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

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 $$n \geq 1$$...)

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 $$0$$.

• Thanks! I see. Can you tell me if everything I did before the "$= 1$" is right though? I think it isn't because it'd imply that the curvature depends only on tangent vectors of $\mathbb{S}^2$, which I'm not sure is correct. – Matheus Andrade Sep 18 '19 at 3:43
• The rest of the argument looks fine to me. It is true in this case that the curvature of a $2$-plane only depends on its projection to $S^2$, but that occurs only because $\mathbb{R}^n$ is flat. – Jason DeVito Sep 18 '19 at 5:34

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 $$0 \ne 1$$. :)

• Because "sectional curvature" is designed to mean "look at the portion of my object that's locally produced by exponentiating curves in this 2-plane in the tangent space at $Q$ and then compute the ordinary curvature of that surface at the point $Q$." And ordinary curvature can be computed as a product of curvatures in the two principal-curvature directions. Of course, you have to believe that the directions I chose in fact represent principal curvature directions on $S^2$ and on $S^1 \times \Bbb R$...but that's in any ugrad diff'l geomerty text, e.g., ONeill. – John Hughes Sep 18 '19 at 11:08