conventions are to have any sensible physical meaning, morphisms must correspond to forward time evolution.

So only typesetters and screen-renderers (and English language readers), who insist and putting (0,0) in the upper left corner of the page, have it backwards. Then again, mathematicians have known that for years.

But, of course, we do live in a democracy. If you tell me that infinitely more people and publications think time flows down, then I'll happily switch my conventions.

Mathcamp High School, to within an order of magnitude

Although I've been a Mathcamp JC, I've carefully stayed away from any talk financial. I've certainly been around, so perhaps have more knowledge than most, but I'm pretty sure that any numerics I might quote are, to within my level of accuracy, publicly available.

Mathcamp tuition for the five-week summer hovers around $3000; half of this goes to the university for room and board. (For comparison, one week without scholarship at a private American university costs more than $1000.) There are 100 students, many of whom do not pay full fair; we can expect an operating budget from tuition around 50 thousand dollars (if fully half of camp is covered by scholarships; perhaps 100 thousand is a good upper estimate). What, then, are expenses? There are roughly 20 staff, who each make, travel and housing included, between three and four thousand dollars for the summer. (This is less, per week, than many camps pay, but certainly well worth it.) That right there eats up most of the operating budget; housing for visitors (at any given time there are perhaps four former staff and perhaps four Famous Professors) already pushes us over any reasonable estimate. And, of course, visiting professors receive around $100 a day, plus travel, far below what many conferences pay. So it's possible, with squeezing, that Mathcamp breaks even. More likely, the Mathematical Foundation of America is doing its job well.


Greg, a camper, brought up one day after lunch the fantasy of a Mathcamp High School. Is it feasible? he asked. Is it good? I responded.

Mathcamp High School, under the present model, would be extremely expensive. College tuition is on par with tuition at elite boarding schools: Exeter, for instance, expects about $37 000 per year. Mathcamp successfully draws students from many socioeconomic backgrounds; it would have to be very careful to continue to do so in a yearlong model. Is MFOA up to the task?

Of course, I would not go to Mathcamp High School. Boarding schools can be extremely valuable, especially for the unfortunately many children who come from less-than-supportive families. But students with good families and good local schools should not, generally, choose boarding schools, even if they are economically feasible.

Mathcamp High School would also have to be extremely careful about keeping the culture of freedom that it currently rides to great success. Can an elite school for gifted youth avoid all quantitative grades and measurements? Can it allow students complete freedom to choose their classes and design their curricula? There are Waldorf schools that succeed. How do they do it?

One thing MHS would have to do is find young, enthusiastic, and brilliant teachers who can commit to years-long tenures. Mathcamp the summer program has a student-teacher ratio of almost five-to-one; advising groups (not all teachers advise) are around size seven. To make MHS work would require JCs and Mentors to be even more involved in helping students pick classes and put together four-year plans, assuring that they cover complete curricula.

Where would the staff come from? Undergrad and graduate students have their own academic careers to see to. But they are who you want: young and enthusiastic and able to directly participate in the culture.

Perhaps you run it at a college. Perhaps, as part of the deal allowing Mathcamp to stay on that campus for the year, the staff can enroll as full-time students. Then, especially if MHS's campus continues to move from year to year, Mathcamp can continue to bring in students from around the country.

I would gladly spend a year on staff at MHS. But I would not give up four years at Stanford to work as an MHS staff advisor, and I had said earlier that helping students plan high-school-long curricula takes staff who are there for the long haul. And what about graduate students?

MHS could no longer provide only math classes. American colleges demand that American high schools provide general liberal educations, and I'm sure that the Mathcamp model would work in other academic areas. What I'm not sure about is how tied Mathcamp's particularly dorky culture is to the math emphasis. The value at Mathcamp of sitting around in the lounge talking math, and math specifically, is astronomical: full immersion in the math jokes and culture and language cannot be replaced.

The most important question, in any discussion of extending Mathcamp from five weeks to forty, is how diminished would be the immediate experience. And how much is a good thing. Campers talk extremely positively about how "intense" a summer they had. Mathcamp levels of intensity are not sustainable, although some colleges come close. Is close good enough?


