## The Alternating Group is Simple III

I will now conclude my series of posts about the alternating group by proving that $A_n$ is simple for $n \geq 5$. Just as with $A_5$ I stole this argument from Dummit & Foote; while I feel I might have been able to come up with the argument for $A_5$ on my own, the argument for $A_n$ is a bit too clever for me. If anyone knows who came up with it, please let me know.

We are going to use again and again the formula for conjugating permutations from my last post, so I will repeat it here for reference:

Lemma 1: Let $\sigma = (a_1 a_2 \ldots a_k)$ be a cycle and let $\tau$ be any permutation. Then $\tau \sigma \tau^{-1} = (\tau(a_1) \tau(a_2) \ldots \tau(a_k))$

Let us jump right into the proof of the main result:

Theorem: $A_n$ is simple for every $n \geq 5$.

Proof: We use induction on $n$. The base case, $n = 5$, was handled in the last post. So assume that $A_{n-1}$ is simple, and let $H$ be a proper normal subgroup of $A_n$, $n \geq 6$. Our aim is to show that $H$ is the trivial group.

Our first step is to prove that no non-identity element of $H$ can fix any symbol. Let $G_i$ denote the subgroup of $A_n$ consisting of all elements that fix the symbol $i$; by Lemma 1 we have $\tau G_i \tau^{-1} = G_{\tau(i)}$ for any permutation $\tau$. Note that $G_i \cong A_{n-1}$ for each $i$, so if $H$ intersects some $G_i$ nontrivially then $G_i \subseteq H$ by the induction hypothesis. Moreover, since any $G_j$ can be obtained from $G_i$ by conjugation and $H$ is normal, we have that $G_j \subseteq H$ for all $H$.

Now, any element of $A_n$ can be written as the product of pairs of transpositions. A pair of transpositions can only permute up to four symbols, so since $n \geq 5$ every pair of transpositions fixes at least one symbol and hence is in some $G_i$. Thus every element of $A_n$ can be written as a product of permutations each of which is in some $G_i$; since $G_i \subseteq H$, it follows that $A_n \subseteq H$, contradicting our assumption that $H$ is a proper subgroup.

So no non-identity element of $H$ can fix any symbol. Consequently, if two elements of $H$ agree on even one symbol then they must be the same, for if $\tau_1(i) = \tau_2(i)$ then $\tau_1 \tau_2^{-1}$ fixes $i$ and hence is the identity. To complete the proof we will use this observation to show that the identity is the only element of $H$.

• No element of $H$ can contain a $k$-cycle for $k \geq 3$:
Suppose $\sigma \in H$ contains a $k$-cycle $(a_1 a_2 a_3 \ldots)$. Since $n \geq 5$ it is possible to choose $\tau$ which fixes $a_1$ and $a_2$ but not $a_3$. By Lemma 1 we have:
$\tau (a_1 a_2 a_3 \ldots) \tau^{-1} = (a_1 a_2 \tau(a_3) \ldots)$
Thus $\sigma$ and $\tau \sigma \tau^{-1}$ are two permutations in $H$ which agree on $a_1$ but not on $a_2$; this is a contradiction.
• No element of $H$ can be the product of disjoint $2$-cycles:
Suppose such an element $\sigma$ were to exist. Since $n \geq 6$ and $\sigma$ can’t fix any symbols, it must be the product of at least three disjoint $2$-cycles:
$\sigma = (a_1 a_2)(a_3 a_4)(a_5 a_6)\ldots$
Let $\tau = (a_1 a_2)(a_3 a_5)$. We have:
$\tau \sigma \tau^{-1} = (a_1 a_2)(a_5 a_4)(a_3 a_6)\ldots$
This time $\sigma$ and $\tau \sigma \tau^{-1}$ agree on $a_1$ and $a_2$ but not on $a_3$, a contradiction.

We conclude that no element of $H$ can have a cycle of length larger than $1$; this means that $H$ is the trivial group.
QED

## The Alternating Group is Simple II

In my last post I described the alternating group and its place in the world of groups. I will now prove that $A_5$ is simple, and in the third and final post of this series I will prove that $A_n$ is simple for $n \geq 5$. The plan of attack is as follows: first I will carry out some preliminary analysis of conjugacy in $S_n$ and $A_n$, and then by identifying all conjugacy classes in $A_5$ I will prove that $A_5$ is simple. I will then prove that $A_n$ is simple for $n \geq 5$ by induction. I’m not sure who invented this argument; all I know is that I learned it in Dummit & Foote.

