On the Berger conjecture for manifolds all of whose geodesics are closed

Abstract

We define a Besse manifold as a Riemannian manifold (M, g) all of whose geodesics are closed. A conjecture of Berger states that all prime geodesics have the same length for any simply connected Besse manifold. We firstly show that the energy function on the free loop space of a simply connected Besse manifold is a perfect Morse–Bott function with respect to a suitable cohomology. Secondly we explain when the negative bundles along the critical manifolds are orientable. These two general results, then lead to a solution of Berger’s conjecture when the underlying manifold is a sphere of dimension at least four.

Appendix A. Small subsets of \({\mathsf {Sp}}(n-1,\omega )\)

Let $(\mathbb {R}^{2(n-1)},\omega )$ be the symplectic vector space, and let ${\mathsf {Sp}}(n-1,\omega )=\{P\in {\mathrm {GL}}(2(n-1),\mathbb {R})\mid \omega (Ax,Ay)=\omega (x,y)\}$ be the real symplectic group. In this appendix we focus on the subspaces of symplectic matrices with real eigenvalues of higher geometric multiplicity.

Lemma A.1

Let $P\in {\mathsf {Sp}}(n-1,\omega )$ and let $\lambda$ be a positive real eigenvalue of P of algebraic multiplicity a.

• (a) If $\lambda \ne 1$ then up to conjugation with an element of ${\mathsf {Sp}}(n-1,\omega )$, the matrix P can be written as $P={\mathrm {diag}}(\lambda U^{{\text {tr}}}, (\lambda U)^{-1}, R)$ where $U\in {\mathrm {GL}}(a,\mathbb {R})$ is unipotent, $R\in {\mathsf {Sp}}(n-a-1,\omega )$, and $U^{{\text {tr}}}$ denotes the transpose of U.

• (b) If $\lambda =1$ then up to conjugation with some element of ${\mathsf {Sp}}(n-1,\omega )$, the matrix P can be written as $P={\mathrm {diag}}(U, R)$ where $U\in {\mathsf {Sp}}(a,\omega )$ is unipotent and $R\in {\mathsf {Sp}}(n-a-1,\omega )$.

Proof

By the so-called refined Jordan decomposition, there are commuting matrices $P_s, P_u\in {\mathsf {Sp}}(n-1,\omega )$ such that $P_s$ is diagonalizable over $\mathbb {C}$, $P_u$ is unipotent, and $P=P_sP_u$. In particular, $P_s$ has the same eigenvalues of P with the same algebraic multiplicities.

Let $E_1$ denote the direct sum of the eigenspaces of $P_s$ of eigenvalues $\lambda$ and $\lambda ^{-1}$, and $E_2$ the sum of the other eigenspaces. Since $E_1$ and $E_2$ are symplectic subspaces (see for example [2]) then, up to conjugation with a symplectic matrix, we can write $P_s={\mathrm {diag}}(\lambda {\mathrm {Id}}_a,\lambda ^{-1}{\mathrm {Id}}_a, R_1)$ (if $\lambda \ne 1$) or $P_s={\mathrm {diag}}({\mathrm {Id}}_{2a},R_1)$ (if $\lambda =1$) for some $R_1\in {\mathsf {Sp}}(n-a-1,\omega )$.

Since $P_u$ commutes with $P_s$, it can be written either as $P_u={\mathrm {diag}}(U^{{\text {tr}}},U^{-1}, R_2)$ for some $U\in {\mathrm {GL}}(a,\mathbb {R})$ unipotent and $R_2\in {\mathsf {Sp}}(n-a-1,\omega )$ (if $\lambda \ne 1$), or $P_u={\mathrm {diag}}(U, R_2)$ for some $U\in {\mathsf {Sp}}(a,\omega )$ unipotent and $R_2\in {\mathsf {Sp}}(n-a-1,\omega )$ (if $\lambda =1$).

Since $P=P_uP_s$ we have proved the lemma. $\square$

Given an algebraic group $G\subseteq {\mathrm {GL}}(N,\mathbb {R})$, recall that a torus $T\subseteq G$ is a connected, abelian subgroup whose elements are diagonalizable over $\mathbb {C}$. Every algebraic group admits at least one torus of maximal dimension, called maximal torus, which is unique up to conjugation by an element of G, and the rank of G, denoted ${\text {rk}}\, G$, is defined as the dimension of a maximal torus of G. We will be mostly concerned with $G={\mathsf {Sp}}(N,\omega )$, in which case ${\text {rk}}\, G=N$. The following Lemma is a consequence of well-known results, but we could not find any reference in the literature.

