50
$\begingroup$

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.)

$\endgroup$
7
  • 51
    $\begingroup$ Quadratic reciprocity. $\endgroup$
    – Terry Tao
    Dec 13, 2013 at 22:55
  • 8
    $\begingroup$ The relation between $\zeta(1-x)$ and $\zeta(x)$ for the Riemann $\zeta$ function. $\endgroup$ Dec 14, 2013 at 2:26
  • 4
    $\begingroup$ 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. $\endgroup$ Dec 14, 2013 at 2:46
  • 11
    $\begingroup$ The Peano definition of addition, even. $\endgroup$
    – Joe Z.
    Dec 14, 2013 at 2:56
  • 4
    $\begingroup$ I saw the title and my first thought was "Littlewood-Richardson coefficients". :) $\endgroup$ Dec 14, 2013 at 20:55

37 Answers 37

1
2
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ BTW, the proof given in EC2 is via (quasi-)symmetric function theory. I don't know if there is a bijective proof of this result. $\endgroup$ Aug 18, 2018 at 15:19
  • 1
    $\begingroup$ I suspect that a bijective proof follows from Brice Huang, Cyclic Descents for General Skew Tableaux, arXiv:1808.04918v1, provided that this preprint is correct (always a risky bet with a bijective combinatorics preprint that has just been on the arXiv for 4 days). $\endgroup$ Aug 18, 2018 at 16:56
2
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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 $a(m,n)=a(n,m)$.

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ The $q,t$ symmetry was also mentioned in this answer of Richard Stanley: mathoverflow.net/a/151790/25028 $\endgroup$ Aug 22, 2018 at 18:22
  • $\begingroup$ ah, yes, i missed that. there are several versions of this - both for Catalan and parking functions, if my memory serves me right. $\endgroup$ Aug 22, 2018 at 20:05
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
5
  • 1
    $\begingroup$ Would be so kind to give a reference for this result ? $\endgroup$ Jan 12, 2014 at 19:00
  • 1
    $\begingroup$ Why would you say it isn't provable? It has been proved. $\endgroup$ Jan 13, 2014 at 11:40
  • $\begingroup$ @DouglasZare: Of course. I suppose Houdini meant "it's not provable without evaluating the very integral", in other terms: it cannot be transformed into a "non trivial" (in whatever sense??) integral over a symmetric function in $\alpha$ and $\beta$. BTW I would doubt such an absolute statement. It might be worth trying to write $f( \alpha,\beta)\sin(\pi \alpha)\sin(\pi\beta)$ as a triple integral, using suitable different integrals with values $\sin(\pi \alpha)$ and $\sin(\pi\beta)$. $\endgroup$
    – Wolfgang
    Jan 13, 2014 at 13:43
  • 3
    $\begingroup$ It is wrong to suggest that this is similar to trisecting an angle, solving the general quintic in radicals, solving the halting problem, or proving AC/CH. How can you say something is unprovable or will never be proved while looking at 3 proofs which simply don't seem fully satisfactory from one perspective? So, $-1$ for that flagrantly incorrect and misleading statement. $\endgroup$ Jan 13, 2014 at 19:13
  • 2
    $\begingroup$ I've removed the unnecessarily hyperbolic/mystical assertions - I think this is an interesting example nonetheless. $\endgroup$
    – Yemon Choi
    Aug 20, 2018 at 2:15
0
$\begingroup$

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.

$\endgroup$
1
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.