Calculating the curvature of product manifold $\mathbb{S}^2 \times \mathbb{R}$

Question

I've read that $\mathbb{S}^2 \times \mathbb{R}$ is one of the model geometries of Thurston which has non constant curvature. I took it to mean that the manifold $(\mathbb{S}^2 \times \mathbb{R}, g)$ has non constant sectional curvature (where $g$ is the standard product metric) and tried to compute its sectional curvature. Here's what I used:

Let $(M, g_1)$ and $(N, g_2)$ be two Riemannian manifolds with curvature tensors $R_1$ and $R_2$. Using that for each $(p, q) \in M \times N$, $T_pM \oplus T_q N \cong T_{(p, q)}(M \times N)$, for each $X, Y, Z \in \Gamma(T(M \times N))$, we have:

• $R(X, Y)Z = R_1(X_1, Y_1)Z_1 + R_2(X_2, Y_2)Z_2$
• $\text{Rm}(X, Y, Z, W) = \langle R(X,Y)Z, W \rangle$

where $X = (X_1, X_2)$, with $X_1 \in \Gamma(TM)$, $X_2 \in \Gamma(TN)$ and analogously for $Y$ and $Z$, and the metric on $M \times N$ is given by:

$$g^{M \times N}_{(p, q)}(X, Y) = g^{M}_{p}(X_1, Y_1) + g^{N}_{q}(X_2, Y_2)$$

Denoting by $R_1$ the curvature tensor for the sphere $\mathbb{S}^2$ and by $R_2$ the one for the real line, it's obvious that $R_2 \equiv 0$. Now, let $X, Y$ be an orthonormal basis for a $2$ plane contained in some tangent space of $\mathbb{S}^2 \times \mathbb{R}$. We have:

$$\begin{align}R(X,Y)Y &= R(X_1 + X_2, Y_1 + Y_2)( X_1 + X_2)\\ &= R(X_1, Y_1)X_1 + R(X_2, Y_1)X_1 + R(X_1, Y_2)X_1 + R(X_2, Y_2)X_1& \\ &+R(X_1, Y_1)X_2 + R(X_2, Y_1)X_2 + R(X_1, Y_2)X_2 + R(X_2, Y_2)X_2 \\ &= R_{1}(X_1, Y_1)X_1 \end{align}$$

since all the other terms disappear, where we're using that $R_2 \equiv 0$. Then:

$$\begin{align}K(X, Y) &= \langle R(X, Y)Y, X \rangle \\ &= \langle R_1(X_1, Y_1)Y_1, X_1 \rangle + \langle R_1(X_1, Y_1)Y_1, X_2 \rangle \\ &= \langle R_1(X_1, Y_1)Y_1, X_1 \rangle = 1 \end{align}$$

because $\mathbb{S}^2$ has constant sectional curvature equal to $1$. So we have that $\mathbb{S}^2 \times \mathbb{R}$ has constant sectional curvature as well.

Where did I make a mistake here? Or does $\mathbb{S}^2 \times \mathbb{R}$ actually have constant curvature?

(I also realized that if my computations are correct, it would imply that $\mathbb{S}^n \times \mathbb{R}$ has constant curvature for all $n \geq 1$...)

Answer 1

The fact that $X$ and $Y$ are orthonormal says nothing about whether or not $X_1$ and $X_2$ are orthonormal. Thus, the condition $\langle R(X_1,Y_1)Y_1, X_1\rangle =1$ does not need to hold. In fact, $X_1$ and $Y_1$ could be linearly dependent, in which case the curvature is $0$.

Answer 2

If you let $u, v$ be orthogonal unit tangent vectors to $S^2$ at a point $P\in S^2$, and $w$ be a tangent vector to $\Bbb R$ at $t \in \Bbb R$, then all three can be regarded as tangent vectors to $S^2 \times \Bbb R$ at $(P, t)$.

The sectional curvature in the plane defined by $(u,v)$ is $1$ (assuming you've got a unit $S^2$).

The sectional curvature in the plane defined by $(u, w)$ is the product of the curvature in the $u$-direction, which is $1$, with the curvature of $\Bbb R$ in the $w$ direction, which is $0$. So The sectional curvature in that plane is $0$. And $0 \ne 1$. :)

This doesn't answer your question of where you went wrong, because I'm too tired to read all you wrote. I'm sorry about this.