In the hope of providing a more year-round Mathcamp presence, I have, instead, a different suggestion. Let's hold Mathcamp more than once a year. Five weeks in the summer, yes, but also two weeks over winter break. These would most reasonably be in the American south or southwest, where it's warmer, but perhaps these two weeks should be in southern Maine, so as to end with four days at MIT's Mystery Hunt. Then another week around spring break: Mathcamp, I feel, has the bravado to encourage students to skip a week of high school if the local spring break falls on a different week from Mathcamp's. Each of these would be run, not as just a reunion, but as a mini camp: JCs and Mentors would live in dorms as RAs, fulfill all the loco-parentis responsibilities of camp counselors, and run activities and teach classes. This would not be a reunion but a week- or two-week-long math conference.

This I could skip school to organize. College final exam periods are largely a laugh; one can find workarounds for missing classes and assignments. I would not give up a year at Stanford for a year JCing, unless I simultaneously had access to a university's classes and undergraduate community. But I would happily — nay, eagerly — give up a few weeks.

Orange Juice: it's what's for breakfast

When I went off to college, my mother strongly encouraged me not to get a credit card. I have a debit/atm card, no debt, and no credit either. Sometime, probably at the end of the summer, I'll sign up for one of those air miles cards, and put all sorts of notes-to-self in my calendar about how much to spend, and when to pay it off, and when to cancel. I need the credit history, because it won't be too long before I actually will want to borrow. And it's possible, given the right circumstances, to make money (or at least air fare) off those cards. But in my case the right circumstances include having parents pay for tuition and bail me out when needed (never more than the cost of books and board bill, which I've been covering out of pocket, although they offered to pay for them).

And they've made it very clear that when I hit grad school, so exactly a year from now, I'm responsible for paying for everything. Rent, food, etc. comes out of whatever salary and stipend I get.

One way to live very well and very cheaply, if you're willing to spend the time and energy on thinking about cooking and eating, is to always cook vegan. Or vegan plus eggs, since eggs are cheaper than soy. Or, rather, it's very easy to spend huge amounts of money on vegan products — soy milks and egg replacers and yummy, unnecessary stuff. But, if you have the pallet, vegetables and beans and soy are cheaper than meat and dairy.

At school, this is a big part of how we cut costs — my quarterly board bill is less than any other eating arrangement on campus. I'll be doing half the ordering for our kitchen, and it's great to buy bulk flours. Our most expensive products are the (organic, free range) dairy: cheese and butter is expensive. It's also high fat, and especially high in saturated fat (is how it stays solid at room temp). It's a constant challenge to try to convince the residents, many of whom have never even tried vegetarianism before, that they don't need cheese and butter to survive. I plan to wow them, early on, with vegan desserts: my brother got me a vegan cookbook that, because of its veganism, is also zero-colesterol, almost zero-saturated fat, and generally very low fat. Silken tofu, my current favorite ingredient, in almost every cake, frosting, and pudding.

Poverty is one of the major causes of American obesity. Eating healthy requires resources: time, energy, education, and money. Whereas McDonalds will sell you all the calories you need in a meal for a dollar and no wait. But if you have the conveniences of, for instance, an academic life, in which the government school provides medical coverage, athletic facilities, something intellectual to do, and a small amount of money, living cheaply and eating well is very easy. The trick is to be a food snob: prefer your own cooking, buy only the very best ingredients, and think carefully about what you eat. And eat vegan. And organic. And local. And, most importantly, be part of the "slow food" push. WholeFoods will happily provide expensive vegan organic premade and packaged products.

Below is my signature dessert, which I usually think of as a vegan gluten-free brownie recipe, but I'll present here as a chocolate raspberry cake (as I had it for my birthday), with commentary on how to modify. As always, check local availability before committing to any particular fresh produce — by varying the fruit, one can make a seasonal cake in almost any season.

Chocolate raspberry cake

Preheat oven 350°F (325 for gluten-free). Grease two nine-inch round cake pans (or one 9x13 pan for brownies), and, for cakes, cut parchment or wax paper into circles to exactly fit on the bottom of the pans (for easier removal), place in, and grease both sides.