### Conjugacy Classes in the Alternating Group

To understand the normal subgroups of a group it is very useful to first think carefully about its conjugacy classes; this is because a normal subgroup is by definition the union of conjugacy classes. Fortunately conjugation in the symmetric group is easy to understand using “cycle notation”. A $k$-cycle in $S_n$ is a permutation which fixes all but $k$ symbols $a_1, \ldots, a_k$ which acts on these symbols as:

$a_1 \to a_2 \to \ldots \to a_k \to a_1$

The notation for this cycle is $(a_1 a_2 \ldots a_k)$. It is not hard to show that every permutation decomposes as the product of disjoint cycles, and the decomposition is unique up to reordering the cycles. Indeed, cycle notation makes it particularly easy to understand conjugation.

Lemma 1: Let $\sigma = (a_1 a_2 \ldots a_k)$ be a cycle and let $\tau$ be any permutation. Then $\tau \sigma \tau^{-1} = (\tau(a_1) \tau(a_2) \ldots \tau(a_k))$

Proof: For $i < k$ we have $\tau \sigma \tau^{-1}(\tau(a_i)) = \tau \sigma(a_i) = \tau(a_{i+1})$ and similarly $\tau \sigma \tau^{-1}(\tau(a_k)) = \tau(a_1)$.
QED

The lemma extends easily to the case where $\sigma$ is the product of cycles, so we see that conjugation by $\tau$ preserves the cycle structure of $\sigma$ while relabelling the symbols in the cycle. In particular, two elements of $S_n$ are conjugate if and only if the number and lengths of cycles are the same. For instance, $(12)(345)$ is conjugate to $(124)(35)$ in $S_5$ but not to $(12345)$.

Note that conjugacy in $A_n$ is a little more subtle. A $k$-cycle is even if and only if $k$ is odd, but not all $k$-cycles are conjugate in $A_n$. For instance the transposition $\tau = (45)$ conjugates $(12345)$ to $(12354)$ in $S_5$, but there is no even permutation which conjugates $(12345)$ to $(12354)$ and hence they are not conjugate in $A_5$.

To prove that $A_5$ is simple, we will need to determine the sizes of all of its conjugacy classes. We will do this using the following tool:

Lemma 2: Let $g$ be an element of a group $G$, let $Z_G(g)$ be the centralizer of $g$ (i.e. the set of all elements of $G$ which commute with $g$) and let $C_G(g)$ denote the conjugacy class of $g$. Then $|Z_G(g)| \cdot |C_G(g)| = |G|$

Proof: Let $G$ act on itself by conjugation. The orbit of $g$ under this action is $C_G(g)$ and the stabilizer is $Z_G(g)$, so the result follows from the orbit-stabilizer theorem.
QED

We will apply this lemma as follows. First we will use our understanding of conjugacy in $S_n$ to identify the centralizer of a cycle. From that it is easy to identify the centralizer of a cycle in $A_n$, and that will allow us to count the conjugates of a cycle in $A_n$.

Proposition 3: Let $\sigma \in S_n$ be a $k$-cycle. Then:
$Z_{S_n}(\sigma) = \{\sigma^i \tau:\: 0 \leq i < k,\, \tau \in S_{n-k}\}$

Proof: By Lemma 1, the conjugates of $\sigma$ in $S_n$ are precisely the $k$-cycles. To specify a $k$-cycle one must specify the symbols in the $k$-cycle and the order in which they appear; there are $\frac{n!}{k!(n-k)!}$ ways to choose $k$ symbols and $k!$ different orders in which they can appear, though $k$ of the orders define the same cyclic permutation. Thus there are $\frac{n!}{k!(n-k)!} \cdot (k-1)! = \frac{n!}{k \cdot (n-k)!}$ conjugates of $\sigma$; by Lemma 2, $|Z_{S_n}(\sigma)| = k \cdot (n-k)!$.

