5
$\begingroup$

Let $M$ be a smooth surface and let $f: M \to \mathbb{R}^3$ be a family of immersions given by

$$ f(t) = f_0 + tuN_0, $$

where $f_0$ is some initial immersion, $N_0$ is the associated Gauss map, and $u: M \to \mathbb{R}$ is an arbitrary smooth function. One can easily work out, e.g., the first variation in the induced area form, but I am having a surprising amount of difficulty deriving (or tracking down known expressions for) variations of higher-order quantities.

Question: What are the first variations of the mean curvature $H$ and the Gauss curvature $K$ with respect to the immersion $f$?

Bonus question: What are the first variations of the principal curvatures?

If these results are not immediately available, I am also appreciative of strategies and techniques for obtaining them. Finally, my apologies if these questions have well-established answers - I am having no luck finding them in the literature!

$\endgroup$
14
$\begingroup$

If you just calculate using the moving frame, you'll get the answer for the variation of the principal curvatures in a few lines: $$ \delta\kappa_i = \mathrm{Hess}(u)(e_i,e_i) + \kappa_i^2\,u . $$ Here, $\delta\kappa_i$ is the first $t$-derivative of $\kappa_i$ at $t=0$ (i.e., the 'first variation of $\kappa_i$'), $\mathrm{Hess}(u)$ is the quadratic form that is the Hessian of $u$ (using the induced metric on the surface), and $e_1$ and $e_2$ are the principal unit vector fields on the surface (i.e., $e_i$ is tangent to the $i$-th principal curve). In particular, since $H=\tfrac12(\kappa_1+\kappa_2)$ and $K = \kappa_1\kappa_2$, you'll get $$ \delta H = -\tfrac12\Delta u + (2H^2{-}K)\,u $$ and $$ \delta K = -2H\,\Delta u - \mathrm{II}{\cdot}\mathrm{Hess}(u) + 2HK\,u, $$ where $\Delta u$ is the Laplacian of $u$ (i.e., minus the trace of $\mathrm{Hess}(u)$ with respect to the first fundamental form), and $\mathrm{II}{\cdot}\mathrm{Hess}(u)$ is the inner product of the second fundamental form $\mathrm{II}$ and the Hessian of $u$ (interpreted as symmetric matrices using the orthonormal basis $e_i$, this inner product is just the trace of the product of the two corresponding matrices).

Added at request of the OP:

Sorry that this took so long. Here is how the calculation goes using the moving frame. (As usual, the work is explaining the notation. The calculation itself is easy.)

I'm only going to treat the case of Darboux surfaces, i.e., surfaces in $\mathbb{E}^3$ for which the principal curvatures are distinct. Let $X:\Sigma\times\mathbb{R}\to\mathbb{E}^3$ be a $1$-parameter family of Darboux immersions of an abstract, simply connected, oriented surface $\Sigma$. Let $N:\Sigma\times\mathbb{R}\to S^2$ be the function that gives the oriented unit normal vector field $N(\cdot,t):\Sigma\to S^2$ along the immersion $X(\cdot,t):\Sigma\to\mathbb{E}^3$ for each $t\in\mathbb{R}$. By reparametrizing the family $X$, I can assume that $\partial X/\partial t = u N$ for some (unique) function $u:\Sigma\times\mathbb{R}\to\mathbb{R}$. Since the individual immersions are Darboux, there will exist a smooth mapping $(e_1,e_2,e_3):\Sigma\times\mathbb{R}\to\mathrm{SO}(3)$ such that $e_3 = N$ and such that $e_i(\cdot,t)$ for $i=1,2$ is a Darboux (i.e., principal) frame field along $X(\cdot,t):\Sigma\to\mathbb{E}^3$. Now, setting $$ \mathrm{d}X = e_1\,\omega_1 + e_2\,\omega_2 + e_3\,\omega_3\,, $$ one has $\omega_3 = u\,\mathrm{d}t$ and the forms $\omega_1,\omega_2,\mathrm{d}t$ are a basis for the $1$-forms on $\Sigma\times\mathbb{R}$. The Darboux framing assumption is that $$ \omega_{3i} = \kappa_i\,\omega_i + \mu_i\,\mathrm{d}t $$ for $i=1,2$, where $\kappa_i$ are the (distinct) principal curvatures.

