From Busemann function to Cheeger-Gromoll splitting
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This post is a part of the memoire of my M1 internship at I2M. The memoire contains, needless to say, less errors than this page.
We will prove the following result by Cheeger and Gromoll by a slightly modified approach of A. Besse.
Let \(M\) be a complete, connected Riemannian manifold with non-negative Ricci curvature. Suppose that \(M\) contains a line then \(M\) is isometric to \(M'\times \mathbb{R}\) with \(M'\) a complete, connected Riemannian manifold with non-negative Ricci curvature.
Note that the notion of geodesic ray or geodesic line used here is rather strict: A geodesic line \(\gamma\) is a geodesic parameterized by \(\mathbb{R}\) such that the distance between two point is exactly the distance on the geodesic, for example, geodesic line, if it passes by \(p\in M\) with velocity \(v\) of norm 1, satisfies \[ d(exp_p(tv), exp_p(-sv)) = s+t,\quad \forall s,t>0 \]
1 Busemann function
Let \(\gamma\) be a geodesic ray. We construct the Busemann function \(b\) associated to the ray as \[ b(x) = \lim_{t\to+\infty}\left( t - d(x,\gamma(t))\right) \] where the limit exists because the sequence \(f_t:\ x\mapsto t - d(x,\gamma(t))\) is non-decreasing and bounded above by \(d(x,\gamma(0))\). The convergence is also uniform in every compact of \(M\).
In Euclidean space for example, the Busemann function is the orthogonal projection on \(\gamma\). We will see that in a Riemannian manifold with non negative curvature, the Busemann function will serve as a projection.
Now with a fixed \(x_0\in M\), the tangent vectors at \(x_0\) of the geodesics connecting \(x_0\) and \(\gamma(t)\) is in the unit sphere of \(T_xM\), which is compact. Let \(X\) be a limit point of these tangents vectors, we defined \[ b_{X,t}(x) = b(x_0) + t - d(x, exp_{x_0}(tX)) \] where \(exp_{x_0}(tX)\) is the geodesic starting at \(x_0\) with velocity \(X\).
- From the construction of \(X\), one has \(b(x_0) + t = b(exp_{x_0}(tX))\), therefore \(b_{X,t} \leq b\) with equality in \(x_0\). We say that \(b\) is supported by \(b_{X,t}\) at \(x_0\). In general a function \(f\) is supported by \(g\) at \(x_0\) if \(f(x_0)=g(x_0)\) and \(f\geq g\) in a neighborhood of \(x_0\).
- \(b_{X,t}\) is smooth and a computation in local coordinate gives \(\Delta b_{X,t} \geq -\frac{\dim M - 1}{t}\)
- \(\|\nabla b_{X,t}\| = 1\)
The estimation given on the second point of Remark \ref{rem:3-calcul-cheeger-gromoll} is established using Jacobi fields:
The function \(f(x) = d(x,x_0)\) satisfies at a point \(x\) out of the cut-locus of \(x_0\): \[ \nabla f(x) \leq \frac{n-1}{l} \] where \(n=\dim M, l = d(x,x_0) = f(x)\) in Riemannian manifold \(M\) with non-negative Ricci curvature.
Let \(N(t), 0\leq t \leq l\) be the velocity of the geodesic \(\gamma\) from \(x_0\) to \(x\), and \(E_1,\dots, E_{n-1},N\) be a parallel frame along \(\gamma\). Let \(J_i\) be the unique Jacobi fields along \(\gamma\) with \(J_i(l) = E_i(l)\) and \(J_i(0)=0\) (existence and uniqueness of \(J_i\) is due to the fact that \(x\) is not in the cut-locus).
