Curvature and Fundamental Groups

hbghlyj

Mock Exam

• Prove each of the following statements for a compact manifold $M$ of dimension $n \geqslant 2$.
  
  • If $M$ admits a Riemannian metric $g$ with sectional curvature $K \leqslant 0$ then $M$ has an infinite fundamental group.
  • If $M$ admits a Riemannian metric with sectional curvature $K>0$ then $M$ has a finite fundamental group.
  
  For the converse of each the above statements, either prove it if it is true or give a counterexample if it is false, justifying your answer in each case.
  [You may assume the Cartan-Hadamard, Bonnet-Myers and Synge theorems, provided they are clearly stated. You may also assume the Hopf-Rinow theorem and that if $M_1, M_2$ are manifolds and $\pi: M_1 \to M_2$ is a covering map of finite degree, then $M_1$ is compact if $M_2$ is compact.]
• Let $(M, g)$ be a complete simply connected Riemannian manifold with $K \leqslant 0$. Let $\gamma: \mathbb{R} \to M$ be a normalised geodesic and suppose that $q \in M \setminus \gamma(\mathbb{R})$.
  
  • For $s \in \mathbb{R}$ let $\delta(s)=d(q, \gamma(s))$ and let $\alpha_s$ be the minimizing geodesic from $q$ to $\gamma(s)$. Show that for the variation $f(s, t)=\alpha_s(t)$ the energy $E_f$ of $f$ satisfies
    \[
    \frac{1}{2} E_f'(s)=g(\gamma'(s), \alpha_s'(\delta(s))) \quad \text { and } \quad E_f''(s)>0
    \]
  • Deduce that there exist a unique minimizing geodesic $\alpha$ from $q$ to a point $p$ on $\gamma$ so that $\alpha$ is orthogonal to $\gamma$ at $p$, i.e. $g(\alpha', \gamma')=0$ at $p$, and that $p$ is then the closest point to $q$ on $\gamma$.
    [You may use the First and Second Variation Formulas and their derivations without proof.]

hbghlyj

Let $(M,g)$ be a Riemannian manifold. The sectional curvature $K(p,\sigma)$ at a point $p \in M$ and a 2-dimensional subspace $\sigma \subset T_pM$ is defined as:
\[
K(p,\sigma) = \frac{g(R(X,Y)Y,X)}{g(X,X)g(Y,Y) - g(X,Y)^2}
\]
where $X, Y$ are linearly independent tangent vectors that span $\sigma$, and $R$ is the Riemann curvature tensor. This measures the Gaussian curvature of the 2-dimensional submanifold obtained by exponentiating $\sigma$ at $p$.
Relevant Theorems:

• Cartan-Hadamard Theorem: If $(M,g)$ is a complete, simply connected Riemannian manifold with $K \leq 0$, then the exponential map $\exp_p: T_pM \to M$ is a diffeomorphism for any $p \in M$. Thus, $M$ is diffeomorphic to $\mathbb{R}^n$.
• Bonnet-Myers Theorem: If $(M,g)$ is a complete Riemannian manifold with Ricci curvature bounded below by $\text{Ric} \geq (n-1)k > 0$ for some constant $k > 0$, then $M$ is compact with diameter $\leq \pi/\sqrt{k}$ and has finite fundamental group.
• Synge's Theorem: Let $M$ be a compact Riemannian manifold with positive sectional curvature. If $M$ is even-dimensional and orientable, then $M$ is simply connected. If $M$ is odd-dimensional, then it is orientable.

• • $K \leq 0$ implies infinite fundamental group
    Proof: Suppose $M$ is a compact manifold with a Riemannian metric $g$ such that $K \leq 0$. Let $\tilde{M}$ be its universal covering space.
    
    Since $M$ has $K \leq 0$, the metric $g$ lifts to a metric $\tilde{g}$ on $\tilde{M}$ with the same sectional curvature. By the Cartan-Hadamard theorem, $(\tilde{M},\tilde{g})$ is diffeomorphic to $\mathbb{R}^n$.
    
    Now, if the fundamental group $\pi_1(M)$ were finite, then the covering map $\pi: \tilde{M} \to M$ would have finite degree. Since $M$ is compact, by the given assumption, $\tilde{M}$ would also be compact. But $\tilde{M} \cong \mathbb{R}^n$, which cannot be compact. This contradiction shows that $\pi_1(M)$ must be infinite.
  • $K > 0$ implies finite fundamental group
    Proof: Let $M$ be a compact manifold with a Riemannian metric $g$ such that $K > 0$. Then, in particular, the Ricci curvature satisfies $\text{Ric} \geq (n-1)K_{\min} > 0$, where $K_{\min} > 0$ is the minimum sectional curvature.
    
    By the Bonnet-Myers theorem, $M$ has finite fundamental group.
  
  Converse Statements
  
  • "If $M$ has an infinite fundamental group, then $M$ admits a metric with $K \leq 0$." is false. Consider $M = S^2 \times S^1$. Then $\pi_1(M) \cong \mathbb{Z}$, which is infinite. However, $M$ does not admit a metric with nonpositive sectional curvature because the Cartan-Hadamard Theorem states that if the sectional curvature is nonpositive, the universal cover is $\mathbb{R}^n$. However, the universal cover of $M$ is $S^2 \times \mathbb{R}$, which is not diffeomorphic to $\mathbb{R}^3$.
  • "If $M$ has a finite fundamental group, then $M$ admits a metric with $K > 0$." is false. Consider $M=\mathbb{RP}^3$, which has finite fundamental group $\mathbb{Z}_2$. However, $M$ does not admit a metric of positive sectional curvature by Synge's Theorem.
• • The variation is defined as $f(s,t) = \alpha_s(t)$ for $t \in [0,\delta(s)]$.
    By the first variation formula for a geodesic with a fixed starting point and moving endpoint, we have:
    \[
    \frac{1}{2} E_f'(s) = g(\gamma'(s), \alpha_s'(\delta(s)))
    \]
    Let $J(t) = \frac{\partial f}{\partial s}(s_0,t)$ be the Jacobi field along $\alpha_{s_0}$ arising from this variation. By Second variation formula:
    \[
    \frac{1}{2} E_f''(s_0) = \int_0^{\delta(s_0)} \left[ \|\nabla_t J\|^2 - g(R(J, \alpha_{s_0}')\alpha_{s_0}', J) \right] dt
    \]In a manifold with $K \leq 0$, we have $g(R(J, \alpha_{s_0}')\alpha_{s_0}', J) \leq 0$. Therefore:
    \[
    \frac{1}{2} E_f''(s_0) \geq \int_0^{\delta(s_0)} \|\nabla_t J\|^2 dt > 0
    \]
    This inequality is strict because $J$ cannot be parallel along $\alpha_{s_0}$ (otherwise, $f$ would be a trivial variation). Thus, $E_f''(s) > 0$, establishing strict convexity.
  • Uniqueness of orthogonal geodesic
    
    From part (i), we have $\frac{1}{2}E_f'(s) = g(\gamma'(s), \alpha_s'(\delta(s)))$ and $E_f''(s) > 0$.
    
    Since $E_f''(s) > 0$, the function $E_f(s)$ is strictly convex and thus has a unique minimum. At this minimum $s_0$, we have $E_f'(s_0) = 0$, which implies:
    \[
    g(\gamma'(s_0), \alpha_{s_0}'(\delta(s_0))) = 0
    \]
    This means that the geodesic $\alpha_{s_0}$ from $q$ to $\gamma(s_0)$ is orthogonal to $\gamma$ at the point $\gamma(s_0)$.