The permutation $\sigma^i$ clearly commutes with $\sigma$. Any permutation $\tau$ which fixes the $k$ symbols that $\sigma$ acts on also commutes with $\sigma$, and the subgroup of all such permutations is isomorphic to $S_{n-k}$. Thus the permutations $\sigma^i \tau$, $\tau \in S_{n-k}$, all commute with $\sigma$; there are $k \cdot (n-k)!$ distinct permutations of this form, so they make up the entire centralizer of $\sigma$.
QED

### Simplicity of $A_5$

We are now ready to prove the main result of this post:

Theorem: $A_5$ is simple.

Proof: The only possible cycle structures of non-identity elements in $A_5$ are $(123)$, $(12345)$, and $(12)(34)$. Recall that in $S_5$ the cycle structure completely determines the conjugacy class; in $A_5$ some of these conjugacy classes may split. Let us analyze each conjugacy class in turn using Proposition 3.

• $(123)$: The centralizer of $(123)$ in $S_5$ consists of the six permutations $(123)^i \tau$ where $i = 0, 1, 2$ and $\tau$ is either the identity or $(45)$, so $(123)$ has $120/6 = 20$ conjugates in $S_5$. If $\tau = (45)$ then $(123)^i \tau$ is odd, so the centralizer in $A_5$ has only three elements and hence the number of conjugates is still $60/3 = 20$. Thus all $3$-cycles are conjugate in $A_5$.
• $(12345)$: The centralizer of $(12345)$ in both $S_5$ and $A_5$ is just the cyclic subgroup $\{(12345)^i\}$, so there are $120/5 = 24$ conjugates in $S_5$ and $60/5 = 12$ conjugates in $A_5$. The other $12$ elements in the $S_5$ conjugacy class are accounted for by the $A_5$ conjugacy class of $(12354)$ which is disjoint from that of $(12345)$.
• $(12)(34)$: It is straightforward to check that $(12)(34)$ commutes with the identity, itself, $(13)(24)$ and $(14)(23)$. If $\tau$ does not fix the symbol $5$ then $\tau (12)(34) \tau \neq (12)(34)$ by Lemma 1, so $(12)(34)$ does not commute with $\tau$. A similar argument shows that $(12)(34)$ does not commute with any $3$-cycle, so the centralizer has exactly $4$ elements and hence $(12)(34)$ has $60/4 = 15$ conjugates in $A_5$.

Including the identity, we have accounted for the conjugacy classes of all $60$ elements of $A_5$: $60 = 1 + 20 + 12 + 12 + 15$. So let $H$ be a normal subgroup of $A_5$. Since $H$ is normal it is the union of conjugacy classes (including the identity), so $|H|$ is the sum of $1$ and some subset of $\{20, 12, 12, 15\}$. But $|H|$ must also divide $|A_5| = 60$; checking cases the only possible choices for $|H|$ are $1$ and $60$.
QED

## 1+2+3+4+5+…

It’s been almost a year since I last blogged, and I’ve spent much of that time feeling guilty about not blogging enough. So here we are. I was lured out of my state of blog-apathy by a recent post by mathbabe; in fact, this will be the second of my very few blog posts inspired by that blog. If you’re not already reading that blog, you really should – it’s a brilliant mix of math, politics, economics, data, and sex.

In any event, mathbabe was commenting on a video which has apparently been making its way around the internet. In this video, some mathematicians (Or perhaps physicists? What are string theorists calling themselves these days?) attempted to explain the mind-boggling “fact” that

$1 + 2 + 3 + 4 + 5... = -1/12$

Watch the video if you like, but by now a number of other mathematicians have rightfully pointed out that most of the fishy manipulations in the video amount to fraudulent nonsense which can be used to justify just about anything. This infuriates me, because the people who made the video could have used the opportunity to legitimately blow people’s minds by placing the equation above (which does make sense, from the right point of view!) in its proper context and explaining some beautiful mathematics.

I don’t have the apparatus to make a cool video, but I do have a blog. So I’m going to make an attempt to do what I think the video should have done (I am not optimistic that my attempt will get picked up by Slate, of course). Instead of adding up all of the positive integers, I’m going to start by adding up all of the powers of two:

$2 + 4 + 8 + 16 +... = -2$

We still get a negative number, so this equation should be just as counter-intuitive as the original one (though admittedly $-1/12$ is pretty bizarre). Our strategy for making sense of both equations will be the same:

1. Write down an equation which makes sense (both logically and intuitively) in a narrow context
2. Observe that the right-hand side of the equation actually makes sense in a much larger context than the left-hand side
3. Use the right-hand side as a proxy for the left-hand side in the larger context

The strange equations that I wrote above are mathematical counterparts of taking a word such as “hyperlink” which only really makes sense in the context of the internet and applying it to real world mail. You would end up with a sentence which looks pretty bizarre, but there would nevertheless be a certain logic to it.

