# Formula for the Hessian on a Riemannian manifold

Let $(M,g)$ be a Riemannian manifold and let us denote the Hessian of a function $u$ by $\mathrm{Hess}(u)$. Denote the gradient of a function $f$ by $\nabla f$. I want to prove that the formula

$2\mathrm{Hess}(u)(\nabla f, \nabla h)= g(\nabla f, \nabla g(\nabla u, \nabla h)) + g(\nabla h, \nabla g(\nabla u, \nabla f)) - g(\nabla u, \nabla g(\nabla f, \nabla h))$

is valid for all functions $f$ and $h$ on $M$.

I am completely stuck. I have tried to compute the gradient of $g(\nabla f, \nabla h)$ for example trying to use the definition of the gradient but nothing seems to work. I would preffer not to put this in coordinates since it seems that it will be a pain, but if it is the only way then, well...

Honestly, I think index notation is less of a pain here: when we apply the product rule/metric compatiblity, it makes it easy to keep track of which slot of the Hessian we're applying the metric to. Since $f,u,h$ are all scalars, we can write all derivatives as postfix indices without getting confused, and raise and lower indices with the metric as usual. In this notation the RHS is

$$f_i \nabla^i (u^j h_j) + h_i \nabla^i (u^j f_j) - u^j \nabla_j (f_i h^i).$$

Expanding the derivatives with the product rule gives

$$f_i u^{ji} h_j + f^i u^j h_{ij} + h_i u_j f^{ij} + h_i f_j u^{ij} - u^j h^i f_{ij} - u_j f_i h^{ij}.$$

Cancelling the 3rd and 5th terms and likewise the 2nd and last we are left with

$$2 f_i h_j u^{ij} = 2 {\rm Hess}(u)(\nabla f, \nabla h)$$ as desired.

Note that we treated $f_i, u_i, h_i$ as vectors throughout - this reflects the fact that this a special case of the identity

$$2 (\nabla \theta) (X,Y) = g(X, \nabla (\theta(Y))) + g(Y, \nabla (\theta (X)) - \theta (\nabla (g(X,Y)))$$

with $\theta = du, X = \nabla f, Y = \nabla g$.

• Thank you very much for your answer. I am not really comfortable with the Einstein summation convention and with indices. Please bear with me: for example $\nabla^i$ denotes the $i$-th coordiante of the gradient right? but then what happened to the gradients, for example in the term $u^jh_j$? wouldn't it have to be $\nabla^j u\nabla ^j h^j$ instead? also I don't see the coefficients of the metric anywhere which i find weird. – Sak Oct 6 '16 at 3:09
• @Sak: as I mentioned in the first paragraph, I'm being super lazy and just writing $u^j$ instead of $\nabla^j u$. I'm writing the metric implicitly by raising/lowering indices: for example $u^j h_j$ is really $g_{ij} u^i h^j = g(\nabla u, \nabla h)$. Since all the derivatives are covariant, you can pretty much pretend the metric isn't there. One way to interpret this calculation would be to fix geodesic normal coordinates and let all indices be lower, at which point it is exactly the same calculation as in Euclidean space. – Anthony Carapetis Oct 6 '16 at 3:30