Examples of Complete Kähler Metrics with Nonnegative Holomorphic Sectional Curvature

Abstract

In this paper, we consider complete noncompact Kähler metrics with nonnegative holomorphic sectional curvature. In particular, we prove some results on existence, non-existence, and function theory on such manifolds.

Appendices

Appendix A: SO(3)-Invariant Kähler Metrics on the Affine Quadric $M_t$

1.1 Appendix A.1: The Adapted Complex Structure

Unless specified we use column vectors.

We assume $t>0$, we may write $X=T\,S^2$ by

$$\begin{aligned} X=\{(x, \xi ) \in {\mathbb {R}}^3 \times {\mathbb {R}}^3 \mid \ |x|=1, x \cdot \xi =0\} \end{aligned}$$

(42)

and $M_t$ by

$$\begin{aligned} M_t=\{(u, v) \in {\mathbb {R}}^3 \times {\mathbb {R}}^3 \mid \ |u|^2-|v|^2=t, u \cdot v=0\} \end{aligned}$$

(43)

and consider a diffeomorphism $\Phi : X \rightarrow M_t$

$$\begin{aligned} (u, v)=\Phi (x, \xi )=\left( \sqrt{t}\cosh (|\xi |)x, \sqrt{t}\frac{\sinh (|\xi |)}{|\xi |}\xi \right) . \end{aligned}$$

(44)

For simplicity, we set $a(|\xi |)=\frac{\sinh (|\xi |)}{|\xi |}$. Let $J_0$ be the complex structure on $M_t$, and J be the pull back of $J_0$ onto X, we may solve

$$\begin{aligned} \mathrm{{d}}\Phi \left( \frac{\partial }{\partial x_i}, \frac{\partial }{\partial \xi _i}\right) =\left( \frac{\partial }{\partial u_j}, \frac{\partial }{\partial v_j}\right) \begin{pmatrix} \sqrt{t} \cosh |\xi | \delta _i^j &{} \sqrt{t} a \, x_j \xi _i \\ 0 &{} \sqrt{t} a \left( \delta _i^j+\frac{a' \xi _j \xi _i}{a\, |\xi |}\right) \end{pmatrix}. \\ J_0 \left( \frac{\partial }{\partial u_i}, \frac{\partial }{\partial v_i}\right) =\left( \frac{\partial }{\partial u_j}, \frac{\partial }{\partial v_j}\right) \begin{pmatrix} 0 &{}\quad -I \\ I &{}\quad 0 \end{pmatrix}. \end{aligned}$$

Since $J=(d\Phi )^{-1} \cdot J_0 \cdot d\Phi$, we may solve

$$\begin{aligned} J \left( \frac{\partial }{\partial x_i}, \frac{\partial }{\partial \xi _i}\right) =\left( \frac{\partial }{\partial x_j}, \frac{\partial }{\partial \xi _j}\right) \begin{pmatrix} \frac{1}{a\sqrt{t}} (\delta _j^i-\frac{a' \xi _i\xi _j}{a|\xi |+a'|\xi |^2}) &{} -\frac{\tanh (|\xi |) }{|\xi |} (\delta _j^i+\frac{a' \xi _j \xi _i}{a|\xi |}) \\ \coth (|\xi |) |\xi | (\delta _j^i-\frac{a' \xi _i\xi _j}{a|\xi |+a'|\xi |^2}) &{} x_j \xi _i \end{pmatrix}. \end{aligned}$$

(45)

We remark that this complex structure J on X is called the adapted complex structure on the complexification of $S^2$, which is much studied by Lempert and Szőke [33], Szőke [44], Patrizio and Wong [38], and Guillemin and Stenzel [21] in a more general context.

The group $G=SO(3, {\mathbb {R}})$ acts on X by

$$\begin{aligned} g \cdot (x, \xi )=(gx, g\xi ). \end{aligned}$$

Following Stenzel ( [43]), we consider $X \setminus X_0$ where $X_0$ denote the zero section of the tangent bundle, then we may identify ${\mathbb {R}}^{+} \times G$ with $X \setminus X_0$ by

$$\begin{aligned} (r, g)=\left( g \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, r \, g \cdot \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\right) \end{aligned}$$

Our goal is to construct a natural global $(1, 0)-$frame on $X \setminus X_0$. For any point $(x, \xi ) \in X \setminus X_0$ where $x=(x_1, x_2, x_3)^{T}$ and $\xi =(\xi _1, \xi _2, \xi _3)^{T}$, we set $r=|\xi |$, then there is a suitable $g \in G$ so that

$$\begin{aligned} g\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}=x, \ \ g\begin{pmatrix} 0 \\ r \\ 0 \end{pmatrix}=\xi . \end{aligned}$$

In fact, we may simply pick $g=(e_w, e_{\xi }, e_x)$ where

$$\begin{aligned} e_{\xi }=\frac{\xi }{|\xi |}=\frac{\xi }{r}, \ \ e_x=x, \ \ e_w=e_{\xi } \times e_x. \end{aligned}$$

Now we take a basis of the Lie algebra of G as

$$\begin{aligned} W_1=\begin{pmatrix} 0 &{}\quad 1 &{}\quad 0\\ -1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \end{pmatrix},\ W_2=\begin{pmatrix} 0 &{}\quad 0 &{}\quad 1\\ 0 &{}\quad 0 &{}\quad 0 \\ -1 &{}\quad 0 &{}\quad 0 \end{pmatrix},\ W_3=\begin{pmatrix} 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad -1 &{}\quad 0 \end{pmatrix}. \end{aligned}$$

