I'm trying to work through a derivation of the stability operator from minimal surface theory. Suppose $\Sigma^k$ is a minimal submanifold of $\mathbb{R}^n$, and suppose $X$ is a normal vector field on $\Sigma$ with compact support vanishing on the boundary.

Part of the derivation involves the integration by parts $\int_{\Sigma} \langle \nabla_{\Sigma}^N X, \nabla_{\Sigma}^N X \rangle = - \int_{\Sigma} \langle X, \Delta^N_{\Sigma} X \rangle$, where $\nabla_{\Sigma}^N$ is the normal projection of the covariant derivative on $\Sigma$, and $\Delta^N_{\Sigma}$ is the normal Laplacian defined by $\Sigma_{i = 1}^k \nabla_{E_i}^N \nabla_{E_i}^N X - \nabla^N_{\left(\nabla_{E^i} E_i\right)^T} X$, with $E_i$ being an orthonormal frame for $\Sigma$. Why does this hold? I'm aware of Green's identity holding for the Laplacian and gradient of scalar functions, but the operators involved here are normal projections, and I want to know why the identity still holds in this case.

I have tried to compute $\nabla_{\Sigma}^N (\nabla_{\Sigma}^N X)$ in hopes that its inner product with $X$ is the same as that with $\Delta_{\Sigma}^N X$, up to some terms which vanish under the integral like a divergence, but I haven't gotten anywhere.

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    $\begingroup$ Your integration by parts intuition is not quite right. What's missing is the boundary term — because $X$ vanishes on the boundary. So what you're integrating over $\Sigma$ is the derivative term. (This is not unlike Green's identities in classical multivariable calculus.) $\endgroup$ Commented Jun 25, 2022 at 18:28
  • $\begingroup$ @maxematician I deleted my answer because I couldn't see a way to fix the problems you pointed out with it. Could you point out the source where you saw this calculation? Does it come up in the proof of the second variation formula? $\endgroup$ Commented Jun 26, 2022 at 3:20
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    $\begingroup$ Colding-Minicozzi’s A Course on Minimal Surfaces. It’s in the first chapter. There’s also Section 6 of these notes by Fernando Coda Marques web.math.princeton.edu/~rcabral/pdfs/minimalsurfaces.pdf $\endgroup$ Commented Jun 26, 2022 at 3:39
  • $\begingroup$ @TedShifrin I was assuming $X$ is compactly supported, so $X$ and all of its derivatives vanish on the boundary. I edited the post to reflect that. $\endgroup$ Commented Jun 26, 2022 at 21:47

1 Answer 1


The normal bundle along the submanifold in the Riemannian background is a special case of the following situation.

Let $M$ be a suitably good manifold of dimension $n$ with a Riemannian metric $g$. It has the Levi-Civita connection $\nabla = \nabla^g$ compatible with $g$, that is $\nabla g = 0$. Taking care with the orientation, there is also the Riemannian volume form $\mathrm{d}M$ on $M$.

Furthermore, let $E \to M$ be a (finite-dimesional) vector bundle over $M$ and forgive me if I sloppily denote the space of its sections by $E$ as well. Let $h: E \otimes E \to \mathbb{R}$ be a vector-bundle metric. Let $\nabla^E$ be some metric connection in $E$, that is $\nabla^E h = 0$.

There is a whole family of tensor products of the tangent, cotangent, and the given vector bundle $E$ on the manifold $E$, which are endowed with the corresponding tensor-product connection. More precisely, on each of such bundles, the connection acts by extending the Levi-Civita connection and the connection $\nabla^E$ using the Leibniz rule and linearity. Each of those connections we are going to denote simply by $\nabla$ and refer to them collectively as the coupled Levi-Civita connection.

The vector-bundle connection Laplacian $\Delta^E$ is defined on any section $X$ of the bundles, described above, by the formula $$ \Delta^E X := g^{a b} \nabla_a \nabla_b X $$ where $\nabla$ is the coupled Levi-Civita connection. The formula for the normal Laplacian in the OP is just a particular case of this.

Now I claim that $$ \big( \nabla^a h(X, \nabla_a X) \big) \mathrm{d}M = \mathrm{d}\omega $$ for some $(n - 1)$-form $\omega$ on $M$. This allows us to use the general Stokes theorem $$ \int_{M} \mathrm{d}\omega = \int_{\partial M} \omega $$ and, if we take into account the OP restrictions on the sections $X$ with regards to the boundary, we can say that $$ \int_{M} \big( \nabla^a h(X, \nabla_a X) \big) \mathrm{d}M = 0 $$

Using the Leibniz rule, we calculate $$ \nabla^a h(X, \nabla_a X) = h(\nabla^a X, \nabla_a X) + h(X, \nabla^a \nabla_a X) = |\nabla X|^2 + h(X, \Delta^E X) $$

From the last display, integrating and using the Stokes theorem, we obtain $$ \int_{M} |\nabla X|^2 \mathrm{d}M = - \int_M h(X, \Delta^E X) \mathrm{d}M $$

Proof of the claim. This is going to be a calculation using an abstract index notation technique for dealing with differential forms.