Then basic manipulation of Jacobi fields gives (without the fact that curvature is non-negative): \[ \Delta f(x) = \int_0^l dt \sum_{i=1}^{n-1}\left(<\nabla_N J_i, \nabla_N J_i> - < R(N,J_i)J_i,N >\right) = \sum_{i=1}^{n-1} I_\gamma(J_i,J_i) \] where \(I_\gamma\) is the index form of \(\gamma\). Note that the Jacobi fields \(J_i\) coincide with the fields \(\frac{t}{l}E(t)\) at \(0\) and \(l\), therefore by the fundamental inequality of index form (see Sakai Takashi, Riemannian geometry for details about Jacobi fields and Fundamental inequality of index form): \[ I_\gamma(J_i,J_i)\leq I_\gamma(\frac{t}{l}E_i,\frac{t}{l}E_i) \] hence \[ \Delta f(x) \leq \int_0^l \sum_{i=1}^{n-1}<\nabla_N \frac{t}{l}E_i, \nabla_N \frac{t}{l}E_i> - < R(N,\frac{t}{l}E_i)\frac{t}{l}E_i,N > \]
The curvature term being \( \frac{t^2}{l^2}Ric(N,N)\) therefore non-negative, one has \[ \Delta f(x) \leq \int_0^l dt \sum_{i=1}^{n-1}<\nabla_N \frac{t}{l}E_i, \nabla_N \frac{t}{l}E_i> = \frac{n-1}{l}. \]
We also note that for Theorem No description for this link it suffices to show that \(b\) is harmonic. In fact, from the smoothness one has \(\nabla b(x_0) =\nabla b_{X,t}(x_0)\), which means \(\|\nabla b\| = 1\) at every point in \(M\). For each point \(y\in M\), there exists a unique \(x\) with \(b(x)=0\) and time \(t\) when the flow of \(\nabla b\) starting at \(x\) arrives at \(y\). \(M\) is therefore homeomorphic to \(\bar M\times \mathbb{R}\) by the map \(F: y\mapsto (x,t)\). In order that \(F\) is an isometry, it suffices to prove that the gradient field \(\nabla b\) is parallel. In fact, \(\bar M\) being equiped with the restriction of the metric on \(M\), the fact that \(F\) is isometric is equivalent to the the fact that the flow \(\Phi^t\) of \(\nabla b\) is isometric for every time \(t\), which means \( \frac{d}{dt} <\Phi_*^t u, \Phi_*^t u>\) vanishes at \(t=0\). But \[ \frac{d}{dt}<\Phi_*^t u,\Phi_*^t u>|_{t=0} = 2<\nabla_{\partial t}\Phi_*^t u, u>|_{t=0} = 2<\nabla_u \nabla b,u> \] where in the second equality we used Schwarz lemma for commuting derivatives of \(\Phi(t,x) = \Phi^t(x)\). The vanishing of \(<\nabla_u \nabla b(x),u>\) for every vector \(u\) is, by bilinearity, equivalent to that of \(\nabla_u \nabla b\) for every \(u\), meaning that \(\nabla b\) is parallel.
The fact that \(\nabla b\) is parallel is due to a simple computation: \[ Ric(N,N) = -N(\Delta b) - \|\nabla N \|^2 \] where \(\|\nabla N\|^2 = \sum_{i=1}^{n-1}\sum_{j=1}^{n-1}<\nabla_{E_i}N, E_j>^2\). We see that \(N = \nabla b\) is parallel if \(\Delta b =0\).
- One can show (see A. Besse) that every gradient field \(\nabla b\) of norm 1 at every point is actually harmonic.
- Using de Rham decomposition, one has directly the splitting of \(M\) if it is simply-connected since \(N\) is parallel and \(M\) is complete.
2 Harmonicity
The Busemann function associated to a geodesic ray is subharmonic, it is a consequence of the following lemma.
In a connected Riemannian manifold, if a continuous function \(f\) is supported at any point \(x\) by a family \(f_\epsilon\) (depending on \(x\)) with \(\Delta(f_\epsilon)\leq \epsilon\), then \(f\) can not attains its maximum (unless when \(f\) is constant).
Given a small geodesic ball \(B\), suppose that we have a function \(h\) on \(B\) with \(\Delta h <0\) in \(B\) and \(f+h\) attains maximum at \(x\) in the interior of \(B\). Then \(f_\epsilon + h\) also attains maximum at \(x\), which means \(\Delta f_\epsilon + \Delta h \geq 0\), which is contradictory.
For the construction of the function \(h\), one suppose that \(B\) is small enough such that \(f|_{\partial B} \leq max_B f=: f(x_0)\) and equality is not attained at every points in \(\partial B\). Then choose \[ h = \eta (e^{\alpha \phi} - 1) \] with and \(\phi(x) = -1\) if \(x\in \partial B\) and \(f(x) = f(x_0)\), \(\phi(x_0) = 0\), \(\nabla \phi \ne 0\) and a large \(\alpha\) such that \[ \Delta h = \eta (-\alpha^2\| \nabla \phi\| + \alpha \Delta \phi)e^{\alpha \phi}. \] is negative.
