Lemma 12.22.10A017BFF9468

Proof. First let $q:M'\to M$ be one blowup along a smooth center, with exceptional divisor $E$, and let $h_ M$ be a Hermitian form on $M$. The restriction of $\mathcal O(-E)$ to each projective fiber of $E$ is $\mathcal O(1)$. A Hermitian metric on the normal bundle of the center therefore induces a metric on $\mathcal O(-E)|_ E$ whose curvature is positive on the tangent spaces of the fibers. Extend this metric to $\mathcal O(-E)$ on $M'$ by extending the smooth logarithm of its ratio to a fixed reference metric from the closed submanifold $E$, and let $\beta $ be its curvature. The semipositive form $q^*h_ M$ is positive off $E$, and at points of $E$ its kernel is exactly the tangent space of the fiber, on which $\beta $ is positive. Hence $Cq^*h_ M+\beta >0$ for every sufficiently large $C$: otherwise, unit tangent vectors violating the inequality for $C\to \infty $ would converge, by compactness, to a vector in the kernel of $q^*h_ M$ on which $\beta $ is nonpositive. With $\eta _ E=-\beta $, the curvature of the induced metric on $\mathcal O(E)$, this reads $Cq^*h_ M-\eta _ E>0$.

For a composition of blowups, apply this step successively, with $h_ M$ the positive form already obtained on the previous blowup. Each step multiplies the previous divisor by a positive constant, pulls it back and adds the new exceptional divisor, so the divisor stays effective and exceptional. This proves (12.22.2). Each exceptional divisor maps into the image of its center, which has dimension at most $n-2$. $\square$