# Structures that turn out to exhibit a symmetry even though their definition doesn't

Sometimes (often?) a structure depending on several parameters turns out to be symmetric w.r.t. interchanging two of the parameters, even though the definition gives a priori no clue of that symmetry.

As an example, I'm thinking of the Littlewood–Richardson coefficients: If defined by the skew Schur function $s_{\lambda/\mu}=\sum_\nu c^\lambda_{\mu\nu}s_\nu$, where the sum is over all partitions $\nu$ such that $|\mu|+|\nu|=|\lambda|$ and $s_{\lambda/\mu}$ itself is defined e.g. by $s_{\lambda/\mu}= \det(h _{\lambda_i-\mu_j-i+j}) _{1\le i,j\le n}$, it is not at all straightforward to see from that definition that $c^\lambda_{\mu\nu} =c^\lambda_{\nu\mu}$.

Granted that this way of looking at it may seem a bit artificial, as I guess that in many of such cases, it is possible to come up with a "higher level" definition that shows the symmetry right away (e.g. in the above example, the usual (?) definition of $c_{\lambda\mu}^\nu$ via $s_\lambda s_\mu =\sum c_{\lambda\mu}^\nu s_\nu$), but showing the equivalence of both definitions may be more or less involved. So I am aware that it might just be a matter of "choosing the right definition". Therefore, maybe it would be better to think of the question as asking especially for cases where historically, the symmetry of a certain structure has been only stated 'later', after defining or obtaining it in a different way first.

Another example that would fit here: the Perfect graph theorem, featuring a 'conceptual' symmetry between a graph and its complement.

What are other examples of "unexpected" or at least surprising symmetries?

(NB. The 'combinatorics' tag seemed the most obvious to me, but I won't be surprised if there are upcoming examples far away from combinatorics.)

• Quadratic reciprocity. – Terry Tao Dec 13 '13 at 22:55
• The relation between $\zeta(1-x)$ and $\zeta(x)$ for the Riemann $\zeta$ function. – Lev Borisov Dec 14 '13 at 2:26
• Number of partitions of $n$ into no more than $k$ terms that are each no larger than $l$. The symmetry between $l$ and $k$ might not be immediately obvious to novices. – Yoav Kallus Dec 14 '13 at 2:46
• The Peano definition of addition, even. – Joe Z. Dec 14 '13 at 2:56
• I saw the title and my first thought was "Littlewood-Richardson coefficients". :) – darij grinberg Dec 14 '13 at 20:55

If $a$ and $b$ are positive integers, and you make the definition $$a \cdot b = \underbrace{a + \cdots + a}_{b \text{ times} }$$ then it's a slightly surprising fact that $a \cdot b$ is actually equal to $b \cdot a$.

• Indeed, this fails in general when $a,b$ are ordinals. – Terry Tao Dec 15 '13 at 4:51
• It's even more surprising if you start with the inductive definitions of plus and times. The proof that $ab=ba$ comes as Proposition 72 in the first development of this theory, by Grassmann in 1861. – John Stillwell Jan 13 '14 at 9:12

A nice example from classical mechanics is this: there is a hidden $SO(4)$ symmetry in the elliptical orbits of a particle in an inverse square potential, ie. the Kepler problem.

The system has an obvious $SO(3)$ symmetry because the inverse square law is invariant under rotations. But there's no a priori clue that an $SO(4)$ symmetry exists in this system.

This carries over to the quantum mechanical case when you solve the Schrödinger equation for an inverse square potential.

The result is that the hidden $SO(4)$ symmetry explains the "coincidence" that many hydrogen atom states have the same energy.

1. I think that if you put yourself back in the position of someone discovering this for the first time, the equality (under suitable hypotheses) $${\partial^2f\over\partial x\partial y}={\partial^2 f\over\partial y\partial x}\quad (1)$$ should count.

2. Here's a surprising application of that suprising equality. Suppose you're a profit-maximizing competitive firm, hiring both labor ($L$) (at a wage rate of $W$) and capital ($K$) (at a rental rate of $R$). Then an increase in $W$ will, in general, lead you to reduce your output and so employ less capital, but at the same time lead you to substitute capital for labor and so employ more capital. On balance, the derivative $dK/dW$ could be either positive or negative. Likewise for the derivative $dL/dR$. It does not seem to me to be at all intuitively obvious that these derivatives even have the same sign, much less that they are equal. But if one takes $f$ in (1) to be profit as a function of $x$ (labor) and $y$ (capital) then one discovers that in fact

$${dK\over dW}={dL\over dR}$$

(Of course this looks more symmetric if you write $X_1$ and $X_2$ for labor and capital, and $P_1$ and $P_2$ for the wage rate and the rental rate.)