Let’s see how this all plays out mathematically. We’ll start with something that isn’t likely to stir up much controversy:

$1/2 + 1/4 + 1/8 +... = 1$

This is the mathematical counterpart of the observation that if you walk across half of a room, then a quarter of the room, then an eigth, and so on then you will have crossed the whole room. (Of course, there are some philosophical questions to be raised by the fact that the phrase “and so on” took the place of an infinite number of actions. Even non-controversial infinite series deserve serious thought.)

You might also convince yourself that

$1/3 + 1/9 + 1/81 +... = 1/2$

It might not be obvious that the answer is $1/2$, but this answer is at least plausible: we start with a number which is smaller than $1/2$ and add increasingly tiny numbers to it. And if you plug numbers into a calculator you will get good numerical evidence that this equation makes sense; the further out you go in the sum, the closer you get to $1/2$. In general, if the absolute value of $s$ is a number smaller than $1$, we have:

$s + s^2 + s^3 +... = \frac{s}{1 - s}$

Of course, the only context in which the left-hand side really makes sense is when $|s| < 1$; this ensures that the powers of $s$ get very small very vast and thus the settles near a particular value. If $|s| \geq 1$ then there is no such guarantee: the powers of $s$ do not get smaller, and you can get a number as large as you want by adding up enough terms in the sum.

The right-hand side, on the other hand makes sense in a much larger context: we can plug in any number except $s = 1$! In particular, we can plug $s = 2$ into $\frac{s}{1-s}$ to obtain $\frac{2}{1-2} = -2$. Since the expressions $s + s^2 + s^3 +...$ and $\frac{s}{1-s}$ agree when $|s| < 1$, it makes sense to use the latter expression as a proxy for the former at other values of $s$, such as $s = 2$. In othe words, it is not entirely stupid to write:

$2 + 4 + 8 +... = -2$

There is some theory which makes this equation even less stupid: $\frac{s}{1-s}$ is (in a sense which can be made precise) the only sensible way to extend $s + s^2 + s^3 +...$ beyond the set $|s| < 1$. Properly justifying this requires techniques coming from one of the most beautiful subjects in all of mathematics: the calculus of complex numbers. It should not be at all obvious, but in the end this whole discussion is really all about the mysterious powers of complex numbers.

The same techniques allow us to analyze the sum $1 + 2 + 3 +...$ which got this post started; this time, our starting point is the Riemann Zeta function:

$\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} +... = \zeta(s)$

This time the sum on the left-hand side makes sense as long as $|s| > 1$, but the same tools described above imply that the Riemann Zeta function can be “analytically continued” to allow any input except $s = 1$, and its value at $s = -1$ can be calculated to be $-1/12$. This calculation could occupy another entire blog post, so I will not go any further than that at this time.

Now that I have explained the sense in which it is not completely stupid to say that the sum of all the positive integers is $-1/12$, I would like to conclude by arguing that it still is pretty stupid. Notice that according to the reasonging described in this post we did not assign the sum a value by thinking about it intrinsically as we can with, for instance $1/2 + 1/4 + 1/8 +...$; instead we related the sum to the Riemann Zeta function and analyzed that function. But there are infinitely many other possible functions which have a similar relationship to $1 + 2 + 3 +...$, and many of them will assign different values to the series following the steps outlined here. In fact, you can use these steps to justify giving the sum any value you want. Still, the Riemann Zeta function enjoys a privileged position in mathematics (and physics) so $-1/12$ is a pretty good choice.

## Gender and the Mathematical Community

I still haven’t posted all that much in this blog, and essentially nothing research-related. I’ve been writing a bit offline, and I’ll probably adapt some of what I’ve been thinking about into blog form fairly soon. In this post I’d like to address some issues related to sexism and gender bias in the mathematical (and perhaps broader scientific) community. I think about these issues rather often, but I’m writing about them now because of recent posts in The Accidental Mathematician (Izabella Laba’s blog) and mathbabe (Cathy O’Neil’s blog).