Let $X_i$ be the corresponding left-invariant vector fields induced by the action of $\exp (tW_i)$. We may solve

$$\begin{aligned}&X_1=r\left( \frac{\partial }{\partial \xi _1}, \frac{\partial }{\partial \xi _2}, \frac{\partial }{\partial \xi _3}\right) e_w=\begin{vmatrix} \frac{\partial }{\partial \xi _1}&\frac{\partial }{\partial \xi _2}&\frac{\partial }{\partial \xi _3}\\ \xi _1&\xi _2&\xi _3 \\ x_1&x_2&x_3 \end{vmatrix};\end{aligned}$$

(46)

$$\begin{aligned}&X_2=\left( \frac{\partial }{\partial x_1}, \frac{\partial }{\partial x_2}, \frac{\partial }{\partial x_3}\right) e_w=\frac{1}{r}\begin{vmatrix} \frac{\partial }{\partial x_1}&\frac{\partial }{\partial x_2}&\frac{\partial }{\partial x_3}\\ \xi _1&\xi _2&\xi _3 \\ x_1&x_2&x_3 \end{vmatrix};\end{aligned}$$

(47)

$$\begin{aligned}&X_3=\left( \frac{\partial }{\partial x_1}, \frac{\partial }{\partial x_2}, \frac{\partial }{\partial x_3}\right) e_{\xi }-r(\frac{\partial }{\partial \xi _1}, \frac{\partial }{\partial \xi _2}, \frac{\partial }{\partial \xi _3})e_x. \end{aligned}$$

(48)

For example, we may get $X_1$ by solving the derivative of the following expression at $\theta =0$

$$\begin{aligned} (e_w, e_{\xi }, e_x)\begin{pmatrix} \cos \theta &{}\quad \sin \theta &{}\quad 0\\ -\sin \theta &{}\quad \cos \theta &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{pmatrix} \left[ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ r \\ 0 \end{pmatrix}\right] . \end{aligned}$$

We may check

$$\begin{aligned}&[X_1, X_2]=-X_3,\ \ [X_1, X_3]=X_2, \ \ [X_2, X_3]=-X_1;\end{aligned}$$

(49)

$$\begin{aligned}&\left[ \frac{\partial }{\partial r}, X_1\right] =0, \ \ \left[ \frac{\partial }{\partial r}, X_2\right] =0,\ \ \left[ \frac{\partial }{\partial r}, X_3\right] =0. \end{aligned}$$

(50)

Note that

$$\begin{aligned} \frac{\partial }{\partial r}=\left( \frac{\partial }{\partial \xi _1}, \frac{\partial }{\partial \xi _2}, \frac{\partial }{\partial \xi _3}\right) e_{\xi }. \end{aligned}$$

(51)

Use (45) we may solve

$$\begin{aligned} J\left( \frac{\partial }{\partial r}\right) =-X_3,\ \ \ \ J(X_1)=-\tanh (r)X_2. \end{aligned}$$

(52)

1.2 Appendix A.2: The Non-existence of SO(3)-invariant Kähler Metrics of $H \ge 0$

We begin with an observation on the Kähler class on any affine quadric in ${}{\mathbb {C}}^{n}$.

Claim: Any Kähler metric on $M_t=\{z_1^2+z_2^2+\cdots +z_{n+1}^2=t\} \subset {\mathbb {C}}^{n+1}$ where $n \ge 2$ has a global Kähler potential.

Note that $M_t$ is smooth affine, hence Stein, we have $H^{0, 1}_{{\bar{\partial }}}(M_t)=0$. As $M_t$ being diffeomorphic to X which is the tangent bundle of $S^n$, $M_t$ is homotopy equivalent to $S^n$. When $n \ge 3$, $H^2(M_t, {\mathbb {R}})$ vanishes, therefore any Kähler form $\omega$ on $M_t$ can be written as $\sqrt{-1}\partial {\overline{\partial }} P$ for some global real-valued function P. If $n=2$, $H^2(M_t, {\mathbb {R}})={\mathbb {R}}$, we may write its generator $\alpha$ as the pull back of $\pi : M_t \rightarrow S^2$ of the generator of $H^2(S^2, {\mathbb {R}})$. Then any Kähler form on $M_t$ can be written as $\omega =a\alpha +\sqrt{-1}\partial {\overline{\partial }} P$ for some $a \in {\mathbb {R}}$ and some real-valued function P on $M_t$. However, a has to be 0. Otherwise, since any Kähler form admits a local potential, so does $\alpha$. It is impossible since $S^2$ is totally real in $M_t$. Hence we proved the claim.

Now we focus on the affine quadric in ${\mathbb {C}}^3$, every SO(3)-invariant Kähler metric on X can be written as

$$\begin{aligned} \omega =\sqrt{-1} \partial {\bar{\partial }} P \end{aligned}$$

(53)

where P is a smooth function on X which only depends on $r=|\xi |$. We may further write

$$\begin{aligned} \omega =\sqrt{-1}(P^{\prime \prime } \partial r \wedge {\bar{\partial }} r +P' \partial {\bar{\partial }} r). \end{aligned}$$

(54)

From (52), we get

$$\begin{aligned} g\left( \frac{\partial }{\partial r}, \frac{\partial }{\partial r}\right) =\frac{P^{\prime \prime }}{2}, g(X_2, X_2)=\frac{P'}{2\tanh (r)}. \end{aligned}$$

and

$$\begin{aligned} 2g=P^{\prime \prime }\left[ (\mathrm{{d}}r)^2+(\mathrm{{d}}\omega ^3)\right] +P'\tanh (r) (\omega ^1)^2+\frac{P'}{\tanh r} (\omega ^2)^2. \end{aligned}$$