• Under the same heading: equality of the mutual inductance $M_{12}$, ratio of the emf induced in coil 1 to the rate of change of current in coil 2, to $M_{21}$. – Yoav Kallus Dec 13 '13 at 23:09
• Maybe it's just sleep deprivation, but I don't see how that second equality works out. It doesn't look like the dimensions match; the left-hand side seems to be in dimensions of capitaltime/labor, while the right-hand side seems to be in dimensions of labortime/capital. – user2357112 supports Monica Dec 14 '13 at 12:49
• @user2357112 : They both have units of capital*labor/output. If output is $F(K,L)$ then profit is $F(K,L)-RK-WL$. Profit maximization implies that $\partial F/\partial K=R$ and $\partial F/\partial L=W$. Use this and the equality of the two cross partials to get the result. – Steven Landsburg Dec 14 '13 at 13:55
• This is only one instance of a much more general fact known as Onsager's reciprocity formula. This is found everywhere there is a thermodynamical formulation. – Denis Serre Aug 22 '18 at 19:04

Higher Homotopy groups $\pi_n(X)$ are abelian. This is quite surprising if you see the defintion for the first time and probably got in touch with the classical fundamental group before, which is not abelian in general.

In fact, higher homotopy groups should serve as a generalization to the fundamental group in contrast to the abelian homology groups, when they were introduced, but as one recognized, that they are abelian too, they seemed to be not a nice generalization.

• I'd like to say something which is well-known by algebraic topologists. This commutativity actually hides some asymmetry. If you look at the usual proof that e.g. $\pi_2(X)$ is abelian, then to prove $ab=ba$ you need e.g. to "move" $b$ over $a$. But if you do it twice, moving $a$ over $b$ back, you have made a loop in the double loop space $\Omega^2 X$, i.e. you get an element of $\pi_3(X)$ that tells you that while $a$ and $b$ commute, you have two choices of making them commute ($a$ over $b$ or $b$ over $a$), and they are not identical. See: Whitehead products (and $E_n$ operads ☺). – Najib Idrissi Aug 18 '18 at 15:59

Rolling one surface on another without slipping binds the velocity of the rolling surface and its angular velocity, giving a rank 2 subbundle in the tangent bundle of the 5-dimensional space of tangential positionings of the 2 surfaces in space. This subbundle, when you roll one sphere on another, has an 8 dimensional symmetry group, unless one sphere has exactly one third the radius of the other sphere, in which case the subbundle is preserved by a 14 dimensional group of diffeomorphisms of the 5-dimensional manifold: the split real form of the simple Lie group $G_2$.

• This subbundle is my favorite example of a non-integrable distribution (if the surfaces are "generic", at least) - you can physically see that rolling a sphere in an "infinitesimal square" on a plane makes the sphere rotate. – Peter Samuelson Dec 14 '13 at 15:29

The joint distribution of IID normal random variables is spherically symmetric.

Although invariance under permutations of the coordinates is obvious for any IID variables, spherical symmetry is rare. In fact, this characterizes the normal distribution.

• This is not a characterization in dimension 1! – KConrad Jan 13 '14 at 5:46
• @KConrad: One-dimensional normal distributions are the subject of the statement: If IID copies $(X_1,...,X_n)$ of a random variable $X$ have a spherically symmetric distribution in $\mathbb{R}^n, n\gt 1,$ then $X$ is normally distributed with mean $0$. Maxwell's Theorem is actually a little stronger than this. – Douglas Zare Jan 13 '14 at 6:21
• What I meant was that spherical symmetry when $n=1$ (i.e., using the group $O(n)$ when $n = 1$) does not characterize the normal distribution. For any fixed $n > 1$, spherical symmetry implies normality. The Wikipedia page for Maxwell's theorem, at the moment I write this, leaves off the condition that $n > 1$ and when I look at Maxwell's theorem in other references it is pretty common to see that the author forgets to say $n > 1$ in the theorem. – KConrad Jan 13 '14 at 7:13
• There are characterizations of normal distributions that work directly in dimension 1, e.g., the characterization using maximum entropy. Maxwell's theorem is a characterization that requires using dimension > 1. – KConrad Jan 13 '14 at 7:17
• @Kconrad: Yes, you are right that you need to use more than $1$ copy (I did say variableS and coordinateS), but this is a hidden symmetry of a $1$-dimensional normal distribution, not just higher dimensional normal distributions. If you really think there is a problem, you are welcome to edit the answer to improve it. The entropy characterization doesn't seem to be a "surprising symmetry" which is what this question asked. – Douglas Zare Jan 13 '14 at 8:01

A pedestrian definition of the rank of a matrix as the maximum number of linearly independent columns equals the maximum number of linearly independent rows.

