Existence and non-existence of constant scalar curvature and extremal Sasaki metrics

Abstract

We discuss the existence and non-existence of constant scalar curvature, as well as extremal, Sasaki metrics. We prove that the natural Sasaki–Boothby–Wang manifold over the admissible projective bundles over local products of non-negative CSC Kähler metrics, as described in [3], always has a constant scalar curvature (CSC) Sasaki metric in its Sasaki-Reeb cone. Moreover, we give examples that show that the extremal Sasaki–Reeb cone, defined as the set of Sasaki–Reeb vector fields admitting a compatible extremal Sasaki metric, is not necessarily connected in the Sasaki–Reeb cone, and it can be empty even in the non-Gorenstein case. We also show by example that a non-empty extremal Sasaki–Reeb cone need not contain a (CSC) Sasaki metric which answers a question posed in [16]. The paper also contains an appendix where we explore the existence of Kähler metrics of constant weighted scalar curvature, as defined in [43], on admissible manifolds over local products of non-negative CSC Kähler metrics.

Appendix A: Constant weighted scalar curvature on admissible Kähler manifolds

In this appendix we will show how the technique we used in the proof of Theorem 3.1 can be adapted to an existence result for Kähler metrics of constant weighted scalar curvature. The notion of constant weighted scalar curvature was introduced more generally by A. Lahdili in [43], see also [39].

Suppose that $\Omega$ is any admissible Kähler class on an admissible manifold $N^{ad}$ which fibers over a local product⁸ of nonnegative CSCK metrics. We will show that when the value of the weight is sufficiently large, there exists an admissible Kähler metric in $\Omega$ with constant weighted scalar curvature. Specifically we prove the following theorem:

Theorem A.1

Suppose $\Omega$ is any admissible Kähler class on the admissible manifold $N^{ad}={\mathbb {P}}(E_0 \oplus E_{\infty }) \longrightarrow N$, where N is a compact Kähler manifold which is a local product of nonnegative CSCK metrics. If $p > \max \{m+1,2d_0+2,2d_\infty +2\}$, then there always exist some $c\in (-1,1)$ such that the corresponding admissible $(c\,{\mathfrak {z}}+1,p)$-extremal metric in $\Omega$ has constant $(c\, {\mathfrak {z}}+1,p)$-scalar curvature.

Note that Lemmas A.2 and A.3 below supply additional technical details for the proof of Theorem  3.1. In Sect. 1 we will touch on the limitations of Theorem A.1, by exploring an example that falls outside of the realm of Theorem A.1.

Proof

Assume that we have an admissible manifold $N^{ad}$ as above with the property that N is a local product of non-negative CSCK metrics and let $\Omega$ be an admissible Kähler class on $N^{ad}$. As we discussed in the beginning of Sect. 3, we know that $\Omega$ admits an admissible $(c\,{\mathfrak {z}}+1,p)$-extremal metric for any choice of $c\in (-1,1)$ and $p\in {\mathbb {R}}$. Recall that the admissible $(c\,{\mathfrak {z}}+1,p)$-extremal metric has $(c\,{\mathfrak {z}}+1,p)$-scalar curvature $Scal_{c\,{\mathfrak {z}}+1,p}=A_1{\mathfrak {z}}+A_2$, where, $A_1$ and $A_2$ are given (8) and (9). Thus admissible $(c\,{\mathfrak {z}}+1,p)$-extremal metric has, g has constant weighted scalar curvature, $Scal_{c\,{\mathfrak {z}}+1,p}$, if and only if $A_1=0$. In turn the equation $A_1=0$ may be re-written as

$$\begin{aligned} \alpha _{1,-(1+p)}\beta _{0,(1-p)}-\alpha _{0,-(1+p)}\beta _{1,(1-p)}=0. \end{aligned}$$

(18)

As already mentioned in the proof of Theorem 3.1 we know that

• $c\, \alpha _{r+1,k}+\alpha _{r,k}=\alpha _{r,k+1}$

• $c\, \beta _{r+1,k}+\beta _{r,k}=\beta _{r,k+1}$.

