# Is the warped product: $dr^{2} + r^{2} \gamma$ always flat in 3 dimensions?

Suppose you have a Riemannian 2-sphere $$(S^{2} , \gamma)$$. Define a metric $$g$$ on $$M = S^{2} \times [1,\infty)$$ in this way:

If $$(x_{1}, x_2)$$ is a coordinate chart on $$S^{2}$$ then $$g = dr^{2} + r^{2}\gamma_{ij}dx^i dx^j$$ Where $$i,j = 1,2$$. Notice that $$\gamma_{ij}$$ does not depend on $$r$$ and so this is a warped product. My question is: is this metric flat? And is this metric the Euclidean metric on $$M$$?

Let us fix a coordinate chart on $$S^{2}$$ call it $$(\theta, \phi) \in U$$ where $$U$$ is some open set of $$\mathbb{R}^2$$. Then we are looking at $$(U\times [1,\infty) , g)$$ where $$g = dr^2 + \tilde{\gamma}$$ where $$\tilde{\gamma} = r^2 \gamma$$.

I think this means that the hypersurfaces $$\{ r = r_{0} \}$$ are all totally umbilical since the second fundamental form $$K$$ satisfies:

$$K_{ij} = -\frac12 \frac{d}{dr} \tilde{\gamma} = -\frac1r \tilde{\gamma}$$

So the principle curvatures are $$-\frac1r$$ and so the Gaussian curvature is $$\frac{1}{r^2}$$ . Using this, I computed the Rici curvature components of M and found them to be $$0$$ (But I don’t trust my computations at all). This implies that the Rieman curvature tensor vanishes on $$M$$ and so $$M$$ is flat (since the Weyl part of Rieman curvature tensor vanishes for 3-dim manifolds, the vanishing of the Ricci curvature implies the vanishing of the Rieman Curvature tensor). Is there any mistake in what I am doing?

• I don't think it is flat unless $\gamma_{ij}dx^idx^j$ is a metric on $S^2$ of constant curvature $1$. – Yu Ding Apr 20 at 4:04
• @YuDing Thank you for your comment. How do you see that? And my computations were wrong :(. – Laithy Apr 20 at 14:53
• Such a metric certainly can be made with $r\in (0, \infty)$, not just $[1, \infty)$. Intuitively, we try to allow $r\in [0, \infty)$, then $r=1$ is the set of points with distance $1$ from the "point" $r=0$; if the space is flat, then locally $r=1$ must be part of the standard sphere in ${\mathbb R}^3$, which has constant curvature $1$. – Yu Ding Apr 21 at 0:54

From OP's comment under the question it appears that $$\gamma$$ is an arbitrary metric on $$S^2$$. The coordinates are local, so this question is more or less just asking about the behavior of the cone metric $$\hat \gamma := dr^2 + r^2 \pi^* \gamma$$ over a Riemannian surface $$(\Sigma, \gamma)$$ on $$\hat\Sigma := \Bbb R_+ \times \Sigma$$. Here, $$\pi : \hat\Sigma = \Bbb R_+ \times \Sigma \to \Sigma$$ is just the canonical projection onto the second factor. Computing directly$$^*$$ gives that the respective Ricci curvatures $$\operatorname{Ric}, \widehat{\operatorname{Ric}}$$ of $$\gamma, \hat\gamma$$ are related by $$\widehat{\operatorname{Ric}} = \pi^*(\operatorname{Ric} - \gamma) ,$$ and then forming traces gives that the respective scalar curvatures $$R, \hat R$$ are related by $$\hat R = r^{-2} \pi^* (R - 2) .$$ These two identities immediately give that $$\widehat{\operatorname{Ric}} = 0 \Leftrightarrow \operatorname{Ric} = \gamma \Leftrightarrow \hat R = 0 \Leftrightarrow R = 2 ,$$ which in particular recovers Yu Ding's observation from the comments.
$$^*$$One can compute these identities efficiently using isothermal coordinates: There are some local coordinates $$(x_1, x_2)$$ on $$\Sigma$$ for which $$\gamma = e^{2 \Upsilon} (dx_1^2 + dx_2^2)$$ for some smooth function $$\Upsilon(x_1, x_2)$$. Applying the Koszul Formula yields relations between $$\hat\nabla$$ and $$\nabla$$, e.g., $$\hat \nabla_{\partial_{x_i}} \partial_{x_j} = \nabla_{\partial_{x_i}} \partial_{x_j} - r \gamma(\partial_{x_i}, \partial_{x_j}) \partial_r ,$$ among simpler relations, and then compute the curvatures directly from these. (In this display equation, we've implicitly used the decompositions $$T_{(r, p)} \hat\Sigma = T_r \Bbb R_+ \oplus T_p \Sigma$$.)
• Thank you very much for your answer. I am getting different answers for the scalar curvature. $$\widehat{R} = R - \frac{2}{r^2}$$. but gives the same result at the end. That warped product is flat iff $\gamma$ has constant curvature 1. If it is flat, does that mean it's the euclidean metric? There should be many spheres with constant curvature 1 right? thank you again. – Laithy Apr 21 at 16:13
• Oh my $R$ is the scalar curvature of ${S^2, r^2\gamma)$. So we do get exactly the same answer. – Laithy Apr 21 at 16:18
• You're welcome, I'm glad you found it helpful. "(Locally) flat" means "(locally) isometric to the Euclidean metric". As for your second question, the answer depends on what you mean by "many spheres". Given the usual round metric $h$ on $S^2$ and any diffeomorphism $\phi : S^2 \to S^2$, the pullback metric $\phi^* h$ (which for all but a finite-dimensional family of $\phi$ does not coincide with $h$) also has Gaussian curvature $K_{\phi^* h} = \phi^* K_h = 1$. – Travis Apr 21 at 19:48
• Thank you again. The following question will be my last: 1) Since our manifold is not simply connected (also not complete) (i think), then flat (vanishing of Riemann curvature tensor) does not imply globally isometric to euclidean. So if $\gamma$ has constant curvature 1, that does not mean that $dr^2 + r^2 \gamma$ is the euclidean metric and so $\gamma$ might not necessarily be the round metric. Is any of that correct? – Laithy Apr 21 at 20:46
• In fact $M$ is simply connected: It is the product of two simply connected spaces. (However, its second homotopy group is nontrivial, so like you say it cannot be homeomorphic to $\Bbb R^3$.) Any metric $\gamma$ on $\Bbb S^2$ of constant curvature $1$ is isometric to the round metric, and so there is an isometric embedding $(M, dr^2 + r^2 \gamma) \hookrightarrow \Bbb (R^3, \bar g)$, where $\bar g$ is the Euclidean metric. – Travis Apr 22 at 15:41