Algebraic Value of A Covariant Derivative I am having trouble with the Notation for a question, and require some help. The question reads:

If $(u_1, u_2)$ are coords on a surface $S$, show that:
  $$\left<\left[\frac{D}{\partial u_1}, \frac{D}{\partial u_2}\right](V), V\right>=0$$

What does the first term in the Inner Product mean? I know that:
$$\left[\frac{DV}{\partial u_1}\right]$$
Stands for the Algebraic Value of the Covariant Derivative. But what does the above mean? The Algebraic Value of Both the Covariant Derivatives combined?
 A: $\newcommand{\dd}{\partial}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$The notation refers to the commutator of second covariant derivatives:
$$
\Brak{\left[\frac{D}{\dd u_{1}}, \frac{D}{\dd u_{2}}\right](V), V}
= \Brak{\left[\frac{D}{\dd u_{1}}\left(\frac{D}{\dd u_{2}}\right) - \frac{D}{\dd u_{2}}\left(\frac{D}{\dd u_{1}}\right)\right](V), V}.
$$
(The vanishing of this inner product is a fundamental symmetry of the Riemann curvature tensor.)
A: I'm guessing $ \langle u,w \rangle = u \cdot w$ and so you need to show that,
$$ \left(\begin{pmatrix} \nabla\omega_1 \cdot u_1 \\ \nabla\omega_2 \cdot u_1 \\ \nabla\omega_3 \cdot u_1 \end{pmatrix}  \begin{pmatrix} \nabla\omega_1 \cdot u_2 \\ \nabla\omega_2 \cdot u_2 \\ \nabla\omega_3 \cdot u_2\end{pmatrix} - \begin{pmatrix} \nabla\omega_1 \cdot u_2 \\ \nabla\omega_2 \cdot u_2 \\ \nabla\omega_3 \cdot u_2\end{pmatrix} \begin{pmatrix} \nabla\omega_1 \cdot u_1 \\ \nabla\omega_2 \cdot u_1 \\ \nabla\omega_3 \cdot u_1 \end{pmatrix}\right) \cdot V = 0  $$  
where $V = \sum_j \left(V \cdot U_j\right) \ U_j$
