Pome391E9AF09F50391E9AF09F50
On a product of $k$ manifolds, the external tensor product of bundles with positive mean curvature has, at every nonzero fiber vector, a positive curvature direction tangent to each factor. Indeed, the curvature of the tensor product metric has no components mixing different factors, so this count applies to every fiber vector, including vectors that are not decomposable tensors. Theorem 11.18.2 then improves the convexity bound by $k$.
Let $k\geq 2$. For $1\leq i\leq k$, let $X_ i$ be a connected compact complex manifold of dimension $n_ i\geq 1$, let $E_ i\to X_ i$ be a holomorphic vector bundle of rank $r_ i\geq 1$, and choose a smooth Hermitian metric $h_ i$ on $E_ i$ and a Hermitian form $\omega _ i$ on $X_ i$. With $p_ i:X=\prod _ iX_ i\to X_ i$ the projections, put
For an endomorphism $A_ i$ of $E_ i$, write $\widehat A_ i$ for its action on the $i$th tensor factor, with the identity on every other factor. We use the Chern curvature, the forms $\alpha _{h,v}$ and the mean curvatures of 11.18.
Lemma 11.20.1.B46ABA0FEC10 The curvature of $h$ is
For every $v\in E_ x$, the coefficient form of $\alpha _{h,v}$ is block diagonal with respect to the decomposition $T_ x^{1,0}X=\bigoplus _ iT_{x_ i}^{1,0}X_ i$, and
Proof. In product holomorphic frames, the matrix of the metric is $M=\bigotimes _ iM_ i$, and its connection matrix is
Each summand depends only on the coordinates of $X_ i$. Applying $\bar\partial $ gives the curvature formula and shows that all components mixing different factors vanish. Pairing with $v$ gives the block diagonal form. Contracting the $i$th block with $\omega _ i$ commutes with the identity actions on the other tensor factors, which gives the trace identity. $\square$
Theorem 11.20.2.0B75DFE662A7 If every $K_ i$ is positive definite, then
is $(n+r-k)$-convex.
Proof. At each point, choose an $h_ i$-orthonormal eigenbasis of $K_ i$. The resulting product basis is $h$-orthonormal, and $\widehat K_ i$ has the positive eigenvalues of $K_ i$, repeated according to the dimensions of the other tensor factors. Hence $h(\widehat K_ i v,v)>0$ for every nonzero $v$, whether or not $v$ is decomposable.
By lemma 11.20.1, each diagonal block of $\alpha _{h,v}$ has positive trace and therefore a positive eigenvalue, so the block diagonal form has at least $k$ positive eigenvalues. Since $k\leq n$, theorem 11.18.2 with $s=k$ gives the assertion. $\square$
Corollary 11.20.3.F8F43BAF4F9A Suppose that each $\omega _ i$ is Gauduchon, that $E_ i$ is $\omega _ i$-slope-stable as in definition 11.18.3, and that $\deg _{\omega _ i}E_ i>0$. Then the complement $\mathbb P(\mathcal O_ X\oplus E)\setminus \mathbb P(\mathcal O_ X)$ is $(n+r-k)$-convex. For $k$ stable ample Hirzebruch examples as in corollary 11.18.5, with their polarizations, this gives $(2^ k+k)$-convexity.
Proof. As in the proof of corollary 11.18.4, choose a Hermitian–Einstein metric on each $E_ i$. Its mean curvature is $\lambda _ i\operatorname {Id}_{E_ i}$, with $\lambda _ i>0$ by the positivity of the degree and the integrated trace identity. Apply theorem 11.20.2. For the Hirzebruch factors, $n_ i=r_ i=2$, so $n+r-k=2k+2^ k-k=2^ k+k$. $\square$
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