Hermite's reciprocity: as representations of $GL_2$, we have $$S^k(S^l\mathbb{C}^2)\simeq S^l(S^k\mathbb{C}^2).$$

Consider the Desargues configuration. It consists of (1) two triangles, say $ABC$ and $A'B'C'$ such that the lines $AA'$, $BB'$, and $CC'$ all meet at a point $P$, and (2) the three points of intersection of corresponding sides $X=(BC)\cap(B'C')$, $Y=(AC)\cap(A'C')$, and $Z=(AB)\cap(A'B')$. Desargues's theorem says that then $X$, $Y$, and $Z$ are collinear. The Desargues configuration consists of the 10 points mentioned above ($A,B,C,A',B',C',P,X,Y,Z$) and the 10 lines mentioned (the three sides of both triangles, the three lines through $P$, and the line $XYZ$). The surprising (to me) symmetry is an action of the cyclic group of order 5. In fact, the graph whose vertices are the 10 points of the Desargues configuration and whose edges join any two points that are not together on any of the configuration's 10 lines is the Petersen graph, which is usually drawn in a way that makes the cyclic 5-fold symmetry visible.

• Have used Desargues for easily a hundred times in my schooldays and never realized this. I actually wasn't aware that the Petersen graph had any deeper meaning than that of a counterexample to some conjectures of days gone by. Nice!! – darij grinberg Dec 14 '13 at 21:01

In fact, the "correct" definition of Littlewood-Richardson coefficients shows a surprising $S_3$-symmetry among all the indices $\lambda,\mu,\nu$. See Thomas and Yong - An $S_3$-symmetric Littlewood–Richardson rule.

A further example related to symmetric functions is the symmetry between the area and bounce statistics of Dyck paths. See for instance Chapter 3 of Haglund - The $q, t$-Catalan numbers and the space of diagonal harmonics. No combinatorial proof of symmetry is known.

There are many enumeration problems with "hidden symmetry." For instance, what is the probability that 1 and 2 are in the same cycle of a (uniform) random permutation of $1,2,\dots,n$? More interesting, suppose that I shuffle an ordinary deck of 26 red cards and 26 black cards. I turn the cards face up one at a time. At any point before the last card is dealt, you can guess that the next card is red. What strategy maximizes the probability of guessing correctly? The surprising answer is that all strategies have a probability of 1/2 of success! There is a very elegant way to see this.

• I see how to solve the card problem by proving a more general result for R red cards and B black cards, and then using induction on the size of the deck. (There are two cases: Either my strategy is to guess before the first card or my strategy is contingent on the first card.) But I wonder if that's the "very elegant way" you have in mind. – Steven Landsburg Dec 14 '13 at 0:32
• @StevenLandsburg: imagine the dealer turns over the bottom card of the deck when you guess, instead of the top one. Clearly this situation is symmetric to the one described above, but also clearly every strategy gives 50/50 odds as the outcome is determined before the game even starts. – Sam Hopkins Dec 14 '13 at 1:00
• Can you fix the first link to point to the abstract rather than directly to the PDF? Thank you! – Harry Altman Dec 14 '13 at 18:11

The outer automorphism of $S_6$.

From school days... Take positive reals x,y,z,w. The following statement is actually symmetric in x,y,z,w:

"there exists an equilateral triangle of side length w, and a point whose distances from the three vertices are x,y,z" A quick proof: Let $ABC$ be equilateral and $P$ arbitrary. Construct $BPQ$ equilateral. Let $AB=AC=BC=w$, $AP=x$, $BP=y$ and $CP=z$. Then $BP=PQ=BQ=y$ by construction, $CP=z$ and $CB=w$ obviously, so it remains to check that $CQ=x$. Now note that triangle $CBQ$ is the $60^\circ$ rotation of $ABP$ around $B$.