In blender, combine wet ingredients until smooth:

• 1 cup (8 oz) silken tofu
  
• 1/2 cup raspberry (or other fruit) jam
  
• 1/4 cup canola oil
  
• 1 Tbsp vanilla

In standing mixer with paddle blade, mix dry ingredients:

• 2 cups sugar (for fudge brownies, use 3 cups)
  
• 1 cup unsweetened cocoa powder (for brownies, use 1 1/2 cups)
  
• optional: up to 1 Tbsp instant coffee powder
  
• 2 cups all-purpose flour (or cake flour, or tapioca flour for gluten-free; if making cake with tapioca flour, supplement with 2 tsp xanthan gum, a gluten substitute derived from bacteria, and 1/4 cup cornstarch)
  
• 1 tsp baking soda (for brownies, use less; baking powder also works, and has less leavening power, because the batter is already acidic)

Pour in wet ingredients. (For brownies, also add

• 3 cups (vegan) dark chocolate chips)

Mix, adding up to

• 1 cup soymilk (be sure, if making gluten-free, to check the brand — Soy Dream and Almond Breeze are both safe, whereas Edensoy and Vitasoy are not)

if you feel like batter is too dry (at school we don't get silken tofu, so I use 1 cup firm tofu and 1 cup soy milk; at home, silken tofu means that with the soy milk it is often too liquidy). Pour into greased pans, and bake until done. (Gluten free at 325 takes a little over an hour; hotter temperatures burn the edges. Glutinous can go faster and hotter.) Enjoy licking the extra batter off the pan: no eggs means no salmonella.

Frosting and assembly

Tofu generally comes in 16-oz packs, and I usually use about 9 oz in this cake. So the rest, rather than trying to keep it, goes into the frosting. (In theory one would have the presence of mind to do the frosting a day ahead, so that the tofu can set. But I never do.)

Wash and clean standing mixer bowl, and fit with wire whisk. Whip

• silken tofu

until smooth. (You might decide instead to puree it in the blender, and then move to the mixer, or do it all in the blender or food processor. I've seen recipes calling for any of the three.) Then add, mostly to taste

• cocoa powder
• powdered sugar
• ground instant coffee
• corn starch and/or tapioca powder to thicken

until you reach a sweet and spreadable consistency. Place in freezer to set (or fridge if you have enough time).

For a raspberry chocolate cake, I also like to acquire fresh raspberries, and to make a raspberry syrup/glaze. This latter is very easy: in a sauce pan, heat raspberry jam with a little water until it dissolves, just before boiling (careful not to overheat and burn the sugar).

Once cakes are done, let cool 10 minutes then remove from pans and let cool completely. To assemble, place one cake face down on plate. Spread a thin layer of frosting, and cover with

• fresh raspberries, cut in half

and sprinkle on a little glaze. Then place second layer on top, and frost sides and top. Cover top with

• whole fresh raspberries

and drizzle with glaze.

Serve, and amaze your friends, after they've commented on how moist and rich it is, by revealing its ingredients.

Two of these things are kind of the same

In november I discussed two famous proofs of the infinitude of primes, arguing that the two proofs are essentially the same. I'd like to play that game again, with a more complicated theorem.

Theorem (Brouwer in two dimensions) Any continuous map $f:D\to D$ from the disk into itself must have a fixed point.

Proof 1 (Algebraic) Assume the contrary. Let $S$ be the circle, and $h:S\to D$ be the embedding identifying $S$ with the boundary of the disk. For each $p\in D$, consider the ray starting at $f(p)$ passing through $p$ (it's unique because $p\neq f(p)$), and let $g(p)$ be the (unique by convexity (and Jordan?)) point at which this ray intersects the boundary $S$. Then $g$ acts as the identity on the boundary; i.e. the diagram

    h       g
S ----> D ----> S
 \_____________/
       id

commutes, and $g$ is continuous (exercise for the reader). But the fundamental groups for $S$ and $D$ are $\Z$ and trivial, respectively, so applying the functor which measures a (path-connected) space's fundamental group (up to isomorphism) gives the patently impossible diagram

    h       g
Z ----> 1 ----> Z
 \_____________/
       id

Contradiction, QED.

Proof 2 (Combinatoric) Instead of the circular disk previously used, consider a triangle with vertices $A = (0,0)$, $B = (1,0)$, and $C = (0,1)$. We need the following lemma:

Lemma (Sperner) Consider a large triangle $ABC$ and a triangulation (which may add points to the boundary of $ABC$, but, being a triangulation, interior triangles never meet corner-to-edge). Color the vertices of this triangulation in the following manner: color $A$ red, $B$ blue, and $C$ yellow; points on the edge $AB$ cannot be yellow, and analogously for the other two edges; interior points can be any of the three colors. Then there must be a red-blue-yellow triangle in the triangulation.