Further, it is easy to check the general formulas below:

• $\frac{d}{dc}\left[ \alpha _{r,k}\right] =k\alpha _{r+1,k-1}$

• $\frac{d}{dc}\left[ \beta _{r,k}\right] =k \beta _{r+1,k-1}$,

These observations are used in the calculations below.

We now turn our attention to the weighted Einstein–Hilbert functional as defined by Futaki and Ono in [32]: In the present case, (the appropriate restriction of) this weighted Einstein–Hilbert functional is given by

$$\begin{aligned} WH_K(c):=\frac{S_c}{V_c^{\frac{p-2}{p}}}, \end{aligned}$$

where $V_c= \int _{N^{ad}}(c\,{\mathfrak {z}}+1)^{-p}\,d\mu _g$ and $S_c= \int _{N^{ad}} Scal_{c\,{\mathfrak {z}}+1,p}(g) (c\,{\mathfrak {z}}+1)^{-p}\,d\mu _g$. Using the formulas from [7], it is not hard to check that, up to an overall positive rescale, $WH_K(c)= \frac{\beta _{0,2-p}}{\left( \alpha _{0,-p}\right) ^{\frac{p-2}{p}}}$. To simplify things a bit, we will actually work with $H_K(c):=[WH_K(c)]^p$, that is,

$$\begin{aligned} H_K(c)= \frac{\left( \beta _{0,2-p}\right) ^{p}}{\left( \alpha _{0,-p}\right) ^{p-2}}. \end{aligned}$$

Note that while $H_S(c)$ from the proof of Theorem  3.1 is analogous to $H_K(c)$, it is NOT a special case of $H_K(c)$ and for $p=m+2$, the two functions are not equal. We easily verify that

$$\begin{aligned} \begin{array}{ccl} H_K'(c) &{} = &{} \frac{p(2-p)\left( \alpha _{0,-p}\right) ^{p-2}\left( \beta _{0,2-p}\right) ^{(p-1)}\beta _{1,1-p} - (p-2)(-p)(\alpha _{0,-p})^{(p-3)}\alpha _{1,-(1+p)}\left( \beta _{0,2-p} \right) ^{p}}{\left( \alpha _{0,-p}\right) ^{2(p-2)}}\\ &{}= &{}\frac{p(p-2)(\beta _{0,2-p})^{p-1} (\alpha _{0,-p})^{p-3}}{\left( \alpha _{0,-p}\right) ^{2(p-2)}}\left[ \alpha _{1,-(1+p)}\beta _{0,2-p}-\alpha _{0,-p}\beta _{1,1-p} \right] \\ &{}= &{}\frac{p(p-2)(\beta _{0,2-p})^{p-1} (\alpha _{0,-p})^{p-3}}{\left( \alpha _{0,-p}\right) ^{2(p-2)}} \left[ \alpha _{1,-(1+p)}\beta _{0,(1-p)}-\alpha _{0,-(1+p)}\beta _{1,(1-p)}\right] . \end{array} \end{aligned}$$

Now $\alpha _{0,-p}>0$ and $\beta _{0,2-p}> 0$ (the latter is due to the fact that we are assuming N is a local product of nonnegative CSCK metrics) and therefore critical points of $H_K(c)$ correspond exactly to solutions of Eq. (18). Showing that for $p > \max \{m+1,2d_0+2,2d_\infty +2\}$, $H_K(c)$ must have a least one critical point $c\in (-1,1)$, would finish the proof.

To that end, first notice that $H_K(c)$ is a smooth positive function for $c\in (-1,1)$. To show that H(c) has a critical point inside $(-1,1)$, we will show that $\displaystyle \lim _{c\rightarrow \pm 1^{\mp }}H_K(c) = +\infty$. We need two lemmas. Let o(x) denote any function which is smooth in a neighborhood of $x=0$ and satisfies that $o(0)=0$. We shall recycle this notation for several functions of this nature.

