On some ${\mathcal {AC}}^{\perp }$ manifolds

Abstract

We give a description of all Gray AC manifolds (M, g) whose Ricci tensor has two eigenvalues of multiplicity 1 and $\dim M-1$.

1 Introduction

In this paper, we study complete Riemannian manifolds whose Ricci tensor $\rho$ satisfies the relation

$$\begin{aligned} \nabla _X\rho (X,X) =\frac{2X \text {Scal}}{n+2} g(X,X) \end{aligned}$$

(1.1)

where ${\text {Scal}}= \text {tr}_g \rho$ is the scalar curvature of (M, g) and $n=\dim M$. This property was studied by A. Gray [4] and by A. Besse [1]. A. Gray called Riemannian manifolds satisfying (1.1) ${\mathcal {AC}}^{\perp }$ manifolds; we shall also call them Gray manifolds. We describe all simply connected and real analytic Gray ${\mathcal {AC}}^{\perp }$ manifolds whose Ricci tensor has two eigenvalues $\lambda ,\mu$ of multiplicity $\dim M-1$ and 1 .

If ([g], D) is an Einstein–Weyl structure on a compact manifold M, and $g_0$ is the Gauduchon metric in the conformal class [g], then $(M,g_0)$ is an $\mathcal {AC}^{\perp }$ manifold (see [6]) with the above property. If $\omega _0$ is the Lee form of $g_0$, i.e., $Dg_0=\omega _0\otimes g_0$, then the Weyl structure determined by the pair $(g_0,-\omega _0)$ is also Einstein–Weyl (see [6]). Hence if ([g], D) is an Einstein–Weyl structure on a compact manifold M, and $g_0$ is the Gauduchon metric of ([g], D), then on M there is another Einstein–Weyl structure $([g],D^-)$ such that $D^-g_0=-\omega _0\otimes g_0$. Thus, $(M,g_0)$ is then a Riemannian manifold which admits a pair of Einstein–Weyl structures determined by the pairs $(g_0,\omega _0)$ and $(g_0,-\omega _0)$. More generally, a Riemannian manifold (M, g) which admits two Einstein–Weyl structures determined by pairs $(g,\omega )$ and $(g,-\omega )$ is an $\mathcal {AC}^{\perp }$ manifold whose Ricci tensor has two eigenvalues $\lambda ,\mu$ of multiplicity $\dim M-1$ and 1. We prove that Gray manifolds with two eigenvalues $\lambda ,\mu$ of multiplicity $\dim M-1$ and 1 always admit a conformal vector field $\xi$ which is an eigenfield of the Ricci tensor of (M, g). The field $\xi$ is a Killing vector field or a conformal closed gradient vector field.

In particular, we give a description of all complete, real analytic, simply connected Riemannian manifolds (M, g) admitting a pair of Einstein–Weyl structures $(g,\omega ),(g,-\omega )$. We show that on such manifolds $\omega (X)=g(\xi ,X)$ where $\xi$ is either Killing vector field or a closed gradient conformal field. In the second case, we give a complete classification of such manifolds. We prove that in this case the two Einstein–Weyl structures are conformally Einstein.

For general $\mathcal {AC}^{\perp }$ manifolds, we prove that the eigenvalues $\lambda ,\mu$ of the Ricci tensor satisfy the equations: $n\mu -2(n-1)\lambda =C_0=\mathrm{const}$ in the case of a non-Killing conformal vector field and $(n-4)\lambda +2\mu =C_0=\mathrm{const}$ in the case of a Killing vector field.

We also give new examples of compact $\mathcal {AC}^{\perp }$ manifolds diffeomorphic to the sphere $S^{n},n>2$.

2 ${\mathcal {AC}}^{\perp }$ manifolds and Einstein–Weyl manifolds

By an ${\mathcal {AC}}^{\perp }$ manifold (see [1, 4]), we mean a Riemannian manifold (M, g) satisfying the condition

$$\begin{aligned} \mathfrak {C}_{X Y Z}\nabla _X\rho (Y,Z)=\frac{2}{\dim M+2}\mathfrak {C}_{X Y Z}X \text {Scal} g(Y,Z), \end{aligned}$$

(2.1)

where $\rho$ is the Ricci tensor of (M, g), $\text {Scal}$ is the scalar curvature and $\mathfrak {C}$ means the cyclic sum.

In [6], it is proved that a Riemannian manifold (M, g) is an ${\mathcal {AC}}^{\perp }$ manifold if and only if the Ricci endomorphism Ric of (M, g) is of the form ${Ric}=S+\frac{2}{n+2} \text {Scal} Id$ where S is a Killing tensor and $n=\,$dimM. Let us recall that a (1, 1) tensor S on a Riemannian manifold (M, g) is called a Killing tensor if $g(\nabla S(X,X),X)=0$ for all $X\in \mathrm{TM}$.

Define the integer-valued function $E_S(x)=($the number of distinct eigenvalues of $S_x$) and set $M_S=\{ x\in M:E_S$ is constant in a neighborhood of $x\}$. The set $M_S$ is open and dense in M, and the eigenvalues $\lambda _i$ of S are distinct and smooth in each component U of $M_S$. Let us denote by $D_{\lambda _i}$ the eigendistribution corresponding to $\lambda _i$. We have (see [6]).

Proposition

Let S be a Killing tensor on M and U be a component of $M_S$ and $\lambda _1,\ldots ,\lambda _k \in C^{\infty }(U)$ be eigenfunctions of S. Then for all $X\in D_{\lambda _i}$ we have

$$\begin{aligned} \nabla S(X,X)=-\frac{1}{2} \nabla \lambda _i\Vert X \Vert ^2 \end{aligned}$$

(2.2)

and $D_{\lambda _i}\subset \ker d\lambda _i$. If $i\ne j$ and $X\in \Gamma (D_{\lambda _i}),Y \in \Gamma (D_{\lambda _j})$ then

$$\begin{aligned} g(\nabla _X X ,Y)=\frac{1}{2}\frac{Y\lambda _i}{\lambda _j-\lambda _i}\Vert X \Vert ^2. \end{aligned}$$

(2.3)

If $T(X,Y)=g(SX,Y)$ is a Killing tensor on (M, g) and c is a geodesic on M, then the function $\phi (t)=T(\dot{c}(t),\dot{c}(t))$ is constant on the domain of c. In fact, $\phi '(t)=\nabla _{\dot{c}(t)}T((\dot{c}(t),\dot{c}(t))=0$.

