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In the article: https://arxiv.org/abs/0906.3217 the authors prove in Lemma 1 a formula which helps compute more easily the integral of the Hessian of a function defined on $\Bbb{S}^2$. More precisely, if $h : \Bbb{S}^2 \to \Bbb{R}$ is a $C^2$ function, $Hess(h)(X,Y) = \langle \nabla_X \nabla h,Y\rangle$ is the Hessian of $H$ and $H(h)$ is the determinant of the Hessian, then $$ \int_{\Bbb{S}^2} H(h) dA = \frac{1}{2} \int_{\Bbb{S}^2} |\nabla h|^2 dA$$ where $dA$ is the area element on $\Bbb{S}^2$. This formula is helpful and simplifies some aspects of my numerical computations. Here and in the following, $\nabla$ and $\Delta$ represent the tangential gradient and Laplace-Beltrami operator on $\Bbb{S}^2$.

The above formula is proved in connection with bodies of constant width, but in the proof in the article they don't seem to use this fact. In my numerical experiments the formula gives the expected result in the general case. However, I work in the case where $H(h)>0$ (I don't know if this is relevant or not...)

I was wondering if it is possible to obtain a similar simplification for the integral $\int_{\Bbb{S}^2} h H(h) dA$? More precisely, is it possible to obtain something of the form $$ \int_{\Bbb{S}^2} h H(h) dA = \int_{\Bbb{S}^2} \mathcal{F}(h,\nabla h,\Delta h)dA $$ where $\mathcal{F}$ is "simple" (polynomial)?

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  • $\begingroup$ At first glance, this looks wrong. In any case, Lemma 1 assumes that the body has constant width. You don't appear to be doing so here. $\endgroup$
    – Deane Yang
    Commented Mar 12, 2018 at 15:19
  • $\begingroup$ I contacted the authors and apparently they don't use the fact that the body has constant width in Lemma 1.... My (numerical) experiments suggest that the formula in the article is OK in the general case. Unfortunately, my level in differential geometry is not enough to check their proof... Nevertheless, I work in the case where $H(h)>0$, so maybe I should add that. $\endgroup$
    – Beni Bogosel
    Commented Mar 12, 2018 at 15:33
  • $\begingroup$ OK, I see. One way, probably not the easiest but a straightforward way, is to write everything in terms of stereographic coordinates. The integrals can now be written explicitly as integrals over $\mathbb{R}^2$. You can first confirm their formula and then carry out a similar calculation for your equation. It'll be a mess, but if you do it carefully a few times, you should be able to figure it all out. $\endgroup$
    – Deane Yang
    Commented Mar 12, 2018 at 17:15
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    $\begingroup$ The Lemma is actually not difficult to prove directly; you don't need to go to coordinates. There is a (Lorentzian) inner product on $S^2(T^*\mathbb{S}^2)$ such that $\nabla^2u\cdot\nabla^2u = H(u)$ for all smooth $u$, and the $L^2$-adjoint of $\nabla^2:C^\infty(\mathbb{S}^2)\to S^2(T^*\mathbb{S}^2)$ turns out to satisfy $$(\nabla^2)^*\nabla^2u = \tfrac12 \Delta u=\tfrac12 \nabla^*\nabla u.$$ (Perhaps surprisingly, it's only second order, because of the Lorentzian nature of the inner product.) Now, integration by parts gives the Lemma. $\endgroup$
    – Robert Bryant
    Commented Mar 12, 2018 at 19:56
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    $\begingroup$ By the way, the formula for a general compact, oriented Riemannian surface $M$ with Gauss curvature $K$ is $$\int_M H(u)\,\mathrm{d}A = \frac12\int_M K\,|\nabla u|^2\,\mathrm{d}A.$$ This (sort of) restores the scaling balance that Deane may have been worried about when he first expressed doubts that the equation could be true. $\endgroup$
    – Robert Bryant
    Commented Mar 12, 2018 at 20:40

2 Answers 2

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There is a formula that is roughly of the kind that the OP desires. On a closed, oriented Riemannian surface $(M,g)$ with Gauss curvature $K$, the formula is $$ \int_MuH(u)\,\mathrm{d}A = \frac12\int_M \left(K u |\nabla u|^2 - \nabla^2(u)(J\nabla u,J\nabla u)\right)\,\mathrm{d}A. $$ Here, $\nabla^2(u)$ is what the OP called the Hessian quadratic form of $u$ and $J:TM\to TM$ is the `rotation by $\pi/2$ operator' on tangent vectors.

I do not know whether the right hand side can be further simplified so that the polynomial depends only on $u$, $\nabla u$, and $\Delta u$.

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  • $\begingroup$ Thank you very much for your answer and for the useful comments which clarified the usage of the Lemma from the article. $\endgroup$
    – Beni Bogosel
    Commented Mar 13, 2018 at 20:31
  • $\begingroup$ I can see that I should have integrated by parts one more time, using the identity $$ \mathrm{d}\left(|\nabla u|^2\, {\ast}\mathrm{d}u\right) = -\left(3\Delta u\,|\nabla u|^2 + 2\,\nabla^2(u)(J\nabla u,J\nabla u)\right)\,\mathrm{d} A.$$ (However, I see that my Laplacian is the negative of Jeffrey Case's. Mine has positive eigenvalues.) $\endgroup$
    – Robert Bryant
    Commented Mar 14, 2018 at 9:28
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You can also derive the lemma from the Bochner formula, which in general dimensions can be written

$$ \frac{1}{2}\Delta\lvert\nabla u\rvert^2 = \lvert\nabla^2u\rvert^2 - (\Delta u)^2 + \delta\left((\Delta u)\,du\right) + \mathrm{Ric}(\nabla u,\nabla u) . $$

If $(M,g)$ is a Riemannian surface with Gauss curvature $K$, then $\mathrm{Ric}=Kg$ and the determinant of the Hessian is $H(u)=\frac{1}{2}\left((\Delta u)^2-\lvert\nabla^2u\rvert^2\right)$. Multiplying the Bochner formula by $u$ and integrating by parts gives

$$ \int u\,H(u)\,dA = \frac{1}{4}\int \left(2Ku - 3\Delta u\right)\lvert\nabla u\rvert^2\,dA . $$

In your situation, $K=1$.

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  • $\begingroup$ Thank you for your answer. This resembles what I was looking for. $\endgroup$
    – Beni Bogosel
    Commented Mar 13, 2018 at 20:33

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