Now, one has the structure equations $\mathrm{d}e_a = e_b\,\omega_{ba}$, where $\omega_{ba} = -\omega_{ab}$ for $1\le a,b\le 3$ and (summation convention assumed) $$ \mathrm{d}\omega_a = -\omega_{ab}\wedge\omega_b \qquad\text{and}\qquad \mathrm{d}\omega_{ab} = - \omega_{ac}\wedge\omega_{ca}\,. $$

Now, $\mathrm{d}u = u_1\,\omega_1 + u_2\,\omega_2 + \dot u\,\mathrm{d}t $, so the equation $\mathrm{d}\omega_3 = -\omega_{31}\wedge\omega_1 -\omega_{32}\wedge\omega_2$ becomes $$ (u_1\,\omega_1 + u_2\,\omega_2)\wedge\mathrm{d}t = - \omega_{31}\wedge\omega_1 -\omega_{32}\wedge\omega_2 = -\mu_1\,\mathrm{d}t\wedge\omega_1 - \mu_2\,\mathrm{d}t\wedge\omega_2\,, $$ so it follows that $\mu_i = u_i$. Moreover, since $\mathrm{d}(\mathrm{d}u)=0$, this gives $$ \mathrm{d}u_1\wedge\omega_1 -u_1\omega_{12}\wedge\omega_2 +\mathrm{d}u_2\wedge\omega_2 -u_2\omega_{21}\wedge\omega_1 \equiv 0 \mod \mathrm{d}t, $$ so this implies, by Cartan's Lemma, that, for some $u_{ij}=u_{ji}$, I have $$ \left. \begin{aligned} \mathrm{d}u_1&\equiv u_2\omega_{21} + u_{11}\,\omega_1 + u_{12}\,\omega_2\\ \mathrm{d}u_2&\equiv u_1\omega_{12} + u_{12}\,\omega_1 + u_{22}\,\omega_2 \end{aligned}\right\} \mod \mathrm{d}t $$ Geometrically, this means that the Hessian of $u(\cdot,t)$ is the quadratic form $$ \mathrm{Hess}\bigl(u(\cdot,t)\bigr) = u_{11}(\cdot,t)\,{\omega_1}^2 +2u_{12}(\cdot,t)\,\omega_1\omega_2 + u_{22}(\cdot,t)\,{\omega_2}^2. $$ One also has $$\mathrm{d}\omega_i = - \omega_{ij}\wedge\omega_j -\omega_{i3}\wedge\omega_3 = -\omega_{ij}\wedge\omega_j + u\kappa_i\,\omega_i\wedge\mathrm{d}t. $$

Finally, computing the exterior derivatives of the $\omega_{3i}$ for $i=1,2$ via the structure equation $\mathrm{d}\omega_{3i} = -\omega_{3j}\wedge\omega_{ji}$ yields $$ \mathrm{d}(\kappa_i\,\omega_i + u_i\,\mathrm{d}t) = -(\kappa_j\,\omega_j + u_j\,\mathrm{d}t)\wedge\omega_{ji} $$ (where $j$ is summed over $1,2$). Expanding this, using the above equations, and comparing the coefficients of the $\omega_i\wedge\mathrm{d}t$ term on the RHS and LHS, one obtains the relation $$ \dot\kappa_i = u_{ii} + {\kappa_i}^2\,u\,,\tag1 $$ where, for $i=1,2$, one has an expansion of the form
$$ \mathrm{d}\kappa_i = \kappa_{ij}\,\omega_j + \dot\kappa_i\,\mathrm{d}t $$ (where $j$ is summed over $1,2$). Then equation (1) is the desired formula.

$\endgroup$
  • $\begingroup$ Thanks Robert. I'm accepting this answer because it gives an explicit expression for the variations, but of course the other answers below are also useful. However, my experience with moving frames is limited and I was unable to re-derive the result myself. The main difficulty is that I do not know how to set up a moving frame on an evolving surface in an intelligent way, so that I don't run into the same difficulties as with other methods of derivation. Can you share at least the ansatz you used for this calculation? Thanks! $\endgroup$ – Omega Tree Oct 18 '14 at 12:56
  • $\begingroup$ @OmegaTree You're welcome. Sure, I'll write a description of the calculation, but I probably won't have time to do it until this evening (I'm on the East Coast of the US). $\endgroup$ – Robert Bryant Oct 18 '14 at 13:36
  • $\begingroup$ Thanks Robert. Whenever you do get a spare moment, I think it would be very helpful to have such a description archived with the results above. $\endgroup$ – Omega Tree Oct 20 '14 at 18:41
  • $\begingroup$ Hi Robert, any chance you might still say something about how to set up moving frames on an evolving surface? Sorry to be a pest, but it appears to be a very useful tool! $\endgroup$ – Omega Tree Nov 1 '14 at 13:31
  • $\begingroup$ @OmegaTree: Sorry, I've been really busy lately and haven't had time to put this in. I'll try to get to it this weekend. $\endgroup$ – Robert Bryant Nov 1 '14 at 13:43
6
$\begingroup$

