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Compute the sectional curvature, Ricci curvature, and scalar curvature of $M:=S^n(1) \times S^m(1)$ where $S^n(1)$ denotes the $n$-dimensional sphere of radius $1$ with standard round metric.

I figured that this question is what has been mentioned in section 4.2.2 of Petersen's Riemannian Geometry and I think the proof should be pretty similar.

So, what we have is that by letting $Y$ be a unit vector field on $S^n(1)$, $V$ a unit vector field on $S^m(1)$, and $X$ a unit vector field on either $S^n(1)$ or $S^m(1)$ that is perpendicular to both $Y$ and $V$, and using the Koszul formula, we get that $$ \begin{aligned} 2 g\left(\nabla_Y X, V\right) & =g([Y, X], V)+g([V, Y], X)-g([X, V], Y) \\ & =g([Y, X], V)-g([X, V], Y) \\ & =0, \end{aligned} $$ as $[Y, X]$ is either zero or tangent to $S^n$ and likewise with $[X, V]$. Thus $\nabla_Y X=0$ if $X$ is tangent to $S^m$ and $\nabla_Y X$ is tangent to $S^n$ if $X$ is tangent to $S^n$. This shows that $\nabla_Y X$ can be computed on $S^n(1)$.

Now, here, what Petersen does is the following:

[…] can be computed on $S_a^n$. We can then calculate $R$ knowing the curvatures on the two spheres we obtain: $$ \begin{aligned} & \mathfrak{R}(X \wedge V)=0, \\ & \Re(X \wedge Y)=a X \wedge Y, \\ & \mathfrak{R}(U \wedge V)=b U \wedge V . \end{aligned} $$ In particular, proposition 4.1.1 shows that all sectional curvatures lie in the interval $[0, \max \{a, b\}]$. It also follows that $$ \begin{aligned} \operatorname{Ric}(X) & =(n-1) a X, \\ \operatorname{Ric}(V) & =(m-1) b V, \\ \text { scal } & =n(n-1) a+m(m-1) b . \end{aligned} $$ Therefore, we conclude that $S_a^n \times S_b^m$ always has constant scalar curvature, is an Einstein manifold exactly when $(n-1) a=(m-1) b$ (which requires $n, m \geq 2$ or $n=m=1$ ), and has constant sectional curvature only when $n=m=1$. Note also that the curvature tensor on $S_a^n \times S_b^m$ is always parallel.

In this proof, $S_a^n$ basically denotes the sphere of radius $1/a$ and in my problem $a, b=1$. So, this should be even easier than what Petersen does. However, I don't really understand what Petersen does and I think what I should be having at the end is that:

  • $M$ has constant scalar curvature.
  • $M$ is Einstein exactly when $n=m$, not for when $n=m=1$ only.
  • $M$ has constant sectional curvature only when $n=m=1$.

I'd appreciate it if someone can help figuring a better thing to solve this or helps me understand what Petersen is doing and what should I change to get the things I think I should get.

Jendrik Stelzner
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1 Answers1

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Let us look at a more general picture. Consider $(M_1,g_1)$ and $(M_2,g_2)$ two Riemannian manifolds, and $(M,g)$ their Riemannian products. Let $X$ and $Y$ be two decomposable vector fields, that is $$ X(p_1,p_2) = X_1(p_1) \oplus X_2(p_2) \quad \in \quad T_{(p_1,p_2)}(M_1\times M_2) = T_{p_1}M_1\oplus T_{p_2}M_2,\\ Y(p_1,p_2) = Y_1(p_1) \oplus Y_2(p_2) \quad \in \quad T_{(p_1,p_2)}(M_1\times M_2) = T_{p_1}M_1\oplus T_{p_2}M_2. $$ From Koszul formula, it follows that we have $ \nabla^g_X Y = \nabla^{g_1}_{X_1}Y_1 \oplus \nabla^{g_2}_{X_2} Y_2. $ This is basically your computations. Note that this is only true for decomposable vector fields: a counter example is given by $X=Y=x\partial_y$ on $\Bbb R^2 = \Bbb R\times \Bbb R$. However, it is enough information in order to prove that $$ R^g(X,Y,Z,T) = R^{g_1}(X_1,Y_1,Z_1,T_1) + R^{g_2}(X_2,Y_2,Z_2,T_2), $$ where $X(p_1,p_2) = X_1(p_1,p_2) + X_2(p_1,p_2)$, etc. This is equivalent to the equality $ R^g = (\pi_1)^*R^{g_1} + (\pi_2)^*R^{g_2} $ with $\pi_i\colon M \to M_i$ the canonical projection.

Now, consider a tangent plane $P\subset T_{(p_1,p_2)}M = T_{p_1}M_1\oplus T_{p_2}M_2$. Three cases can occur:

  1. $P\subset T_{p_1}M_1$,
  2. $P\subset T_{p_2}M_2$,
  3. $P \not \subset T_{p_i}M_i$, $i\in \{1,2\}$.