Proof of Lemma Let $ABC$ be colored as above, and for each red-yellow edge place two dots: one on either side of the edge. Then there are definitely an even number of dots in the picture. How many are outside triangle $ABC$? For a dot to be outside, it must come from some edge along $AC$, and as you move from $A$ to $C$, each dot corresponds to a parity change between $A$ and $C$; as the total number of such changes must be odd, there are an odd number of dots outside $ABC$. But an interior triangle without both red and yellow vertices will not contain any dots, and red-yellow-yellow and red-yellow-red triangles contain two dots each. So there must be an odd number of interior red-yellow-blue triangles. qed.

Now finish the proof of Brouwer. Let $f$ be a continuous map without a fixed point from triangle $ABC$ to itself, and color each point by considering the vector $v(p)$ from point $p$ to $f(p)$: If $v(p)$ points into the first quadrant or along the positive $x$- or $y$- axes, color $p$ red; if $v(p)$ is in the fourth quadrant including the negative $y$-axis but excluding the positive $x$-axis, color $p$ yellow; and if $v(p)$ is in the second of third quadrants (i.e. if the $x$-component of $v$ is strictly negative), color $p$ blue. I leave it to the reader to check that every triangulation of $ABC$ colored like this satisfies the statement of Sperner's lemma.

Now consider some triangulation of $ABC$ where every small triangle has diameter less than $1$. This must contain a red-blue-yellow subtriangle $A_1,B_1,C_1$ (labeled so that letters correspond to colors in the obvious way). Now triangulate so that every subtriangle has diameter less than $1/2$; this contains red-blue-yellow subtriangle $A_2,B_2,C_2$. Now again so that all diameters are less than a third, and find $A_3,B_3,C_3$. Continuing in this manner, we get three sequences $A_i$, $B_i$, and $C_i$, so that, for each $j$, the pairwise distances between $A_j$, $B_j$, and $C_j$ are all less than $1/j$. But the triangle is compact, so $A_i$ must have a limit point, i.e. it contains some convergent subsequence $A_{i(j)}$, converging to point $P$. But the corresponding subsequences $B_{i(j)}$ and $C_{i(j)}$ must also converge to $P$, since they get arbitrarily close to the $A_{i(j)}$s.

What color is $P$? Well, $v(A_i)$ is in the first quadrant (or on the boundary) for each $i$, so $v(P)$ must, by continuity of $f$, be in the (closure of) the first quadrant --- i.e. $P$ is a red point. But by the same argument, $v(P)$ must be in the closure of the yellow fourth quadrant, so it must be on the boundary between yellow and red --- i.e. $v(P)$ is on the $x$-axis. But yet again it must be in the closure of the blue half-plane. So the only possibility is for $v(P)$ to be the zero-vector. But we assumed that $f$ had no fixed points, contradiction, QED.



To compare and understand these two proofs, let's run them at the same time. We might as well consider both to take place in a triangle (although, for that matter, Sperner's lemma was combinatorial, and all Proof 2 needed really was that the disk is compact and convex).

The first hint to the proofs' similarity is that both care about the line connecting $p$ with $f(p)$. Sure, the first one uses the ray from $f(p)$ to $p$, the second the vector in the other direction, but by taking negations we might as well care about the vector from $f(p)$ to $p$. And in the second proof we don't care how long the vector $v(p)$ is; we might as well normalize to get $w(p)$ on the unit circle. Of course, $g(p)$ (from the first proof) and $w(p)$ are patently different functions, but letting them act on the boundary $S$ gives two loops in the circle that are, at least, homotopically equivalent. (Intuitively, $g$ asks where the ray meets the unit circle; $w$ asks where it will meet the circle of infinite radius.)