(55)

where $dr, \omega ^1, \omega ^2, \omega ^3$ denote the dual form of $\frac{\partial }{\partial r}, X_1, X_2, X_3$. Indeed it is possible to write down these forms explicitly

$$\begin{aligned}&\mathrm{{d}}r=(\mathrm{{d}}\xi _1, \mathrm{{d}}\xi _2, \mathrm{{d}}\xi _3)e_{\xi };\ \ \omega ^1=\frac{1}{r}(\mathrm{{d}}\xi _1, \mathrm{{d}}\xi _2, \mathrm{{d}}\xi _3)e_w;\\&\omega ^2=(\mathrm{{d}}x_1, \mathrm{{d}}x_2, \mathrm{{d}}x_3)e_w;\ \ \omega ^3=(\mathrm{{d}}x_1, \mathrm{{d}}x_2, \mathrm{{d}}x_3)e_{\xi }; \end{aligned}$$

From (52), we solve

$$\begin{aligned} J \mathrm{{d}}r=\omega ^3, \ J \omega ^1=\frac{1}{\tanh (r)}\omega ^2. \end{aligned}$$

(56)

Then we get the unitary $(1, 0)-$frame $\{e_1, e_2\}$ and coframe $\{f^1, f^2\}$:

$$\begin{aligned}&e_1=\frac{1}{\sqrt{P^{\prime \prime }}} \left( \frac{\partial }{\partial r} -\sqrt{-1} J \frac{\partial }{\partial r}\right) , \,\,\, e_2=\frac{1}{\sqrt{P^{\prime } \tanh (r)}} (X_1 -\sqrt{-1} J X_1); \end{aligned}$$

(57)

$$\begin{aligned}&f^1=\frac{\sqrt{P^{\prime \prime }}}{2}(dr-\sqrt{-1}Jdr),\,\,\,\, f^2=\frac{\sqrt{P^{\prime }\tanh (r)}}{2}(\omega ^1-\sqrt{-1}J\omega ^1). \end{aligned}$$

(58)

Next we solve the structural equation

$$\begin{aligned} \begin{pmatrix} \nabla e_1\\ \nabla e_2 \end{pmatrix}=\begin{pmatrix} \theta _1^1 &{}\quad \theta _1^2 \\ \theta _2^1 &{}\quad \theta _2^2 \end{pmatrix} \begin{pmatrix} e_1\\ e_2 \end{pmatrix}. \end{aligned}$$

(59)

Here the connection form $\theta _i^j$ is skew-Hermitian. If we write

$$\begin{aligned} \theta =\gamma +\rho \end{aligned}$$

(60)

where $\gamma$ is (1, 0) part and $\rho$ is (0, 1) part. Then, we have

$$\begin{aligned} \gamma _i^j=(e_i, [\overline{e_j}, e_k]) f^k, \ \ \overline{\gamma _i^j}+\rho _j^i=0. \end{aligned}$$

(61)

We may check

$$\begin{aligned}{}[\overline{e_1}, e_1]=&-\frac{P^{\prime \prime \prime }}{2(\sqrt{P^{\prime \prime }})^3}(e_1-\overline{e_1}),\\ [\overline{e_1}, e_2]=&\frac{1}{\sqrt{P^{\prime \prime }}} \Big [-\frac{P^{\prime \prime }}{2P'}-\frac{1}{\sinh (2r)}+\frac{1}{\tanh (r)} \Big ]e_2\\&+\frac{1}{\sqrt{P^{\prime \prime }}}\frac{-2}{\sinh (2r)} \overline{e_2}.\\ [\overline{e_2},e_1]=&\frac{1}{\sqrt{P^{\prime \prime }}}\Big [-\frac{1}{\tanh (r)}+ \frac{P^{\prime \prime }}{2P'}+\frac{1}{\sinh (2r)}\Big ]\overline{e_2}\\&+\frac{1}{\sqrt{P^{\prime \prime }}}\frac{2}{\sinh (2r)} e_2.\\ [\overline{e_2}, e_2]=&-\frac{\sqrt{P^{\prime \prime }}}{P'}(e_1-\overline{e_1}), \end{aligned}$$

Using (61), we may conclude

$$\begin{aligned} \theta =\begin{pmatrix} \frac{P^{\prime \prime \prime }}{2(\sqrt{P^{\prime \prime }})^3}(f^1-\overline{f^1}) &{} \frac{\sqrt{P^{\prime \prime }}}{P'}f^2+\frac{2}{\sqrt{P^{\prime \prime }}\sinh (2r)}\overline{f^2}\\ -\frac{2}{\sqrt{P^{\prime \prime }}\sinh (2r)}f^2-\frac{\sqrt{P^{\prime \prime }}}{P'}\overline{f^2} &{} \frac{1}{\sqrt{P^{\prime \prime }}} (-\frac{1}{\tanh (r)}+\frac{P^{\prime \prime }}{2P'}+\frac{1}{\sinh (2r)})(f^1-\overline{f^1}) \end{pmatrix} \end{aligned}$$

(62)

As an immediate consequence, we see $\nabla _{e_1+\overline{e_1}} (e_1+\overline{e_1})=0$, i.e., $\frac{\partial }{\partial r}$ generates a geodesic.

