PomeLemma 11.27.1.00DBBF153DBA At a point $x$, the metric $h$ is RC-positive if and only if $A_ x$ is invertible.
11.27 Ricci-flat Kähler surfacesFFC070E9C8C2
FFC070E9C8C2
The tangent bundle of a Ricci-flat Kähler surface gives an example for the conformal construction of section 11.26. Ricci-flatness makes the mean curvature of the tangent metric zero and its determinant metric flat, but it does not make every directional curvature zero. In real dimension four, the remaining curvature acts on the three-dimensional space of anti-self-dual two-forms, and its invertibility gives exactly the positive directions needed for the conformal construction.
In this section, $X$ is a compact complex surface, so $n=2$, and $\omega $ is a Ricci-flat Kähler form on $X$. Write $h$ for the Hermitian metric on $T^{1,0}X$ defined by $\omega $, and $g$ for the associated Riemannian metric, so that $\omega (S,T)=g(JS,T)$.
Use $g$ to identify bivectors and two-forms. For the complex orientation, the Hodge star gives
\[ \Lambda ^2=\Lambda ^+\oplus \Lambda ^-,\qquad \Lambda ^\pm =\{ \zeta :*\zeta =\pm \zeta \} . \]
Each summand has real rank three. The first is spanned by $\omega $ and the real parts of $(2,0)$-forms, and the second consists of the real primitive $(1,1)$-forms; both descriptions follow by writing the star in a unitary real frame.
Let $R^\nabla $ be the curvature of the Levi–Civita connection, with $R^\nabla (S,T)=\nabla _ S\nabla _ T-\nabla _ T\nabla _ S-\nabla _{[S,T]}$, and define the self-adjoint curvature operator $\mathscr R$ on $\Lambda ^2$ by
\[ \langle \mathscr R(S\wedge T),U\wedge V\rangle =g(R^\nabla (S,T)V,U). \]
Since the curvature of a Kähler metric has type $(1,1)$, the operator $\mathscr R$ annihilates the real parts of $(2,0)$-forms. Its value on $\omega $ is the Ricci form, which vanishes. Thus
\[ \mathscr R=0\oplus A,\qquad A:\Lambda ^-\longrightarrow \Lambda ^-. \]
The endomorphism $A$ is symmetric and trace free, since the trace of $\mathscr R$ is half the scalar curvature; in this Ricci-flat setting it is the anti-self-dual Weyl operator. Norms of curvature tensors below are Hilbert–Schmidt norms of these endomorphisms.
Proof. For a real tangent vector $V$, set $v=\tfrac 12(V-\mathrm iJV)$. The Chern and Levi–Civita connections agree on a Kähler manifold. Their curvature pairings give
\[ \alpha _{h,v}=\tfrac 12\bigl (\mathscr R(V\wedge JV)\bigr )^\flat . \]
Indeed, $h(v,w)=\tfrac 12(g(V,W)-\mathrm i g(JV,W))$ for $w=\tfrac 12(W-\mathrm iJW)$, and $R^\nabla $ commutes with $J$. Since $R^\nabla $ is skew-adjoint, the real part of $h(F_ hv,v)$ vanishes, which gives the formula.
The anti-self-dual projection of $V\wedge JV$ is
\[ \Pi (V)=V\wedge JV-\tfrac 12|V|^2\omega . \]
It is nonzero for $V\ne 0$, and every nonzero element of $\Lambda ^-$ is of this form. To see this, diagonalize a nonzero real primitive $(1,1)$-form in a unitary real frame. It has the form $c(e_1\wedge Je_1-e_2\wedge Je_2)$, with $c>0$ after interchanging the two complex lines. Taking $V=\sqrt{2c}\, e_1$ gives exactly this form as $\Pi (V)$. Also $|\Pi (V)|^2=|V|^4/2$.
Consequently, since $\mathscr R$ vanishes on $\Lambda ^+$, we have $\alpha _{h,v}\ne 0$ for every nonzero $v$ if and only if $\ker A_ x=0$. Since $K_{\omega ,h}=0$, lemma 11.26.2 gives the assertion. $\square$
Lemma 11.27.2.594B40F9CD8C If $g$ is not flat, the open set $\{ x:\det A_ x\ne 0\} $ is nonempty.
Proof. Suppose that $\det A=0$ everywhere. On the open set where $A\ne 0$, its three eigenvalues are $\lambda _0,-\lambda _0,0$ with $\lambda _0>0$; they are distinct, so $A$ has smooth local eigenframes there.
Step 1: a Bochner identity. On an open set where $A$ has distinct eigenvalues, we compute $\Delta |A|^2$, keeping track of the normalizations.
Choose an oriented local eigenframe $\xi _1,\xi _2,\xi _3$ of $\Lambda ^-$ with $|\xi _ i|^2=2$, and denote the corresponding eigenvalues by $\lambda ,\mu ,\nu $. As skew-adjoint endomorphisms, these forms satisfy the multiplication rules of the quaternions. Write
\[ \nabla \xi _1=c\otimes \xi _2-b\otimes \xi _3,\quad \nabla \xi _2=-c\otimes \xi _1+a\otimes \xi _3,\quad \nabla \xi _3=b\otimes \xi _1-a\otimes \xi _2. \]
Here $a,b,c$ are real one-forms. The Ricci identity and the contracted Bianchi identity, using $\operatorname {Ric}(g)=0$, give
\[ \mathrm da+b\wedge c=\lambda \xi _1,\qquad \mathrm d\lambda =(\lambda -\mu )\xi _3c+(\lambda -\nu )\xi _2b, \]
and their simultaneous cyclic permutations. In the second formula the two-forms act as skew-adjoint endomorphisms on one-forms. Put $D=\mathrm d^*\mathrm d=-\operatorname {div}_ g\nabla $ on functions. Applying $\mathrm d^*$ to the second formula yields
\[ D\lambda =2\lambda ^2+4\mu \nu +2(\nu -\lambda )|b|^2+2(\mu -\lambda )|c|^2. \]
For example, the calculation uses
\begin{align*} \mathrm d^*(\xi _3c)& =-\langle b,\xi _1c\rangle +\langle a,\xi _2c\rangle +\langle a,\xi _3b\rangle -2\nu ,\\ \mathrm d^*(\xi _2b)& =-\langle b,\xi _1c\rangle -\langle a,\xi _2c\rangle -\langle a,\xi _3b\rangle -2\mu , \end{align*}
which follow from the connection and curvature formulas above; substitution and the cyclic formulas for $\mathrm d\mu ,\mathrm d\nu $ cancel the mixed pairings.