A conformal vector field $\xi$ on a Riemannian manifold satisfies the relation $L_{\xi }g=\alpha g$. This is equivalent to $\nabla _X\omega (Y)+\nabla _Y\omega (X)=\alpha g(X,Y)$ where $\omega =g(\xi ,.)$.

Finally, let us recall that a Riemannian manifold with Killing Ricci tensor is called after A. Gray an $\mathcal A$ manifold ( it is an ${\mathcal {AC}}^{\perp }$ manifold with constant scalar curvature).

We start with some basic facts concerning Einstein–Weyl geometry. For more details, see [10, 12], [11].

Let M be an n-dimensional manifold with a conformal structure [g] and a torsion-free affine connection D. This defines an Einstein–Weyl (E–W) structure if D preserves the conformal structure, i.e., there exists a 1-form $\omega$ on M such that

$$\begin{aligned} Dg=\omega \otimes g \end{aligned}$$

(2.4)

and the Ricci tensor $\rho ^D(X,Y)=\text {tr}\{Z\rightarrow R^D(Z,X)Y\}$ of D, where $R^D$ is the curvature tensor of (M, D), satisfies the condition

$$\begin{aligned} \rho ^D(X,Y)+\rho ^D(Y,X)=\bar{\Lambda }g(X,Y) \text { for every } X,Y\in \mathrm{TM} \end{aligned}$$

for some function $\bar{\Lambda }\in C^{\infty }(M)$. P. Gauduchon ([2]) proved the fundamental theorem that if M is compact, then there exists a Riemannian metric $g_0\in [g]$ for which the Lee form $\omega _0$ associated with $g_0$ is co-closed-$\delta \omega _0=0$ where $Dg_0=\omega _0\otimes g_0$, and $g_0$ is unique up to homothety. The metric $g_0$ is called Gauduchon or the standard metric of the E–W structure ([g], D) on M. Note that if M is not compact, a standard metric may not exist.

Let $\rho$ be the Ricci tensor of (M, g) and let us denote by Ric the Ricci endomorphism of (M, g), i.e., $\rho (X,Y)=g(X, Ric Y)$. We recall an important theorem (see [10]):

Theorem

A metric g and a 1-form $\omega$ determine an E–W structure if and only if there exists a function $\Lambda \in C^{\infty }(M)$ such that

$$\begin{aligned} \rho +\frac{1}{4}(n-2) \mathcal D\omega =\Lambda g \end{aligned}$$

(2.5)

where $\mathcal D\omega (X,Y)=(\nabla _X\omega )Y+(\nabla _Y\omega )X+\omega (X)\omega (Y)$ and $n=\dim \ M$. If (1.5) holds then

$$\begin{aligned} \bar{\Lambda }=2\Lambda +\text {div}\omega -\frac{1}{2} (n-2)\parallel \omega ^{\sharp }\parallel ^2 \end{aligned}$$

(2.6)

Compact Einstein–Weyl manifolds with the Gauduchon metric are Gray manifolds. By a Riemannian manifold with a pair of Einstein–Weyl structures, we mean a manifold (M, g) for which there exists a 1-form $\omega$ such that $(g,\omega ),(g,-\omega )$ determine two Einstein Weyl structures $([g],D),([g],D^-)$ with $Dg=\omega \otimes g, D^-g=-\omega \otimes g$. Note that every Einstein–Weyl structure (D, [g]) on a compact manifold M determines a Riemannian manifold (M, g) with two Einstein–Weyl structures, namely $([g], D),([g], D^-)$ where g is the Gauduchon metric for the conformal manifold (M, [g]) and $D^-g=-\omega _0\otimes g$ where $\omega _0$ is the Lee form of g with respect to D (see [6]).

It is not difficult to prove that any Riemannian manifold admitting a pair of E–W structures is an $\mathcal {AC}^{\perp }$ manifold with two eigenvalues $\lambda ,\mu$ of the Ricci tensor of multiplicities $n-1,1$ and such that $\lambda \ge \mu$ (see [3]). In this case, $\omega (X)=g(\xi ,X)$ where $\xi$ is a conformal vector field. We will prove the converse: if (M, g) is a simply connected, real analytic $\mathcal {AC}^{\perp }$ manifold with two eigenvalues $\lambda ,\mu$ of multiplicities $n-1,1$ and $\lambda >\mu$ at least at one point, then (M, g) admits a pair of Einstein–Weyl structures $( g, \omega )$ and $( g,-\omega )$.

3 ${\mathcal {AC}}^{\perp }$ manifolds (M, g) whose Ricci tensor has two eigenvalues of multiplicity $\dim M-1$ and 1

We start with:

Theorem 1

Let (M, g) be a real analytic complete, simply connected ${\mathcal {AC}}^{\perp }$ manifold whose Ricci tensor has two eigenvalues $\lambda ,\mu$ of multiplicity $n-1,1$ in the set $V=\{x\in M:\lambda (x)\ne \mu (x)\}$. Then, the set $N=\{x\in M:\lambda (x)=\mu (x)\}$ has an empty interior and there exists a conformal vector field $\xi$ such that $g(\xi ,\xi )=|\lambda -\mu |$ and $(Ric-\mu Id)\xi =0$. If $\xi$ is a Killing vector field, then $\lambda '$ is constant, while if $\xi$ is conformal not Killing, then $\mu '$ is constant. Here $\lambda ',\mu '$ are eigenvalues of the Killing tensor $S=Ric-\frac{2\text {Scal}}{n+2 }Id$ corresponding to $\lambda ,\mu$ respectively.

