NB: I have edited this question to clarify what the OP is asking – Robert Bryant

Problem: Find a holomorphic function $f$ where where $f(x+iy) = u(x,y) + i\,v(x,y)$, such that the graph $\Gamma_u = \left\{\bigl(x,y,u(x,y)\bigr)\mid (x,y)\in\mathrm{dom}(f)\right\}\subset\mathbb{R}^3$ has Gauss curvature $K=-1$ at every point.

Because the Gauss curvature of such a graph $\Gamma_u$ is given by the formula $$ K(x,y) = \frac{u_{xx}u_{yy} - u_{xy}^2}{ (1 + u_x^2 + u_y^2)^2}, $$ the equation $K(x,y) = -1$ is a PDE of Monge-Ampere type. However, in this case, because $u$ is the real part of a holomorphic function and hence is harmonic, the function $u$ must also satisfy $u_{xx}+u_{yy} = 0$.

One wants to solve this combined system or else find a way to reduce it to an ODE so that one can investigate numerical solutions.

  • $\begingroup$ I'm not convinced that you have the equation correctly stated. Why does $v$ not show up in the denominator? Maybe you should give a bit more detail on how $f$ determines a metric of Gauss curvature $-1$ so that we can believe you have the right equation. $\endgroup$ – Robert Bryant Oct 11 '14 at 10:30
  • $\begingroup$ A natural way of associating a Riemannian metric to a function $f:\mathbb R^2\to\mathbb R^2$ is to take the pullback or pushforward of the Euclidean metric over this map. Is one of these the metric whose curvature you are interested in? $\endgroup$ – Joonas Ilmavirta Oct 11 '14 at 10:49
  • $\begingroup$ @RobertBryant: Gauss curvature either at (x,y,Re(f(z)=u(x,y)) or at (x,y,Im(f(z)) is the same,by virtue of CR relations reduces to Monge form's K above ( r t - s^2)/ (1 +p^2 +q^2)^2 .It is convenient to deal either with u or with v. Also please see Joseph Shomberg's article referenced by SetHead given here. $\endgroup$ – Narasimham Oct 11 '14 at 13:24
  • $\begingroup$ @RobertBryant: Answered by Orangeskid in: math.stackexchange.com/questions/967020/… $\endgroup$ – Narasimham Oct 11 '14 at 13:33
  • 1
    $\begingroup$ @Narasimham: I have edited the question to clarify what you are asking, since the original question didn't have all of the definitions that you needed in order for the question to make sense. $\endgroup$ – Robert Bryant Oct 12 '14 at 11:45

Now I understand the OP's question, which I would state this way:

Is there a holomorphic function $f(x,y) = u(x,y) + \mathrm{i}\,v(x,y)$ on a domain in the $xy$-plane such that the induced metric on the graph of $u$ has Gauss curvature $-1$? (Note: The OP needs $f$ to be holomorphic, not just complex, if the given formula for the Gauss curvature is going to be correct.)

The answer to this question is 'no'; here is why: The hypotheses can be expressed in terms of $u$ as follows: Find the common (local) solutions to the pair of equations $$ u_{xx} + u_{yy} = 0\qquad\text{and}\qquad {u_{xx}}^2 + {u_{xy}}^2 = \bigl(1 + {u_x}^2+{u_y}^2\bigr)^2. $$ I claim that there are no such solutions. To see this, note that for any solution of these equations on a simply connected domain in the $xy$-plane, there must exist a function $\theta$ on that domain such that $$ u_{xx} = \cos\theta\,\bigl(1 + {u_x}^2+{u_y}^2\bigr),\quad u_{xy} = \sin\theta\,\bigl(1 + {u_x}^2+{u_y}^2\bigr),\quad u_{yy} = -\cos\theta\,\bigl(1 + {u_x}^2+{u_y}^2\bigr). $$ Differentiating these equations and expanding out $(u_{xx})_y = (u_{xy})_x$ and $(u_{xy})_y = (u_{yy})_x$ then shows that the function $\theta$ must satisfy the differential equations $$ \theta_x = 2(u_x\,\sin\theta -u_y\,\cos\theta),\quad\text{and}\quad \theta_y = -2(u_y\,\sin\theta +u_x\,\cos\theta). $$ Finally, using these formulae and the above formulae for $u_{xx}, u_{xy}, u_{yy}$ to expand out the identity $(\theta_x)_y-(\theta_y)_x = 0$ then yields $4=0$, a contradiction.

Thus, there are no solutions of this kind.


There are some results in the literature reducing some Monge-Ampre and Gauss curvature type equations (and others) into an 'ODE'; see A NOTE ON SURFACES WITH RADIALLY SYMMETRIC NONPOSITIVE GAUSSIAN CURVATURE and the related article HOLOMORPHIC METHODS IN PDE


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.