Why image of curvature is a Lie subalgebra?

Question

In the red line of picture below, why it is Lie algebra ? $M_{\alpha\beta}$ is the Lie bracket ? But $M_{\alpha\beta}$ is symmetric .

Picture below is from the 216 page of this paper.

$\ \ \ \ $ $\text{T}\scriptstyle{\text{HEOREM}}\ $ $2.2.1\ $ (Hamilton's strong maximum principle). Let $M_{\alpha\beta}$ be a smooth solution of the equation $(2.2.1)$. Suppose $M_{\alpha\beta}\geq0$ on $\Sigma\times[0,T]$. Then there exists a positive constant $0<\delta\leq T$ such that on $\Sigma\times(0,\delta)$, the rank of $M_{\alpha\beta}$ is constant, and the null space of $M_{\alpha\beta}$ is invariant under parallel translation and invariant in time and also lies in the null space of $N_{\alpha\beta}$.

Answer 1

Remember that the curvature operator is a symmetric bilinear form on the Lie algebra $$\Lambda^2T_pM\simeq \mathfrak{so}(n). $$

The structure coefficients $C^{\alpha \beta}_\gamma$ are just the components of the Lie bracket in our basis $v^\alpha$ for this algebra. By the choice of diagonalizing basis, the image of $M$ is just $\underset{\alpha>k}{\operatorname{span}} v^\alpha$, and thus the condition for this to be closed under the Lie bracket is $C^{\alpha \beta}_\gamma = 0$ whenever $\alpha,\beta>k$ but $\gamma\le k$.

Answer 2

For $\gamma,\ \xi >k$ and $\alpha \leq k$, assume that $$ C_\alpha^{\gamma\xi } := (v^\alpha,[v^\gamma,v^\xi ]) =0 $$ This implies that $[v^\gamma,v^\xi ]$, Lie bracket of elements in $V$ where $V$ is an orthogonal complement of the kernel, is in $V$. That is $V$ is closed under Lie bracket