Proof

Let $V_+=\{x\in M:\lambda (x)>\mu (x)\}$, $V_-=\{x\in M:\lambda (x)<\mu (x)\}$. Note that $M=V_+\cup V_-\cup N$. Let $\mathcal D$ be the distribution on V defined by $D_{\mu }=\mathcal D=\ker (Ric-\mu I)$. Let us define on V the tensor $m=g(p_{\mathcal D}.,p_{\mathcal D}.)$ where $p_{\mathcal D}:\mathrm{TM}\rightarrow \mathcal D$ is the orthogonal projection onto $\mathcal D$. Then, $\rho =\lambda g+(\mu -\lambda )m$. Let us assume that $V_+\ne \emptyset$. The distribution $\mathcal D_{|V_+}$ is locally spanned by a unit vector field $\xi _0$. We have $Ric\xi _0=\mu \xi _0$. Hence on $V_+$,

$$\begin{aligned} \rho =\lambda g- \omega \otimes \omega \end{aligned}$$

where $\omega =g(\xi , .)$ and $\xi =\sqrt{\lambda -\mu }\xi _0$.

Now we shall check when the tensor $S=Ric-\frac{2}{n+2}\text {Scal} g$ is a Killing tensor. Note that for $T(X,Y)=g(SX,Y)$ we have $T=\lambda 'g-\omega \otimes \omega$ where $\lambda '=\lambda -\frac{2}{n+2}\text {Scal}$. Hence

$$\begin{aligned} \nabla _XT(X,X)=X\lambda 'g(X,X)-2\nabla _X\omega (X)\omega (X)=X\lambda 'g(X,X)-h(X,X)\omega (X) \end{aligned}$$

where h is the symmetric (2, 0) tensor defined by $h(X,Y)=\nabla _X\omega (Y)+\nabla _Y\omega (X)$. Consequently, T is a Killing tensor if and only if

$$\begin{aligned} X\lambda 'g(X,X)-h(X,X)\omega (X)=0. \end{aligned}$$

(3.1)

If $\omega (X)=0$, then from (3.1) it follows that $d\lambda '(X)=0$. Thus, $d\lambda '=\alpha \omega$ for some $\alpha \in C^{\infty }(V_+)$. Consequently

$$\begin{aligned} \omega (X)(\alpha g(X,X)-h(X,X))=0 \end{aligned}$$

for every $X\in \mathrm{TM}_{|V_+}$.

Let $F(X,Y)=\alpha g(X,Y)-h(X,Y)$. Then, F is a symmetric (2, 0) tensor and

$$\begin{aligned} \omega (X)F(X,X)=0 \end{aligned}$$

(3.2)

for every $X\in \mathrm{TM}_{|V_+}$. Let us write $X=tY+\xi$ where $Y\in D_{\lambda }$, $\omega (Y)=0$, $t\in {\mathbb {R}}$. Then, $\omega (X)=g(\xi ,\xi )\ne 0$, and from (2.2) we get $F(tY+\xi ,tY+\xi )=0$. Consequently, $t^2F(Y,Y)+2tF(Y,\xi )+F(\xi ,\xi )=0$. Since $t\in {\mathbb {R}}$ is arbitrary, we obtain $F(Y,Y)=F(\xi ,Y)=F(\xi ,\xi )=0$. Thus, $F=0$. Hence, $h(X,Y)=\alpha g(X,Y)$ and $\xi$ is a local conformal field. Since $h(X,Y)=\nabla _X\omega (Y)+\nabla _Y\omega (X)$, we get $n\alpha =-2\delta \omega$. We also have

$$\begin{aligned} d\lambda '=\alpha \omega \end{aligned}$$

(3.3)

and $0=d\alpha \wedge \omega +\alpha d\omega$. Hence in the set $U_+=\{x\in V_+:\alpha (x)\ne 0\}$, we obtain $d\omega =-d\ln |\alpha |\wedge \omega$. It follows that in $U_+$ the distribution $D_{\lambda }=\ker \omega$ is integrable. Thus for $X,Y\in D_{\lambda }$, we get $\nabla _X\omega (Y)=g(\nabla _X\xi ,Y)=-g(\xi ,\nabla _XY)=-g(\xi ,\nabla _YX)=g(\nabla _Y\xi ,X)=\nabla _Y\omega (X)$. Consequently, $\nabla _X\omega (Y)=\frac{1}{2}\alpha g(X,Y)$ and $p_{D_{\lambda }}(\nabla _X\xi )=\frac{1}{2}\alpha X$. Hence for a vector field $\xi$ and $X\in D_{\lambda }$, we get $\nabla _X\xi =\frac{\alpha }{2}X+\phi (X)\xi$ for $\phi =\frac{1}{2}d\ln g(\xi ,\xi )$. Now for $X,Y\in D_{\lambda }$, we have

$$\begin{aligned} 2\nabla _X\nabla _Y\xi= & {} X\alpha Y+\alpha \nabla _XY+2X\phi (Y)\xi +\phi (Y)(\alpha X+2\phi (X)\xi ),\\ 2\nabla _Y\nabla _X\xi= & {} Y\alpha X+\alpha \nabla _YX+2Y\phi (X)\xi +\phi (X)(\alpha Y+2\phi (Y)\xi ),\\ 2\nabla _{[X,Y]}\xi= & {} \alpha [X,Y]+2\phi ([X,Y])\xi \end{aligned}$$

