# 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

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.

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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
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