Lemma A.2

For any $k \ge m+1$,

$$\begin{aligned} \alpha _{0,-k}=\displaystyle \frac{\delta _0(c) + o(1-c)}{(1-c)^{k-1-d_0}}=\frac{\delta _\infty (c) + o(1+c)}{(1+c)^{k-1-d_\infty }}, \end{aligned}$$

where $\delta _0$ and $\delta _\infty$ are smooth functions defined in open neighborhoods of $c=1$ and $c=-1$ respectively, such that $\delta _0(1)>0$ and $\delta _\infty (-1)>0$.

Proof

By repeated integration by parts, we can see that $\alpha _{0,-k}= \int _{-1}^1 (c\,t +1)^{-k} \overbrace{p_c(t)}^{\text {degree}\,m-1}dt$ is a rational function of c which is positive and has no vertical asymptotes within the interval $(-1,1)$.

First notice that

$$\begin{aligned} \alpha _{0,-k} = \int _{-1}^1 (c\,t +1)^{-k} (1+t)^{d_0} p_0(t)dt = \int _{-1}^1 (c\,t +1)^{-k} (1-t)^{d_\infty } p_{\infty }(t)dt, \end{aligned}$$

where $p_0(t) = p_c(t)/(1+t)^{d_0}$ and $p_\infty (t) = p_c(t)/(1-t)^{d_\infty }$ are polynomials of degree $m-1-d_0$ and $m-1-d_\infty$, respectively. Keep in mind that the complex dimension $m=\sum _{a\in {\hat{\mathcal {A}}}}d_a+1$ (with $d_a>0$ for $a\in {\mathcal {A}}$ and $d_a\ge 0$ for $a\in \hat{\mathcal {A}}$) and note that $p_0(-1)>0$ and $p_{\infty }(1)>0$. From the definition of $p_c(t)$ we also know that $p_0(t)$ and $p_\infty (t)$ are positive for $-1<t<1$. To understand the behavior of $\alpha _{0,-k}$ near $c=\pm 1$, we consider repeated integration by parts and observe that

$$\begin{aligned} \alpha _{0,-k}=\displaystyle \frac{\left( \frac{(k-2-d_0)! \, d_0! \, p_0(-1)}{(k-1)! \, c^{d_0+1}} + o(1-c)\right) }{(1-c)^{k-1-d_0}}=\frac{\left( \frac{(-1)^{d_\infty +1}(k-2-d_\infty )! \, d_\infty ! \, p_\infty (1)}{(k-1)! \, c^{d_\infty +1}} + o(1+c)\right) }{(1+c)^{k-1-d_\infty }} \end{aligned}$$

This proves the lemma. $\square$

Lemma A.3

For any $l \ge m$,

$$\begin{aligned} \beta _{0,-l}=\displaystyle \frac{\gamma _0(c) + o(1-c)}{(1-c)^{l-d_0}}=\frac{\gamma _\infty (c) + o(1+c)}{(1+c)^{l-d_\infty }}, \end{aligned}$$

where $\gamma _0$ and $\gamma _\infty$ are smooth functions defined in open neighborhoods of $c=1$ and $c=-1$ respectively, such that $\gamma _0(1)>0$ and $\gamma _\infty (-1)>0$.

Proof

By repeated integration by parts, we can see that

$$\begin{aligned} \begin{array}{ccl} \beta _{0,-l} &{}= &{} \int _{-1}^1\overbrace{\Big (\sum _{a\in {\hat{\mathcal {A}}}} \frac{x_ad_as_a}{1+x_at}\Big ) p_c(t)}^{\text {degree}\, m-2} (c\,t +1)^{-l} dt \\ &{} + &{} \big ((1-c)^{-l}p_c(-1) + (1+c)^{-l} p_c(1)\big ) \end{array} \end{aligned}$$

is a rational function of c which is non-negative and has no vertical asymptotes within the interval $(-1,1)$. Moreover, it is easy to see that if $d_0=0$,

$$\begin{aligned} \beta _{0,-l} = \frac{\left( p_c(-1)+o(1-c)\right) }{(1-c)^l}, \end{aligned}$$

with $p_c(-1)>0$ in this case, and if $d_\infty =0$,

$$\begin{aligned} \beta _{0,-l} = \frac{\left( p_c(1)+o(1+c)\right) }{(1-c)^l} \end{aligned}$$

with $p_c(1)>0$ in this case.