The thrust of Izabella Laba’s post (entitled “Gender Bias 101 for Mathematicians”) is that gender bias in the mathematical community is not limited to a few grouchy old codgers, but rather that it is a systematic cultural and psychological phenomenon which afflicts everybody. There are two potentially controversial assertions implicit in this statement:

1. Gender bias in the mathematical community exists.
2. Gender bias in the mathematical community is pervasive and systematic.

The first assertion is pretty hard to argue with, though I’m sure some people still try. Every math department with which I have been affiliated is *massively* male dominated, and there is ample evidence that hiring practices, salaries, journals, etc. are stacked in favor of men. I’m not going to try to document or justify this in any detail because I don’t have the facts available at my fingertips and because the issue has been argued to my satisfaction elsewhere (e.g. in the Accidental Mathematician).

The second assertion might be more surprising to some, and it’s the one I want to discuss here. Izabella Laba’s post quotes a recent study in which faculty from research oriented universities were presented with applications for a lab manager position with randomly assigned male or female names. The study found that a given application with a male name at the top was consistently rated more highly than the same application with a female name. Interestingly enough, the pattern was independent of the gender of the faculty evaluator: female professors were just as biased as male professors. Cathy O’Neil contributes another study which shows that 15-year-old girls outperform 15-year-old boys in science exams in some countries but not others (not in the United States), indicating that gender gaps in science are cultural rather than biological.

Both of these studies are quite compelling, and I’m sure there are others which point to the same conlusion. My intention is to participate in this discussion subjectively rather than objectively. In short, I am going to use the rest of this post to analyze my own gender-oriented biases. Something feels a bit self-indulgent about this exercise, but I think it will be healthy for me even if it isn’t useful for anyone else.

I will begin by admitting outright that I am biased against women. I consider myself to be a pretty progressive guy – perhaps even more progressive than most – and I think that most people who know me would say that overall I do a good job of treating women with the same respect with which I treat men. But this is not because I don’t have biases, it’s because I work very hard to identify them and eliminate them or at least minimize their impact on my behavior. I am unqualified to generalize my own psychological observations to everyone else, but I suspect that it is neurologically almost impossible for a person socialized in 20th or 21st century American society to avoid gender biases: we are bombarded with overt and covert messages about gender constantly and starting at a very young age. Given what I have been learning lately about how insignificant our conscious thought processes are in comparison to our subconscious psychological machinery, these messages must take their toll.

What forms do gender biases take? There are many answers with varying applicability to me. Here is a non-comprehensive unordered list that I have assembled from reading things online, talking to people, and making my own observations.

• Intelligence and Competence Bias: This is simply the assumption that women are less intelligent or less competent than men. I have heard numerous stories in which Andrew launches into a lengthy explanation to Barbara about a subject in which Barbara is more of an expert than Andrew. Here is a particularly cringe-inducing example of this. I tend not to offer unsolicited explanations to men or women very often, and when providing solicited explanations I usually make an effort to identify my audience’s background, so I don’t think I am terribly guilty of this particular behavior. Instead, I notice this bias in myself when I am seeking an expert on a particular subject and I am presented with a male option and a female option. Sometimes I catch myself behaving or thinking according to the assumption that the male expert is more knowledgable or more adept than the female expert even if I have no particular reason to make such a judgement. I have to force myself to think deliberately about what I know and don’t know when making these sorts of comparisons.
• Experience Bias: Lately I have started noticing a disturbing pattern in my judgements about a person’s age, experience, education level, etc.: my estimates are consistently lower than reality for women and higher than reality for men. I have heard many stories from women in which they are demoted from faculty member to graduate student or from graduate student to undergraduate by a male interlocutor, and I am embarrassed to admit that I have done this before. I have also heard stories in which a female graduate student or faculty member has been assumed to be a secretary or staffperson; I don’t necessarily consider this to be a “demotion,” but I doubt I would appreciate it if it happend to me. These days I try to avoid guessing somebody’s position or experience level at all, and if I do make a guess it’s generally “faculty” regardless of gender (in a university setting). Still, it requires conscious effort on my part.
• Common Ground Bias: This is the assumption that, all things being equal, I will have more in common with a male than a female. This bias is fairly understandable – there are, after all, real biological and social differences between men and women – but I think it has unfortunate consequences in an academic setting. Few of my mathematical conversations with my peers begin, “Hi, my name is Paul. Would you like to have a conversation about elliptic cohomology?” Instead, they typically begin with the typical introductory social graces and lead into mathematical territory after a basic rapport has been established. This rapport is more difficult to establish with a person with whom I assume I will have a harder time identifying with before the conversation even begins, and consequently I am more likely to engage in mathematical conversations with my male colleagues than my female colleagues. I don’t know how socially isolated women in math departments feel, but I suspect that it’s more of a problem than I realize. My plan for reducing the impact of this bias is to simply be more bold and less awkward about engaging people in conversation, but this isn’t always easy.
• Sexual Biases: I am a heterosexual man who is attracted to intelligent and ambitious women, and the women that one finds employed in a math department often fit this description. Sexual attraction is firmly rooted in extremely powerful subconscious processes, and I am certain it affects my interactions with my female colleagues in ways that I don’t fully understand. If nothing else, it consumes some measure of my mental energy that is liberated when I’m interacting with men. It seems very hard to deal with the subconscious aspects of this bias, but I long ago adopted a mechanism which at least helps me manage the factors that are under my control. I decided early on in graduate school that I would categorically avoid romantically pursuing anyone in my own department. This allows me to sidestep the hazards associated with workplace romances in general, but mainly it helps me ensure that I treat all of my colleagues as professionally as possible. I don’t know how often the average woman in a math department is forced to deal with romantic overtures from her male colleagues, but given the highly skewed gender ratios I’m guessing it’s more than I imagine. I am also largely ignorant of the consequences of this behavior.

