I think of a snapshot of a single period of a doubly periodic function as one parallelogramshaped tile in a tessellation. Could a function have a period that repeats like a honeycomb or some other not rectangular tessellation?
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What kind of functions are we talking about here? – Qiaochu Yuan Apr 28 '11 at 19:38

C> C, complex – futurebird Apr 28 '11 at 19:50

1The fact that there exist functions that "repeat like a honeycomb" does not contradict the fact that the fundamental domain is a parallelogram, and in fact a rectangle. Suppose two parallel sides of the hexagon are horizontal. Go from the center of one hexagon in a horizontal direction until you reach the next center of a hexagon (thus NOT one of the ones that are adjacent to the one you started in) and that's one side of a rectangle. Then go from that same center in a vertical direction until you reach the next center (of a hexagon that IS adjacent to the one where you started$\,\ldots\qquad$ – Michael Hardy Aug 25 '16 at 22:23

$\ldots\,$and that's another side of the rectangle. $\qquad$ – Michael Hardy Aug 25 '16 at 22:23
2 Answers
(many literature searches and Mathematica experiments later...)
The usual Jacobi and Weierstrass elliptic functions have as their "repeating unit" a parallelogram (which can be made rhomboidal or square through appropriate choices of parameters). It is known that apart from parallelograms, hexagons can tile the plane by translation; so, why can't there be a doubly periodic function that has a hexagonal repeating unit?
It turns out that A.C. Dixon (the guy whose book on elliptic functions Hans linked to), in a long 1890(!) paper, studied a class of elliptic functions (now named after him) based on the inversion of the Abelian integral
$$\int\frac{\mathrm dt}{\left(1t^3\right)^{2/3}}=t {}_2 F_1\left({{\frac13\quad \frac23}\atop{\frac43}}\mid t^3\right)$$
where ${}_2 F_1\left({{a\quad b}\atop{c}}\mid x\right)$ is a Gaussian hypergeometric function.
There are two of these Dixon elliptic functions, $\operatorname{sm}(z,0)=\operatorname{sm}(z)$ and $\operatorname{cm}(z,0)=\operatorname{cm}(z)$, corresponding to the usual sine and cosine respectively. Both functions have a real period $\pi_3=B\left(\frac13,\frac13\right)$ (where $B(a,b)$ is the beta function) and a complex period $\pi_3\exp(2i\pi/3)$, and satisfy the following relations (reminiscent of usual trigonometric identities):
$$\begin{align*} &\operatorname{sm}\left(\frac{\pi_3}{3}z\right)=\operatorname{cm}(z)\\ &\operatorname{sm}^3(z)+\operatorname{cm}^3(z)=1\\ &\operatorname{sm}^\prime(z)=\operatorname{cm}^2(z),\quad \operatorname{cm}^\prime(z)=\operatorname{sm}^2(z) \end{align*}$$
and, most relevant to the purposes of this question, a rotational invariance:
$$\exp(2i\pi/3)\operatorname{sm}(z\exp(2i\pi/3))=\operatorname{sm}(z),\quad \operatorname{cm}(z\exp(2i\pi/3)) =\operatorname{cm}(z)$$
Plots of the Dixon functions on the real line don't look very interesting:
but, as with the usual elliptic functions, the fun starts in the complex plane:
These contour plots clearly display the hexagonal structure of the Dixon functions. Here is a single "fundamental period hexagon" for $\operatorname{sm}(z)$:
Note that a section of the real line (in the plots above, $\left(\frac{\pi_3}3,\frac{2\pi_3}{3}\right)$) corresponds to a chord of the period hexagon.
Both Dixon elliptic functions possess three poles (once you've identified the congruent poles in the period hexagon) and three zeros within the fundamental hexagon. Of course, one could go the usual route and consider the "repeating unit" of the Dixon function to be a particular rhombus; this is equivalent; since the rhombus can be appropriately dissected into a regular hexagon, and viceversa.
The Dixon elliptic functions can also be expressed in terms of Weierstrass elliptic functions:
$$\operatorname{sm}(z)=\frac{6\wp\left(z;0,\frac1{27}\right)}{13\wp^\prime\left(z;0,\frac1{27}\right)}$$
$$\operatorname{cm}(z)=\frac{3\wp^\prime\left(z;0,\frac1{27}\right)+1}{3\wp^\prime \left(z;0,\frac1{27}\right)1}$$
(there are also expressions for Dixon functions in terms of Jacobi elliptic functions, but they are rather complicated.)
Finally, if you're interested in knowing more about the Dixon elliptic functions (including combinatorial applications), this paper is a good starting point.
A Mathematica notebook for those interested in exploring the topic further is available from me upon request.
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4So this makes me wonder: what about the other 15 wallpaper groups? – graveolensa May 02 '11 at 21:40