Lemma A.2

Given an algebraic group $G\subseteq {\mathrm {GL}}(N,\mathbb {R})$, the set $G_u$ of unipotent elements in G is an affine variety of codimension equal to the rank of G.

Proof

The set $G_u$ is invariant under the action of G by conjugation. Fixing B a Borel (i.e., maximal connected solvable) subgroup of G, let $B_u$ denote the subset of unipotent elements in B. Every G-orbit meets B at least once by [4, 11.10], and therefore the map $G\times B_u\rightarrow G_u$ sending (g, A) to $gAg^{-1}$ is surjective. The normalizer $N(B_u)=\{g\in G\mid gB_ug^{-1}\subseteq B_u\}$ coincides with B by a theorem of Chevalley [4, 11.16] and therefore $\dim B_u+\dim G-\dim B=\dim G_u$. By [4, 10.6] $B_u$ is normal in B and $B/B_u$ is isomorphic to a maximal torus, in particular $\dim B=\dim B_u+{\text {rk}}\,G$. With the equation before, we obtain $\dim G_u=\dim G-{\text {rk}}G$. $\square$

Let ${\mathsf {Sp}}_1(n-1,\omega )$ denote the space of symplectic matrices whose positive real eigenvalues have geometric multiplicity 1. The next result shows that the complement of ${\mathsf {Sp}}_1(n-1,\omega )$ in ${\mathsf {Sp}}(n-1,\omega )$ has codimension at least 3.

Proposition A.3

The set of matrices $P\in {\mathsf {Sp}}(n-1,\omega )$ with some eigenvalue $\lambda \in (0,1]$ of geometric multiplicity $>1$ has codimension $\ge 3$ in ${\mathsf {Sp}}(n-1,\omega )$.

Proof

Given $\lambda$ in (0, 1] let ${\mathcal {M}}_{\lambda }$ denote the space of matrices in ${\mathsf {Sp}}(n-1,\omega )$ with eigenvalue $\lambda$ of geometric multiplicity at least 2. This set can be also described as

$$\begin{aligned} {\mathcal {M}}_{\lambda }=\left\{ X\in {\mathsf {Sp}}(n-1,\omega )\mid {\text {rk}}(X-\lambda I)\le 2(n-2)\right\} \end{aligned}$$

from which it follows that ${\mathcal {M}}_\lambda$ is an algebraic variety, and we can talk about its dimension. To prove the lemma, it is enough to prove that the codimension of ${\mathcal {M}}_{\lambda }$ is $\ge 4$ for $\lambda \ne 1$, and $\ge 3$ if $\lambda =1$.

We also define ${\mathcal {M}}_\lambda (n_1,n_2)$ to be the subspace of matrices P in ${\mathcal {M}}_\lambda$ such that the generalised eigenspace with eigenvalue $\lambda$ can be written as a sum of two P-invariant subspaces of dimension $n_1,n_2$. The set ${\mathcal {M}}_\lambda$ consists of a finite union of the ${\mathcal {M}}_\lambda (n_1,n_2)$ and it suffices to show that each of them has the required codimension.

Suppose first that $\lambda \ne 1$. Fixing one ${\mathcal {M}}={\mathcal {M}}_\lambda (n_1,n_2)$, let $\Sigma \subseteq {\mathcal {M}}_{\lambda }$ denote the subset of matrices P that, in some fixed basis, can be written as $P={\mathrm {diag}}(P_1,R)$, with $P_1, R$ both symplectic, and $P_1$ decomposing further as ${\mathrm {diag}}(\lambda U_1^{{\text {tr}}}, (\lambda U_1)^{-1},\lambda U_2^{{\text {tr}}}, (\lambda U_2)^{-1})$ where $U_1\in {\mathrm {GL}}(n_1,\mathbb {R})$, $U_2\in {\mathrm {GL}}(n_2,\mathbb {R})$ have the form

$$\begin{aligned} \left( \begin{array}{cccc}1 &{} &{} &{} \\ 1&{} 1 &{} &{} \\ &{} \ddots &{} \ddots &{} \\ &{} &{} 1 &{} 1\end{array}\right) \end{aligned}$$

