Leaf space isometries of singular Riemannian foliations and their spectral properties

Abstract

In this note, the authors show by example that an isometry between leaf spaces of singular Riemannian foliations need not induce an equality of the basic spectra. If the leaf space isometry preserves the mean curvature vector fields, then it is proved that the basic spectra are equivalent, i.e. that the leaf spaces are isospectral. As a corollary to the main result, the authors identify geometric conditions that ensure preservation of the mean curvature vector fields, and therefore isospectrality of the leaf spaces.

Appendix

Here, we include an alternate proof of Corollary 1.6 using the characterization of the mean curvature vector field in terms of the trace of the shape operator of the leaves. In particular, we show that the projections of the mean curvature vector fields are preserved using Jacobi fields and special properties of foliations on spheres to calculate the eigenvalues and their multiplicities of the shape operators. This proof starts by proving the result in the case of a regular Riemannian foliation in the first lemma below, before generalizing to the singular case. This is essentially just a modification of Proposition 4.1.1 of [8].

Lemma 4.1

If $M_1$ and $M_2$ are space forms of the same non-negative curvature $\kappa$ that admit regular Riemannian foliations ${\mathcal {F}}_1$ and ${\mathcal {F}}_2,$ and $\varphi :M_1/{\mathcal {F}}_1 \rightarrow M_2/{\mathcal {F}}_2$ is a smooth SRF isometry, then $d\varphi (H_{1*})=H_{2*}$.

Proof

Let $p_1\in M_1$ and $p_2\in M_2$. Assume further that ${\bar{p}}_1\in M_1/{\mathcal {F}}_1$ and ${\bar{p}}_2=\varphi ({\bar{p}}_1)\in M_2/{\mathcal {F}}_2$ denote their images in the respective quotient spaces–i.e. $\pi _1(p_1)={\bar{p}}_1$ where $\pi _1:M_1\rightarrow M_1/{\mathcal {F}}_1$ and similarly for ${\bar{p}}_2$. Let $x_i\in (V_i^\perp )_{p_i}$ denote horizontal vectors for $i=1,\,2$ and let ${\bar{x}}_i =d\pi _i(x_i)$ denote the images of these vectors in the quotient space. Suppose further that ${\bar{x}}_2:=d\varphi ({\bar{x}}_1).$ Let $\gamma _i(t)=exp_{p_i}(tx_i)$ for $i=1,\,2$. Finally, let $S_{x_i}$ denote the shape operators at $x_i$ for $M_i$, $i=1,\,2$.

Let $u_1\in (V_1)_{p_1}$ be the eigenvector with eigenvalue $\lambda$ for $S_{x_1}$. Recall from [8] that a Jacobi field J defined along a curve c is said to be projectable if and only if it satisfies $J'^{\varvec{v}}=-S_{{\dot{c}}}J^{\varvec{v}}-A_{{\dot{c}}}J^{\varvec{h}}$. (Here, the superscripts denote the vertical and horizontal components of the vector fields.) And if J is vertical and projectable, $J'^{\varvec{v}}=-S_{{\dot{c}}}J.$

Now consider the Jacobi field $J_1(t)$ along $\gamma _1(t)$ defined by the following initial value problem: $J_1''(t)+\kappa J_1(t)=0,$ $J_1(0)=u_1$ and $J_1'(0)=-S_{x_1}u_1=-\lambda u_1$. Since $u_1$ is vertical, $J_1$ is vertical. The initial condition on the first derivative implies that $J_1$ is projectable, a fact which we will use shortly.

Returning to the initial value problem, we observe that $J_1(t)=f(t)E(t)$ where E is the parallel field along $\gamma _1$ with $E(0)=u_1$, and f satisfies the initial value problem $f''+\kappa f=0,$ $f(0)=1,$ $f'(0)=-\lambda$. By solving, we see that when $\kappa =0$, $f(t)=1-\lambda t$, and when $\kappa >0$ we have $f(t)= \cos (t\sqrt{\kappa })-\frac{\lambda }{\sqrt{\kappa }}\sin (t\sqrt{\kappa }).$

Assume for the moment that $\lambda \not =0$ if $\kappa =0$. Then $J_1(t_0)=0$ for some $t_0\in {\mathbb {R}}$. If $\kappa =0,$ $f(t_0)=0$ implies that $\lambda =1/t_0.$ If $\kappa >0$, then $f(t_0)=0$ imples that $\lambda =\frac{\sqrt{\kappa }}{\tan (t_0\sqrt{\kappa })}.$