A $(n)$-form $\xi$ using the index notation can be presented as $\xi_{a_1 \dots a_n}$, where the sequentially numbered indexes are assumed to be antisymmetrized: $$ \xi_{a_1 \dots a_n} = \xi_{[a_1 \dots a_n]} $$

The exterior derivative $\mathrm{d}$ on the differential form $\xi$ as above is defined by the formula: $$ (\mathrm{d} \xi)_{a_0 a_1 \dots a_n} := (n + 1) \nabla_{a_0} \xi_{a_1 \dots a_n} $$

Recall that the volume form $\mathrm{d} M$ is a ${n}$-form, so we can write it as $(\mathrm{d} M)_{a_1 \dots a_n}$ to emphasize that.

Now we want $\xi$ to be a $1$-form (a covector field). We can introduce a notation, just for the convenience sake, for the interior product of the covector $\xi$ with the volume form $\mathrm{d} M$ as so: $$ (\iota_{\xi} \mathrm{d} M)_{a_2 \dots a_n} := g^{a a_1} \xi_a (\mathrm{d} M)_{a_1 \dots a_n} $$ Clearly, $\iota_{\xi} \mathrm{d} M$ is a $(n-1)$-form.

Let us compute the exterior derivative of $\iota_{\xi} \mathrm{d} M$ to see what happens. $$ \big( \mathrm{d} (\iota_{\xi} \mathrm{d} M) \big)_{a_0 a_2 \dots a_n} = n \nabla_{a_0 } (\iota_{\xi} \mathrm{d} M)_{a_2 \dots a_n} = n \nabla_{a_0 } g^{a a_1} \xi_a (\mathrm{d} M)_{a_1 \dots a_n} = \\ = n g^{a a_1} \big( \nabla_{a_0 } \xi_a (\mathrm{d} M)_{a_1 \dots a_n} \big) = n g^{a a_1} \big( (\nabla_{a_0 } \xi_a) (\mathrm{d} M)_{a_1 \dots a_n} + \xi_a \nabla_{a_0 } (\mathrm{d} M)_{a_1 \dots a_n} \big) = \\ = n g^{a a_1} \big( (\nabla_{a_0 } \xi_a) (\mathrm{d} M)_{a_1 \dots a_n} $$

The last expression in the display above is a $(n)$-form on $M$, so it must be proportional to $\mathrm{d} M$: $$ g^{a a_1} \big( (\nabla_{a_0 } \xi_a) (\mathrm{d} M)_{a_1 \dots a_n} = f (\mathrm{d} M)_{a_0 a_2 \dots a_n} $$

Contracting with $\mathrm{d} M$ and using its properties (see R.Wald, General Relativity, p.433), we can find that $$ \nabla^b \xi_b = n f $$

We see that for any covector $\xi$ the so-called divergence term $(\nabla^a \xi_a) \mathrm{d} M$ is an exact differential form: $$ (\nabla^a \xi_a) \mathrm{d} M = \mathrm{d}(\iota_{\xi} \mathrm{d} M) $$

Taking $\xi_a = h(X, \nabla_a X)$ we obtain the claimed property.

  • $\begingroup$ Thanks for the post. I can see that Green's identity holds for generalized gradients and Laplacians on a vector bundle, which I intuitively guessed was the case. All that remains is an explanation for why the normal Laplacian takes the particular form given in the OP. It shouldn't be too hard to work out, and I'll try to do it myself if I have time, but I'd appreciate seeing the details. $\endgroup$ Commented Jun 30, 2022 at 15:52
  • $\begingroup$ @maxematician To clarify the expression for the normal Laplacian in an ONF you can start with the formula for the second covariant derivative of a bundle section $s$ in the tangent directions $X$ and $Y$: $\nabla^2_{X, Y} s = \nabla_{X} (\nabla_{Y} s) - \nabla_{\nabla_{X} Y} s$ and calculate the contraction. $\endgroup$ Commented Jul 1, 2022 at 1:13
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    $\begingroup$ With regards to the details about the second covariant derivative, one can find this answer useful. $\endgroup$ Commented Jul 2, 2022 at 12:28
  • $\begingroup$ @YuriVyatkin I am a little confused with the definition of $\triangle^E$. It seems more fair to define it by $\triangle^EX=g^{ab}(\nabla_a\nabla_b X - \nabla_{\nabla_ab}X)$. $\endgroup$
    – Y.Wayne
    Commented Aug 27, 2022 at 2:00
  • $\begingroup$ @Y.Wayne Nope, this is not how the abstract index notation works. In particular, $\nabla_a b$ makes no sense. Keep in mind that $\nabla_a \nabla_b X$ is a traditional (and confusing!) notation for the second covariant derivative, which should be better written as $(\nabla \nabla X)_{a b}$ and then there is a formula for it, which you are alluding to. $\endgroup$ Commented Aug 27, 2022 at 2:44

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