The set ${\mathcal {M}}$ is preserved under the action of ${\mathsf {Sp}}(n-1,\omega )$ on itself by conjugation and, by Lemma A.1, every orbit meets $\Sigma$ in at least one point. Therefore, the map $\Sigma \times {\mathsf {Sp}}(n-1,\omega )\rightarrow {\mathcal {M}}$ is surjective and, letting $N(\Sigma )=\{A\in {\mathsf {Sp}}(n-1,\omega )\mid A\Sigma A^{-1}\subseteq \Sigma \}$ denote the normalizer of $\Sigma$, we have

$$\begin{aligned} \dim {\mathcal {M}}=\dim {\mathsf {Sp}}(n-1,\omega )+ \dim \Sigma -\dim N(\Sigma ). \end{aligned}$$

(A.1)

Clearly a matrix $P={\mathrm {diag}}(P_1,R)$ in $\Sigma$ is uniquely determined by $R\in {\mathsf {Sp}}(n'-1,\omega )$, $n'=n-n_1-n_2$, and therefore $\dim \Sigma =\dim {\mathsf {Sp}}(n'-1,\omega )$.

We now compute $N(\Sigma )$. Suppose that $n_1\le n_2$, and let ${\mathcal {A}}\subseteq {\mathrm {GL}}(n_1+n_2,\mathbb {R})$ be the set of matrices such that

Clearly $\dim {\mathcal {A}}=3n_1+n_2\ge 4$. For any matrix $A\in {\mathcal {A}}$ and $B\in {\mathsf {Sp}}(n'-1,\omega )$, the matrix ${\mathrm {diag}}(A^{{\text {tr}}},A^{-1},B)$ lies in $N(\Sigma )$. In particular, $\dim N(\Sigma )\ge \dim {\mathcal {A}}+{\mathsf {Sp}}(n'-1,\omega )$ and therefore $\dim {\mathcal {M}}\le \dim {\mathsf {Sp}}(n-1,\omega )-\dim {\mathcal {A}}\le {\mathsf {Sp}}(n-1,\omega )-4$.

If $\lambda =1$ then any $P\in {\mathcal {M}}_{\lambda }$ can be written, in a suitable symplectic basis, as ${\mathrm {diag}}(U,R)$ where $U\in {\mathsf {Sp}}(n_0,\omega )$, for some $n_0$, is unipotent with at least two linearly independent eigenvectors, and $R\in {\mathsf {Sp}}(n-n_0-1,\omega )$. We now define $\Sigma$ to be the set of matrices that, under the same fixed basis, can be written as ${\mathrm {diag}}(U',R')$ for some $R'\in {\mathsf {Sp}}(n'-1,\omega )$ and some unipotent matrix $U'\in {\mathsf {Sp}}(n_0,\omega )$. If we let ${\mathsf {Sp}}(n_0,\omega )_u$ denote the set of unipotent matrices in ${\mathsf {Sp}}(n_0,\omega )$, we have

$$\begin{aligned} \dim \Sigma = \dim {\mathsf {Sp}}(n_0,\omega )_u+\dim {\mathsf {Sp}}(n-n_0-1,\omega ). \end{aligned}$$

where ${\mathsf {Sp}}(n_0, \omega )_u$ denote the unipotent matrices in ${\mathsf {Sp}}(n_0, \omega )$. The normalizer $N(\Sigma )$ contains the matrices of the form ${\mathrm {diag}}(P_1,R')$ with $P_1\in {\mathsf {Sp}}(n_0,\omega )$ and $R'\in {\mathsf {Sp}}(n-n_0-1,\omega )$, thus

$$\begin{aligned} \dim N(\Sigma )\ge \dim {\mathsf {Sp}}(n_0,\omega )+\dim {\mathsf {Sp}}(n-n_0-1,\omega ). \end{aligned}$$

Once again ${\mathsf {Sp}}(n-1,\omega )$ acts on ${\mathcal {M}}_\lambda$ and by Lemma A.1 every orbit meets $\Sigma$. In particular, ${\mathcal {M}}_\lambda$ is contained in the space spanned by the orbits of $\Sigma$, and $\dim {\mathcal {M}}_\lambda \le \dim {\mathsf {Sp}}(n-1,\omega )-(\dim N(\Sigma )-\dim \Sigma )$. By the computation above and Lemma A.2, we have

$$\begin{aligned} \dim {\mathcal {M}}_\lambda&\le \dim {\mathsf {Sp}}(n-1,\omega )- {\mathrm {rk}}\,{\mathsf {Sp}}(n_0,\omega )\\&= \dim {\mathsf {Sp}}(n-1,\omega )-n_0. \end{aligned}$$

The codimension of ${\mathcal {M}}_\lambda$ is then $\ge 3$, unless possibly when $n_0=1$ or 2.