• I have problems with this when w=10x=10y=10z. You might add an inequality to show when a triangle might exist. Gerhard "Not Doubting The Equivalence, However" Paseman, 2013.12.18 – Gerhard Paseman Dec 18 '13 at 19:44
• @GerhardPaseman I think the statement includes cases like yours where a triangle inequality is violated. It just says that the structure exists iff it exists for any one permutation of $x,y,z,w$. And in your case, it exists for none. :) – Wolfgang Dec 19 '13 at 8:07
• Right. I am not disagreeing with the argument or the statement. I am disagreeing with the presentation. Even if it gives the game away, I would posit "Let there be x,y,z,w satisfying the following inequalities:...", then follow up with the supposedly asymmetrical statement of the existence of an object. I agree that the proof convinces me the statement has a hidden symmetry. Gerhard "Ask Me About System Design" Paseman, 2013.12.19 – Gerhard Paseman Dec 19 '13 at 19:31

The combinatorial definition of the Schur functions is $$s_\lambda(x) = \sum_{T \in SSYT(\lambda)} x^{cont(T)}$$ where $SSYT(\lambda)$ is the set of semi-standard Young tableaux of shape $\lambda$ and $x^{cont(T)}$ is the product over all $i$ of $x_i^{\# i\text{'s in }T}$. This is not manifestly a symmetric function. The Bender-Knuth involution proves that $s_\lambda(x)$ is invariant after swapping $x_i$ with $x_{i+1}$, and thus $s_\lambda(x)$ is, indeed, symmetric.

• And more startlingly (or at least far less obviously), the Stanley symmetric functions and their generalizations. – darij grinberg Jan 22 '14 at 17:43
• And the LLT polynomials. And the Eulerian quasisymmetric functions (which are symmetric - (i did not name them...). – Per Alexandersson Aug 20 '18 at 8:47

The Jacobson radical of a ring $R$ is defined to be the intersection of all maximal left ideals in $R$. It turns out that the Jacobson radical is the intersection of all maximal right ideals in $R$ as well, so the Jacobson radical does not depend on whether one considers left or right ideals. In particular, the Jacobson radical of a ring is a two-sided ideal. In fact, there are several characterizations of the Jacobson radical that do not appear to be symmetric with respect to "leftness" and "rightness" including the following.

1. The intersection of all maximal left ideals.

2. $\bigcap\{\textrm{Ann}(M)|M\,\textrm{is a simple left}\,R-\textrm{module}\}$

3. $\{x\in R|1-rx\,\textrm{has a left inverse for each}\,r\in R\}$

4. $\{x\in R|1-rx\,\textrm{has a two-sided inverse for each}\,r\in R\}$

Let $G$ be a finite group with order $n$. For each $d$ dividing $n$, the number of subgroups of $G$ of order $d$ equals the number of subgroups of order $n/d$ if $G$ is abelian. More broadly, the lattice of subgroups of a finite abelian group looks the same if you flip it around by 180 degrees.

This is not at all obvious at the level at which the statement can first be understood, essentially because there is no natural way to construct subgroups of index $d$ from subgroups of order $d$ in a general finite abelian group with order divisible by $d$. It is not clear at a beginning level how the commutativity of the group leads to such conclusions.

Consider a differential inequality, like the Hardy-Sobolev inequality $$\left|\int\int_{{\mathbb R}^N\times{\mathbb R^N}}\frac{\overline{f(x)}g(y)}{|x-y|^\lambda}dxdy\right|\leq C\|f\|_r\|g\|_s.$$ Even if you put the sharp constant $C$ in this inequality, for most functions the inequality is strict. Now look for maximizers, i.e., functions for which the LHS is equal to the RHS: they are highly symmetric functions, actually spherically symmetric and very smooth. This is a general phenomenon, connected with monotonicity of $L^p$ and Sobolev norms with respect to symmetrization procedures.

I always found $\mathrm{Tor}_R\left(M,N\right) \cong \mathrm{Tor}_R\left(N,M\right)$ for a commutative ring $R$ and two $R$-modules $M$ and $N$ to be mysterious. Then again I have no idea about homology and thus wouldn't be surprised if this is a triviality from an appropriate viewpoint.

Volker Strehl's generalized cyclotomic identity (Corollary 6 in Volker Strehl, Cycle counting for isomorphism types of endofunctions states that $\prod\limits_{k\geq 1} \left(\dfrac{1}{1-az^k}\right)^{M_k\left(b\right)} = \prod\limits_{k\geq 1}\left(\dfrac{1}{1-bz^k}\right)^{M_k\left(a\right)}$ in the formal power series ring $\mathbb Q\left[\left[z,a,b\right]\right]$, where $M_k\left(t\right)$ denotes the $k$-th necklace polynomial $\dfrac{1}{k}\sum\limits_{d\mid k} \mu\left(d\right) t^{k/d}$. I recall this being not particularly difficult, but quite useful.