Since $J_1$ is projectable, by Theorem 1.6.1 of [8], we have that $d\pi _1(J_1)$ is a Jacobi field along $\pi _1(\gamma _1)$. It follows easily that $d\varphi (d\pi _1(J_1))$ is a Jacobi field along $\varphi (\pi _1(\gamma _1(t)))=\pi _2(\gamma _2(t))$. By Lemma 1.6.1 of [8], there exists a unique Jacobi $J_2$ along $\gamma _2$ with $d\pi _2(J_2)=d\varphi (d\pi _1(J_1)),$ $J_2'^{\varvec{v}}(0)=-S_{x_2}J_2(0),$ and $J_2(t_0)=0$. In particular, $J_2$ must be vertical, because $J_1$ is vertical. Since $J_2(t_0)=0$, and $J_2$ is vertical, $J_2=fE_2$ for some parallel field $E_2$. Hence, $J_2'(0)=-\lambda J_2(0),$ and thus $J_2(0)$ is an eigenvector with eigenvalue $\lambda$ of $S_{x_2}$.

Now consider multiplicities. Suppose $\lambda$ is an eigenvalue of $S_{x_1}$ with multiplicity k. Let $E_1(\lambda )$ denote the $\lambda$ eigenspace of $S_{x_1}.$ As noted in the proof of Proposition 4.1.1 of [8], for any Jacobi field ${\bar{J}}_1$ along the projected geodesic $\pi _1\circ \gamma _1$ with ${\bar{J}}_1(0)={\bar{J}}_1(t_0)=0$ then, by Lemma 1.6.1 of [8], there exists a unique projectable Jacobi field $J_1$ along $\gamma _1$with the $J_1(0)\in E_1(\lambda )$. This implies that $t_0$ is a conjugate point of $\pi _1\circ \gamma _1$ of multiplicity (as a conjugate point) of at most k.

On the other hand, if $J_1$ is a projectable Jacobi field along $\gamma _1$ with $J_1(0)=-S_{x_1}J_1(0),$ with $J_1(t_0)=0,$ $J(0)\not =0$ then $d\pi _1(J_1)$ is a Jacobi field that vanishes at $t=0$ (because $J_1(0)\in E_1(\lambda )\subset V_1$) and at $t=t_0.$ We may assume that $d\pi _1(J_1)$ is non-trivial, since otherwise $J_1$ is vertical, and thus a holonomy field (See Definition 1.4.3, of [8]). Such fields vanish identically at any point; thus, the condition that $J_1(t_0)=0,$ implies that $J_1$ is trivial, a contradiction. Thus, $t_0$ is a conjugate point of multiplicity exactly k for $\pi _1\circ \gamma _1$.

By an identical argument, if $\lambda$ is an eigenvalue of $S_{x_2}$ with multiplicity $k'$, then $t_0$ is a conjugate point of multiplicity exactly $k'$ for $\pi _2\circ \gamma _2$.

But, $\varphi \circ \pi _1\circ \gamma _1=\pi _2\circ \gamma _2$, and $d\varphi (d\pi _1(J_1))=d\pi _2(J_2)$. Hence, $J_1(t_0)=0$ if and only if $J_2(t_0)=0$, so conjugate points are preserved, and therefore $k=k'$, and thus the eigenvalues of $S_{x_1}$ are the same as those of $S_{x_2}$, including multiplicities. Hence, $d\varphi (H_{1*})({\bar{p}}_1)=H_{2*}({\bar{p}}_2)$.

If $\kappa =0$, some care must be taken with the zero eigenvalues. Suppose now that $\kappa =0$ and $\lambda =0$, then $J_1(t)\not =0$ for finite t. In this case, we can take $t_0=\infty$ and $\lambda =1/t_0=0.$ By the argument above, $J_2(t)\not =0$ for all t, and hence $J_2(0)$ will be an eigenvector of $S_{x_2}$ with eigenvalue zero.

