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It can be shown that a helicoid and catenoid are flexible surfaces in that that can be isometrically deformed (parametrically) into each other (simply, a helicoid and catenoid are isometric). The parametric equations on the coordinates are:

\begin{equation} x(u,v) = \cos \alpha \sinh v \sin u + \sin \alpha \cosh v \cos u, \\ y(u,v) = -\cos \alpha \sinh v \cos u + \sin \alpha \cosh v \sin u, \\ z(u,v) = u \cos \alpha + v \sin \alpha \end{equation}

I'm curious if the flexibility of this surface is a Ricci flow.

The only instance where the Ricci flow of a metric is a family of isometric metrics is for steady Ricci solitons. For the ricci soliton equation given by:

$$ Ricc(g) = \lambda g - \frac{1}{2} \mathcal{L}_V g $$

The case of a steady ricci soliton is when $\lambda = 0$. My work is the following:

The induced metric is $\cosh^2 v \ (du^2 + dv^2)$ and the Ricci scalar is $R = -2 \DeclareMathOperator{\sech}{sech} \sech^4 v$. Notice the deformation portions (the sines and cosines in alpha) don't appear at all. Since helicoids/catenoids are 2d surfaces, I found it a tad bit simpler to use the relation $Ricc(g) = \frac{1}{2}Rg$ so the steady ricci soliton relation becomes: $(R + \mathcal{L}_v)g = 0$. Solving for the vector field $V$ s.t. we can prove we have steady ricci flow gives me the following PDE's on its components:

$$ \partial_1 V^2 + \partial_2 V^1 = 0, \ \ \partial_1 V^1 = \partial_2 V^2, \ \ V^2 \tanh v + \partial_1 V^1 = 1 $$

Here, 1 corresponds to the polar angle $u$ and 2 to the vertical coordinate $v$. I find for the lie derivative of my metric:

$$ \mathcal{L}_V g_{12} = \mathcal{L}_V g_{vu} = (\cosh^2 v) (\partial_1 V^2 + \partial_2 V^1) \\ \mathcal{L}_V g_{ii} = 2 [V^2 (\sinh v \cosh v) + \partial_i V^i (\cosh^2 v)] $$

The first equation for the off diagonal components happens to be what the steady ricci flow equation reduces to since the off diagonal components of $g$ are zero.

For the first two equations I listed on $V^1$ and $V^2$, finding a nice solution that actually has some interesting implications isn't hard. It's the one with the $\tanh v $ that's giving me trouble. I think there should be a solution here, since if one let's alpha in the original coordinates increase forever, the helicoid will continue to deform into the catenoid and the catenoid back into the helicoid forever.

DingleGlop
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    What is the actual question here? – Matthew Cassell Aug 18 '24 at 03:37
  • @MatthewCassell I can't figure out the components of $V$. I can't figure out how to reconcile the first two equations I got on $V^1$ and $V^2$ with the one with $\tanh^2 v$. I'm asking for a nudge in the right direction to find V – DingleGlop Aug 18 '24 at 04:24
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    You mean 'the one with $\color{red}{\tanh v}$'? – Matthew Cassell Aug 18 '24 at 04:26
  • @MatthewCassell yes I'm trying to find solutions – DingleGlop Aug 18 '24 at 04:41
  • Yes but is the third PDE in terms of $\tanh v$ or $\tanh^{2} v$? Your comment above doesn't match the body of your question. – Matthew Cassell Aug 18 '24 at 04:42
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    @MatthewCassell you are correct, my apologies. it's only in terms of $\tanh v$. it is not squared. my post is what's correct. – DingleGlop Aug 18 '24 at 04:51
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    The first two equations can be solved by characteristics. Following here gives $$V^{1}=f_{1}(u+iv)+f_{2}(u-iv),\quad V^{2}=-if_{1}(u+iv)+if_{2}(u-iv)$$ The second and third equations decouple $V^{1}, V^{2}$ so that \begin{align} V^{2}{v}+\tanh vV^{2}&=1&\implies V^{2}&=\tanh v+(\cosh v)^{-1}g{1}(u)\V^{1}{v}+V^{2}{u}&=0&\implies V^{1}&=-\arctan(\sinh v)g_{1}'(u)+g_{2}(u)\end{align} It seems clear that the sets of solutions are incompatible. – Matthew Cassell Aug 18 '24 at 04:53
  • @MatthewCassell this means there's no Ricci flow here yes? – DingleGlop Aug 18 '24 at 04:56
  • I've never studied Ricci flow so I have no idea. – Matthew Cassell Aug 18 '24 at 04:57
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    FWIW, Ricci flow is an intrinsic local concept, depending only on the metric and Ricci curvature. The catenoid and helicoid are locally isometric; particularly, they're not distinguished by local intrinsic geometry. The extrinsic deformation from helicoid to catenoid is not Ricci flow: In fact, if we view it as a family of metrics on the plane induced by immersions, the family is constant. – Andrew D. Hwang Aug 18 '24 at 11:44
  • @AndrewD.Hwang is it permittable for an isometric deformation to change the curvature of a surface? I would think so thinking on it but now I'm a bit unsure. if so, given two surfaces that have an isometry between them such that their metrics differ due to the only an isometric deformation, would it then be a Ricci flow? I believe the shrinking of a sphere into an infinitesimally small sphere is an example of a $C^1$ continuous deformation of a closed surface in 3d space, which happens to be one of the simplest examples of Ricci flow – DingleGlop Aug 18 '24 at 13:13
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    Ricci flow does change the metric, and the (Gaussian) curvature of a surface metric. That includes a round sphere: The circumference shrinks exponentially, while the curvature grows like the inverse square of the circumference. <> By contrast, the helicoid-catenoid deformation preserves Gaussian curvature, and if set up suitably in the plane actually preserves the metric itself. – Andrew D. Hwang Aug 18 '24 at 15:09
  • @AndrewD.Hwang even in the case of steady Ricci flow for solitons? – DingleGlop Aug 19 '24 at 16:16
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    It sounds as if you're asking whether an arbitrary Riemannian metric, viewed as a constant path in the space of metrics, is a Ricci flow soliton...? – Andrew D. Hwang Aug 20 '24 at 12:25
  • @AndrewD.Hwang I think I don't understand what a steady Ricci flow soliton would correspond to intuitively in regards to what the flow is – DingleGlop Aug 21 '24 at 08:20
  • Not an expert, but it ought to mean a metric equal (or proportional) to its Ricci tensor $\rho$, so the scaled Ricci flow becomes something like $\dot{g} = g - \rho = 0$. – Andrew D. Hwang Aug 21 '24 at 15:32

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