Every nontrivial commutativity of some family of operators probably qualifies as an unexpected symmetry. Here are three examples:

1. Consider the group ring $\mathbb Z\left[S_n\right]$ of the symmetric group $S_n$. For every $i\in \left\{1,2,...,n\right\}$, define an element $Y_i \in \mathbb Z\left[S_n\right]$ by $Y_i = \left(1,i\right) + \left(2,i\right) + ... + \left(i-1,i\right)$ (a sum of $i-1$ transpositions). Then, $Y_i Y_j = Y_j Y_i$ for all $i$ and $j$ in $\left\{1,2,...,n\right\}$. This is a simple exercise, and the $Y_i$ are called the Young-Jucys-Murphy elements.

2. Consider the group ring $\mathbb Z\left[S_n\right]$ of the symmetric group $S_n$. For every $i\in \left\{0,1,...,n\right\}$, define an element $\mathrm{Sch}_i \in \mathbb Z\left[S_n\right]$ as the sum of all permutations $\sigma \in S_n$ satisfying $\sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(i\right)$. (Note that $\mathrm{Sch}_0 = \mathrm{Sch}_1$ when $n\geq 1$.) Then, $\mathrm{Sch}_i \mathrm{Sch}_j = \mathrm{Sch}_j \mathrm{Sch}_i$ for all $i$ and $j$ in $\left\{0,1,...,n\right\}$. In fact, $\mathrm{Sch}_i \mathrm{Sch}_j = \sum\limits_{k=0}^{\min\left\{n,i+j-n\right\}} \dbinom{n-j}{i-k} \dbinom{n-i}{j-k} \left(n+k-i-j\right)! \mathrm{Sch}_k$, which makes the symmetry maybe not that surprising (no similar equalities hold in cases 1 and 3!). See Manfred Schocker, Idempotents for derangement numbers, Discrete Mathematics, vol. 269 (2003), pp. 239-248 for a proof. (This is also proven in my answers to Is this sum of cycles invertible in QSn? now, except that instead of the condition $\sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(i\right)$ I require $\sigma\left(n-i+1\right) < \sigma\left(n-i+2\right) < ... < \sigma\left(n\right)$ in that thread. But the two conditions can be transformed into one another by the automorphism $S_n \to S_n,\ \sigma \mapsto w \circ \sigma \circ w$ of $S_n$, where $w$ is the permutation in $S_n$ that sends each $i$ to $n+1-i$.)

3. Consider the group ring $\mathbb Z\left[S_n\right]$ of the symmetric group $S_n$. For every $i\in \left\{1,2,...,n\right\}$, define an element $\mathrm{RSW}_i \in \mathbb Z\left[S_n\right]$ as

$\sum\limits_{1\leq u_1 < u_2 < ... < u_i\leq n} \sum\limits_{\substack{\sigma\in S_n, \\ \sigma\left(u_1\right) < \sigma\left(u_2\right) < ... < \sigma\left(u_i\right)}} \sigma$.

Then, $\mathrm{RSW}_i \mathrm{RSW}_j = \mathrm{RSW}_j \mathrm{RSW}_i$ for all $i$ and $j$ in $\left\{1,2,...,n\right\}$. This is Theorem 1.1 in Victor Reiner, Franco Saliola, Volkmar Welker, Spectra of Symmetrized Shuffling Operators, arXiv:1102.2460v2, and a nice proof remains to be found.

• The Tor symmetry is basically just that $M \otimes N \cong N \otimes M$, and you take the derived functors of both sides. Generalizing, any and all nice properties of (co)homology groups would seem to be mysterious symmetries if you consider the definition to be messing around with projective or injective modules, and not something more intrinsic like derived functors. – Ryan Reich Dec 15 '13 at 5:14
• Regarding Volker Strehl's identity, it seems to be true for any function, not just $\mu$, although presumably taking $\mu$ has some application. Thus let $f:\mathbb{N}\to\mathbb{Q}$ be any function. Then in $\mathbb Q[[a,b,z]]$, we have the formal identity $$\prod_{k\ge1} \left(\frac{1}{1-az^k}\right)^{\frac{1}{k}\sum_{d\mid k} f(d)b^{k/d}} = \exp\left( \sum_{d=1}^\infty\frac{f(d)}{d} \sum_{i,j=1}^\infty \frac{a^ib^j}{ij} z^{ijd}\right).$$ so symmetry in $a$ and $b$ is clear. Proof: take logs of both sides, use the series for $\log(1-t)^{-1}$, and flip the order of series. – Joe Silverman Aug 18 '18 at 17:40
• @JoeSilverman: Nice observation! – darij grinberg Aug 18 '18 at 18:12

Morley's trisector theorem allows you to build a triangle which is maximally symmetric out of one which has no symmetry at all.

