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Given two Riemannian manifolds $(M,g^M)$ and $(N,g^N)$ is there a natural way to combine them to be a Riemannian manifold? Some kind of $(M \times N, g^{M \times N})$.

Willie Wong
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Yrogirg
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1 Answers1

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Yes. Using the natural isomorphism $T(M \times N) \cong TM \times TN$, define the metric on $T(M \times N)$ as follows for each $(p,q) \in M \times N$.

$$g^{M\times N}_{(p,q)} \colon T_{(p,q)}(M \times N) \times T_{(p,q)}(M \times N) \to \mathbb{R},$$ $$((x_1,y_1),(x_2,y_2)) \mapsto g^M_p(x_1,x_2) + g^N_q(y_1,y_2).$$

Alternately, if you think of the metrics as vector bundle isomorphisms $g^M \colon TM \to T^* M$ and $g^N \colon TN \to T^* N$, then the metric on $M \times N$ is just the induced vector bundle isomorphism

$$g^M \oplus g^N \colon TM \oplus TN \to T^*M \oplus T^* N,$$ $$(x,y) \mapsto (g^M(x), g^N(y))$$

Here, $TM \oplus TN$ is just $TM \times TN$ as a set, and the base manifold of $TM \oplus TN$ is $M \times N$.

Yrogirg
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Victor Dods
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    Another way to write what Victor wrote is that there exists natural projection maps $\pi_M:M\times N\to M$ and $\pi_N: M\times N \to N$. The object $g^{M\times N} := \pi_M^g^M + \pi_N^ g^N$ consisting of pullbacks of the constituent metrics is obviously symmetric as a bilinear form. It suffices to check that it is positive definite (which uses the natural isomorphism of tangent bundles ). This also leads to the notion of warped products. – Willie Wong Jul 20 '12 at 07:43
  • Nice, that's a concise way to put it. – Victor Dods Jul 20 '12 at 08:41
  • Is there a nice book which discusses this? – ABIM Jul 03 '18 at 16:56