Thus, $2R(X,Y)\xi =X\alpha Y-Y\alpha X-\phi (X)\alpha Y+\phi (Y)\alpha X$ for $X,Y\in \ker \omega$. Let $\beta =d\alpha -\alpha \phi$. Consequently, taking the trace over vectors perpendicular to $\xi$ in the previous formula we get $0=\rho (Y,\xi )=\frac{1}{2}(\beta (Y)-(n-1)\beta (Y))=-\frac{n-2}{2}\beta (Y)$. Hence, $\beta (X)=0$ for $X\in D_{\lambda }$ and

$$\begin{aligned} R(X,Y)\xi =0 \end{aligned}$$

for $X,Y\in D_{\lambda }$. We have $d\ln |\alpha |(X)=\phi (X)$ for $X\in D_{\lambda }$. Hence, for $X\in D_{\lambda }$, $\nabla _X\xi =\frac{\alpha }{2}X+d\ln |\alpha |(X)\xi$ and $\nabla _X\omega (\xi )=d\ln |\alpha |(X)\omega (\xi )$. Thus, $\nabla _{\xi }\omega (X)=-d\ln |\alpha |(X)\omega (\xi )$ and $d\omega (X,\xi )=2d\ln |\alpha |(X)\omega (\xi )$. On the other hand, $d\omega =-d\ln |\alpha |\wedge \omega$ and $d\omega (X,\xi )=-d\ln |\alpha |(X)\omega (\xi )$. Consequently $d\ln |\alpha |(X)=0$ for $X\in D_{\lambda }$. It follows that $d\omega =0$ and $\nabla _X\xi =\frac{1}{2}\alpha X$ for all $X\in \mathrm{TM}$. The formula $\nabla _X\xi =\frac{1}{2}\alpha X$ is clear if $g(X,\xi )=0$. We will show that it also holds for $X=\xi$. In fact, if X is any vector perpendicular to $\xi$ then $0=\alpha g(X,\xi )=g(\nabla _X\xi ,\xi )+g(\nabla _{\xi }\xi ,X)=g(\nabla _{\xi }\xi ,X)$. Hence $\nabla _{\xi }\xi =\gamma \xi$ for some function $\gamma$. On the other hand, we have $\alpha g(\xi ,\xi )=g(\nabla _{\xi }\xi ,\xi )+g(\nabla _{\xi }\xi ,\xi )=2\gamma g(\xi ,\xi )$. It follows that $\gamma =\frac{1}{2}\alpha$. Thus, $R(X,Y)\xi =\frac{1}{2} d\alpha (X)Y-\frac{1}{2}d\alpha (Y)X$, and consequently, $\rho (Y,\xi )=-\frac{n-1}{2}d\alpha (Y)$. Hence, $d\alpha =-\frac{2\mu }{n-1}\omega$.

We also have $d g(\xi ,\xi )(X)=2g(\nabla _X\xi ,\xi )=\alpha \omega (X)=d\lambda '(X)$. Note that $d\mu '=0$ in $M-N$ where $\mu '$ is an eigenvalue of the Killing tensor associated with Ric corresponding to the one-dimensional distribution $D_{\mu }$ [see (2.3)]. Note that $g(\xi ,\xi )=\lambda '-\mu '$.

Analogously, we construct a conformal local vector field $\xi$ in $V_-$. In particular in $V_-$, we have $d\lambda '=-\alpha \omega$ where $n\alpha =-2\delta \omega$ and $\nabla _X\xi =\frac{\alpha }{2}X$. Note that if $\xi ,\xi '$ are two local fields constructed as above, then $\xi =\pm \xi '$ in the intersection of their domains. Now let us define $M_-=\text {int}\{x\in M-N:\alpha (x)=0\}$ and $M_+=\{x\in M-N:\alpha (x)\ne 0\}$.

We now show that if $M_-\ne \emptyset$, then $M_+=\emptyset$ hence $\xi$ is a Killing vector field (see [6]). Let $x_0\in M_-$ and $x_1\in M_+$. If there exists a smooth curve c joining $x_0$ and $x_1$ and such that $\text {im }c\cap N=\emptyset$, then by analytic continuation we see that $\lambda '$ is constant along this curve, i.e., $\lambda '$ is constant on a certain neighborhood $U_x$ of any $x\in \text {im} c$. Analogously, $\mu '$ is constant, and hence, $\alpha (x_1)=0$, a contradiction. A curve as above indeed exists: if U is a connected component of $x_1$ in $M_+$, then there exists a geodesic c with $c(0)=x_0$ and $c(1)\in U$ which doesn’t meet N (see [6]).

If $N=\emptyset$, then $\xi$ exists and is smooth in a neighborhood of any point $x\in M$. Note that in the case of a Killing vector field the constructed vector field is smooth also if $N\ne \emptyset$, and then one of the sets $V_+,V_-$ vanishes. Next we show that a similar situation holds for a conformal vector field.

Let us assume now that $\text{ int }\{x:\alpha (x)=0\}=\emptyset$. In every component of the set $M-N$ and in the interior of N, the function $\mu '$ is constant. Since $\mu '$ is continuous, it is constant on M. The set $\{x:\alpha (x)\ne 0\}$ is dense in $M-N$. In this set, $\nabla _X\xi =\frac{\alpha }{2}X$. If $x_0\in M- N$, then in a neighborhood of $x_0$ the local field $\xi$ is smooth and $\nabla _X\xi =\frac{\alpha }{2}X$ also if $\alpha (x_0)=0$ (we take the limit). Hence in $M-N$, we have $d\alpha =-\frac{2\mu }{n-1}\omega$. Note that the function $\alpha ^2$ is independent of the choice of the local field $\xi$, and $\alpha ^2=\frac{||\nabla \lambda '||^2}{|\lambda -\mu |}$; hence,$\alpha$ is continuous in $M-N$. Note that $\alpha d\alpha =-\frac{2\mu }{n-1}\alpha \omega =-\frac{2\mu }{n-1}d\lambda '$. On the other hand, since $\mu '$ is constant, it follows that $n\mu -2(n-1)\lambda =C_0=\mathrm{const}$ and $d\lambda =\frac{n}{2(n-1)}d\mu$. Since $\lambda -\mu =\lambda '-\mu '$ and $\mu '$ is constant, we get $d\lambda '=d\lambda -d\mu$ and $(n-1) \alpha d\alpha =-2\mu ( d\lambda - d\mu )=\frac{2-n}{n-1}\mu d\mu$. Hence in any component of $M-N$, we have $\pm (n-1)^2\alpha ^2+(n-2)\mu ^2=C_1=\mathrm{const}$ (the minus sign is in the components contained in $V_-$). Since $\mu '$ is constant, it follows that $\lambda ',\lambda ,\mu$ are real analytic functions on the whole of M (we have $(n-1)\lambda +\mu =\text {Scal}, n\mu -2(n-1)\lambda =C_0=\mathrm{const}$ and the scalar curvature $\text {Scal}$ of (M, g) is real analytic).