Next we may solve the curvature form by

$$\begin{aligned} \Theta _{i}^j=\mathrm{{d}}\theta _i^j-\theta _i^p \wedge \theta _p^j. \end{aligned}$$

(63)

We solve the curvature form

$$\begin{aligned} \Theta _{1}^1=&\Big (-\frac{P^{(4)}}{(P^{\prime \prime })^2} +\frac{[P^{(3)}]^2}{(P^{\prime \prime })^3}\Big )f^1\wedge \overline{f^1}+ \Big (-\frac{P^{(3)}}{P'P^{\prime \prime }}+\frac{P^{\prime \prime }}{(P')^2}-\frac{4}{P^{\prime \prime } [\sinh (2r)]^2}\Big )f^2\wedge \overline{f^2},\end{aligned}$$

(64)

$$\begin{aligned} \Theta _{1}^2=&\Big (\frac{4}{P'\sinh (2r)}-\frac{2P^{(3)}}{(P^{\prime \prime })^2\sinh (2r)} -\frac{8\cosh (2r)}{P^{\prime \prime }[\sinh (2r)]^2} \Big )f^1\wedge \overline{f^2}\nonumber \\&+\Big (-\frac{P^{(3)}}{P^{\prime \prime }P'}+\frac{P^{\prime \prime }}{(P')^2}-\frac{4}{P^{\prime \prime }[\sinh (2r)]^2}\Big )f^2\wedge \overline{f^1}, \end{aligned}$$

(65)

$$\begin{aligned} \Theta _{2}^1=&-\overline{\Theta _{1}^2},\end{aligned}$$

(66)

$$\begin{aligned} \Theta _{2}^2=&\big (\frac{-2\alpha '(r)}{\sqrt{P^{\prime \prime }}}-\alpha \frac{P^{(3)}}{(P^{\prime \prime })^{\frac{3}{2}}}\Big ) f^1 \wedge \overline{f^1}+\Big (-\frac{2\alpha \sqrt{P^{\prime \prime }}}{P'}-\frac{P^{\prime \prime }}{(P')^2} +\frac{4}{P^{\prime \prime }[\sinh (2r)]^2}\Big )f^2\wedge \overline{f^2}. \end{aligned}$$

(67)

Here $\alpha (r)=\frac{1}{\sqrt{P^{\prime \prime }}} \Big (\frac{P^{\prime \prime }}{2P'}+\frac{1}{\sinh (2r)}-\frac{1}{\tanh (r)}\Big )$. We may further simplify

$$\begin{aligned} \Theta _{2}^2=&\Big ( -\frac{P^{(3)}}{P^{\prime \prime }P'}+\frac{P^{\prime \prime }}{(P')^2}- \frac{4}{P^{\prime \prime } [\sinh (2r)]^2} \Big ) f^1 \wedge \overline{f^1}\nonumber \\&+\Big (-\frac{2P^{\prime \prime }}{(P')^2}+\frac{2\coth (2r)}{P'}+\frac{4}{P^{\prime \prime }[\sinh (2r)]^2}\Big )f^2\wedge \overline{f^2}. \end{aligned}$$

(68)

From now on, we write $R_{1 \bar{1} 1 \bar{1}}=R(e_1, \overline{e_1}, e_1, \overline{e_1})$ and set similarly

$$\begin{aligned} A=R_{1 \bar{1} 1 \bar{1}}, \ B=R_{1 \bar{1} 2 \bar{2}}, \ C=R_{2 \bar{2} 2 \bar{2}}, \ D=R_{1 \bar{2} 1 \bar{2}}. \end{aligned}$$

(69)

These are all the non-zero components of curvature tensor after we ignore those obtained from obvious symmetries. We may solve the bisectional curvature formed by two unit (1, 0) vectors $U=u_i e_i$ and $V=v_i e_i$.

$$\begin{aligned} R(U, {\overline{U}}, V, {\overline{V}})=A|u_1 v_1|^2+C|u_2 v_2|^2+B|u_1v_2+u_2v_1|^2+2D{\text {Re}} (u_1 \overline{u_2} v_1 \overline{v_2}). \end{aligned}$$

(70)

Lemma A.1

(53) defines a complete SO(3)-invariant Kähler metric on X if and only if the function $P: (0, \infty ) \rightarrow {\mathbb {R}}$ satisfies

• (i) $P'(r)>0$ and $P^{\prime \prime }(r)>0$ on $(0, +\infty )$.

• (ii) P extends to a smooth function function on $[0, +\infty )$ and there exists $c_1>0$ such that $P'(r)=c_1 r+O(r^2)$ near $r=0$.

• (iii) $\int _0^{\infty } \sqrt{P^{\prime \prime }(r)} \mathrm{{d}}r=\infty$.

Using the Cohn–Vossen inequality, some analysis of the curvature components A, C enables us to exclude the existence of any complete invariant Kähler metric with nonnegative holomorphic sectional curvature on $M_t$.

Proof of Theorem 1.2

Step 1: We consider the affine quadric $M_t \subset {\mathbb {C}}^3$.