1@deoxy: That's an interesting question, too. Maybe you should ask it as a separate question? – J. M. ain't a mathematician May 03 '11 at 05:11


@Raskolnikov: I wouldn't know how to define an elliptic function whose behavior would mimic other geometries, so I can't say there. Sorry. :( – J. M. ain't a mathematician May 03 '11 at 07:55

1@J.M. this question: http://math.stackexchange.com/questions/36737/ellipticfunctionsonthe17wallpapergroups – graveolensa May 03 '11 at 17:45

Is there any relationship between the Dixonian functions and the cubic theta functions mentioned here: http://www.math.illinois.edu/REGS/reports10/Schultz.pdf  I ask because the identity $\mathrm{sm}^{3}(z)+\mathrm{cm}^{3}(z) = 1$ is reminiscent of $a(q)^{3}+b(q)^{3}=c(q)^{3}$. – graveolensa May 03 '11 at 18:55

@deoxy: That's a good question also. :) We know that the Weierstrass functions certainly can be expressed in terms of theta functions; that might pave a way towards seeing if the Dixon functions relate to cubic theta functions. – J. M. ain't a mathematician May 03 '11 at 19:02

None of this should be taken to mean you don't get a parallelogram as a fundamental region. Take one regular hexagon and part of another, forming a rectangle, and tile the plane appropriately with those rectangles, and you'll see the tiling by hexagons. $\qquad$ – Michael Hardy Aug 23 '16 at 15:46

When you write \mathrm{sm} instead of \operatorname{sm} then you don't automatically get proper spacing in $a\operatorname{sm} b$ and $a\operatorname{sm}(b)$. (I give both examples to show the contextdependent nature of the space to the right of $\operatorname{sm}.$) Hence my edit to this answer. $\qquad$ – Michael Hardy Aug 23 '16 at 15:47

I have created this new Wikipedia article based on information in this answer. Whoever wants to can contribute to it. "J.M.", could you perhaps add some suitable graphics to it, and other information? https://en.wikipedia.org/wiki/Dixon%27s_elliptic_functions $\qquad$ – Michael Hardy Aug 23 '16 at 18:31

1@Michael, "None of this should be taken to mean you don't get a parallelogram as a fundamental region."  yes, I did mention the dissection into a *rhombus* in this answer. – J. M. ain't a mathematician Dec 17 '16 at 20:16

You can put one period of this function in terms of [Inverse Beta Regularized](https://reference.wolfram.com/language/ref/InverseBetaRegularized.html). Should this identity be added? – Tyma Gaidash May 19 '22 at 22:53
I don't think so. If there are exactly two periods (not parallel), then we have a parallelogram, so in your case there must be at least three independent periods. But that implies that the function is constant (or multivalued), as proved for example in this old book by Dixon on elliptic functions (see §32 on p. 19).
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On the other hand, Hans displays something almost, but not quite hexagonal [here](http://www.mai.liu.se/~halun/complex/elliptic/) (scroll down to the equianharmonic case of the Weierstrass $\wp$ function). – J. M. ain't a mathematician Apr 28 '11 at 19:40

1I think it depends on how the tesilation works, I just realized that if you create you honeycomb bu translation it can be repartitioned in to parallelograms. Now if we ad a rotation, then it would not work. I think that this needs to be restricted to tesilations where you can pick up the whole plane and map it to itself after translation and rotation.  still, it's not obvious to me that there are more than three periods... Can you expand on that? – futurebird Apr 28 '11 at 19:47

Well, as you say, you can have a honeycomb pattern, but that's really just a doubly periodic pattern in disguise (with periods 1 and $\exp(i\pi/3)$, say). If you really want to go beyond the case where there is a fundamental region in the shape of a parallelogram lurking somewhere, you will have to look for something with more than two periods, and that is apparently not possible. – Hans Lundmark Apr 28 '11 at 20:59