Now we show that $\nabla \lambda '=0$ in N. It is clear that $\nabla \lambda '=0$ in the interior of N. Let $x_0\in \partial N$. Then, there exists a sequence of points $x_m$ from one component of $M-N$ such that $lim x_m=x_0$. This component, for example, is contained in $V_+$. In that component, $(n-1)^2\alpha ^2+(n-2)\mu ^2=C_0=\mathrm{const}$. In particular, $lim_{m\rightarrow \infty }\alpha ^2(x_m)=\frac{1}{(n-1)^2}(C_0-(n-2)\mu (x_0)^2)$ is finite which means that $\nabla \lambda '=0$ at $x_0$.

Now let assume that $N\ne \emptyset$ and let $x_0\in N$. Let c(t) be a geodesic with $c(0)=x_0,\Vert \dot{c}(0)\Vert =1$. If $S=Ric-\frac{2\text {Scal}}{n+2}g$, then $T(.,.)=g(S.,.)$ is a Killing tensor and $T-\lambda 'g=\pm \omega \otimes \omega$. Note that $\lim _{t\rightarrow 0}T(\dot{c}(t),\dot{c}(t))=\mu '$, and since $T(\dot{c}(t),\dot{c}(t))$ is constant, it follows that $g(\xi (c(t)),\dot{c}(t))^2=\pm (\lambda '-\mu ')$. Hence, $g(\frac{1}{\sqrt{|\lambda '-\mu '|}}\xi (c(t)),\dot{c}(t))=\pm 1$. Since both vectors have length 1, it follows that $\xi (c(t))=\pm \sqrt{|\lambda '-\mu '|}\dot{c}(t)$. If we introduce geodesic polar normal coordinates centered at $x_0$, then we can assume that $\xi =\sqrt{|\lambda '-\mu '|}\frac{\partial }{\partial t}$ where t is a radial coordinate. In particular, $\nabla \lambda '=\alpha \sqrt{|\lambda -\mu |}\frac{\partial }{\partial t}$ on $M-N$ and $\nabla \lambda '=0$ on N. Hence, $d\lambda '(X)=0$ for any X tangent to the geodesic sphere and $\lambda '$ is constant on geodesic spheres with center $x_0$. Thus, $\lambda '$ is a function of the radial coordinate t only.

If there existed a hypersurface $\Sigma \subset B(x_0,r)$ with $x_0\in \Sigma$ such that the intersection of $\Sigma$ with every geodesic sphere $S(x_0,s),s<r,$ is not empty, and on which $\lambda '-\mu '=0$, then we would have $\lambda '=\mu '$ on the whole of B. In fact, if for $x_1\ne x_0$ we have $\lambda '(x_1)=\mu '$, then $\lambda '(x)=\mu '$ on the sphere $S(x_0,s)$ containing the point $x_1$. Hence, the sphere $S(x_0,s)$ would be a hypersurface $\Sigma$ in a ball $B(x_1,t)$, where t is the injectivity radius at the point $x_1$, which has the above property. It follows that the set N has a non-empty interior U. Since $\lambda ,\mu$ are real analytic and $\lambda =\mu$ on U, it follows that $\lambda =\mu$ on M, a contradiction. It follows that in a geodesic ball $B(x_0,r)$ where r is the injectivity radius of M at $x_0$, the set $\{x\in B(x_0,r): \lambda '(x)=\mu '\}$ consists of only the point $x_0$. Hence, the set N consists of isolated points. This implies that $M-N$ has only one component. It follows that $\lambda '-\mu '$ has a constant sign and one of the sets $V_+,V_-$ is empty.

For $x_0\in N$, let us define a local field $\xi$ on the geodesic ball $B(x_0,r)$ where r is the injectivity radius at $x_0$ by $\xi =\sqrt{|\lambda '-\mu '|}\frac{\partial }{\partial t}$ where t is the radial coordinate. Then since M is simply connected, there exists a continuous vector field $\xi$ on M such that $\rho =\lambda g -\omega \otimes \omega$ if $V_-=\emptyset$ or $\rho =\lambda g +\omega \otimes \omega$ if $V_+=\emptyset$ and $\omega =g(\xi ,.)$ (we take the germs of local fields $\xi$ as points of a two-sheeted covering of M). The field $\xi$ is smooth on $M-N$ and defines a function $\alpha$ on $M-N$. The function $\alpha$ is smooth on $M-N$. We shall show that $\xi$ is in fact smooth.

First we show that the function $\alpha$ has a continuous extension to the points $x_0\in N$. We assume, for example, that $V_-=\emptyset$. Then, the function $\alpha ^2$ is real analytic and $(n-1)^2\alpha ^2+(n-2)\mu ^2=C_1=\mathrm{const}$ in $M-N$; hence, $\alpha ^2$ has a real analytic extension on the whole of M. If $\alpha (x_0)^2>0$, then $\alpha$ has a constant sign in a certain neighborhood of $x_0$ and $\alpha =\pm \frac{1}{n-1}\sqrt{C_1-(n-2)\mu ^2}$ is a smooth extension of $\alpha$ on the neighborhood of $x_0$. If an extension of $\alpha ^2$ satisfies $\alpha ^2(x_0)=0$, then certainly $\mathrm{lim}_{x\rightarrow x_0}\alpha (x)=0$ and taking $\alpha (x_0)=0$ we get a function continuous at $x_0$.