First we discuss the non-existence of complete SO(3)-invariant Kähler metric with nonnegative bisectional curvature. Assume g is such a metric, we consider the intersection of $M_t$ and the complex plane $\{z_1=0\}$ in ${\mathbb {C}}^3$. Let Z be the corresponding complex submanifold in X after the identification $\Phi : X \rightarrow M_t$. It is direct to check that the restriction of tangent vectors $\frac{\partial }{\partial r}$ and $X_3$ on X along Z are indeed tangent vectors on Z, Consider the induced complex structure on Z, we have the same expression of $e_1$ as in (57). From the connection form (62), we see that Z is totally geodesic. Hence, the induced metric on Z has nonnegative holomorphic sectional curvature (Guass curvature), However, Z is biholomorphic to a cylinder. Applying the Cohn–Vossen inequality on complete surfaces ([10, 26]), we get that the integral of the Gauss curvature on Z is bounded by $2\pi \chi (Z)=0$. We conclude Z has zero Gauss curvature, i.e., $A=R(e_1, \overline{e_1}, e_1, \overline{e_1})$ vanishes along Z. Since $A=-\frac{P^{(4)}}{(P^{\prime \prime })^2} +\frac{[P^{(3)}]^2}{(P^{\prime \prime })^3}=0$, we solve $P^{\prime \prime }=c_1 e^{c_2r}$. Combined with Lemma A.1, we may write $P'(r)=c_1 (e^{c_2r}-1)$ with both $c_1, c_2>0$.

From the formula of A, B, C, D in (69), we see they are all homogeneous with respect to $c_1$, so we simply set $c_1=1$. Now we check B, after plugging into $P'(r)=e^{c_2r}-1$.

$$\begin{aligned} B&=-\frac{c_2}{e^{c_2r}-1}+\frac{c_2e^{c_2r}}{(e^{c_2r}-1)^2}-\frac{4}{c_2 e^{c_2r} [\sinh (2r)]^2} \nonumber \\&=\frac{e^{c_2r} \{ [c_2 \sinh (2r)]^2-16 [\sinh (\frac{c_2}{2}r)]^2 \}}{c_2e^{c_2r} [\sinh (2r)]^2 (e^{c_2r}-1)^2} \end{aligned}$$

(71)

Since $B \ge 0$, we get $E(r):=c_2 \sinh (2r)-4\sinh (\frac{c_2}{2}r) \ge 0$. Let r be large enough, we conclude that $c_2 \le 4$. After we solve $E'(r)$, it is easy to see that $c_2 \le 4$ is also sufficient to get $E(r) \ge 0$ for any $r \ge 0$.

Now we turn to the formula of C:

$$\begin{aligned} C&=-\frac{2c_2e^{c_2r}}{(e^{c_2r}-1)^2}+\frac{2\coth (2r)}{(e^{c_2r}-1)} +\frac{4}{c_2e^{c_2r}[\sinh (2r)]^2}\nonumber \\&=\frac{-2[c_2e^{c_2r}]^2[\sinh (2r)]^2+c_2e^{c_2r}(e^{c_2r}-1)\sinh (4r)+4(e^{c_2r}-1)^2}{c_2e^{c_2r}(e^{c_2r}-1)^2 [\sinh (2r)]^2 } \end{aligned}$$

(72)

Let $F(r):=-2[c_2e^{c_2r}]^2[\sinh (2r)]^2+c_2e^{c_2r}(e^{c_2r}-1)\sinh (4r)+4(e^{c_2r}-1)^2$, letting r large enough, the term $(-\frac{c_2^2}{2}+\frac{c_2}{2}) e^{(2c_2+4)r}$ dominates. So if $C \ge 0$ for any $r \ge 0$, we must have $c_2 \le 1$. On the other hand, near $r=0$, $F(r)=(-10c_2^3-\frac{32}{3}c_2)r^3+O(r^4)$. Therefore, F(r) cannot be nonnegative near $r=0$. This contradiction shows that g can not have nonnegative bisectional curvature.

In fact the above argument only makes use of the nonnegativity of A and C, we have proved that $M_t$ does not admit any complete SO(3)-invariant Kähler metric with $H \ge 0$.

Step 2: In general, given the smooth affine quadric $\{z_1^2+z_2^2+\cdots +z_{n+1}^2=t\} \subset {\mathbb {C}}^{n+1}$, we may consider its intersection with the hyperplane $\{z_1=0\}$. It is a totally geodesic (as it is fixed by the isometry subgroup SO(n)) complex submanifold. Moreover, the induced metric on the intersection is also complete as the restriction of $\frac{\partial }{\partial r}$ generates a geodesic tending to infinity. Therefore, by a dimension reduction, we reduce the problem to the case of smooth affine quadric in ${\mathbb {C}}^3$. $\square$

1.3 Appendix A.3: Other SO(3)-Invariant Kähler Metrics on the Affine Quadric

Proposition A.2

(Stenzel [43]) $P(r)=2\sqrt{t}\cosh (r)$ defines the Eguchi–Hanson metric on X.

In [43], the author observed that

$$\begin{aligned} \sinh (r) (\mathrm{{d}}r-\sqrt{-1}J\mathrm{{d}}r) \wedge (\omega ^1-\sqrt{-1}J\omega ^1) \end{aligned}$$

is a $SO(3, {\mathbb {C}})$-invariant holomorphic 2-form on X. Indeed, we may first check $\frac{\mathrm{{d}}z_1 \wedge \mathrm{{d}}z_2}{z_3}$ is a global holomorphic 2-form on $M_t$, then verify

$$\begin{aligned} \frac{\mathrm{{d}}z_1 \wedge \mathrm{{d}}z_2}{z_3}=\sinh (r) (\mathrm{{d}}r-\sqrt{-1}Jdr) \wedge (\omega ^1-\sqrt{-1}J\omega ^1). \end{aligned}$$