Now assume that $d_0 \ge 1$ and define

$$\begin{aligned} q_0(t):= \sum _{a\in {\hat{\mathcal {A}}}\setminus \{0\} } \frac{x_ad_as_a}{1+x_at}\frac{p_c(t)}{(1+t)^{d_0-1}} + x_0 d_0 s_0 \frac{p_c(t)}{(1+t)^{d_0}}. \end{aligned}$$

The degree of the polynomial $q_0(t)$ is $m-d_0-1$ and

$$\begin{aligned} \displaystyle q_0(-1) = x_0 d_0 s_0 \prod _{a\in {\hat{\mathcal {A}}}\setminus \{0\} } (1-x_a)^{d_a} =d_0 (d_0+1) \prod _{a\in {\hat{\mathcal {A}}}\setminus \{0\} } (1-x_a)^{d_a} >0. \end{aligned}$$

By repeated integration by parts we then have

$$\begin{aligned} \begin{array}{ccl} \beta _{0,-l} &{}= &{} \int _{-1}^1 (c\,t +1)^{-l} (1+t)^{d_0-1} q_0(t) dt + (1+c)^{-l} p_c(1) \\ &{} =&{} \displaystyle \frac{\left( \frac{(l-2-(d_0-1))! \, (d_0-1)! \, q_0(-1)}{(l-1)! \, c^{d_0}} + o(1-c)\right) }{(1-c)^{l-d_0}}. \end{array} \end{aligned}$$

Similarly if $d_\infty \ge 1$ we define

$$\begin{aligned} q_\infty (t):= \sum _{a\in {\hat{\mathcal {A}}}\setminus \{\infty \} } \frac{x_ad_as_a}{1+x_at}\frac{p_c(t)}{(1-t)^{d_\infty -1}} + x_\infty d_\infty s_\infty \frac{p_c(t)}{(1-t)^{d_\infty }}. \end{aligned}$$

The degree of the polynomial $q_\infty (t)$ is $m-d_\infty -1$ and

$$\begin{aligned} \displaystyle q_\infty (1) = x_\infty d_\infty s_\infty \prod _{a\in {\hat{\mathcal {A}}}\setminus \{\infty \} } (1+x_a)^{d_a} =d_\infty (d_\infty +1) \prod _{a\in {\hat{\mathcal {A}}}\setminus \{\infty \} } (1+x_a)^{d_a} >0. \end{aligned}$$

By repeated integration by parts we now have

$$\begin{aligned} \begin{array}{ccl} \beta _{0,-l} &{}= &{} \int _{-1}^1 (c\,t +1)^{-l} (1-t)^{d_\infty -1} q_\infty (t) dt + (1-c)^{-l} p_c(-1) \\ &{} =&{} \displaystyle \frac{\left( \frac{(-1)^{d_\infty }(l-2-(d_\infty -1))! \, (d_\infty -1)! \, q_\infty (1)}{(l-1)! \, c^{d_\infty }} + o(c+1)\right) }{(c+1)^{l-d_\infty }}. \end{array} \end{aligned}$$

From the above observations we have now proved Lemma A.3 regardless of the values of $d_0$ and $d_\infty$. $\square$