If $n_0=1$, then every matrix in ${\mathcal {M}}_{\lambda }$ can be written, under some basis, as

$$\begin{aligned} {\mathrm {diag}}({\mathrm {Id}}_2,R),\qquad R\in {\mathsf {Sp}}(n-2,\omega ) \end{aligned}$$

(A.2)

Fixing a basis and letting $\Sigma$ denote the space of matrices that, in the fixed basis, can be written as in (A.2), we have that $\dim \Sigma =\dim {\mathsf {Sp}}(n-2,\omega )$ and the normalizer $N(\Sigma )$ contains ${\mathsf {Sp}}(1,\omega )\times {\mathsf {Sp}}(n-2,\omega )$. Therefore, $\dim N(\Sigma )-\dim \Sigma \ge 3$. Using Eq. (A.1), we obtain $\dim {\mathcal {M}}_\lambda \le \dim {\mathsf {Sp}}(n-1,\omega )-3$.

If $n_0=2$, then every matrix in ${\mathcal {M}}_{\lambda }$ can be written, under some basis, as

$$\begin{aligned} U= {\mathrm {diag}}(U_1,U_2)\qquad U_i=\left( \begin{array}{cc}1 &{} \\ \sigma _i &{} 1\end{array}\right) ,\, \sigma _i\in \{-1,0,1\}. \end{aligned}$$

(A.3)

Fixing a basis and letting $\Sigma$ denote the space of matrices that, in the fixed basis, can be written as in (A.3), we have $\dim \Sigma =\dim {\mathsf {Sp}}(n-3,\omega )$. If for example $\sigma _1=\sigma _2=1$ then $N(\Sigma )$ contains all the matrices of the form ${\mathrm {diag}}(P_1,R)$ where $R\in {\mathsf {Sp}}(n-3,\omega )$ is any matrix, and $P_1\in {\mathsf {Sp}}(2,\omega )$ has the form

where a, b, c, d satisfy the linear equation $(a+b)\cos \theta =(c-d)\sin \theta$. Therefore $\dim N(\Sigma )\ge \dim {\mathsf {Sp}}(n-3,\omega )+4$ and, by Eq. (A.1), we obtain $\dim {\mathcal {M}}_\lambda \le \dim {\mathsf {Sp}}(n-1,\omega )-4$. The same computations can be checked for the other values of $\sigma _1$ and $\sigma _2$. $\square$

Let ${\mathcal {G}},{\mathcal {G}}_1$ denote the subspaces of ${\mathsf {Sp}}(n-1,\omega )\times \mathbb {R}_+$ given by

$$\begin{aligned} {\mathcal {G}}&=\left\{ (P,\lambda )\in {\mathsf {Sp}}(n-1,\omega )\times \mathbb {R}_+\mid \dim \ker (P-\lambda \, {\mathrm {Id}})\le 1 \right\} \\ {\mathcal {G}}_1&=\left\{ (P,\lambda )\in {\mathsf {Sp}}(n-1,\omega )\times \mathbb {R}_+\mid \dim \ker (P-\lambda \, {\mathrm {Id}})= 1 \right\} \end{aligned}$$

Clearly ${\mathcal {G}}$ is open and dense in ${\mathsf {Sp}}(n-1,\omega )\times \mathbb {R}_+$, ${\mathcal {G}}_1\subseteq {\mathcal {G}}_0$ and, by construction, we have ${\mathsf {Sp}}_1(n-1,\omega )\times \mathbb {R}_+\subseteq {\mathcal {G}}$.

Proposition A.4

The map

$$\begin{aligned} \chi :{\mathcal {G}}&\longrightarrow \mathbb {R}\\ (P,\mu )&\longmapsto \det (P-\mu \,{\mathrm {Id}}) \end{aligned}$$

is a submersion in a neighbourhood of ${\mathcal {G}}_1$.

Proof

We are going to prove a stronger statement. In fact, we prove that for any $(S,\lambda )\in {\mathcal {G}}_0$ we can find a vector $v_{\scriptscriptstyle (S,\lambda )}\in T_S{\mathsf {Sp}}(n-1,\omega )$ such that $d_{\scriptscriptstyle (S,\lambda )}\chi (v_{\scriptscriptstyle (S,\lambda )})> 0$.

