On a Riemannian manifold $(M,g)$ let $u = u(t,x)$ the solution to the heat equation $\partial_t u = \frac 12 \Delta u$. The Laplace-Beltrami operator etc. are taken with respect to the metric $g$.
I'd like to prove that the following equality holds:
$$\left( \frac 12 \Delta - \partial_t\right) \frac{\vert \nabla u \vert^2}{u} = \frac 1u \left\vert \ \mathrm{Hess}\,u - \frac{\nabla u \otimes \nabla u}{u}\right\vert^2 + \frac{\mathrm{Ric}(\nabla u, \nabla u)}{u}$$
My idea was to use a direct calculation (product rule for the Laplacian) and Bochner's formula
$$\frac 12 \Delta \vert \nabla u \vert^2 = \vert \ \mathrm{Hess}\,u \ \vert^2 + \mathrm{Ric}(\nabla u, \nabla u).$$
Finally some terms will vanish by assumption that $(\Delta - \partial_t)u = 0$. My problem is that I don't understand the notation:
$\nabla u \otimes \nabla u \overset?=\sum_i \partial_{x_i} u \cdot \sum_j \partial_{x_j} u$
Thanks.
$\nabla u$ is a $(1,0)$ tensor $\nabla u= u^i\frac{\partial}{\partial x^i}$, where $u^i = g^{ij} \frac{\partial u}{\partial x^j}$ , so $\nabla u \otimes \nabla u$, just like $\text{Hess } u$, is a $(2,0)$ tensor given by
$$(\nabla u\otimes \nabla u)^{ij} = u^i u^j = g^{ik}g^{jl} \frac{\partial u}{\partial x^k}\frac{\partial u}{\partial x^l}$$