doubly periodic functions as tessellations (other than parallelograms)

(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

$$\int\frac{\mathrm dt}{\left(1-t^3\right)^{2/3}}=t {}_2 F_1\left({{\frac13\quad \frac23}\atop{\frac43}}\mid t^3\right)$$

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 vice-versa.

The Dixon elliptic functions can also be expressed in terms of Weierstrass elliptic functions:

$$\operatorname{sm}(z)=\frac{6\wp\left(z;0,\frac1{27}\right)}{1-3\wp^\prime\left(z;0,\frac1{27}\right)}$$

$$\operatorname{cm}(z)=\frac{3\wp^\prime\left(z;0,\frac1{27}\right)+1}{3\wp^\prime \left(z;0,\frac1{27}\right)-1}$$

(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.

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 multi-valued), as proved for example in this old book by Dixon on elliptic functions (see §32 on p. 19).