From Lemmas A.2 and A.3 we now have that for $k \ge m+1$ and $l\ge m$,

$$\begin{aligned} \frac{\left( \beta _{0,-l}\right) ^{k}}{\left( \alpha _{0,-k}\right) ^{l}}= \frac{\left( \gamma _0(c) + o(1-c)\right) ^k}{\left( \delta _0(c)+o(1-c)\right) ^l}\left( 1-c\right) ^{d_0(k-l)-l} = \frac{\left( \gamma _\infty (c) + o(1+c)\right) ^k}{\left( \delta _\infty (c)+o(1+c)\right) ^l}\left( 1+c\right) ^{d_\infty (k-l)-l}. \end{aligned}$$

In particular, for $p\ge m+2$, we see that (letting $l=p-2$ and $k=p$),

$$\begin{aligned} H_K(c)= & {} \frac{\left( \beta _{0,2-p}\right) ^{p}}{\left( \alpha _{0,-p}\right) ^{p-2}}=\frac{\left( \gamma _0(c) + o(1-c)\right) ^p}{\left( \delta _0(c)+o(1-c)\right) ^{p-2}}\left( 1-c\right) ^{2d_0 +2-p}\\= & {} \frac{\left( \gamma _\infty (c) + o(1+c)\right) ^p}{\left( \delta _\infty (c)+o(1+c)\right) ^{p-2}}\left( 1+c\right) ^{2d_\infty +2-p}. \end{aligned}$$

Therefore, for $p > \max \{m+1,2d_0+2,2d_\infty +2\}$, $\displaystyle \lim _{c\rightarrow \pm 1^{\mp }}H_K(c) = +\infty$. This, together with the fact that $H_K(c)$ is smooth and bounded from below over the interval $(-1,1)$, allows us to conclude that $H_K(c)$ has at least one critical point $c\in (-1,1)$. For this value of c, we have $H_K'(c)=0$ and therefore c solves (18). $\square$

Recall that $\displaystyle m=\sum \nolimits _{a\in {\hat{\mathcal {A}}}}d_a+1$. Assuming (as we should) that ${\mathcal {A}}\ne \emptyset$, we have that for $p=2m$ the condition $p > \max \{m+1,2d_0+2,2d_\infty +2\}$ is automatic. As a special case, we thus get the following corollary to Theorem  A.1, which in particular confirms Conjecture 1 in [7]. Note that the corollary is also a special case of an existence theorem stated in [35].

Corollary A.4

Suppose $\Omega$ is any admissible Kähler class on the admissible manifold $N^{ad}={\mathbb {P}}(E_0 \oplus E_{\infty }) \longrightarrow N$, where N is a compact Kähler manifold which is a local product⁹ of nonnegative CSCK metrics. Then there always exist some $c\in (-1,1)$ such that the corresponding admissible $(c\,{\mathfrak {z}}+1,2m)$-extremal metric g in $\Omega$ has the property that $h=(c\,{\mathfrak {z}}+1)^{-2}g$ has constant scalar curvature and therefore is conformally Kähler Einstein–Maxwell.

1.1 Limitation of Theorem A.1

On a more pessimistic note, if we instead work with $p=m+2$, we cannot assume that $p > \max \{m+1,2d_0+2,2d_\infty +2\}$ is true and thus we cannot appeal to Theorem A.1. The following example illustrates that, in this case, existence is more unpredictable—even if the underlying admissible manifold is fixed.

Example A.1

Let ${\mathcal {A}}=\{1\}$, $N=N_1={\mathbb {C}}{\mathbb {P}}^1$, and $s_1=2$ (so $g_1$ is the Fubini-Study metric on ${\mathbb {C}}{\mathbb {P}}^1$). If we then let $d_\infty =1$ and $d_0=3$, we have $m=6$ and $p=8$. Clearly $p=2d_0+2$ and hence $p > \max \{m+1,2d_0+2,2d_\infty +2\}$ is false. Setting $x_1=x$ we have that each value $0<x<1$ determines a certain admissible Kähler class. We will see how the existence of a solution $c\in (-1,1)$ to (18) depends on the value of x.

