Concave function of Weingarten operator and convex cone of matrices

Question

Picture below is from 143 page of Huisken, Gerhard; Sinestrari, Carlo, Mean curvature flow with surgeries of two-convex hypersurfaces, Invent. Math. 175, No. 1, 137-221 (2009). ZBL1170.53042.

$A=\{h_{ij}\}$ is the second fundamental form . Weingarten operator is $W=\{h^i_j\}$. Denote by $\lambda_1\le\cdot\cdot\cdot\le\lambda_n$ the principal curvature.

First, I understand Weingarten operator as a matrix or a linear map from tangent space to tangent space. So, how the Weingarten operator applied to two tangent vectors?

Second, why $\lambda_1 +\lambda _2$ is concave function of the Weingarten operator ? I just know what is concave function, but don't know what is concave function of a operator.

Third, what is the convex cone of matrices ? I just know convex cone is a subset of a vector space that is closed under linear combination.

Answer 1

1. How do you turn a linear map in to a bilinear form? By lowering the index with the metric, as always; i.e. $W(v_1,v_2) = \langle W(v_1), v_2 \rangle.$ Under this identification, $W$ is essentially the same thing as the second fundamental form (maybe up to a sign).
2. The given formula shows that $\lambda_1 + \lambda_2$ is a minimum of linear functions $W \mapsto \sum_i W(e_i,e_i)$. Linear functions are concave, and minimums of concave functions are concave. Here concavity has its usual meaning, since the set of matrices of a given size has a natural vector space structure.
3. Again, the set of matrices is a vector space, so the usual notion of convex cone makes sense (and this is what is meant).