This is a rather specialized example, but dear to my heart.

Consider the set of "Richardson subvarieties" of the flag manifold $GL_n/B$, intersections of Schubert and opposite Schubert varieties. The only part of the Weyl group that preserves this set is $\{1,w_0\}$ where the $w_0$ exchanges Schubert and opposite Schubert varieties.

Now project these varieties to a $k$-Grassmannian, obtaining "positroid varieties". This includes the Richardson varieties in the Grassmannian, and many new varieties.

Now the part of the Weyl group that preserves this collection is the dihedral group $D_n$! The symmetry has gotten bigger by a factor of $n$.

Maxwell's equations were originally formulated for Newtonian physics. However, special relativity has found that these equations have a surprising symmetry to Lorentz transformations. The equations remain true in a moving reference frame. The transformation of the values is such that (loosely speaking) what looks like pure electric charge in one reference frame can be electric current and charge in another reference frame; and what looks like pure electric field from one reference frame can be magnetic and electric field in another reference frame.

See https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism for a precise formulation.

Here is an example from potential theory where symmetry is a not-so-obvious property: the Green function of a bounded open subset $\Omega \subset \mathbb{C}$. More precisely, having specified a point $a \in \Omega$, one defines the classical Green function for $\Omega$ with pole at $a$, , as a function on $\mathbb{C}$ with the following properties: (i) $G_\Omega(\cdot; a)$ is harmonic in $\Omega \setminus \{a\}$; (ii) $z \mapsto G(z;a) + \log |z-a|$ extends to a harmonic function on $\Omega$; (iii) for each $w \in \partial \Omega$, $\lim_{z \to w} G_\Omega(z;a)=0$.

The symmetry property says that $G_\Omega(z;w)=G_\Omega(w;z)$ for any $z,w \in \Omega$ such that $z \ne w$. Note that the functions on either side of the equation are different: one has a pole at $w$ and the other at $z$. It is not very hard to prove the symmetry property, but it is not obvious either.

The existence of such a function is related to the solution of a Dirichlet problem for the Laplace equation in $\Omega$. Analogous functions can be considered for domains in $\mathbb{R}^n, \ n>2$ or in $\mathbb{C}^n, n > 1$, and they also enjoy the symmetry property.

A couple very disparate answers that spring to mind (fortunately, this is community wiki, and actual experts should feel very free to improve my exposition of either):

The negative gradient flow for the Chern-Simons functional on a 3-manifold $M$ naturally satisfies a four-dimensional symmetry. Namely, if one has a principal $G$-bundle on $M$ and some connection $A$ on this $G$-bundle (which I'll carelessly think of as a $\mathfrak{g}$-valued $1$-form on $M$), the Chern-Simons functional $CS(A) = \int_M \Big( dA + \frac{2}{3} A \wedge A \Big) \wedge A$ is a perfectly well-defined function on the space of connections, and one can attempt to perform the negative gradient flow with respect to a natural metric on this space of connections (this being a very natural thing to do from the point of view of Morse theory, for example). If you want, you can interpret the solution to this flow as a connection on the bundle pulled back to $M \times \mathbb{R}$, and while this connection clearly transforms nicely under $Diff(M)$, there's no particular reason to think it's a well-behaved object under the diffeomorphism group of the four-manifold $M \times \mathbb{R}$. However, this negative gradient flow equation turns out to be exactly the anti-self dual equation $F^+ = 0$, where the curvature $F = dA + A \wedge A$ and its self-dual part is $F^+ = \frac{1}{2}(F + *F)$. This equation manifestly respects the symmetries of the entire four-manifold, and this point of view is a very effective one for proving even basic things, like gauge invariance, of the Chern-Simons functional. Witten is very fond of making this point and my understanding is that this insight allowed him to extend his QFT description of the Jones polynomial to a QFT description of its categorification, Khovanov homology.