We have now demonstrated that the non-zero eigenvalues of $S_{x_1}$ are the same as those of $S_{x_2}$, counting with multiplicities. Hence, their traces are the same, and it follows that $d\varphi (H_{1*})({\bar{p}}_1)=H_{2*}({\bar{p}}_2),$ and thus $d\varphi (H_{1*})=H_{2*}.$ $\square$

Alternative proof of Corollary 1.6

Let $p_1\in M_1$ be a point belonging to the regular region, and let $x_1\in (V_1^\perp )_{p_1}$, and let X be a horizontal vector field with $X_{p_1}=x_1$. Define $p_2$ and $x_2$ similarly with $p_2=\varphi (p_1)$ and $x_2=d\varphi (X_{p_1}).$ Note that $p_2$ necessarily lies in the regular region as well because of the leaf codimension preservation condition. Let $u_1\in (V_1)_{p_1}$ be the eigenvector with eigenvalue $\lambda$ for $S_{x_1}$, as in the proof of Lemma 4.1, and let $\gamma _1(t)=exp_{p_1}(tx_1).$ Now consider the Jacobi field $J_1(t)$ along $\gamma _1(t)$ defined by the following initial value problem: $J_1''(t)+\kappa J_1(t)=0,$ $J_1(0)=u_1$ and $J_1'(0)=-S_{x_1}u_1=-\lambda u_1$. As above, $J_1(t)$ will vanish at some $t_0\in {\mathbb {R}}$ unless $\kappa =0$. For the moment, we will assume that such a $t_0<\infty$ exists. If $\gamma _1(t_0)$ belongs to the regular region of $M_1$, then we may use the argument of the previous lemma to show that $d\varphi (H_{1*})({\bar{p}}_1)=H_{2*}({\bar{p}}_2)$.

If $\gamma _1(t_0)$ belongs to the singular strata of $M_1$ such that $\pi _1\circ \gamma _1(t)$ lies in the orbifold part of $M_1/{\mathcal {F}}_1$, denoted by $(M_1/{\mathcal {F}}_1)_{orb}$, then we will show that the non-zero eigenvalues of $S_{x_1}$ are the same as the non-zero eigenvalues of $S_{x_2}$ when $\kappa =0,$ and all eigenvalues of $S_{x_1}$ are the same as the eigenvalues of $S_{x_2}$ when $\kappa >0.$

Note that the complement of $(M_1/{\mathcal {F}}_1)_{orb}$ has codimension at least two, thus almost every projected horizontal geodesic stays in $(M_1/{\mathcal {F}}_1)_{orb}$ for all time. Further $\varphi ((M_1/{\mathcal {F}}_1)_{orb})=(M_2/{\mathcal {F}}_2)_{orb},$ and $\varphi \circ \pi _1\circ \gamma _1=\pi _2\circ \gamma _2.$ From this we will conclude that the trace of $S_{x_1}$ equals the trace of $S_{x_2}$ whenever $\pi _1\circ \gamma _1$ is contained in $(M_1/{\mathcal {F}}_1)_{orb},$ which is open and dense. By continuity of the mean curvature form, we will then conclude that $d\varphi (H_{1*})=H_{2*}.$

We now take cases, supposing first $\kappa >0$, then $\kappa =0$.

If $\kappa >0$, then we may use the following argument from the thesis of M. Radeschi, [14]. We will first suppose that $M_1$ is a round $n_1$-sphere of curvature $\kappa$, and $M_2$ is similarly a round $n_2$-sphere of curvature $\kappa$, each admitting singular Riemannian foliations ${\mathcal {F}}_1$ and ${\mathcal {F}}_2,$ respectively. At the conclusion of this case, we will show how to generalize the result from spheres to positive curvature space forms.

From [12], Lemma 6.1, we know that if $L_s$ is a singular leaf through $p_0:=\gamma _1(t_0)$, and $\sigma$ is a horizontal geodesic starting at $p_0$ that leaves the stratum containing $p_0$, then for some small $\varepsilon$ there is a neighborhood $U_{\sigma (\varepsilon )}\subset L_{\sigma (\varepsilon )},$ (contained in the regular region), and a neighborhood $U_{p_0}\subset L_s$ such that the closest-point map $\Pi : U_{\sigma (\varepsilon )}\rightarrow U_{p_0}$ is a submersion with non-trivial kernel and $d \Pi _{\sigma {\epsilon }} v = J_v(0)$ where $J_v$ is the holonomy Jacobi field along $\sigma$ such that $J_v{\epsilon }=v.$ We can then prove the following claim (generalizing Lemma 3.0.6 of [14]) which we include for the sake of completeness of the exposition: Let $(M,{\mathcal {F}})$ be a singular Riemannian foliation on a sphere M of curvature $\kappa$, and suppose $L_s$ is a singular leaf, $\sigma$ is a horizontal geodesic starting at $p_0\in L_s$, and $\Pi : U_{\sigma (\varepsilon )}\rightarrow U_{p_0}$ the local submersion, then

$$\begin{aligned} Ker\, d\,\Pi _{\sigma (\varepsilon )}=\Bigl \{ v\in V_{\sigma (\varepsilon )}\,|\,A^*_xv=0,\, S_xv=\frac{\sqrt{\kappa }}{\tan (\varepsilon \sqrt{\kappa })}v\Bigr \} \end{aligned}$$