Let c(t) be a unit speed geodesic with $c(0)=x_0\in N$. We take $\xi$ which in geodesic coordinates around $x_0$ is $\xi =\sqrt{|\lambda '-\mu '|}\frac{\partial }{\partial t}$. Note that $\sqrt{|\lambda '-\mu '|}=\pm g(\dot{c},\xi )$. The function $\phi (t)=\sqrt{|\lambda '-\mu '|}\circ c(t)$ if $t\ge 0$ and $\phi (t)=-\sqrt{|\lambda '-\mu '|}\circ c(t)$ if $t<0$ is differentiable at 0 and smooth in a neighborhood of 0. Indeed note that $\phi (t)=g(\dot{c}(t),\xi (c(t)))$ and $\phi '(t)=g(\dot{c},\nabla _{\dot{c}}\xi )=\frac{1}{2}\alpha (c(t)$ for $t\ne 0$. Since $\alpha$ is continuous, it follows that $\phi$ is differentiable at 0 and has continuous derivative. The function $\alpha (t)=\alpha (c(t))$ is differentiable at 0 and $\alpha '(0)=0$. In fact, $d\alpha =-\frac{2}{n-1}\mu \omega$ and $d\alpha (\dot{c})=-\frac{2}{n-1}\mu \phi (t)$. Hence, $\alpha '(t)=-\frac{2}{n-1}\mu \circ c(t)\phi (t)$. Note that $\frac{1}{2}\alpha (t)=\phi '(t)$. The last equation holds also for $t=0$ since $\alpha$ is continuous. Hence, we get an equation $\phi ''(t)=-\frac{\mu (c(t))}{n-1}\phi (t)$ with initial conditions $\phi (0)=0,\phi '(0)=\frac{1}{2}\alpha (x_0)$ where the function $\mu (c(t))$ is real analytic. If $\alpha (x_0)=0$, then we get $\phi (t)=0$, a contradiction. Hence, $\alpha (x_0)\ne 0$ and $\xi =\frac{1}{\alpha }\nabla \lambda '$ is a smooth vector field in a neighborhood of $x_0$. Hence, $\xi$ is a smooth vector field.$\diamondsuit$

Remark

$\mathcal {AC}^{\perp }$ manifolds for which $\xi$ is a Killing vector field are described in [6]. In the rest of the paper, we shall assume that $\dim M=n+1$.

Theorem 2

Let us assume that (M, g) is a real analytic, simply connected $\mathcal {AC}^{\perp }$ manifold, whose Ricci tensor has two eigenvalues $\lambda ,\mu$ of multiplicities n, 1 respectively, and such that $\mu -\frac{2\text {Scal}}{n+3}$ is constant and $\lambda -\frac{2\text {Scal}}{n+3}$ is non-constant. Then, $\lambda -\mu \ge 0$ on the whole of M or $\lambda -\mu \le 0$ on the whole of M. If $|\lambda -\mu |>0$ on M, then M is a warped product $M={\mathbb {R}}\times _{f^2} M_*$ where $M_*$ is an Einstein manifold. If $N\ne \emptyset$, then $N=\{x_0\}$ or $N=\{x_0,x_1\}$ and $M={\mathbb {R}}_+\times _{f^2} M_*$ in the first case, while $M=(0,\epsilon )\times _{f^2} M_*$ in the second case, where $M_*=(S^n,g_\mathrm{can})$ is a sphere with the standard metric of constant sectional curvature. In all cases, the function f satisfies an equation

$$\begin{aligned} -(n-1)\left( \frac{f''}{f}-\frac{(f')^2}{f^2}\right) =\tau f^{-2}-Cf^2 \end{aligned}$$

(3.4)

where $C\in {\mathbb {R}}-\{0\}$ and $Ric_{M_*}=\tau g_{M_*}$. If $N\ne \emptyset$ and $M_*$ is a sphere with sectional curvature 1, then the function f satisfies the initial conditions $f(0)=0,f'(0)=1$ in the first case and in the second case $f(0)=0,f'(0)=1, f(\epsilon )=0,f'(\epsilon )=-1$.

Proof

If $N=\emptyset$, the proof is given in [7] (see also [5]). Let us assume first that $\sharp N\ge 2$. Let $x_0\in N$ and $r=d(x_0,N-\{x_0\})=d(x_0,x_1)$ where $x_1\in N$. We show that the map $\mathrm{exp}_{x_0}:B(0,r)\rightarrow M$, where $B(0,r)=\{X\in T_{x_0}M:||X||<r\}$, is a diffeomorphism onto its image.

If $\mathrm{exp}_{x_0}X=\mathrm{exp}_{x_0}Y$, then we consider two cases $X=0,Y\ne 0$ and $X,Y\ne 0$. In the first case, the geodesic $c(t)=\mathrm{exp}_{x_0}t\frac{Y}{||Y||}$ is a loop of length $s=||Y||<r$. In the second case, let $c(t)=\mathrm{exp}_{x_0}t\frac{X}{||X||}, c_1(t)=\mathrm{exp}_{x_0}t\frac{Y}{||Y||}$. Let $t_0=||X||,t_1=||Y||$. Then, $c(t_0)=c_1(t_1)$ and $\dot{c}(t_0)=\dot{c}_1(t_1)$ since $\xi (c(t_0))=\sqrt{|\lambda -\mu |}\dot{c}(t_0)=\sqrt{|\lambda -\mu |}\dot{c}_1(t_0)$. It follows that $c(t_0+s)=c_1(t_1+s)$. Let us assume that $t_1>t_0$. Then, $c(0)=x_0=c_1(t_1-t_0)$. Hence, $c_1$ is a geodesic loop at $x_0$ of length $t_1-t_0<r$. This contradicts $\xi (c_1(t))=\sqrt{|\lambda -\mu |}\dot{c}_1(t)$ which is valid up to the first value of $t>0$ for which $\lambda (c_1(t))-\mu (c_1(t))=0$, if we start from $t=0$ (note that $|\lambda (c_1(t))-\mu (c_1(t))|=0$ only if $c_1(t)=x_0$ for $t\in (0,r)$). Thus, $t_0=t_1$ and $c=c_1$, which means that $X=Y$. Similarly, we obtain $Y=0$ if $X=0$. The mapping $\mathrm{exp}_{x_0}:B(0,r)\rightarrow M$ is a diffeomorphism. In fact, otherwise we would have a conjugate point on a geodesic $c(t)=\mathrm{exp} tX$. Hence, we would have two different geodesics starting at $x_0$ and joining $x_0$ to $x_1=\mathrm{exp} sX$, where $s>1, s||X||<r$. It leads to a contradiction as above.