Now for subharmonicity of \(b\), given a harmonic function \(h\) that coincides with \(b\) in the boundary \(\partial B\) of a geodesic ball \(B\), then \(b-h\) is supported by \(b_{X,t} - h\) with \(\Delta (b_{X,t}-h) \to 0\) as \(t\) tends to \(+\infty\), therefore \(b-h \leq (b-h)|_{\partial B} = 0\) in \(B\). We have just proved the following lemma:
The Busemann function of a geodesic ray in a Riemannian manifold \(M\) with non-negative Ricci curvature is subharmonic.
Now let \(b_+\) be the function previously constructed for the ray \(\gamma|_{[0,+\infty[}\) and \(b_-\) the Busemann function for the ray \(\tilde\gamma|_{[0,+\infty[}\) where \(\tilde\gamma(t) = \gamma(-t)\). Note that \(b_+ + b_-\leq 0\) with equality on the line \(\gamma\), but the sum is subharmonic therefore by maximum principle \(b_+ + -b_- = 0\) and \(b\) is harmonic therefore smooth. The splitting theorem of Cheeger-Gromoll follows.
3 Application
A consequence of Theorem No description for this link is the following result from J.Cheeger- D.Gromoll, The splitting theorem for manifold of nonnegative Ricci curvature (Theorem 2)
Let \(M\) be a compact Riemannian manifold with non-negative Ricci curvature, then the universal covering space of \(M\) is of form \( \tilde M = \mathbb{R}^n\times \bar M\) where \(\bar M\) does not contain any lines. Then \(\bar M\) is compact.
It suffices to prove that if \(\bar M\) is not compact, then it contains a line. In fact, it is easy to see that such \(\bar M\) must contains a (strict) geodesic ray. In fact it is obvious that with a fixed \(p\in M\) the function \[ F: v\mapsto \inf \{t>0: d(p,exp_p(tv)) < t\} \] defined on the unit ball \(U_p\) of \(T_p\bar M\) is upper semi-continuous. Therefore if \(F(v)<\infty\) for all unit tangent vector \(v\) at \(p\) then \(F\) is bounded above in \(U_p\) by a constant \(c\). Therefore \(\bar M\subset exp_p(cU_p)\) which is compact (contradiction). Therefore there exists a minimal ray at every point \(p\in \bar M\).
The existence of a line in general might not be true, the only extra property of \(\bar M\) that we will need is that it has a (fundamental) domain \(K\) compact and a family \(\sigma_i\) of isometries such that \(\bar M = \cup_i \sigma_i K\).
Let us first prove that such domain \(K\) exists. Remark that every isometry of \( \mathcal{M}\) acts separately on \( \mathcal{M}\), i.e. of form \(\sigma(u) = (\sigma_1(x), \sigma_2(y))\) for \(u=(x,y)\in \mathcal{M}\) with \(g_1, g_2\) isometries of \( \mathbb{R}^n\) and \(\bar M\). This can be seen by the uniqueness part of de Rham decomposition or simply by noticing that a tangent vector in the \(T_x\mathbb{R}^n\) component is characterized by the fact that its geodesics is a line. As the action of \(\pi_1(M)\) on \(\tilde M\) is free and proper, it has a fundamental domain \(H\). We then can choose \(K\) to be the projection of \(H\) on \(\bar M\) and \(\{\sigma_i\}\) to be the projection of \(\pi_1(M)\) on \(Isom(\bar M)\).
Now let \(\gamma\) be a minimal ray starting from \(p\in M\), for each \(x\in \gamma\) there exists an isometry \(\sigma\) of \(\bar M\) such that \(\sigma(x)\in K\). By compactness of \(K\), there exists a sequence \(t_n \to +\infty\) with \(x_n =\gamma(t_n)\), \(v_n = \dot\gamma(t_n)\) that satisfies \(y_n = \sigma_n(x_n) \to y\in K\) and even more, \((\sigma_n)_* v_n \to v\in T_y\bar M\) in the tangent bundle \(T\bar M\). Then the geodesic of \(\bar M\) starting at \(y\) with vector \(v\) is a line. In fact it suffices to prove that \(d(exp_y(t v), exp_y(-s v)) = s+t\) for \(s,t>0\), but for \(n\) large enough that \(t_n > s\) one has \[ d(exp_{y_n}(tv_n), exp_{y_n}(-sv_n)) = s+t \] then let \(n\to +\infty\), one sees that \(\bar M\) contains a geodesic line, which is contradictory.