(12)

where $A^*$ is the adjoint of the O’Neill tensor, A,  and $x=-\sigma '(\varepsilon ).$ Observe that $Ker\, d\,\Pi _{\sigma (\varepsilon )}$ is a subspace of dimension $dim(L_{\sigma (\varepsilon )})-dim(L_s)$.

This claim is proved as follows. Let $v\in Ker\, d\Pi _{\sigma (\varepsilon )}$ and let J(t) be the unique holonomy Jacobi field along ${\tilde{\sigma }}(t):=\sigma (\varepsilon -t)$ such that $J(0)=v$. Then, since J(t) is a holonomy Jacobi field, $J'(0)= -A^*_xv-S_xv$ with $x={\tilde{\sigma }}'(0)$. We may assume $J(t)=\cos (\sqrt{\kappa }t)E_1(t)+\frac{\sin (\sqrt{\kappa }t)}{\sqrt{\kappa }}E_2(t)$ where $E_1(t)$ and $E_2(t)$ are parallel fields, $E_1(0)=v$, $E_2(0)=-A^*_xv-S_xv$ since M is a sphere of curvature $\kappa$. Noting that $\langle E_i(t),E_j(t)\rangle$ for $i, j\in {1,2}$ are constant, since $E_1(t)$ and $E_2(t)$ are parallel fields, one easily calculates that $\Vert J(t)\Vert ^2$ goes to zero precisely when $A^*_xv=0$ and $S_xv=\frac{\sqrt{\kappa }}{\tan (\varepsilon \sqrt{\kappa })}v,$ proving the claim.

Let $E_\lambda (x_1)$ and $E_\lambda (x_2)$ denote the $\lambda$ eigenspaces of $S_{x_1}$ on $M_1$ and $S_{x_2}$ on $M_2,$ respectively. We wish to show that $E_\lambda (x_1)$ and $E_\lambda (x_2)$ have the same dimension. By the previous claim taking $\varepsilon =t_0$, there are subspaces $K_{p_1}:=E_\lambda (x_1)\cap ker A^*_{x_1}$ and $K_{p_2}:=E_\lambda (x_2)\cap ker A^*_{x_2}$. These subspaces have the same dimension precisely when $codim(L_{\gamma _1(t_0)})=codim(L_{\gamma _2(t_0)}).$ It is now enough to show that the dimensions of the quotient spaces $dim(E_\lambda (x_1)/K_{p_1})=dim(E_\lambda (x_2)/K_{p_2})$.

We do so by constructing a bijection between the quotient spaces, as in the proof of Proposition 3.0.7 of [14], which we include for the sake of completeness.

We begin by showing that there is a well-defined injective map

$$\begin{aligned} E_\lambda (x_1)/K_{p_1}\rightarrow E_\lambda (x_2)/K_{p_2}. \end{aligned}$$

(13)

The existence of a corresponding map in the other direction will follow from reversing the roles of $p_1$ and $p_2$ and using $\varphi ^{-1}$ in place of $\varphi$. It is helpful to define the following. Let

$$\begin{aligned} {\mathcal {K}}_1:= & {} \{J_1\,|\,J_1 \text { is a Jacobi field along }\gamma _1 \text { with } J_1(t_0)=0\} \end{aligned}$$

(14)

$$\begin{aligned} {\mathcal {K}}_2:= & {} \{J_2\,|\,J_2 \text { is a Jacobi field along }\gamma _2 \text { with } J_2(t_0)=0\}. \end{aligned}$$

(15)

Let $ev_0$ be the evaluation map at $t=0$. From the argument above, we have $ev_0({\mathcal {K}}_1)=K_{p_1}$ and $ev_0({\mathcal {K}}_2)=K_{p_2}.$

