Product of metrics of constant sectional curvature

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Mock Exam
Let $(M_j, g_j)$ be complete Riemannian manifolds of dimension $n_j \geqslant 2$ for $j=1,2$ and let $M=M_1 \times M_2$.
Given that $T_{(p_1, p_2)}(M_1 \times M_2) \cong T_{p_1} M_1 \times T_{p_2} M_2$ for all $(p_1, p_2) \in M_1 \times M_2$, we define a section $g$ of $S^2 T^*(M_1 \times M_2)$ by
\[
g_{(p_1, p_2)}((X_1, X_2),(Y_1, Y_2))=(g_1)_{p_1}(X_1, Y_1)+(g_2)_{p_2}(X_2, Y_2)
\]
for all $X_1, Y_1 \in T_{p_1} M_1$ and $X_2, Y_2 \in T_{p_2} M_2$, for all $(p_1, p_2) \in M_1 \times M_2$.

• • Show that $g$ is a complete Riemannian metric on $M_1 \times M_2$.
  • Determine the Ricci curvature tensor and scalar curvature of $(M, g)$ in terms of the Ricci curvature tensors and scalar curvatures of $(M_1, g_1)$ and $(M_2, g_2)$.
    [You may assume that if $\nabla$ is the Levi-Civita connection of $g$ and $\nabla_j$ is the Levi-Civita connection of $g_j$ for $j=1,2$ then
    \[
    \nabla_{(X_1, X_2)}(Y_1, Y_2)=(\nabla_1)_{X_1} Y_1+(\nabla_2)_{X_2} Y_2
    \]
    for all vector fields $X_1, Y_1$ on $M_1$ and $X_2, Y_2$ on $M_2$. You may also use any standard results about existence and uniqueness of geodesics.]
• Suppose further that $(M_j, g_j)$ have constant sectional curvature $K_j$ for $j=1,2$.
  
  • Show that if $n_1=n_2$ then $(M, g)$ is Einstein if and only if $K_1=K_2$.
  • Show that $(M, g)$ is scalar-flat and Einstein if and only if $K_1=K_2=0$.
  • Show that if $(M, g)$ is scalar-flat and $M_j$ are simply connected for $j=1,2$ then $M$ is non-compact.

hbghlyj

The Ricci curvature tensor $\mathrm{Ric}$ of a Riemannian manifold $(M, g)$ is defined as the trace of the Riemann curvature tensor $R$ on its first and third indices:$$\mathrm{Ric}(X, Y) = \mathrm{Tr}(Z \mapsto R(Z, X)Y)$$In local coordinates:$$\mathrm{Ric}_{ij} = R^k_{ikj}$$The scalar curvature $S$ is the trace of the Ricci tensor with respect to the metric:$$S = g^{ij} \mathrm{Ric}_{ij}$$

• • Completeness of $g$:
    The metric $g = g_1 + g_2$ on $M_1 \times M_2$ is Riemannian (symmetric, positive definite, smooth).
    A geodesic in $M_1 \times M_2$ is a pair of geodesics in $M_1$ and $M_2$. Since both $(M_1, g_1)$ and $(M_2, g_2)$ are complete, their geodesics exist for all time, by Hopf–Rinow theorem $(M, g)$ is complete.
  • Ricci and Scalar Curvatures:
    The Ricci tensor of $(M, g)$ splits as:$$\mathrm{Ric}_g((X_1, X_2), (Y_1, Y_2)) = \mathrm{Ric}_{g_1}(X_1, Y_1) + \mathrm{Ric}_{g_2}(X_2, Y_2)$$The scalar curvature is additive:$$S_g = S_{g_1} + S_{g_2}$$
• Now suppose that $(M_j,g_j)$ has constant sectional curvature $K_j$ for $j=1,2$. Recall that a manifold of dimension $n$ and sectional curvature $K$ has Ricci tensor
  $$
  \operatorname{Ric}_{g_j}=(n_j-1)K_jg_j,
  $$
  and its scalar curvature is
  $$
  \operatorname{scal}_{g_j}= n_j(n_j-1)K_j.
  $$
  
  • Einstein Condition When $n_1=n_2$
    A Riemannian metric $g$ is said to be Einstein if there is a constant $\lambda$ such that
    $$
    \operatorname{Ric}_g = \lambda g.
    $$
    In our product situation, if we evaluate on vectors tangent to $M_1$ (and similarly for $M_2$) we have:
    $$
    \operatorname{Ric}_g((X_1,0),(Y_1,0)) = \operatorname{Ric}_{g_1}(X_1,Y_1) = (n_1-1)K_1 g_1(X_1,Y_1).
    $$
    On the other hand, Einstein's condition requires that
    $$
    \operatorname{Ric}_g((X_1,0),(Y_1,0)) = \lambda\, g\bigl((X_1,0),(Y_1,0)\bigr) = \lambda g_1(X_1,Y_1).
    $$
    Thus, for all $X_1,Y_1$ we must have
    $$
    (n_1-1)K_1 = \lambda.
    $$
    A similar computation using vectors tangent to $M_2$ yields
    $$
    (n_2-1)K_2 = \lambda.
    $$
    If we assume $n_1=n_2$ (say, equal to $n$), the two conditions force
    $$
    (n-1)K_1 = (n-1)K_2 \quad \Longrightarrow \quad K_1 = K_2.
    $$Thus, when $n_1=n_2$ the product $(M,g)$ is Einstein if and only if $K_1=K_2$.
  • Scalar–Flat and Einstein Metrics
    A metric is scalar-flat if $\operatorname{scal}_g = 0$. In our situation, we have:
    $$
    \operatorname{scal}_g = n_1(n_1-1)K_1 + n_2(n_2-1)K_2.
    $$
    If in addition $(M,g)$ is Einstein (so that as above $\lambda = (n_1-1)K_1 = (n_2-1)K_2$), then its Einstein constant is given by
    $$
    \lambda=\frac{\operatorname{scal}_g}{n_1+n_2}.
    $$
    Thus, if $\operatorname{scal}_g=0$ then $\lambda=0$ and consequently
    $$
    (n_1-1)K_1 =0\quad \text{and}\quad (n_2-1)K_2=0.
    $$
    Since each $n_j\ge2$ (hence $n_j-1 > 0$) it follows that $K_1=K_2=0$. Conversely if $K_1=K_2=0$ then clearly $\operatorname{Ric}_{g_j}=0$ and the product metric is Ricci-flat (thus Einstein with $\lambda=0$) and scalar flat. Hence, $(M,g)$ is scalar–flat and Einstein if and only if $K_1=K_2=0$.
  • Noncompactness When Scalar–Flat and Factors are Simply Connected
    Under the assumption that $(M,g)$ is scalar–flat and that each $M_j$ is simply connected, we already saw in part (ii) that scalar-flat together with the Einstein condition forces $K_1=K_2=0$. But a complete, simply connected Riemannian manifold with constant sectional curvature $0$ is isometric to Euclidean space $\mathbb{R}^{n_j}$ (by the standard classification of space forms). Therefore, each $M_j$ is isometric to $\mathbb{R}^{n_j}$ and in particular is non–compact. Finally, the product of non–compact manifolds is non–compact. (In fact, one may observe that the metric $g$ is isometric to the Euclidean metric on $\mathbb{R}^{n_1+n_2}$.) Thus, $M=M_1\times M_2$ is non–compact.