I’m sure there are other biases worth mentioning, but this list feels like a good start. One interesting supplementary observation about biases in general is that thinking about them leads to an unfortunate feedback loop: worrying about biases against women affects my behavior toward women. I think this effect is fairly minimal in comparison to the consequences of ignoring my biases and failing to monitor my behavior at all, but it’s there all the same.

My final remark about this subject is that there are many other bias issues which are also largely ignored by the mathematical and scientific community. I have encountered some discussion of racial bias in science, but I have heard almost no discussion about biases related to sexual orientation. If anyone reading this is aware of any studies or references about these issues, I would be interested in seeing them. Also, in this post I have focused on the effects of bias on my interactions with my colleagues, but the way my biases manifest themselves in my teaching is a whole other subject which I might take up in the future.

## The Coaster Roller

My first blog post was inspired by my visits to the Museum of Mathematics, and it appears that my second will be as well.  Like the first post this one will be suitable for a general audience, but I’ll write a follow-up which gets a little more mathematically serious.

I started volunteering at MoMath as an “integrator” this past Sunday; the primary role of the integrator is to float around the museum helping people who seem confused by an exhibit.  A secondary role is to operate the few exhibits which require staff supervision, and I was taught how to operate an exhibit called the “coaster roller”.  The coaster roller works as follows.  There is a small pit full of peculiar-looking objects and a sort of “raft” which sits on top of them (I’ll take a picture next time I visit the museum).  The objects are about the size of basketballs, but they are not spherical.  Here’s a picture of what they might look like, stolen from the internet:

A person (or several children) can sit in the raft and pull themselves along using ropes, rolling over the strange shapes at the bottom.

So what’s so special about the shape of these objects?  The important property that they each possess is that they have constant width, meaning that if one of the objects fits in a vise when it is pointing in one direction then it fits in the same vise when it is pointing in any other direction.  This is important because if the width varied then the distance between the raft and the floor would vary as it rolled along and the ride would get pretty bumpy!

The constant width property is possessed by a sphere but not by a cube: the width of a cube is larger when measured between opposite corners than when measured between opposite faces.  If you think about it for a moment, you might find it hard to convince yourself that bodies of constant width which aren’t spheres can even exist!  But in fact they exist in abundance; I will explain a procedure which allows you to construct infinitely many different ones.

To begin, note that given any plane curve of constant width one can obtain a corresponding surface of constant width by rotating the curve in three-dimensional space (you may have already noticed that each of the surfaces in the picture above is rotationally symmetric).  Not all surfaces of constant width arise this way, but in any event this shows that it is enough to look at curves of constant width if we are happy just finding a few basic examples.