One nice source for such computations is contained in the following notes of Schnurer: https://www2.math.hu-berlin.de/gradkoll/Schnuerer_alpbach%5B1%5D.pdf, Section 3.

To be pedantic, technically the notes seem to not address your question, as they state that they assume that $u$ is a function of the principal curvatures, but you may easily check that the computations (at least up to Lemma 3.4) work for any normal flow.

$\endgroup$
3
$\begingroup$

Have a look at this paper. Chapter 4 contains all that you are asking for, in a more general setting.

$\endgroup$
2
$\begingroup$

I just remembered that there might another way using Jacobi fields. Suppose you have a hypersurface in an $(n+1)$-dimensional Riemannian manifold. You take advantage of the fact that along each geodesic normal to an $n$-dimensional hypersurface, there are $n$ linearly independent Jacobi fields $J_1, \dots, J_n$ that are tangent to the level hypersurfaces of the distance from the original hypersurface, where $g_{ij} = J_i\cdot J_j$ is the induced metric of the level hypersurface at each time $t$. The second fundamental form for each level hypersurface is equal to $h_{ij} = J_i\cdot\nabla_t J_j$. From there you can compute its first variation.

A good paper to study is the classic by Heintze and Karcher.

$\endgroup$
1
$\begingroup$

One (but probably not the easiest) way to proceed is this: Just for the heck of it, I'll explain as much as I can in arbitrary dimensions and codimensions. Work with a fixed set of local co-ordinates on the submanifold. Let $f: M^n \rightarrow \mathbb{R}^{N}$. Let $g_{ij}$ be the induced metric given by $$ \partial_i f \cdot \partial_j f = g_{ij} $$ Using this it's easy to compute the first variation of $g$. Let $\Gamma^k_{ij}$ be the Christoffel symbols. Verify that $$ g_{kl}\Gamma^l_{ij} = \partial_kf \cdot \partial^2_{ij}f. $$ Use this to compute the first variation of $\Gamma^k_{ij}$. Next, verify that $$ \partial_{ij}f = \Gamma^k_{ij}\partial_kf + H_{ij}, $$ where $H_{ij}$ is normal to the tangent space of $M$ (as a subspace of $\mathbb{R}^N$. From this you can compute the first variation of $H_{ij}$. The curvature tensor is given by the Gauss equations: $$ R_{ijkl} = H_{ik}\cdot H_{jl} - H_{il}\cdot H_{jk}. $$ From this you can compute the first variation of the curvature tensor. If $n = 2$, you can take advantage of the fact that Gauss curvature is equal to a constant times scalar curvature $S = g^{ik}g^{jl}R_{ijkl}$ to calculate its variation.

Now let's assume $N = n+1$. To get the first variation of the second fundamental form, you need a formula for the unit normal. If $n = 2$, then you can use the cross product of $\partial_1f$ and $\partial _2f$ divided by its norm. In higher dimensions, the Gauss map is the unit vector dual to $*df/|*df|_g$, where $*$ is the Hodge star. Using this formula you can get a first variation of the normal vector. The second fundamental form of a hypersurface is now given by $$ h_{ij} = \nu\cdot H_{ij}. $$ The mean curvature is easy to get from this. As for the principal curvatures, the first variation of one particular principal curvature is guaranteed to exist only if it is different from all of the others. You can get its variation by implicitly differentiating $$ 0 = \det (H_{ij} - \lambda g_{ij}). $$

$\endgroup$
  • 1
    $\begingroup$ For the mean curvature alone, in a much more general setting (arbitrary codimension embeddings of pseudo-Riemannian manifolds), I computed the first variation of the mean curvature vector in section A.5 of this paper following a variation of what Deane described (using Christoffel symbols instead of frames) here. $\endgroup$ – Willie Wong Oct 8 '14 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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