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

Question

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.)

Answer 1

This example is Proposition 7.19.9 of volume 2 of Stanley's "Enumerative Combinatorics." Define a descent of a (skew) Standard Young Tableau $T$ of shape $\lambda/\mu$ to be an index $i$ such that $i+1$ is in a lower row than $i$. Let $D(T)$ denote the set of descents of $T$. Then for any $|\lambda/\mu|=n$ and for any $1 \leq i \leq n-1$, the number of SYTs $T$ of shape $\lambda/\mu$ such that $i \in D(T)$ is independent of $i$.

As Darij points out in the comments, this unexpected symmetry is closely related to the recent theory of cyclic descents for SYT. For an overview of this developing theory, see for instance https://arxiv.org/abs/1710.06664.

Answer 2

Let $r_4(n)$ be the number of $4$-tuples $a,b,c,d\in \bf Z$ satisfying $a^2+b^2+c^2+d^2=n$. Then $\sum_{n\geq 0}r_4(n)e^{2\pi i\, nz}dz$ is a holomorphic differential form on the upper half-plane that is invariant by a subgroup of finite index in ${\rm SL}_2(\bf Z)$ (acting by $\frac{az+b}{cz+d}$).

The same is true if you replace $r_4(n)$ by $a_n(E)$ where:

-- $E$ is an elliptic curve defined over $\bf Q$,

-- if $p$ is a prime number, $a_p(E)=p+1-N_p(E)$ and $N_p(E)$ is the number of points of $E$ in ${\bf Z}/p{\bf Z}$,

-- $a_n(E)$, for $n\in\bf N$, is defined by $\sum_n a_n(E)n^{-s}=\prod_p(1-a_p(E)p^{-s}+p^{1-2s})^{-1}$ (the product has to be taken over the prime numbers $p$ such that $E$ remains an elliptic curve modulo $p$ which excludes finitely many of them).

Answer 3

Let $a(m,n)$ be the number of partitions with no more than $m$ parts, each part (strictly) less than $n$, and the sum a multiple of $n$. Then $a(m,n)=a(n,m)$.

Answer 4

The $qt$-catalan numbers, which encode the area and bounce statistic of Dyck paths.

These encode a bigraded Frobenius/Hilbert series in diadonal harmonics, and with this interpretation, there is an obvious qt-symmetry. However, it is still an open problem to find a combinatorial proof of the symmetry, by exhibiting a bijection that interchanges the area and bounce statistics.

Answer 5

This answer follows from a bijection of De Mèdicis and Viennot (1994, Adv. Appl. Math.) Let $\mathcal{M}_n$ denote the set of perfect matchings of $[2n]$, i.e. the set of partitions of $[2n] := \{1,2,\ldots,2n\}$ into pairs. Let $M \in \mathcal{M}_n$. For $p = \{a,b\}, q = \{c,d\} \in M$ with $a<b$, $c<d$, and $a<c$, we say that $p$ and $q$ cross if $a < c < b< d$ and we say they nest if $a<c<d<b$. Finally, we say they are aligned if they neither cross nor nest, i.e., $a<b<c<d$. Define:

$\mathrm{ne}(M):= |\{\{p,q\}\subset M\colon \textrm{$p$ and $q$ nest}\}|;$

$\mathrm{cr}(M):= |\{\{p,q\}\subset M\colon \textrm{$p$ and $q$ cross}\}|;$

$\mathrm{al}(M):= |\{\{p,q\}\subset M\colon \textrm{$p$ and $q$ are aligned}\}|.$

Then $\sum_{M \in \mathcal{M}_n}x^{\mathrm{ne}(M)}y^{\mathrm{cr}(M)}=\sum_{M \in \mathcal{M}_n}x^{\mathrm{cr}(M)}y^{\mathrm{ne}(M)}$. However, crossings and alignments (or nestings and alignments) are not equidistributed: $\sum_{M \in \mathcal{M}_n}x^{\mathrm{al}(M)}y^{\mathrm{cr}(M)} \neq \sum_{M \in \mathcal{M}_n}x^{\mathrm{cr}(M)}y^{\mathrm{al}(M)}$.

Here we see a symmetry between crossings and nestings that is at least to some degree nonobvious. In fact, the symmetry between "non-crossing" objects and "non-nesting" objects is an important and somewhat mysterious phenomenon in modern research in algebraic/enumerative combinatorics, especially in the context of "Coxeter-Catalan combinatorics." For an introduction to the basic story, see the classic monograph by Armstrong: https://arxiv.org/abs/math/0611106v2.

Answer 6

Let $$f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$
defined inside the unit square. Then we have $f(\alpha, \beta)=f(\beta, \alpha)$. But why?

Reference: https://math.stackexchange.com/questions/268789/symmetry-of-function-defined-by-integral

Answer 7

Let $H(x,p) + \alpha U(x)$ be a Hamiltonian system in $2n$-dimensional phase space with canonical coordinates $x_i,p_i$. Thus the Hamilton-Jacobi equation would take the form $H(x,p) + \alpha U(x) = E$. Assume that for every value of the parameter $\alpha$ the system admits a constant of the motion $K(\alpha)$ analytic in $\alpha$.

Coupling constant metamorphosis: The Hamiltonian $H'= \frac{H-E}{U}$ admits the constant of the motion $K' = K(-H')$, where now $E$ is a parameter.

So we can switch between coupling constant and energy levels and (super)integrability of the problem stays the same.