Reading a proof of the Schwarz lemma for the Kähler-Ricci flow from p22 of [these lecture notes](https://arxiv.org/pdf/1212.3653.pdf). 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?



Schwarz lemma and bisectional curvature lower bound
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 $\hat g$.