To construct a curve of constant width, begin by drawing an equilateral triangle with side length $w$.  Then draw the circle of radius $w$ centered at each of the three vertices of the triangle.  The boundary of the intersection of the disks enclosed by these circles is then a curve of constant width $w$.  Can you figure out why?  Here’s a picture, stolen from Wikipedia:

In fact, the same procedure works if you start with any regular polygon with an odd number of sides (thanks to Ben Levitt from MoMath who corrected my original claim that it works for any regular polygon).  The curves of constant width obtained in this way are often called Reuleaux polygons for their discoverer (even though they aren’t technically polygons).

There are all sorts of interesting mathematical questions one can ask about bodies of constant width.  Here are some useful facts:

1. All curves of the same constant width have the same perimeter.
2. Among all curves with a given constant width, the circle encloses the largest area.  Similarly, among all surfaces with a given constant width, the sphere encloses the largest volume.
3. Among all curves with a fixed constant width, the Reuleaux triangle encloses the smallest area.
4. Among all surfaces with a fixed constant width, nobody knows which one encloses the smallest volume!

In my next blog post, I plan to discuss some of these facts in greater detail.  If you want to read more in the meantime, I recommend The Enjoyment of Math by Rademacher and Toeplitz, two great 20th century mathematicians. Actually, I recommend that book even if you don’t care to read more about bodies of constant width!

## The 15 Game

New York city recently opened the Museum of Mathematics, the United States’ first museum dedicated exclusively to mathematics.  It opened on December 15 (my birthday!) and it’s tons of fun.  If “museum of mathematics” conjures the image in your mind of a framed painting of the quadratic formula or the bust of a dead Greek intellectual, you would be very pleasantly surprised.  The museum is full of puzzles, games, toys, computer simulations, and innovative hands-on exhibits that you probably never imagined.  Nearly every exhibit is interactive in some way, and I don’t recall a single equation in the whole building.

For my inaugural blog entry, I’m going to write about one of my favorite exhibits in the museum.  It’s based on a seemingly simple game which conceals some surprising mathematical secrets.  I first heard about the game from Prof. Mel Hochster during the summer after I graduated from the University of Michigan in 2007.  He was teaching a summer course on the Fibonacci numbers to talented high school students (I was a course assistant) and he used the game to illustrate the notion of isomorphism, a term which mathematicians use to describe seemingly different phenomena which are secretly the same.

Anyway, here’s the game.  All you need is a friend, a piece of paper, and a pencil.  Write the numbers 1 through 9 at the top of the page, and take turns with your friend choosing a number, crossing out each number once it has been chosen.  The object of the game is to be the first person to select exactly three numbers which add up to 15.

Example:
You pick 5, I pick 9.
You pick 4, I pick 6.
You pick 8, I pick 3.
You pick 2, and you win: 5 + 8 + 2 = 15! (Note that I didn’t win even though I picked 9 and 6 because I needed *exactly* three numbers which add up to 15.)

If you have a friend nearby, give the game a try. It’s a pretty challenging game, and it’s structure isn’t particularly obvious. Does either player have a winning strategy? Can either player force a draw? What is the best starting move? If you think about the game for long enough, you might eventually be able to provide answers to some or all of these questions, but it will probably take some effort.

The beautiful thing about the game is that it is completely equivalent (isomorphic, in fact!) to a much simpler game that almost everybody understands. To see what’s going on, we’ll use a so-called magic square:

The property possessed by this table of numbers which makes it “magic” is that each row, column, and diagonal adds up to 15. Moreover, every triplet of numbers from 1 to 9 which add up to 15 is represented as some row, column or diagonal. Let’s go through the example game above one more time, only this time we’ll draw an “X” through each number that you picked and an “O” through each number that I picked. Here’s what it looks like at the end:

As you can hopefully see, the game is nothing more than tic-tac-toe in disguise!

At the museum, they came up with a clever way to turn this game into an exhibit. The two players play the game on a computer screen, but one player is seated inside a concealed booth with a magic square. The poor player without the magic square has to labor through a lot of arithmetic, while the player in the booth just has to play tic-tac-toe.

I think this game (and the corresponding exhibit) is one of the best non-technical illustrations of what mathematics is all about.

1. It demonstrates that two seemingly very different phenomena can be secretly “the same”.
2. It shows how interesting abstract constructions (like the magic square) can be useful in unexpected ways.
3. It shows how hard problems can become much simpler if you find the right way to look at them.

These are all extremely important themes in mathematics, and I hope to explore each of them further in future blog posts.