Schwarz lemma and bisectional curvature lower bound Reading a proof of the Schwarz lemma for the Kähler-Ricci flow from p22 of these lecture notes. I am confused as to what they mean by taking $$\inf _{x \in M} \{\hat{R}_{i \bar i j \bar j}(x) \mid \{\partial_{z^{1}}, \ldots, \partial_{z^{n}} \} \text{ is orthornormal w.r.t. } \hat{g} \text{ at } x, \ i, j = 1, \ldots, n\} \ .$$ Is this lower bound even invariantly defined on the manifold?
 A: Their idea is correct, but its formulation (and the chosen notation) is indeed a bit sloppy. What they seem to do, in reality, is to define
$$ - \hat C = \inf _{x \in M} \inf \{ \hat R (e_i, \bar {e_i}, e_j, \bar {e_j}) \mid \{e_1, \bar {e_1}, \ldots, e_n, \bar {e_n} \} \text{ is orthornormal w.r.t. } \hat{g} \text{ at } x, \ i, j = 1, \ldots, n\} \ .$$
The inner infimum exists, because $i,j$ take a finite number of values, and the orthonormal frames $\{ e_1, \bar {e_1}, \dots, e_n, \bar {e_n} \}$ at $x$ form a space homeomorphic to $O(2n)$ which is compact.
They tacitly accept that this inner infimum will be a continuous function of $x$ (up to the reader to prove this!), therefore the outer infimum will be finite ($M$ is assumed compact).
Finally, in equation (2.22) they just work at some $x$, taking $\{ e_1, \bar {e_1}, \dots, e_n, \bar {e_n} \}$ to be precisely the frame $\{ \partial_1, \bar \partial_1, \dots, \partial_n, \bar \partial_n \}$ corresponding to the normal coordinates at $x$ with respect to $\hat g$.