Sperner's lemma hides a lot in Proof 2. To elucidate it, I'll make some small perturbations in the proof and the statement. First, in the proof: instead of placing two dots by each red-yellow edge, I could have placed a $+$ and a $-$, so that, aligning the edge with yellow at the "north" and red at the "south", $+$ is on the "east" side and $-$ is on the "west". Then, summing the pluses and minuses, each region (including the infinite one outside the triangle) has some weight, and the total weight of all the regions is zero. As before, pluses and minuses correspond to changes between red and yellow: traveling counterclockwise around the perimeter, a plus is a move from red to yellow, a minus from yellow to red. So the outside region has total weight $-1$. But the only interior triangles with non-zero weight are the red-blue-yellow (in the counterclockwise direction) triangles with weight $+1$ and the red-yellow-blue triangles with weight $-1$. So there must be exactly one more of the former than the latter.

I can perturb even more: instead of focusing only on triangulations, I can write a "Sperner's lemma" for any partition of the big triangle into (non-overlapping polygonal) cells. If I place pluses and minuses as above, then the entire plane has total weight zero, but the exterior has weight $-1$. But the cool thing about these weights is what happens when you combine two adjacent cells: the weights exactly add. And a cell can only have non-zero weight if it has all three colors along its edges.

What are these weights really? Imagine a complete graph on three vertices $K_3$, where I've labeled the vertices Red, Blue, and Yellow. Then each cell determines a closed loop in $K_3$: as you travel around the perimeter of the cell, call off the colors you see, and I'll travel around $K_3$ by moving to whatever color you call out. Then the weight of your cell is exactly the homotopy class of my loop. Why? Because the number of times you go around a circle is exactly the same as the number of (directed) times you cross any given point (and I've given a point between Red and Yellow; in my original Proof 2 coloring, this point might as well be "the $x$-axis).

Of course, in Proof 2, I only care about the parity of the homotopy class, and now I'm using the full homotopy class, in large part because I'm trying to perturb the proof towards Proof 1, which used $\Z$. Except that it didn't really: all Proof 1 actually used was the fact that the circle has non-trivial fundamental group. The two proofs do indeed cite and prove more and different things — in this way, the truly are different proofs — but strictly as proofs of Brouwer's theorem?

I hid a lot in Proof 1, also, because I assumed that you understand fundamental groups. Which isn't really fair: I certainly don't deeply understand them, although I know how to define, calculate, and prove things about them. Perhaps this is the best possible: von Neumann says that one will never move past "I'm used to" to "I understand". But I think that, at least in this case, a careful study will get pretty close to the ideal.

The standard definition of the fundamental group of a (non-empty path-connected) space $X$ goes like this: Imagine that $X$ has some special point, which I'll call the "origin" $O$, and consider the set of all parameterized paths that start and end at the origin. I.e. all functions $f: [0,1] \to X$ that send both $0$ and $1$ to $O$. There's a natural binary operation on paths: I can concatenate by running one and then the other, at twice the speed. This operation is not associative, sadly, but it is associative up to homotopy-equivalence: two paths are homotopy-equivalent if I can continuously transform one into the other. (More formally, $f$ and $g$ are homotopy-equivalent if there's a map $h: [0,1]^2 \to X$ that sends all $(0,s)$ and $(1,s)$ to $O$, and sends $(t,0)$ to $f(t)$ and $(t,1)$ to $g(t)$. The $t$-component parameterizes the paths, and the $s$ component parameterizes the transformation.) This is, as the reader can check, an equivalence relation, and concatenation is not only associative on equivalence classes, but admits a group structure. A priori, this "fundamental" group depends on choice of origin $O$, and there is no canonical isomorphism between the fundamental group at $O$ and the one at some other point $P$. But, as $X$ is path connected, there is some path between $O$ and $P$, and going along this path, then around a loop, then back provides an isomorphism between said groups. So the fundamental group is well-defines up to isomorphism: it maps the category of path-connected topological spaces to the category of isomorphism classes of groups.

So far so good: I'm guessing that, at least if you've seen this before, the previous paragraph was completely followable. You might even have a pretty deep understanding of what the fundamental group measures. Essentially, it measures how many "holes" are in a given space. Every graph has a free fundamental group, and spaces we care about are generally graphs with disk-shaped patches glued on in weird ways; if you know where the disks are, then you can write down the fundamental group as a presentation (each glued-on disk yields a relation by traversing its perimeter). But what's really interesting for the Brouwer proof is not what fundamental groups are (we only care about the fundamental groups of the disk and the circle), but what their relationships are.

