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 $(M \times N, g^{M \times N})$.
Product of Riemannian manifolds?
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riemannian-geometry
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1You know, that $T_{(p,q)}M\times N \cong T_pM \oplus T_qN$? – 2012-07-20
1 Answers
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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 $M \times N$.
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0Is there a nice book which discusses this? – 2018-07-03