And now for something completely different: associativity of the quantum cup product. A familiar object to many people is the cohomology ring $H^*(X)$ of a space $X$, which is associative, (graded) commutative, and just generally great. If $X$ is a symplectic manifold, there's an interesting way to deform the multiplication on this ring using counts of $J$-holomorphic curves passing through various cycles. In effect, one picks a compatible almost-complex structure on the symplectic manifold, and then if one writes $\alpha * \beta = \sum_{\gamma} c_{\alpha \beta \gamma} \gamma$, where we think of $\alpha, \beta, \gamma$ as cycles in $X$ (using Poincare duality), the coefficient $c_{\alpha \beta \gamma}$ is a generating function in some formal variables, the coefficients of which are counts of holomorphic curves of fixed genus and homology class intersecting our three cycles $\alpha, \beta, \gamma$. Using this deformed multiplication gives the quantum cohomology ring $QH^*(X)$. Now, some properties of this ring, like graded commutativity, are fairly easy to see from the definition, but associativity is really quite tricky! (I realise this isn't exactly what you asked in your question as it's not just a symmetry of some coefficient, but you can phrase associativity as a symmetry of something or other—if you want to be technical, a four-point Gromov-Witten invariant—so I think it qualifies.) The associativity is somehow not so bad to see in the algebro-geometric case (or perhaps this is just my bias as an algebraic geometer), but in symplectic geometry you really need some nontrivial analytic estimates at some point in the proof. And you get a lot out of it! Associativity of this quantum cohomology ring encapsulates a wealth of information on enumerative geometry counts associated to $M$; indeed, it was basically this idea that allowed Kontsevich to find his recursion for the number of degree $d$ curves through $3d + 1$ general points in $\mathbb{P}^2$.

Finally, I kind of want to mention strange duality, even though that now really isn't an answer to the question, as you have to modify one side or the other; I'll just copy a very quick summary from the abstract to Belkale - The strange duality conjecture for generic curves: "For $X$ a compact Riemann surface of positive genus, the strange duality conjecture predicts that the space of sections of certain theta bundle on moduli of bundles of rank $r$ and level $k$ is naturally dual to a similar space of sections of rank $k$ and level $r$." The paper itself is a great place to learn more about it if you're interested!

Characters of affine Kac-Moody Lie algebras and Virasoro Lie algebra are modular forms. These modular symmetries are not that much evident from the definitions.

In number theory, Terry Tao already mentioned Quadratic Reciprocity in his first comment, but there's also the reciprocity formula $$s(b,c) + s(c,b) = \frac1{12}\left( \frac{b}{c} + \frac1{bc} + \frac{c}{b} \right) - \frac14$$ for Dedekind sums, symmetrized further in Rademacher's formula $$D(a,b;c) + D(b,c;a) + D(c,a;b) = \frac1{12} \frac{a^2+b^2+c^2}{abc} - \frac14.$$ [Here $D(a,b;c) = \sum_{n\,\bmod\,c} ((an/c)) ((bn/c))$, where $((\cdot))$ is the sawtooth function taking $x$ to $0$ if $x \in {\bf Z}$ and to $x - \lfloor x \rfloor - 1/2$ otherwise; and the Dedekind sum is the special case $s(b,c) = D(1,b;c)$.]

• But I don't understand what is so special about this, at least in terms of symmetry: for about any function $s(\cdot,\cdot)$, including the Legendre symbol, $s(b,c)+s(c,b)$ or $s(b,c)s(c,b)$ is symmetric in $b$ and $c$. Where is the surprise? – Wolfgang Dec 18 '13 at 18:01
• I think the point is that each of $s(b,c)$ and $s(c,b)$ is complicated, but once added together, one obtains an extremely simple formula. It's the simplicity of the right hand side rather than the symmetry. – Matt Young Dec 18 '13 at 20:11
• @Wolfgang asks a fair question. To add to Matt Young's answer, we can define $s'(b,c) = s(b,c) + 1/8 - b/12c - 1/24bc$, and then the reciprocity formula says that $s'(b,c)$ is antisymmetric: $s'(b,c) = -s'(c,b)$. – Noam D. Elkies Dec 18 '13 at 20:25
• @Matt: yes, that is exactly the point, and I guess that is also why Terry Tao's mention of Quadratic Reciprocity got so many "great comment" votes... Now if we started a thread about this kind of "simplicity", that one would be endless (not in a mathematical sense). – Wolfgang Dec 19 '13 at 7:54
• @NoamD.Elkies Granted. That reminds me of the relation between $\zeta(1-s)$ and $\zeta(s)$, cast as $\Xi(1-s)=\Xi(s)$ with appropriate $\Xi$. – Wolfgang Dec 19 '13 at 7:56

Betti numbers: the symmetry $\dim(H^k(M^n))=\dim(H^{n-k}(M^n))$ does not immediately follow from the definition.