Now $\lambda +\mu +\nu =0$ and $|A|^2=\lambda ^2+\mu ^2+\nu ^2$. Multiplying the three scalar equations by their eigenvalues and adding gives
\[ \lambda D\lambda +\mu D\mu +\nu D\nu =18\lambda \mu \nu -2\bigl ((\mu -\nu )^2|a|^2 +(\nu -\lambda )^2|b|^2+(\lambda -\mu )^2|c|^2\bigr ). \]
The connection expression also gives
\[ |\nabla A|^2=|\mathrm d\lambda |^2+|\mathrm d\mu |^2+|\mathrm d\nu |^2 +2\bigl ((\mu -\nu )^2|a|^2+(\nu -\lambda )^2|b|^2 +(\lambda -\mu )^2|c|^2\bigr ). \]
The product rule therefore gives, with $\Delta =\operatorname {div}_ g\nabla =-D$,
D65B8F2D9646
11.27.3
\begin{equation} \Delta |A|^2=2|\nabla A|^2-36\det A=2|\nabla A|^2. \end{equation}
This identity also holds on the interior of $\{ A=0\} $. Both sides are smooth functions, and the two open sets have dense union in $X$, so the identity holds everywhere. Integration over $X$ gives $\nabla A=0$.
Step 2: a parallel operator $A$ forces flatness. If $g$ is not flat, then $A\ne 0$ somewhere, and since $A$ is parallel and $\det A=0$, it is nonzero everywhere and has three distinct constant eigenvalues. Its eigenlines are parallel; local unit sections of them are parallel as well. The induced connection on $\Lambda ^-$ therefore has zero curvature. Under $\mathfrak {so}(4)=\Lambda ^+\oplus \Lambda ^-$, the two summands commute, while the action of $\Lambda ^-$ on itself is faithful. These facts follow directly from the bases $e_1\wedge e_2\pm e_3\wedge e_4$, $e_1\wedge e_3\mp e_2\wedge e_4$, $e_1\wedge e_4\pm e_2\wedge e_3$. Thus zero induced curvature forces $\mathscr R(S\wedge T)$ to lie in $\Lambda ^+$ for every $S,T$. It already lies in $\Lambda ^-$, so $\mathscr R=0$, a contradiction. $\square$
Theorem 11.27.4.8C2D7F0F7808 If $X$ has a nonflat Ricci-flat Kähler metric, then $T^{1,0}X$ admits an RC-positive Hermitian metric whose associated $(1,1)$-form is nonclosed. It admits neither a uniformly RC-positive Hermitian metric nor a metric with positive definite mean curvature relative to any Hermitian form.
Proof. By lemma 11.27.2, lemma 11.27.1, the original tangent metric $h$ is RC-positive on a nonempty open set. Its mean curvature is zero, so theorem 11.26.3 gives a metric $h_ t=\mathrm e^{-tf}h$ that is RC-positive everywhere.
In that construction we may take the Gauduchon form to be $\omega $ itself, and the closed coordinate ball to be a proper subset of $X$. Since $P_\omega f=1$ off that ball, $f$ is not constant. The associated form is $\omega _ t=\mathrm e^{-tf}\omega $, with
\[ \mathrm d\omega _ t=-t\mathrm e^{-tf}\mathrm df\wedge \omega . \]
On a real four-dimensional symplectic vector space, $\zeta \mapsto \zeta \wedge \omega $ is injective on one-forms, as a unitary real frame shows. Thus $\mathrm d\omega _ t\ne 0$ somewhere.
Finally, the induced determinant metric of $h$ is flat because its Chern curvature is the Ricci form of $\omega $, which vanishes. The two obstructions follow from proposition 11.26.7. $\square$
The construction does not change the complex surface: it starts from a Ricci-flat Kähler form and produces a nonclosed Hermitian form whose tangent metric is RC-positive. Nonflat Ricci-flat Kähler metrics exist, for example, on K3 surfaces, by Yau's solution of the Calabi conjecture.
Combined with theorem 11.18.2, the RC-positive tangent metric of theorem 11.27.4 bounds the convexity of the complement of the zero section in the projective completion of the tangent bundle. This bundle admits no metric with positive definite mean curvature, so corollary 11.18.4 does not apply to it.
Corollary 11.27.5.9A2FC61929F1 For a connected compact complex surface $X$ admitting a nonflat Ricci-flat Kähler metric, the complement
\[ \mathbb P(\mathcal O_ X\oplus T^{1,0}X)\setminus \mathbb P(\mathcal O_ X) \]
is $3$-convex. The tangent bundle admits no Hermitian metric with positive definite mean curvature relative to any Hermitian form.
Proof. By theorem 11.27.4, the tangent bundle has an RC-positive Hermitian metric. Its directional curvature form has a positive eigenvalue for every nonzero fiber vector. Apply theorem 11.18.2 with $n=r=2$ and $s=1$. The mean-curvature obstruction is the second assertion of theorem 11.27.4. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.