If X admits a Ricci-flat Kähler metric g, then such a form must be parallel, hence has constant norm with respect to g. Using this fact, one may solve $P(r)=2\sqrt{t}\cosh (r)$. It is also direct to check this Kähler metric has flat Ricci curvature by the curvature formula (69). In general, the distance $s=\int \sqrt{\frac{P^{\prime \prime }}{2}}\mathrm{{d}}r$, we may check

$$\begin{aligned} A=C=-B \sim -\frac{1}{s^6}, \ \ D \sim \frac{1}{s^6}. \end{aligned}$$

Now we study the asymptotics of this Kähler metric. From (44), we know

$$\begin{aligned} f_i(x, \xi )=\Phi ^{*}(u_i+\sqrt{-1}v_i)=\sqrt{t} \cosh (|\xi |) x_i+\sqrt{-1} \frac{\sinh (|\xi |)}{|\xi |}\xi _i \end{aligned}$$

(73)

are holomorphic functions on X, while the distance $\mathrm{{d}}s=\sqrt{\frac{P^{\prime \prime }}{2}}\mathrm{{d}}r$. This shows $|f_i| \sim s^2$, hence each $f_i$ is of quadratic growth. Next we solve the asymptotic growth of volume of geodesic balls, we recall the identification of ${\mathbb {R}}^{+} \times SO(3)$ with $X \setminus X_0$. Let $d{\text {Vol}}_{SO(3)}$ be the normalized Haar measure on SO(3) so that its volume is $\pi ^2$ which is one half of that of the unit sphere $S^3 \subset {\mathbb {R}}^4$. We may identify $dr \wedge \omega ^3 \wedge \omega ^1 \wedge \omega ^2=8\, \mathrm{{d}}r \wedge d{\text {Vol}}_{SO(3)}$.

$$\begin{aligned} {\text {Vol}}({\{0 \le r \le r_0\} \cap X})\,&=\, \int _{0 \le r \le r_0} \frac{1}{4} P'(r)P^{\prime \prime }(r) \mathrm{{d}}r \wedge \omega ^3 \wedge \omega ^1 \wedge \omega ^2 \nonumber \\&=\, \int _0^{r_0} \frac{1}{4} P' P^{\prime \prime } \mathrm{{d}}r \cdot 8{\text {Vol}}(SO(3)) \nonumber \\&=\, [P'(r_0)]^2 {\text {Vol}}(SO(3)) \nonumber \\&\, \sim t \pi ^2 e^{2r_0}. \end{aligned}$$

(74)

While the distance $s=\sqrt{\frac{P^{\prime \prime }}{2}}\mathrm{{d}}r \sim \sqrt{2} e^{\frac{r}{2}} t^{\frac{1}{4}}$, for any given point $p \in X_0$, we obtain

$$\begin{aligned} \lim _{s \rightarrow +\infty } \frac{{\text {Vol}}(B(p, s))}{\omega _{4} s^4}=\frac{1}{2}, \end{aligned}$$

where $\omega _{4}=\frac{\pi ^2}{2}$ is the volume of unit ball in the Euclidean space ${\mathbb {R}}^4$. This is merely a manifestation that the Eguchi–Hanson metric is asymptotically locally Euclidean and modeled on the flat cone ${\mathbb {C}}^2 /{\mathbb {Z}}_2$.

Lemma A.3

Suppose P satisfies the conditions in Lemma A.1. Then the corresponding complete Kähler metric has non-positive bisectional curvature if $A, B, C \le 0$ and $AC \ge |D|^2$.

Proposition A.4

The complete Kähler metric defined by $P(r)=t(2\sinh ^2(r)+1)$ on X is induced from the Euclidean metric on ${\mathbb {C}}^3$. It has non-positive bisectional curvature with quadratic decay. The asymptotic volume growth satisfies

$$\begin{aligned} \lim _{s \rightarrow +\infty } \frac{{\text {Vol}}(B(p. s))}{\omega _{4} s^4}=2. \end{aligned}$$

We note that the bisectional curvature of Kähler metric in Proposition A.4 has a lot of null directions. Indeed, $A \sim -\frac{1}{s^6}, C \sim -\frac{1}{s^2}, B=0$, and $D=-\sqrt{AC}$, For any two (1, 0) vectors $U=u_ie_i$ and $V=v_ie_i$, we get from (70) that

$$\begin{aligned} R(U, {\overline{U}}, V, {\overline{V}})=-|\sqrt{-A} u_1 v_1 +\sqrt{-C} u_2 v_2|^2. \end{aligned}$$

(75)

In other words, for any given unit (1, 0) vector U, we may find a unique unit vector V such that $R(U, {\overline{U}}, V, {\overline{V}})=0$.

Appendix B: Some Auxiliary Results in the proof of Theorem 1.3

Proof of Lemma 4.4

Step 1 We show that $2D+\sqrt{C_{I}\,C_{II}}>0$ for $U \in [-ar, U_{*}]$.

Since $D<0$, we check $P_1=(C_{I}\,C_{II}-4D^2)(ar)^2(n-r+1+kU)^2$. From (32) and (33)

$$\begin{aligned} P_1(U)= & {} k^2 U^2+2karU+2(n-r+1)ar-k^2 a^2 r^2-k^2(ar-U)^2\\= & {} 2kar(1+k)U+2(n-r+1)ar-2k^2a^2r^2. \end{aligned}$$

It is direct to check that $P(-ar)=\frac{(n-r+1)^2}{2k(k+1)}<0$. Hence, $P_1(U)>0$ on $[-ar, U_{*}]$.