In the first two cases, it follows that $ \sec^g(P) = \sec^{g_i}(P) $ where $i$ is such that $P\subset T_{p_i}M_i$. In the last case, consider an orthonormal basis $\{X_1,X_2\}$ of $P$ with $X_i \in T_{p_i}M_i$. This is possible since $T_{p_1}M_1 \perp T_{p_2}M_2$. Then $ \sec^g(P) = R^{g_1}(X_1,0,X_1,0) + R^{g_2}(0,X_2,0,X_2) = 0. $ This is basically what Petersen says in the first part of your citation. In other words, the sectional curvature is supported in each component, and there is no mixed curvature.

Now, let us look at the Ricci curvature. Let us pick an orthonormal basis $\{E_1,\ldots,E_n,E_{n+1},\ldots,E_{n+m}\}$ of $T_{(p_1,p_2)}M$, with the constraint $$ E_1,\ldots, E_n \in T_{p_1}M_1,\quad E_{n+1},\ldots,E_{n+m}\in T_{p_2}M_2. $$ Again, this is possible because $T_{p_1}M_1\perp T_{p_2}M_2$ in $T_{(p_1,p_2)}M$. Let $X=X_1+X_2 \in T_{p_1}M_1\oplus T_{p_2}M_2$ be a tangent vector. Then \begin{align} \DeclareMathOperator{\Ric}{Ric} \DeclareMathOperator{\span}{span} \Ric^g(X) &= \Ric^g(X_1) + \Ric^g(X_2)\\ &= \sum_{i=1}^{n+m} \sec^g(\span\{X_1,E_i\}) g(X_1,E_i) E_i+ \sum_{i=1}^{n+m} \sec^g(\span\{X_2,E_i\}) g(X_2,E_i) E_i\\ &= \sum_{i=1}^{n} \sec^g(\span\{X_1,E_i\}) g(X_1,E_i) E_i+ \sum_{i=n+1}^{n+m} \sec^g(\span\{X_2,E_i\}) g(X_2,E_i) E_i\\ &= \sum_{i=1}^{n} \sec^{g_1}(\span\{X_1,E_i\}) g_1(X_1,E_i) E_i+ \sum_{i=n+1}^{n+m} \sec^{g_2}(\span\{X_2,E_i\}) g_2(X_2,E_i) E_i\\ &= \Ric^{g_1}(X_1) + \Ric^{g_2}(X_2). \end{align} Thus, the Ricci operator $\Ric^g$ respects the splitting $T_{(p_1,p_2)}M = T_{p_1}M_1\oplus T_{p_1}M_2$. Tracing this last equality yields $ \DeclareMathOperator{\scal}{scal} \scal^g(p_1,p_2) = \scal^{g_1}(p_1) + \scal^{g_2}(p_2). $

To summarize, we have: \begin{align} R^g &= (\pi_1)^*R^{g_1} + (\pi_2)^*R^{g_2},\\ \Ric^g &= (\pi_1)^*\Ric^{g_1} + (\pi_2)^*\Ric^{g_2},\\ \scal^g &= (\pi_1)^*\scal^{g_1} + (\pi_2)^*\scal^{g_2}. \end{align}

Now, if we specialize with $M_1 = S^n(r_1)$ and $M_2=S^m(r_2)$, whose respective sectional and scalar curvatures are $\frac{1}{{r_1}^2}$ and $\frac{1}{{r_2}^2}$, and $\frac{n(n-1)}{{r_1}^2}$ and $\frac{m(m-1)}{{r_2}^2}$, we have \begin{align} \DeclareMathOperator{\Id}{Id} \Ric^{S^n(r_1)\times S^m(r_2)} &= \frac{n-1}{{r_1}^2}\Id_{T S^n(r_1)} + \frac{m-1}{{r_2}^2}\Id_{TS^m(r_2)},\\ \scal^{S^n(r_1)\times S^m(r_2)} &= \frac{n(n-1)}{{r_1}^2} + \frac{m(m-1)}{{r_2}^2}. \end{align} Therefore:

  • $S^n(r_1)\times S^m(r_2)$ always have constant scalar curvature.
  • $S^n(r_1)\times S^m(r_2)$ is Einsten if and only if the Ricci operator is a constant multiple of the identity map of the tangent bundle, which happens if and only if $\frac{n-1}{{r_1}^2} = \frac{m-1}{{r_2}^2}$. In your specific case, $r_1=r_2=1$, so that it happens if and only if $n=m$.
  • First of all, when $n=m=1$, $S^1(r_1)\times S^1(r_2)$ is a flat torus, so it has constant (zero) sectional curvature. If $n\geqslant 2$, then there is a plane in $S^n(r_1)$ with sectional curvature $\frac{1}{{r_1}^2}$. But if $P$ is a plane which isn't contained in one factor (that is, it has component in each of the factors), we have already shown that it has vanishing sectional curvature. In particular, $S^n(r_1)\times S^m(r_2)$ does not have constant sectional curvature.

Remark: More generally:

  • The Riemannian product of two Riemannian manifolds with constant scalar curvature has constant scalar curvature.
  • The Riemannian product of two Riemannian manifolds is Einstein if and only if both factors are Einstein with the same Einstein constant.
  • The Riemannian product of two Riemannian manifolds have constant sectional curvature if and only if both factors are flat.
Didier
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