# components of normal field are Jacobi fields?

Given a minimal surface $\Sigma$ in $R^3$ with associated normal field $N$, I am told that each of the components of $N$ is a Jacobi field, meaning that $Lu=0$ where L is the stability (Jacobi) operator and $u$ is the function $u(x) = \langle N(x), e_i \rangle$, with $e_1 = (1,0,0)$, etc.

Supposedly this follows from looking at a one parameter family of isometries and the second variation formula, as given in e.g. Minicozzi and Colding's course on minimal surfaces, which I am studying from.
I am just exploring this material for the first time, I guess I'm missing something obvious but I just can't see the conclusion. As in the reference, the motivation is in obtaining a proof that any minimal graph is stable.

Intuition and rigor both welcome!

The other answer has given a good intuitive explanation. Let me just give a computational one. The Jacobi operator is (1.147 in the book)

$$L u = \Delta u + |A|^2 u + \text{Ric}_M(N, N)u = \Delta u + |A|^2 u .$$

as $M = \mathbb R^3$ has $\text{Ric}_M = 0$. To see this $u = \langle N, v\rangle$ satisfies $Lu_ = 0$ (where $v$ is a fixed vector in $\mathbb R^3$), note that if we fix a normal coordinate at $x$ in the minimal surface, then (note that the $e_i$'s orthonormal basis of the minimal surface, not that of $\mathbb R^3$)

$$\begin{split} \Delta u &= \nabla_{e_i} \nabla_{e_i} u \\ &= \nabla_{e_i} \nabla_{e_i} \langle N, v\rangle \\ &= \nabla_{e_i} \langle \nabla_{e_i}N, v\rangle \\ &= - \nabla_{e_i}\langle A_{ij} e_j, v\rangle \\ &= -\langle \nabla_{e_i}A_{ij} e_j + A_{ij} \nabla^{\mathbb R^3} _{e_i } e_j, v\rangle \\ &= -\langle \nabla_{e_j}A_{ii} e_j + A_{ij}A_{ij} N, v\rangle \end{split}$$

Note that in the last equaliy we used the Codazzi equation $\nabla_{e_k} A_{ij} = \nabla_{e_i} A_{kj}$ and the fact that $\nabla^{\mathbb R^3} _{e_i } e_j = A_{ij} N$. Then the first term is zero as the surface is minimal, thus

$$\Delta u = - A_{ij}^2 \langle N, v\rangle = -|A|^2 u.$$

This is the same as $Lu= 0$.

• @user3499756 : I have just deleted the "intuitive" one as Anthony's explanation is better. I just leave the computational one here. – user99914 Oct 13 '15 at 1:49
• \You can think of all of them are usual directional derivative. From the second to third I use the product rule and the fact that $\nabla_{e_i} v = 0$. @user3499756 – user99914 Oct 13 '15 at 2:05
• You are almost correct, $A_{ij} = \langle A(e_i, e_j) , N\rangle$. @user3499756 – user99914 Oct 13 '15 at 3:35
• $A_{ii} = H = 0$ (note that I am using Einstein summation and the fact that $\{e_i\}$'s are orthonormal, so $H = A_{11} + A_{22}$. @user3499756 – user99914 Oct 13 '15 at 4:27
• That is also just directional derivative of a vector (that $e_i$) in the direction. This vector $\nabla_{e_i}^{\mathbb R^3} e_j$ has no tangential part as we assume that $\nabla_{e_i} e_j = 0$ at $x$ and this vector is by definition the tangenet component of $\nabla_{e_i}^{\mathbb R^3} e_j$. Thus it has only normal part, which is $\langle \nabla_{e_i}^{\mathbb R^3} e_j , N\rangle N = A_{ij} N$. @user3499756 – user99914 Oct 13 '15 at 5:36

Consider the variation of $\Sigma$ generated by translation in the direction $e_i$; i.e. $X(x,t) = X_0(x) + te_i$ for $X_0$ a parametrization of $\Sigma$. This is a family of isometries so the area element is constant in time, and it has normal speed $u = \langle e_i, N\rangle$; so the second variation formula tells us $Lu = 0$.

• Could you elaborate on the last sentence? This approach is the same as in Minicozzi and Colding, which is great because it's what I'm trying to understand, but I am just not seeing the implication $Lu$ = 0. I am getting to the point where we see that the integral of $<X_t, LX_t>$ vanishes, i.e. the integral of the $i$th component of $Le_i$ vanishes. Then I get stuck. – Badam Baplan Oct 13 '15 at 1:33
• @user3499756: Since the area element is preserved (i.e. area is preserved locally, not just globally) you get this equality for the integral over any subset of $\Sigma$. Continuity of the integrand then implies that it must in fact be identically zero. – Anthony Carapetis Oct 13 '15 at 1:41
• Though that's assuming the formula is true for surfaces with boundary; I'm actually unsure whether or not this is the case... there may be some integration by parts involved in the derivation that would break this. It's been a while since I looked at this stuff. – Anthony Carapetis Oct 13 '15 at 1:51
• @AnthonyCarapetis : For the first variation, there's actually no integration by part involved. You need only that the variation field $V$ is normal so that you can have $\partial_t g_{ij} = -\langle V, (A_{ij} +A_{ji}) N\rangle.$ – user99914 Oct 13 '15 at 1:56
• There willl be a divergence term if $V$ is not normal, something like $div(V^\top)$. – user99914 Oct 13 '15 at 2:03