Step 2 We show that $-2B_I \cdot C_{II}+2B_{II} \cdot 2D$ is negative at $U=-ar$. Indeed from (32) and (33), at $U=-ar$, we have

$$\begin{aligned} B_I&=\frac{1}{2ar},\ \ C_{II}=\frac{1}{n-r+1-kar},\ D=-\frac{k}{n-r+1-kar}, \\ B_{II}&=-\frac{k}{n-r+1-kar}. \end{aligned}$$

Plugging into $ar=\frac{n-r+1}{2k(2k+1)}$, we get if $U=-ar$ then $-2B_I \cdot C_{II}+2B_{II} \cdot 2D=-\frac{k}{(n-r+1-kar)^2}<0$.

Step 3 We show that $P_2(U)=(-2B_I \cdot C_{II}+2B_{II} \cdot 2D)(ar)^2(n-r+1+kU)^3$ is decreasing on $[-ar, U_{*}]$. To that end, we compute

$$\begin{aligned} P_2(U)=&-(k^2 U^2+2karU+2(n-r+1)ar-k^2 a^2 r^2)(n-r+1+kU)\\&-k(ar-U)(k^2U+k(n-r+1)2U+k^2a^2r^2)\\ =&\,[k^2(n-r+1)-k^3ar-2k^2ar]U^2+[-4(n-r+1)kar+2k^3a^2r^2]U\\&+[-2(n-r+1)ar+k^2a^2r^2(n-r+1)-k^3a^3r^3]\\ =&\,\frac{4k^3+k^2-2k}{2(2k+1)}(n-r+1)U^2+(-4+\frac{k}{2k+1})\frac{(n-r+1)^2}{2(2k+1)}+\text {constant term}. \end{aligned}$$

Obviously, the coefficient of $U^2$ is positive, while that of U is positive. It implies $P_2(U)$ is decreasing on $[-ar, U_{*}]$. $\square$

Proof of Lemma 4.5

Step 1 We show that $2D+\sqrt{C_{I}\,C_{II}}>0$ for $U \in [U_{*}, +\infty )$.

From Lemma 4.2 and our definition of $\rho$, we see all of $C_I, C_{II}, D$ are continuous on $[-ar, +\infty )$. In particular, we get that $\,2D+\sqrt{C_{I}\,C_{II}}>0$ at $U=U_{*}$ from Lemma 4.4.

Let $P_3(U)=(C_I\, C_{II}-4D^2)(16k)^2 (ar+U)^2(n-r+1+kU)^2$, then we show that $P_3(U)$ is increasing on $[U_{*}, +\infty )$. It follows from (36), (37), and (39) that

$$\begin{aligned} P_3(U)\,=\,&16^2 k^2 \left[ \left( 2-\frac{1}{16k}\right) U+\left( 2ar-\frac{n-r+1}{16k^2}\right) -\frac{1}{16k^2}\ln (n-r+1+kU)\right] \\&\cdot \left[ \left( 2-\frac{1}{16}\right) (n-r+1+kU)-\frac{1}{16}\ln (n-r+1+kU)\right] \\&-[(n-r+1+kU)+\ln (n-r+1+kU)]^2.\\ \frac{\mathrm{{d}}\,P_3(U)}{\mathrm{{d}}\,U}\,=\,&16^2k^2\left[ \left( 2-\frac{1}{16k}\right) -\frac{1}{16k}\frac{1}{n-r+1+kU}\right] \\&\cdot \left[ \left( 2-\frac{1}{16}\right) \left( n-r+1+kU\right) -\frac{1}{16}\ln (n-r+1+kU)\right] \\&+16^2k^2 \left[ \left( 2-\frac{1}{16k}\right) U+\left( 2ar-\frac{n-r+1}{16k^2}\right) -\frac{1}{16k^2}\ln \left( n-r+1+kU\right) \right] \\&\cdot \left[ \left( 2-\frac{1}{16}\right) k-\frac{k}{16\left( n-r+1+kU\right) }\right] \\&-2\left[ k+\frac{k}{n-r+1+kU}\right] \left[ (n-r+1+kU)+\ln (n-r+1+kU)\right] . \end{aligned}$$

Since $C_I>0$ on $[U_{*}, +\infty )$ by (40), we see that the second summand in the right hand side of $\frac{d\,P_3(U)}{d\,U}$ is positive. Making repeated use of (30), we get

$$\begin{aligned} \frac{d\,P_3(U)}{d\,U} \,\ge \,&16^2k^2\left[ \left( 2-\frac{1}{16k}\right) -\frac{1}{16k}\frac{1}{n-r+1+kU}\right] \\&\cdot \left[ \left( 2-\frac{1}{16}\right) (n-r+1+kU)-\frac{1}{16}\ln (n-r+1+kU)\right] \\&-2\left[ k+\frac{k}{n-r+1+kU}\right] [(n-r+1+kU)+\ln (n-r+1+kU)].\\ \,\ge \,&16^2 k^2 \left( 2-\frac{1}{8k}\right) \left( 2-\frac{1}{8}\right) (n-r+1+kU)-2(k+k) \cdot 2(n-r+1+kU)\\ \,\ge \,&\left[ 32 k \left( 2-\frac{1}{8k}\right) \left( 2-\frac{1}{8}\right) -1\right] 8k(n-r+1+kU)>0 \end{aligned}$$

Step 2 Let

$$\begin{aligned} P_4(U) \doteq (-2B_I \cdot C_{II}+2B_{II} \cdot 2D)\cdot 16k^2(ar+U)^2 (n-r+1+kU)^3 \end{aligned}$$

and first we will show that $P_4(U_{*})<0$.

