# Metric trace of third derivative

Let $(M,\langle \cdot, \cdot\rangle)$ be a Riemannian manifold, and $\varphi: M\to \Bbb R$ be a smooth map. Denoting the gradient of $\varphi$ by $\nabla \varphi$ and the covariant Hessian by $\nabla^2\varphi$ (that is, $\nabla^2\varphi(X,Y) = \langle \nabla_X(\nabla\varphi), Y\rangle$), I would like to compute the metric trace $${\rm tr}_{\langle \cdot,\cdot\rangle}\big((X,Y)\mapsto \nabla^2\varphi(\nabla_X(\nabla\varphi), Y)\big).$$I'm a bit stumped. I tried to do it in coordinates using $({\rm d}x^i)^\sharp = g^{ij}\partial_j$ but I couldn't simplify anything. Help?

Writing everything in local coordinates. First of all, we have

$$(\nabla_{\cdot} \nabla \phi)_i^j = g^{jk}\phi_{ik} = \phi_i^j.$$

Let $A(X, Y) = \nabla^2 \phi(\nabla_X \nabla \phi, Y)$. Then

\begin{align} \text{tr} A &= g^{ij} A(\partial_i, \partial_j) \\ &= g^{ij} \nabla^2 \phi (\nabla_{\partial_i} \nabla \phi, \partial _j) \\ &= g^{ij} \nabla^2 \phi ( \phi_i^k \partial_k , \partial_j)\\ &= g^{ij} \phi_{kj}\phi_i^k = \phi_k^i \phi_i^k. \end{align}

Remark: For any symmetric bilinear form $V\times V\to \mathbb R$, let $\tilde B : V\to V$ be given by $$B(X, Y) = g(X, \tilde B Y)\ \ \text{ for all }X, Y\in V.$$

Then with $B = \nabla^2 \phi$,

$$A(X, Y) = B ( \tilde B X, Y) = g(\tilde BX, \tilde B Y).$$

One might assume that $\tilde B$ is diagonalized at one point, then $$\text{tr}A = \sum_{i=1}^n g (\tilde B e_i, \tilde Be_i) = \sum_{i=1}^n \lambda_i^2.$$

• I can follow your computation, but is there a way to describe that result without coordinates? – Ivo Terek May 6 '18 at 0:10
• I am not sure what you want, @IvoTerek . I added something anyway. – user99914 May 6 '18 at 3:48
• @IvoTerek: In its full ugliness, it's the contraction $$\def\tr{\operatorname{tr}}\tr_{18}\tr_{23}\tr_{45}\tr_{67}\nabla^2 \phi \otimes g^{-1} \otimes \nabla^2 \phi \otimes g^{-1}.$$ Since $\nabla^2 \phi$ is symmetric, this can be simply written $|\nabla^2 \phi|^2$. – Anthony Carapetis May 6 '18 at 7:34