# The relation between $\nabla\nabla f$ and $H_f$

I've been reading Lounesto's text on Clifford Algebras and Spinors, and I came across the following in section 1.11:

Using the multiplication table of ${C\mathcal{l}}_2$ we notice that the Clifford product \begin{align}\left(a_1e_1+a_2e_2\right)\left(b_1e_1+b_2e_2\right)&=a_1b_1+a_2b_2+\left(a_1b_2-a_2b_1\right)e_{12}\end{align} of two vectors $a=a_1e_1+a_2e_2$ and $b=b_1e_1+b_2e_2$ and a bivector $a\wedge b=(a_1b_2-a_2b_1)e_{12}$.

This part makes sense. But now, extrapolating a bit, is it possible that the Clifford product

\begin{align}\nabla\nabla&=\begin{bmatrix}\partial_1\\\partial_2\end{bmatrix}\begin{bmatrix}\partial_1\\\partial_2\end{bmatrix}\\[1em]&=\color{green}{\partial_{11}^2e_{11}+\partial^2_{12}e_{12}+\partial^2_{21}e_{21}+\partial^2_{22}e_{22}}\\[1em] &=\partial^2_{11}+\partial^{2}_{22}+e_{12}\left(\partial^{2}_{12}-\partial^{2}_{21}\right),\\[1em] &\text{but also }\\ \partial^2_{11}+\partial^{2}_{22}+e_{12}\left(\partial^{2}_{12}-\partial^{2}_{21}\right)&=\partial^2_{11}+\partial^{2}_{22}+\begin{vmatrix}\partial_{1}&\partial_{2}\\[0.5em] \partial_{1}&\partial_{2}\end{vmatrix}e_{12},\end{align}

has anything to do with the Hessian? Just looking it over the green step reminds me of it; for instance, if we organized each entry (although this probably doesn't make any sense) as \begin{align}\begin{bmatrix}e_{11}&e_{12}\\[0.5em]e_{21}&e_{22}\end{bmatrix},\end{align} then we might organize the second-partials in front as \begin{align}\begin{bmatrix}\partial_{11}&\partial_{12}e_{12}\\[0.5em] \partial_{21}e_{21}&\partial_{22}\end{bmatrix}\overset{?}{=}H.\end{align} I'm afraid this might be totally confusing though; I couldn't help but ask.

## 1 Answer

If you know the Hessian you can compute $\nabla^2$, but if you know only $\nabla^2$, you can't compute the Hessian, since you only know the trace and the antisymmetric parts of the Hessian. The traceless symmetric part is information that $\nabla^2$ does not have.