# Complex-doubly periodic function in two variables?

I am looking for a function $$f:\mathbb C^2 \rightarrow \mathbb C^2$$ that satisfies the two equations

$$\partial_{z_2}f_1(z_1,z_2) + \partial_{z_1} f_2(z_1,z_2)=0 \text{ and }$$ $$\partial_{\bar z_1}f_1(z_1,z_2) - \partial_{\bar z_2} f_2(z_1,z_2)=0$$

and in addition, is doubly-periodic in both its complex variables $$z_1,z_2$$. Does such a function exist and if not, why? I would not even know how to start building such a function.

In particular, I would like to have

$$f_1(z_1+1,z_2)=f_1(z_1,z_2+1)=f_1(z_1,z_2)$$ and

$$f_1(z_1+i,z_2) = e^{2\pi i k_1}f_1(z_1,z_2)$$ and

$$f_1(z_1,z_2+i) = e^{2\pi i k_2}f_1(z_1,z_2)$$

for some fixed $$k_1,k_2 \in \mathbb R.$$ Please let me know if you do have any questions. I had some typos in there, but hopefully everything is coherent now.

• You, probably, want more because as written, you can just take any function $f=(f_1,f_2)$ in which $f_1$ depends only on $z_1$ and $f_2$ only on $z_2$ with any periodicity in each variable you like. Jan 28 at 21:59
• @fedja you are right, there were some additional constraints missing to give this question a meaning. Jan 28 at 22:31
• By (an analog of) Cauchy-Riemann, $f_1+if_2$ only depends on $z_1+iz_2$ and $f_1-if_2$ only on $z_1-iz_2$, is this intended? Jan 31 at 15:39
• @მამუკაჯიბლაძე what does it mean for $f_1+if_2$ to only depend on $z_1+iz_2$? What kind of independent coordinate system are you considering here? Feb 1 at 19:53
• @PritamBemis If I am not confused (and I believe something similar is done in the answer by Robert Bryant below), if you change variables to $z_1+iz_2=z$, $z_1-iz_2=w$, and denote $f_1+if_2=g$, $f_1-if_2=h$, then $\partial g/\partial w=\partial h/\partial z=0$, no? Feb 1 at 19:55

The answer is that the only solutions have the form $$f = (f_1,f_2) = \bigl(c, h(\,\overline{z}_1, z_2)\bigr)$$ where $$h:\mathbb{C}^2\to\mathbb{C}$$ is holomorphic and $$c$$ is a constant, which must equal zero unless $$k_1$$ and $$k_2$$ are integers.
The argument is as follows: The first equation implies that there exists a function $$g:\mathbb{C}^2\to\mathbb{C}$$ such that $$f_1 = \frac{\partial g}{\partial z_1} \quad\text{and}\quad f_2 = -\frac{\partial g}{\partial z_2}.$$ Substituting this into the second equation implies that $$g$$ must satisfy $$\frac{\partial^2 g}{\partial z_1\partial\overline{z}_1} + \frac{\partial^2 g}{\partial z_2\partial\overline{z}_2} = 0.$$ In other words $$g$$ is a harmonic function on $$\mathbb{C}^2$$. Since $$g$$ is harmonic, so is its derivative with respect to $$z_1$$, i.e., $$f_1$$.
The periodicity conditions imposed on $$f_1$$ imply that $$f_1$$ is bounded, and a bounded harmonic function on $$\mathbb{C}^2$$ is constant. Thus, $$f_1 = c$$ for some constant $$c\in\mathbb{C}$$. Obviously, $$c$$ must be zero unless $$k_1$$ and $$k_2$$ are integers.
Since $$f_1$$ is constant, the given equations on $$f_2$$ reduce to $$\frac{\partial f_2}{\partial z_1} = \frac{\partial f_2}{\partial\overline{z}_2} = 0.$$ Hence $$f_2 = h(\overline{z}_1,z_2)$$ for some homorphic function $$h:\mathbb{C}^2\to\mathbb{C}$$.
Remark: It wasn't clear from the OP's question whether the OP wanted $$f$$ to be 'doubly-periodic' or just $$f_1$$, nor was it clear exactly what the OP meant by 'doubly-periodic' because, normally, the 'doubly-periodic' condition wouldn't have the exponential factors in its definition.