Now let $[v]\in E_\lambda (x_1)/K_{p_1},$ and let $v\in E_\lambda (x_1)\subset V_{p_1}$ be any representative of [v]. Let $J_v(t)$ be the projectable Jacobi field along $\gamma _1$ with $J_v(0)=v$, $J'_v(0)=-S_{x_1}(v)=-\lambda v.$ Consider an interval of the form $I=(t_0-\varepsilon , t_0)$ for small $\varepsilon >0$ such that $\gamma _1|_I$ is on the regular part of the foliation. On this interval $J_v$ is a projectable Jacobi field over $\gamma _1$ and we may use Theorem 1.6.1 of [8] to conclude that $d\pi _1(J_v(t))$ is a Jacobi field along the projected geodesic $\pi _1\circ \gamma _1$ that vanishes at $t=0$. Furthermore, $\lim _{t\rightarrow t_0}\Vert J_v(t)\Vert =0$, so it vanishes at $t_0$ as well. Note that $d\pi _1(J_v(t))$ is defined independently of the choice of representative v.

Consider $\gamma _2$ a geodesic in $M_2$ such that $\varphi \circ \pi _1\circ \gamma _1=\pi _2\circ \gamma _2$, restricted to $I=(t_0-\varepsilon ,t_0)$. We note that $d\varphi (d\pi _1(J_v))$ is a Jacobi field along $\pi _2\circ \gamma _2$, and $\gamma _2|_I$ lies in the regular region of $M_2$. Thus, we may use Lemma 1.6.1 of [8] to find the projectable Jacobi field ${\tilde{J}}_v(t)$ that projects to $d\pi _2({\tilde{J}}_v)=d\varphi (d\pi _1(J_v))$, and such that ${\tilde{J}}_v(t_0)=0.$ This ${\tilde{J}}_v$ is uniquely defined up to a Jacobi field in ${\mathcal {K}}_2$, and ${\tilde{J}}_v(0)$ is an eigenvector of $S_{x_2}$ with eigenvalue $\lambda =\frac{\sqrt{\kappa }}{\tan (\sqrt{\kappa }t_0)},$ which is well-defined up to an element of $K_{p_2}=ev_0({\mathcal {K}}_2)$. This defines the map $[v]\mapsto [{\tilde{J}}_v(0)]$ in (13).

Note by assumption the codimensions of the corresponding leaves are the same. By the above argument, we see that the corresponding dimensions of the eigenspaces are the same. Since the eigenvectors span the tangent spaces of the leaves, we see that $n_1=n_2.$ Now suppose that $M_1=S^{n_1}/\Gamma _1$ and $M_2=S^{n_2}/\Gamma _2$ are space forms of the same positive constant curvature $\kappa$, where $\Gamma _1$ and $\Gamma _2$ are finite subsets of the isometries of $S^{n_1}$ and $S^{n_2}$, respectively, acting properly discontinuously. Let ${\tilde{p}}_i: S^{n_i}\rightarrow M_i$ be the usual covering maps, for $i=1, \,2$. By means of Lemma 2.4 applied to each covering map, $\varphi : M_1/{\mathcal {F}}_1\rightarrow M_2/{\mathcal {F}}_2$ lifts to a smooth SRF isometry ${\tilde{\varphi }} :S^{n_1}/\tilde{{\mathcal {F}}}_1\rightarrow S^{n_2}/\tilde{{\mathcal {F}}}_2$ by ${\tilde{\varphi }}={\tilde{p}}_2^{-1}\circ \varphi \circ {\tilde{p}}_1,$ and we may conclude as in the preceding paragraph that $n_1=n_2.$ By Lemma 2.4, the map ${\tilde{\varphi }}$ preserves the mean curvature vector field if and only if $\varphi$ does. By the previously covered case for spheres, ${\tilde{\varphi }}$ preserves mean curvature, and we have the desired result for these space forms as well.

If $\kappa =0$, then all the non-zero eigenvalues of $S_{x_1}$ are the same as the non-zero eigenvalues of $S_{x_2}$ and have the same multiplicities for $M_1={\mathbb {R}}^{n_1}$ and $M_2={\mathbb {R}}^{n_2}$ by Proposition 3.1 of [3].

Suppose now $M_1={\mathbb {R}}^{n_1}/\Gamma _1$ and $M_2={\mathbb {R}}^{n_2}/\Gamma _2$ where $\Gamma _1$ and $\Gamma _2$ are finite subsets of the isometries of the appropriate Euclidean spaces, acting properly discontinuously. The covering map argument of the $\kappa >0$ case can then be modified to show the result follows for compact curvature zero space forms as well. $\square$