# 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.)

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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

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

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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.

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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? A mystery so deep it will never be uncovered. Its not provable. We are looking into the symmetric eyes of God.

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Would be so kind to give a reference for this result ? – Alexander Chervov Jan 12 '14 at 19:00
Why would you say it isn't provable? It has been proved. – Douglas Zare Jan 13 '14 at 11:40
@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 $\sin(\pi\beta)$. – Wolfgang Jan 13 '14 at 13:43
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. – Douglas Zare Jan 13 '14 at 19:13