By (35), (37), (38), and (39) we get

$$\begin{aligned} P_4(U)\,=\,&-\left[ (n-r+1)-kar+\ln (n-r+1+kU)-\frac{k(ar+U)}{n-r+1+kU})\right] \\&\cdot \left[ (2-\frac{1}{16})(n-r+1+kU)-\frac{1}{16}\ln (n-r+1+kU)\right] (n-r+1+kU)\\&-\frac{k}{16} [(n-r+1+kU)+\ln (n-r+1+kU)][\ln (n-r+1+kU)-1](ar+U). \end{aligned}$$

It follows from (30) that

$$\begin{aligned} P_4(U_{*})&< -\left[ (n-r+1)\frac{4k+1}{4k+2}-\frac{1}{4k+1}\right] \\&\quad \cdot \left[ \left( 2-\frac{1}{8}\right) (n-r+1)\frac{4k+1}{4k+2}\right] (n-r+1)\frac{4k+1}{4k+2}\\&\quad +\frac{k}{16} \left[ (n-r+1+kU_{*})+\ln (n-r+1+kU_{*})\right] \left[ 1-\ln (n-r+1+kU_{*})\right] \\&\quad \times (ar+U_{*}). \end{aligned}$$

If $\ln (n-r+1+kU_{*})>1$, then obviously $P_4(U_{*})<0$ holds. Otherwise, we use $U_{*}<0$ to get

$$\begin{aligned} P_4(U_{*})<&-\left[ (n-r+1)\frac{4k+1}{4k+2}-\frac{1}{4k+1}\right] \cdot \left[ \left( 2-\frac{1}{8}\right) (n-r+1)\frac{4k+1}{4k+2}\right] (n-r+1)\frac{4k+1}{4k+2}\\&+\frac{k}{16} [2(n-r+1)]\frac{(n-r+1)}{2k(2k+1)}.\\<&-\frac{(n-r+1)^2}{4k+2} \left[ \left[ (n-r+1)(4k+1)-1-\frac{1}{4k+1}\right] \left( 2-\frac{1}{8}\right) \left( \frac{4k+1}{4k+2}\right) ^2-\frac{1}{8}\right] \\<&-\frac{(n-r+1)^2}{4k+2} \Big [ (n-r+1)4k \left( 2-\frac{1}{8}\right) \frac{25}{36} -\frac{1}{8} \Big ]<0. \end{aligned}$$

Step 3 We show that $P_4(U)$ is decreasing on $[U_{*}, +\infty )$. We have

$$\begin{aligned} \frac{d P_4(U)}{dU}\,=\,&I_1+I_2+I_3+I_4+I_5. \\ I_1\,=\,&-[(n-r+1)\frac{4k+1}{4k+2}k+k\ln (n-r+1+kU)]\\&\cdot \left[ (2-\frac{1}{16})(n-r+1+kU)-\frac{1}{16}\ln (n-r+1+kU)\right]<0\\ I_2\,=\,&-2 B_I (n-r+1+kU)\left[ (2-\frac{1}{16})k-\frac{k}{16(n-r+1+kU)}\right]<0\\ I_3\,=\,&-\frac{k}{16}\left[ k+\frac{k}{n-r+1+kU}\right] \left[ \ln (n-r+1+kU)-1\right] (ar+U)\\ I_4\,=\,&-\frac{k}{16}\left[ (n-r+1+kU)+\ln (n-r+1+kU)\right] \frac{k(ar+U)}{n-r+1+kU}<0\\ I_5\,=\,&-\frac{k}{16}\left[ (n-r+1+kU)+\ln (n-r+1+kU)\right] \left[ \ln (n-r+1+kU)-1\right] . \end{aligned}$$

If $\ln (n-r+1+kU_{*})-1>0$, then both $I_3$ and $I_5$ are negative on $[U_{*}, +\infty )$, there is nothing to prove. Otherwise for any $U \in [U_{*}, +\infty )$ so that $\ln (n-r+1+kU)-1<0$, we estimate

$$\begin{aligned} \frac{d P_4(U)}{dU}<\,&I_1+I_3+I_5\\<\,&-\left[ (n-r+1)\frac{4k+1}{4k+2}+\ln (n-r+1+kU)\right] k\left( 2-\frac{1}{8}\right) (n-r+1+kU)\\&+\frac{k}{16}\left[ k+\frac{k}{n-r+1+kU}\right] \left[ 1-\ln (n-r+1+kU)\right] (ar+U)\\&+\frac{k}{16}[(n-r+1+kU)+\ln (n-r+1+kU)][1-\ln (n-r+1+kU)]\\<\,&-(n-r+1)\frac{4k+1}{4k+2}k\left( 2-\frac{1}{8}\right) (n-r+1+kU)+\frac{k}{16}\cdot 2k(ar+U)\\&\,+ \frac{k}{16}\cdot 2(n-r+1+kU)\\<\,&k(n-r+1+kU)\left[ -\frac{15}{8}(n-r+1)\frac{4k+1}{4k+2}+\frac{k(ar+U)}{8(n-r+1+kU)}+\frac{1}{8}\right] \\<\,&k(n-r+1+kU) \left[ -\frac{15}{8}(n-r+1) \cdot \frac{5}{6}+\frac{1}{8}+\frac{1}{8}\right] <0. \end{aligned}$$

This completes the proof of the lemma. $\square$