Which is to say, what's interesting is that the function measuring a fundamental group is in fact a functor (what isn't, these days?).

One can prove this in general — the image of a loop under a continuous map is again a loop, and continuous maps respect concatenation and homotopy, but continuous maps can also identify equivalence classes if the range has more ways to transform one loop into another. Even more importantly, composition of continuous maps leads to composition of homomorphisms; I'll leave this (elementary) result to the reader.

But what does this mean in the very specific context of Proof 1? Certainly, after unpacking some definitions, we don't need all the machinery of functors, categories, and commuting diagrams. We don't, for instance, need to show that the fundamental group of the disk is trivial (by convexity, I can continuously retract any loop to a point), or that the fundamental group of the circle is not ($\R$ is a covering space of $S$, and I can declare the preimage of the origin in $S$ to be the set of integers $\Z \subset \R$; each loop, then, lifts to a path starting at $0\in\R$ and ending at some integer, and, moreover, homotopy-equivalence lifts). All that really matters is that the loop traversing the perimeter of the disk is homotopic-equivalent to the trivial loop in the disk, but not in the circle.

So what's Proof 1 actually doing? It says, "Here's these two loops in $S\subset D$: one, which I'll label $S(t)$, traces $S$ (for instance, as $S(t) = e^{2\pi i t}$ if we embed everything into the complex plane); the other, which I'll label $1(t)$, sits at a point ($1(t) = 1$). These are homotopic-equivalent in the disk, but not in the circle." Expanding even more, we can write down an explicit homotopic equivalence and examine it: $h(t,s) = s*(e^{2\pi i t} - 1) + 1$. The question, now, is to ask what happens in the image of this contraction in $S$.

Remember how Proof 1 gets from $D$ to $S$. It says, "I have this continuous function $f:D\to D$ without a fixed point, so let me make a new function $g:D\to S$ that sends $p$ to the intersection of the ray through $f(p)$ and $p$ with the circle." I often like to think of functions as dynamic things: I ought to be able to watch a function change something, imagining, in this case, pushing the points of $D$ out towards the perimeter. This image makes it clear why it can't work, since, intuitively, to do this I'd need to punch a hole in the disk — this is what we're trying to prove. In this case, however, I think it will be more elucidating to think of $g$ as labeling each point in $D$ with some point in $S$. Moreover, I can make this labeling a very visual experience: identify $S$ with the color wheel, moving continuously from red to purple to blue to green to yellow to orange to red, and color each point in $D$ based on what color it ends up after hitting it with $g$.

All of a sudden this is looking more like Proof 2. When perturbing Proof 2 towards Proof 1, we started introducing more and more homotopy-ish ideas. Now we're coloring things!

By watching the colors of the gradually shrinking circle, we can tell how the corresponding loop in $S$ transforms. We start with a loop that hits all the colors in order — how do small perturbations effect this? To visualize: do the same coloring with your paths. Instead of thinking of a time-parameterized loop $Q(t)$ in $S$, think of a continuous coloring of some other circle along which $t$ varies. Perturbations of the path correspond to perturbations of the coloring. And the intuitive point, which I haven't proved (it's roughly equivalent to Brouwer, so yes, I have, a couple times, but bear with me), is that no continuous perturbation, no matter how large, of the original spectrum-coloring of the $t$-circle could ever fail to include all the colors. Think of this as an intermediate-value-theorem result: as I travel around the circle, I start at red, say; I might hit red a lot of times, but there's a first orange, and a last red before the first orange, and I certainly hit all the colors between red and orange just between that "last red before the first orange" and that "first orange"; then there's a "first yellow after the first orange" and a "last orange before that first yellow"; etc. But now go back to the disk. The colors on the shrinking loop must be the same as the colors we just explored, except that at the end we have a loop that stays right near a single color (continuity of our coloring says that a small enough ball around any given point can't include all the colors). This contradiction gives us yet another proof (strongly motivated by Proof 1, sure), or Brouwer's Fixed-Point Theorem.

What happened is that as we blithely retracted out loop, we must have passed through some snag — some "brown" point around which any arbitrarily small loop must run through the entire spectrum. If our coloring is continuous everywhere, no such brown point exists. But Sperner's lemma is a way of finding one.

