Let $M$ be a level set of a function in $\mathbb R^3$. Then the mean curvature of $M$ is given by the trace of the second fundamental form which is a divergence term involving the Levi-Civita connection. My question is, why is this the same as the "usual" divergence of the normal vector we learned in early multivariable calculus?


Write $n$ for the normal vector of $M$ in $\mathbb R^3$. Recall that the second fundamental form of $M$ is just $II(X,Y) = <D_Xn,Y>$, where $X, Y \in \mathbb R^3$ are tangent vectors to $M$. Here $< \cdot, \cdot>$ is the Euclidean inner product, and $D_X n$ represents the directional derivative of $n$ in the direction $X$. It is true that the second fundamental form is defined in terms of the Levi-Civita connection, but in $\mathbb R^3$ recall that $\nabla_X Y = D_X Y.$ Now, by the definition of divergence, we see that the mean curvature $H$ satisfies $$ H = \mathbb{tr} II = \mathbb{div} \,n$$

which is just the usual version of divergence you learn in multivariable calc.

To be a little bit more precise, the general definition of divergence is $$\mathbb{div} X = \mathbb{tr} (X \to \nabla X),$$ i.e. the trace of the map sending $X$ to $\nabla X$. Since $\nabla = D$ in $\mathbb R^3$, you can see that this agrees with the usual definition of divergence for a vector in $\mathbb R^3$.

  • $\begingroup$ Hi thanks for your reply. I just have one issue. The mean curvature is defined as taking the trace with an orthonormal basis e_1, e_2 of T_pM. To get the "standard divergence" i need to work in the standard orthobasis of R^3, which is of course very different to taking the divergence with e_1, e_2. My idea was to use the normal vector to extend this to an orthonormal basis of R^3 (since the sum involving the normal contributes nothing), and then say that the divergence doesnt depend on the choice of orthonormal basis. and then that would be that. is that correct? $\endgroup$ – user78410 Jun 3 '13 at 9:49
  • $\begingroup$ Yes, that's correct. The trace is independent of the basis used to calculate it. $\endgroup$ – treble Jun 3 '13 at 14:19
  • $\begingroup$ Okay great. Thanks alot for your help! $\endgroup$ – user78410 Jun 3 '13 at 18:43

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