Let a denote the algebraic multiplicity of $\lambda$ in S. By Lemma A.1, there is a symplectic basis such that S can be written as $S={\mathrm {diag}}(S_1,S_2)$ where $S_1\in {\mathsf {Sp}}(a,\omega )$ only contains the eigenvalues $\lambda , \lambda ^{-1}$ and $S_2\in {\mathsf {Sp}}(n-1-a,\omega )$ has eigenvalues different from $\lambda$ and $\lambda ^{-1}$.

If $\lambda \ne 1$ then by Lemma A.1 we can write $S_1={\mathrm {diag}}(\lambda U^{{\text {tr}}}, (\lambda U)^{-1})$, where

$$\begin{aligned} \lambda U^{{\text {tr}}}=\left( \begin{array}{cccc}\lambda &{} 1 &{} &{} \\ &{} \lambda &{} \ddots &{} \\ &{} &{} \ddots &{} 1 \\ &{} &{} &{} \lambda \end{array}\right) . \end{aligned}$$

If $\lambda >1$ let $v={\mathrm {diag}}(-E_{a,1},E_{1,a})\in \mathfrak {sp}(a,\omega )$ otherwise let $v=(-1)^{a}{\mathrm {diag}}(-E_{a,1},E_{1,a})$. In either case, let $v'={\mathrm {diag}}(v,0)\in \mathfrak {sp}(n-1,\omega )$ and $v_{\scriptscriptstyle (S,\lambda )}={L_S}_*v'\in T_S{\mathsf {Sp}}(n-1,\omega )$, one can compute

$$\begin{aligned} d_{\scriptscriptstyle (S,\lambda )} \chi (v_{\scriptscriptstyle (S,\lambda )}) =\lambda \left| \lambda -\lambda ^{-1}\right| ^a>0. \end{aligned}$$

If $\lambda =1$, then $S_1$ can be written in the following block form

$$\begin{aligned} S_1=\left( \begin{array}{c c} U^{-1} &{} \\ U^{{\text {tr}}}T &{} U^{{\text {tr}}}\end{array}\right) \end{aligned}$$

where T is a symmetric matrix, and

$$\begin{aligned} U^{{\text {tr}}}=\left( \begin{array}{ccc} 1 &{} \cdots &{} 1 \\ &{} \ddots &{} \vdots \\ &{} &{} 1\end{array}\right) ,\qquad U^{-1} =\left( \begin{array}{ccc} 1 &{} &{} \\ -1 &{} \ddots &{} \\ &{} -1 &{} 1 \end{array}\right) \end{aligned}$$

In order for $S_1$ to have geometric multiplicity 1, it must be $(UT)_{a,a}=c\ne 0$. Without loss of generality we can suppose that the sign of c is $(-1)^{a-1}$. Define

Letting $v'={\mathrm {diag}}(v,0)\in \mathfrak {sp}(n-1,\omega )$ and $v_{\scriptscriptstyle (S,\lambda )}={L_S}_*v'\in T_S{\mathsf {Sp}}(n-1,\omega )$ we can easily compute that

$$\begin{aligned} d_{\scriptscriptstyle (S,\lambda )} \chi (v_{\scriptscriptstyle (S,\lambda )})= (-1)^{a-1}c> 0. \end{aligned}$$

$\square$

From Proposition A.4 we obtain the following stronger, more global result.

Proposition A.5

There exists a vector field V on ${\mathsf {Sp}}_1(n-1,\omega )$ such that for every $S\in {\mathsf {Sp}}_1(n-1,\omega )$ and every real eigenvalue $\lambda$ of S, $d_{\scriptscriptstyle (S,\lambda )}\chi (V)>0$.