• Poincare duality (in the form you've stated it) comes from the local symmetries of $n$-manifolds (any point has a neighborhood homeomorphic to $\mathbb{R}^n$) and the global symmetry of $M$ (orientability)--this is not a property of Betti numbers, but rather of the underlying space. – Daniel Litt Dec 13 '13 at 21:18
• @DanielLitt, I know, I just don't want to deal with torsion, and for the purpose of this question Betti numbers' symmetry is sufficient. – Michael Dec 13 '13 at 21:23
• My point is that the symmetry does not come from the Betti numbers, but from the space $M$; I don't think this is an example of what the question asks for. – Daniel Litt Dec 13 '13 at 23:40
• There is a philosophy that the functional equation of a zeta function should be a consequence of Poincare duality on some exotic space. For zeta functions of varieties over finite fields, this was made rigorous in the 1960s, but over number fields it's still just a philosophy. So we have two non-obvious symmetries that are the same, but not obviously the same. In other words, we have a non-obvious symmetry between non-obvious symmetries. – JBorger Jan 12 '14 at 19:01

In the definition of "Latin square" there is complete symmetry between the roles of "row", "column" and "symbol", so that any of the 6 permutations of that role produces another Latin square.

I would like to add an example coming from the area of additive theory known as Freiman's structure theory. If I am not (too) blind, this has not been mentioned yet, and hopefully it qualifies as an appropriate answer.

Assume that $\mathbb{A} = (A, +)$ is a (possibly non-commutative) semigroup, and let $X$ be a non-empty subset of $A$. Given an integer $n \ge 1$, we write $nX$ for $\{x_1+\cdots + x_n: x_1, \ldots, x_n \in X\}$. In principle, we have $1 \le |nX| \le |X|^n$, and for all $k \in \mathbb{N}^+$ and $i \in \{1, \ldots, k\}$ we can actually find a pair $(\mathbb{A}, X)$ such that $|X| = k$ and $|nX| = i$, with the result that, in general, not much can be concluded about the "structure" of $X$. However, if $|nX|$ is sufficiently small with respect to $|X|$ and $\mathbb{A}$ has suitable properties, then "surprising" things start happening, and for instance we have the following:

Theorem. If $\mathbb{A}$ is a linearly orderable semigroup (i.e., there exists a total order $\preceq$ on $A$ such that $x + z \prec y + z$ and $z + x \prec z + y$ for all $x,y,z \in A$ with $x \prec y$) and $|2X| \le 3|X|-3$, then the smallest subsemigroup of $\mathbb{A}$ containing $X$ is abelian.

This implies at once an analogous result by Freiman and coauthors which is valid for linearly ordered groups; see Theorem 1.2 in [F] (a preprint can be found here). I don't know of any similar result for larger values of $n$.

References.

[F] G. Freiman, M. Herzog, P. Longobardi, and M. Maj, Small doubling in ordered groups, to appear in J. Austr. Math. Soc.

The "Little Prince" problem, which I learned from Greg Kuperberg, is a geometric answer to your question.

Here is the problem: the Little Prince stands in (I do mean in, not on) the plane and wants to shape its planet from a given quantity of matter (of given density) in order to maximize the gravity he feels. The most efficient way to go is to shape the planet as a round disk.

The problem has a particular point, the position of the Little Prince, but turns out to have a symmetric solution. Note that the same problem in higher dimension does not have a symmetric solution.

Let me add two points that make this example all the more interesting: first, the results still stands if the Little Prince is also authorized to shape the space (rather the surface) he lives in, with the constraint that it should have nonpositive curvature and be simply connected: he should still make the planet a round flat disk. Second, if one takes a general domain and integrates the inequality between the felt gravity and the optimal gravity, one gets the isoperimetric inequality.

Some categories are self-dual in ways not obvious from their definitions.

One good example is Pontryagin duality, which states that the category of locally compact Hausdorff abelian groups is self-dual, via the taking of continuous character groups.

Another is Connes' cyclic category. It is not obvious that this particular melding of the simplex category (of nonempty finite ordinals) and cyclic groups would result in a self-dual category, and in fact this property would fail if the definition were tweaked just slightly (say by working with all finite ordinals).

The Jordan-Kronecker function is defined by the infinite sum

$$F(x, y) = \sum_{n=-\infty}^\infty \frac{y^n}{1 - x q^n}, \quad |q|<|y|<1$$

and, obviously, restrictions on $x$ to avoid poles. Surprisingly, $$F(x, y) = F(y, x) = -F(-1/x,1/y).$$