We calculate that

$$\begin{aligned} \begin{array}{ccl} f_x(c)&{}:=&{} \alpha _{1,-9}\beta _{0,-7}-\alpha _{0,-9}\beta _{1,-7}\\ &{}=&{} -12 x + (24 + x + 13 x^2)c+ (-17 - 52 x + x^2)c^2+(-1 + 15 x + 28 x^2)c^3. \end{array} \end{aligned}$$

Thus solutions $c\in (-1,1)$ to (18) correspond to roots in $(-1,1)$ of the cubic $f_x(c)$. This cubic (with real coefficients) has discriminant $D_x$ given by

$$\begin{aligned} D_x=-5 (1 + x)^3 (-44352 + 155904 x - 159125 x^2 + 115249 x^3 - 90591 x^4 + 49179 x^5). \end{aligned}$$

We can check numerically that there exists a specific value $\tilde{x} \in (0,1)$ such that $D_{\tilde{x}}=0$ and for $0<x<\tilde{x}$, $D_x>0$, while for $\tilde{x}<x<1$, $D_x<0$. Note that $\tilde{x} \approx 0.429$.

For $\tilde{x}<x<1$, we know that the cubic $f_x(c)$ has exactly one real root $\hat{c}$. One can check that for this range of x values, $f_x(1)<1$ and $\displaystyle \lim _{c\rightarrow +\infty }f_x(c)=+\infty$, so it is clear that $\hat{c}>1$ and thus for $\tilde{x}<x<1$, $f_x(c)$ has no roots in $(-1,1)$. Therefore, the admissible Kähler classes corresponding to values $\tilde{x}<x<1$ admit no admissible Kähler metrics with constant $(c\,{\mathfrak {z}}+1,8)$-scalar curvature. The question remains whether there could be any Kähler metric in those Kähler classes with constant (f, 8)-scalar curvature, where f is some positive killing potential.

For $0<x< \tilde{x}$, we know that the cubic $f_x(c)$ has three distinct real roots. We will see that at least one of them will be inside $(-1,1)$.

• In the case where $0<x< 1/7$, this is completely straightforward since $f_x(-1)=-40 (1 + x)^2<0$ and $f_x(1)=6 (1 - x) (1 - 7 x)>0$.

• In the case where $x=1/7$, we have $f_{1/7}(c) = \frac{4}{49} (c-1) \left( 21 c^2-278 c+21\right)$ and it is easy to check that $21 c^2-278 c+21$ has a root in $(-1,1)$.

• In the case where $1/7<x<\tilde{x}$, we have $f_x(-1)<0$ and $f_x(1)<0$, but we also have that $f_x'(-1) =5 (1 + x) (11 + 19 x)>0$, $f_x'(1)=-13 - 58 x + 99 x^2 <0$, and $\displaystyle \lim _{c\rightarrow +\infty }f_x(c)=+\infty$. Considering the options for the cubic with three distinct real roots, it is clear that two of the roots must be inside $(-1,1)$ and that the third root is in the interval $(1,+\infty )$.

Finally, for $x=\tilde{x}$, we know that the cubic $f_x(c)$ has a multiple root (and all the roots are real). In this case we still have that $f_{\tilde{x}}(-1)<0$, $f_{\tilde{x}}'(-1)>0$, $f_{\tilde{x}}(1)<0$, $f_{\tilde{x}}'(1)<0$, and $\displaystyle \lim _{c\rightarrow +\infty }f_x(c)=+\infty$. It is therefore clear that in this case $f_{\tilde{x}}(c)$ has a double real root in the interval $(-1,1)$ and a single real root in the interval $(1,+\infty )$. In conclusion, for each admissible Kähler class, $\Omega _x$, corresponding to a value $0<x \le \tilde{x}$, there exists at least one value $c \in (-1,1)$ such that $\Omega _x$ admits an admissible Kähler metric with constant $(c\,{\mathfrak {z}}+1,8)$-scalar curvature.