Proof

Given $S\in {\mathsf {Sp}}_1(n-1,\omega )$ and $\lambda \in (0,1]$ a real eigenvalue of S, let $v_{\scriptscriptstyle (S,\lambda )}\in T_S{\mathsf {Sp}}_1(n-1,\omega )$ be the vector constructed in the previous proposition, so that $d_{\scriptscriptstyle (S,\lambda )}\chi (v_{\scriptscriptstyle (S,\lambda )})>0$. It can be easily checked that, at the point $(S,\lambda ^{-1})$, the differential of $\chi$ is

$$\begin{aligned} d_{\scriptscriptstyle (S,\lambda ^{-1})} \chi (v_{\scriptscriptstyle (S,\lambda )}) =-\lambda ^{1-2a}(\lambda -\lambda ^{-1})^a> 0 \end{aligned}$$

and moreover for any other eigenvalue $\mu$ of S different from $\lambda$ and $\lambda ^{-1}$, one has $d_{\scriptscriptstyle (S,\mu )}\chi (v_{\scriptscriptstyle (S,\lambda )})=0$. In particular, letting $v_S=\sum _{\lambda } v_{\scriptscriptstyle (S,\lambda )}$, where the sum is taken over all the real eigenvalues of S in (0, 1], the $v_S$ satisfies

$$\begin{aligned} d_{\scriptscriptstyle (S,\lambda )}\chi (v_S)>0 \end{aligned}$$

for every real eigenvalue $\lambda$ of S. By continuity, we can find a neighbourhood $U_S$ of S and an extension $V_S$ of $v_S$ such that for every $S'\in U_S$ and $\lambda '$ real eigenvalue of $S'$, we have $d_{\scriptscriptstyle (S',\lambda ')}\chi (V_S)>0$.

The open sets $\{U_S\}_{S\in {\mathsf {Sp}}_1(n-1,\omega )}$ form an open cover of ${\mathsf {Sp}}_1(n-1,\omega )$. Choosing a countable subcover $\{U_{S_i}\}_i$ with a subordinate partition of unity $\{\lambda _i\}_i$, the vector field

$$\begin{aligned} V=\sum _i\lambda _i V_{S_i} \end{aligned}$$

has the required properties. $\square$

Proposition A.4 implies that ${\mathcal {G}}_1$ is a smooth hypersurface in ${\mathcal {G}}$. Consider now the projection $\pi :{\mathcal {G}}_0\rightarrow \mathbb {R}$, sending $(Q,\lambda )$ to $\lambda$, and let ${\mathcal {G}}_0=\pi ^{-1}(1)$.

Lemma A.6

The map $\pi :{\mathcal {G}}_1\rightarrow \mathbb {R}$ is a submersion around ${\mathcal {G}}_0$.

Proof

It is enough to find, for every point $(Q,1)\in {\mathcal {G}}_1$, a vector $v\in T_{\scriptscriptstyle (Q,1)}{\mathcal {G}}_1$ such that $d_{\scriptscriptstyle (Q,1)}\pi (v)\ne 0$. Fixing (Q, 1), we know in particular that 1 is an eigenvalue of Q and therefore Q can be written, in some basis, as $Q={\mathrm {diag}}(P,R)$ where $P\in {\mathsf {Sp}}(a,\omega )$, $R\in {\mathsf {Sp}}(n-1-a,\omega )$ and moreover

$$\begin{aligned} P=\left( \begin{array}{c@{\quad }c}U^{-1} &{} 0 \\ U^{{\text {tr}}}T &{} U^{{\text {tr}}}\end{array}\right) \end{aligned}$$

with T symmetric and U unipotent such that

$$\begin{aligned} U^{{\text {tr}}}=\left( \begin{array}{ccc}1 &{} \cdots &{} 1 \\ &{} \ddots &{} \vdots \\ &{} &{} 1\end{array}\right) . \end{aligned}$$

For some t small, let

$$\begin{aligned} P(t)=\left( \begin{array}{c@{\quad }c}e^{-t}U^{-1} &{} 0 \\ e^tU^{{\text {tr}}}T &{} e^tU^{{\text {tr}}}\end{array}\right) , \quad Q(t)={\mathrm {diag}}(P(t),R). \end{aligned}$$

Then the path $(Q(t),e^{-t})$ lies in ${\mathcal {G}}_0$ for small t, and $\pi (Q(t),e^{-t})=e^{-t}$. In particular, letting

$$\begin{aligned} v={d\over dt}\bigg |_{t=0}(Q(t),e^{-t}) \end{aligned}$$

we obtain $d_{\scriptscriptstyle (Q,1)}\pi (v)\ne 0$ thus proving the Lemma. $\square$

The following Corollary is straightforward

Corollary A.7

The subset $\pi ^{-1}\big ((0,1]\big )\subseteq {\mathcal {G}}_1$ is a smooth manifold, with boundary ${\mathcal {G}}_0$.