Remember we ended up describing Sperner's lemma as assigning a weight to each cell in the triangulation based on how many times the colors on the perimeter of the cell move around $K_3$. We could have tried to continuously color the perimeter — for the hypotheses of the lemma, say that the outside edge between the red and blue corners takes values only in the purple range — and measured what are essentially winding numbers for each little cell. Complex analysis, it seems, would give us a proof right away. Except that we're actually begging the question: there's no reason our function need be analytic (continuous functions generally aren't), and for these values to be well-defined we need Brouwer's theorem (or something equivalent to it).

Indeed, there's no particular reason that said brown point should be inside any given red-blue-yellow triangle. You can easily draw two adjacent triangles, one red-blue-yellow and the other red-blue-blue, with the brown point inside the red-blue-blue one. Or even farther away. But there is a sense in which the brown point is "close by".

To whit: I want to measure how colorful each point is, and the way I'm going to do it is by asking (i) what color is the point? (ii) where is the closest point with the color opposite the point's color on the color-wheel? (iii) What is the reciprocal of that distance? This reciprocal I'll call the "colorfulness" of the point; large numbers mean that very differently colored points are close by. When measuring "the distance to the nearest oppositely-colored point", I more properly mean the infimum of such distances, but continuity guarantees me that this infimum is never zero (as I argued in a brief parenthetical four paragraphs above: a small enough ball around any given point can't include all the colors that aren't close to the color of the point). Now here's the coup de grace: this "colorfulness" is a continuous function from the compact disk to the reals, and as such it must be bounded. Which is to say that there is some small epsilon so that any region of diameter less than epsilon cannot include all the colors.

Now it's clear how Proof 2 works. Remember how we took successively finer triangulations? Imagine a triangulation so that every subtriangle has diameter less than epsilon. This triangulation is fine enough to measure all the interesting variations in the coloring of the disk. And, if there are no brown points, then it can't contain a red-yellow-blue subtriangle, by the argument in the previous paragraph (I really ought not to ask for the distance to the nearest opposite-colored point, but to a point that is a third away around the color wheel, but no matter). And yet it has to. (Another proof of Brouwer!) In fact, we can say even more. Let's say that there are a couple brown points, but that we've placed little bubbles, little event horizons around them. Tiny little things, of radius roughly delta. And subtract them out of the disk. Now it's still compact, so the argument goes through, and we can use a fine triangulation that captures all the interesting physical features. Then the only way for a subtriangle to be all of red, blue, and yellow is if it encloses a brown point (er, encloses a point within the event horizon, so it's within delta of the brown point; then make delta, and with it epsilon, smaller). A fine enough triangulation will find your holes for you.

So really Proofs 1 and 2 are measuring much the same things, just coming at the system from slightly different points of view. Proof 1 embraces the continuum, using from the start a continuous coloring, and machinery like the intermediate value theorem and the real numbers. Proof 2 shows how to do the calculations, to whatever accuracy you desire, by using only discreet combinatorial arguments at any given stage. Proof 1 thinks that the universe really exists as a manifold, and Proof 2 counters that you have to be able to measure things. Proof 2 has been reading too much popular quantum gravity: perhaps, it suggests, it's good enough of the universe really is discreet and combinatorial.

I won't claim that Proofs 1 and 2 are really the same. Not only do the use and prove different subsidiary and auxiliary results, but the underlying philosophies are really quite different. I used to prefer Proof 1 as the more demonstrative — the idea that a continuous function can't introduce a hole, but that the circle has one, and that the way you measure "holes" is with these (rather abstract) "fundamental groups" all made sense to me. Proof 2, on the other hand, never seemed particularly elucidative; Sperner's lemma seemed like a trick pulled out of a hat. But, properly explored and explained, I've come to like Proof 2 better. It's much more elementary, eschewing such difficulties of covering spaces and categories. As generally written it's a terrible proof — it's elegant, a la Aigner and Ziegler, but it doesn't explain the underlying physics. I hope, however, that I've delved far enough into it for you to agree with me that it's ultimately more explanatory. Why can't the disk map to itself without a fixed point? Because somewhere any coloring that traverses the spectrum on the boundary must admit a whirlpool, a brown point. And we even have a device for locating and measuring the whirlpools.

No, I won't claim that the two proofs are the same, but they can be perturbed, massaged, and transformed into each other. Not the same — homotopically equivalent.