Note that $\lambda -\mu$ depends only on the distance from $x_0$. Hence, $\mathrm{exp}_{x_0}rS\subset N$ where S is a unit sphere in $T_{x_0}M$. Since $\mathrm{exp}_{x_0}$ is a continuous function, $\mathrm{exp}_{x_0}rS$ is connected and N consists of isolated points, it follows that $\mathrm{exp}_{x_0}rS=\{x_1\}$. Hence, $N=\{x_0,x_1\}$ and every unit geodesic c starting from $x_0$ satisfies $c(r)=x_1$. It follows that $\mathrm{exp}_{x_0}:B(0,r)\rightarrow M-\{x_1\}$ is a diffeomorphism. In the same way, $\mathrm{exp}_{x_1}:B(0,r)\rightarrow M-\{x_0\}$ is a diffeomorphism onto $M-\{x_0\}$.

Note that by the Gauss lemma, the leaves of $D_{\lambda }$ are compact spheres $S^n$ via the diffeomorphism $\mathrm{exp}_{x_0}:S(0,t)\rightarrow M$ where $S(0,t)=\{X\in T_{x_0}M:\Vert X\Vert =t$ where $t\in (0,r)$. Hence, $M=[0,r]\times _{f^2}S^n$. From [B, Lemma 9.114, page 269], it follows that on $S^n$ there is a metric of constant sectional curvature $\lambda ^2g_\mathrm{can}$ and $f(0)=0=f(r),f'(0)=\frac{1}{\lambda }=-f'(r)$. Replacing f by a function $\lambda f$, we see that we can assume that $f(0)=0,f'(0)=1,f(r)=0,f'(r)=-1$.

Now consider the case $N=\{x_0\}$. Then, as above one can prove that $\mathrm{exp}_{x_0}:T_{x_0}M\rightarrow M$ is a diffeomorphism. Hence in this case, $M={\mathbb {R}}_+\times _{f^2} S^n$ where $S^n$ has the a canonical metric. The function f satisfies the initial conditions $f(0)=0,f'(0)=1$.$\diamondsuit$

Now we consider the equation (see [7, p.27])

$$\begin{aligned} -(n-1)(\frac{f''}{f}-\frac{(f')^2}{f^2})=\tau f^{-2}-Cf^2 \end{aligned}$$

with initial conditions $f(0)=0,f'(0)=1$ ; hence, we consider the warped product $M=(0,\epsilon )\times _{f}S^{n}$ or $M={\mathbb {R}}_+\times _{f}S^{n}$ and $\tau =n-1$.

If we write $g=\ln f$, then $(g)''=-\frac{\tau }{n-1}e^{-2g}+\frac{C}{n-1}e^{2g}$. We get $\frac{1}{2}((g')^2)'=(\frac{\tau }{2(n-1)}e^{-2g}+\frac{C}{2(n-1)}e^{2g})'$ and $(g')^2=\frac{\tau }{(n-1)}e^{-2g}+\frac{C}{(n-1)}e^{2g}+A$ for some constant $A\in {\mathbb {R}}$. Consequently, if we assume that $\tau =n-1$, we get an equation for f

$$\begin{aligned} (f')^2=1+Af^2+\frac{C}{n-1}f^4. \end{aligned}$$

Using homothety, we can assume that $C=\pm (n-1)$. Thus, we get $(f')^2=1+Af^2+f^4$ or $(f')^2=1+Af^2-f^4$.

First let us assume that the quadratic polynomial $1+At+\epsilon t^2$ where $\epsilon \in \{-1,1\}$ has real roots. This happens for $\epsilon =-1$, and for $|A|\ge 2$ if $\epsilon =1$. Hence, if $\epsilon =1$ and $A<-2$, we get $(f')^2=(a^2-f^2)(\frac{1}{a^2}-f^2)$ for some $a\in (0,1)$. Let f be the solution of this equation with initial conditions $f(0)=0,f'(0)=1$.

We shall show that f increases until it attains its maximum $a=f(t_0)$ and then decreases, and $f(2t_0)=0,f'(2t_0)=-1$. In fact, f satisfies an equation $f'=\sqrt{\left( a^2-f^2\right) \left( \frac{1}{a^2}-f^2\right) }$ if $f<a$. It follows that $\arcsin (\frac{f}{a})'=\sqrt{ \frac{1}{a^2}-f^2}\ge \sqrt{\frac{1}{a^2}-a^2}$ if $f<a$. Consequently, $\arcsin (\frac{f}{a})\ge t\sqrt{\frac{1}{a^2}-a^2}$ and f increases until it attains the value $f(t_0)=a$. The solution exists on an interval $[0,t_0]$. Now we define $f(t_0+s)=f(t_0-s)$ for $s\in (0,t_0]$. Then, f is defined on an interval $[0,2t_0]$ and is a smooth function on an open neighborhood of this interval.

Similarly if $\epsilon =-1$, then we get an equation $f'=\sqrt{\left( a^2-f^2\right) \left( \frac{1}{a^2}+f^2\right) }$ where $a>0$. Thus, $\arcsin (\frac{f}{a})\ge t\sqrt{\frac{1}{a^2}}$, and again there exists $t_0>0$ such that $f(t_0)=a$. We extend the solution on the interval $[0,2t_0]$ and obtain a solution satisfying $f(2t_0)=0,f'(2t_0)=-1$.

