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Here are two equivalent definitions of Hessian. The trace of a (0,2)-tensor $h$ can be defined in coordinates by
$$\text{tr}_g(h)=\sum_{i,j}g^{ij}h_{ij},$$ where $g^{ij}$ is the component of the inverse matrix $(g_{ij}).$

We know that in calculus, the trace of Hessian of a function $u$ is clearly the Laplacian $\Delta u$(In Riemannian geometry, it is defined to be the divergence of a gradient field). But I don't know how to prove this for general Riemannian manifolds(coordinate free or not).

No One
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1 Answers1

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$\def\tr{\mathrm{tr}}\def\dv{\mathrm{div}}\def\gr{\mathrm{grad}}\def\o{\circ}\def\n{\nabla}$You have a small error in your question: the trace you are describing is the trace of a $(0,2)$-tensor $h$ with respect to the metric $g$, which I write $\tr_gh$, reserving $\tr$ for the invariant trace $\tr\, T = T^i_i$ of a $(1,1)$-tensor $T$.

The equality of these two definitions of the Laplacian essentially just falls out of the definitions. The divergence operator is defined on vector fields as the contraction of the covariant derivative, so $\dv = \tr \o \n$. Thus $$\Delta := \dv \o \gr = \tr \o \n \o \gr.$$ Since the gradient is just what we get by raising an index of the covariant derivative using the metric, this becomes $$\Delta = \tr_g \o (\n \o \n) = \tr_g \o \mathrm{Hess},$$ where we used the metric-compatibility of $\nabla$ to commute the metric past the derivative. In abstract index notation this whole calculation is just $$\Delta f = \nabla_i (g^{ij} \nabla_j f)=g^{ij}\nabla_i\nabla_j f.$$

Anthony Carapetis
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  • Thank you! But can you show me how to prove $\mathrm{div} = \mathrm{tr} \circ \nabla$? – No One Sep 27 '17 at 16:55
  • If that's not your definition of divergence, what is? – Anthony Carapetis Sep 27 '17 at 22:28
  • Thanks! Could you please explain why $g^{ij}\nabla_i\nabla_j f$ is the same as any of those two definitions of Hessian given in the link? – No One Sep 30 '17 at 21:17
  • Also, are you claiming that $\nabla_i g^{ij}=0$ in $\nabla_i (g^{ij} \nabla_j f)=g^{ij}\nabla_i\nabla_j f$? How does it follow from the metric-compatibility? – No One Sep 30 '17 at 21:41
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    I'm using the usual convention for index notation where derivatives are taken before the basis vectors are plugged in; i.e. $\nabla_i g^{ij}$ means $(\nabla g)(\partial_i, dx^i, dx^j)$, not $\nabla_{\partial_i} (g(dx^i,dx^j)).$ Thus this follows immediately from the metric-compatibility $\nabla g = 0.$ The expression $\nabla_i \nabla_j f$ thus represents $(\nabla \nabla f)(\partial_i, \partial_j) = \nabla_i(\nabla f(\partial_j)) - (\nabla f)(\nabla_i \partial_j) = \partial_i(\partial_j f) - (\nabla_i \partial_j) f,$ matching the second formula in your link. – Anthony Carapetis Oct 01 '17 at 00:01
  • I have got more confused now. I am not sure what you mean by $(\nabla g)(\partial_i, dx^i, dx^j)$ and I don't see why you can complete the proof without introducing the Christoffel symbols. – No One Oct 01 '17 at 16:28