We have to show that f is an odd function at $t=0$ and $t=2t_0$. Note that f satisfies the equation $f''=Af+2\epsilon f^3$. Since f is real analytic, it suffices to show that $f^{(2k)}(0)=0$ for $k\in {\mathbb {N}}$. We use induction. We have $f(0)=0,f''(0)=0$. Let us assume that $f^{(2k)}(0)=0$ for $k\le n$ where $n\ge 1$. We will show that $f^{(2n+2)}(0)=0$. Note that

$$\begin{aligned} f^{(2n+2)}(0)=Af^{(2n)}(0)+2\epsilon \sum _{r=0}^{2n}\sum _{s=0}^rC^{2n}_rC^r_sf^{(s)}f^{(r-s)}f^{(2n-r)}, \end{aligned}$$

(3.5)

where $C^n_k=\frac{n!}{(n-k)!k!}$. If s and $r-s$ are odd numbers, then $r=r-s+s$ is an even number and $f^{(2n-r)}(0)=0$. Hence in all cases, the left-hand side of (3.5) equals 0. It follows that in both cases considered above we get a compact $\mathcal {AC}^{\perp }$ manifold diffeomorphic to $S^{n+1}$. If $\epsilon =1$ and $A=-2$, then one can easily check that $f(t)=\tanh t$. In that case, we obtain a complete non-compact $\mathcal {AC}^{\perp }$ manifold diffeomorphic to ${\mathbb {R}}^{n+1}$ (see [7]). If $A>-2$, then we can obtain a solution which tends to infinity which may not be defined on the whole of ${\mathbb {R}}_+$. For example, for $A=2$ we get $f(t)=\tan t$. Summarizing we have

Theorem 3

On the sphere $S^{n+1}$, there exists a one-parameter family of $\mathcal {AC}^{\perp }$ metrics $g_A$ such that $g_A=dt^2+f^2g_\mathrm{can}$ and $(S^{n+1},g_A)=[0,2t_0]\times _{f^2}S^{n}$ where f satisfies the equation

$$\begin{aligned} (f')^2=1+Af^2+f^4 \end{aligned}$$

(3.6)

with $A\in (-\infty ,-2)$, or

$$\begin{aligned} (f')^2=1+Af^2-f^4 \end{aligned}$$

(3.7)

for $A\in {\mathbb {R}}$. In the case of Eq. (3.6), the corresponding $\mathcal {AC}^{\perp }$ manifold admits a pair of Einstein–Weyl structures both conformally Einstein and conformal to the standard sphere with constant sectional curvature.

Proof

Note that in the case (3.6) the corresponding $\mathcal {AC}^{\perp }$ manifold has a Ricci tensor with eigenvalues $\lambda ,\mu$ such that $\lambda -\mu =Cf^2\ge 0$. Hence from (2.5), it determines a pair of Einstein–Weyl structures $(g,\omega '), (g,-\omega ')$ where $\omega '=\frac{2}{\sqrt{n-1}}\omega$ and $\omega (X)=g(\xi ,X)$ was defined in the first part of the paper. Since $d\omega =0$, both Einstein–Weyl structures are conformally Einstein. Let $\omega =d\phi$, then the metrics $g_1=\exp (\phi )g, g_2=\exp (-\phi )g$ are Einstein metrics on $S^{n+1}$. They are conformally equivalent to each other $g_1=\exp (2\phi )g_2$. Thus, it follows from [8, 9] that $g_i$ are standard metrics of constant sectional curvature on $S^{n+1}$. $\square$

Remark

In the case of Eq. (3.6) with $A=-2$, we obtain a solution $f(t)=\tanh t$. The corresponding $\mathcal {AC}^{\perp }$ manifold $M={\mathbb {R}}_+\times _{f^2} S^n$ is complete (see [7]) and diffeomorphic to ${\mathbb {R}}^{n+1}$. It also admits a pair of Einstein–Weyl structures which are conformally Einstein. At least one of them has to be non-complete since any complete Einstein manifold which is conformal to a complete Einstein manifold has to be a sphere with standard metric of constant sectional curvature (see [8]).

Theorem 4

On a manifold $M={\mathbb {R}}\times _{f^2} M_*$ where $(M_*,g_{M_*})$ is an Einstein manifold such that $Ric_{M_*}=\tau g_{M_*}$ with $\tau <0$ there exists a one-parameter family of complete $\mathcal {AC}^{\perp }$ metrics $g_A$ such that $g_A=dt^2+f^2g_{M_*}$, where f satisfies the equation

$$\begin{aligned} (f')^2=\frac{\tau }{n-1}+Af^2-f^4 \end{aligned}$$

(3.8)

with $A\in {\mathbb {R}}, A^2>-4\tau ,\tau <0$. The eigenvalues $\lambda ,\mu$ of the Ricci tensor of $(M,g_A)$ satisfy $\lambda \le \mu$; hence, these manifolds do not admit a pair of Einstein–Weyl structures.

Proof

Equation (3.8) is equivalent to the equation

$$\begin{aligned} f''=Af-2f^3 \end{aligned}$$

(3.9)

with appropriate initial conditions. Hence, we can apply [B,Lemma 16.37 p.445 ]. Note that $(f')^2=-(f^2-a^2)(f^2+\frac{\tau '}{a^2})$ if $A^2>-4\tau ',A>0,\tau '<0$ where $\tau '=\frac{\tau }{n-1}, a>0$. We can assume that $a<\frac{\sqrt{-\tau '}}{a}$. Hence for $P(f)=-(f^2-a^2)(f^2+\frac{\tau '}{a^2})$, we have $P(a)=P(\frac{\sqrt{-\tau '}}{a})=0$ and $P'(a)P'(\frac{\sqrt{-\tau '}}{a})\ne 0$. Consequently, the equation $(f')^2=P(f)$ admits a periodic solution $f:{\mathbb {R}}\rightarrow [a,\frac{\sqrt{-\tau '}}{a}]$. In that way, we get complete manifolds $(M,g_A)$. Note that these manifolds admit compact quotients.$\diamondsuit$

Remark

Note that an equation of type (3.9) ( for a function $\frac{1}{f}$) was studied in [1] and complete and compact manifolds constructed in Theorem 4 are known (as manifolds with $Dr\in C(Q)$ in [1]). However, the $\mathcal {AC}^{\perp }$ metrics on spheres $S^{n+1}$ were not found by A. Besse.