# Product of Riemannian manifolds?

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 $(M \times N, g^{M \times N})$.

• You know, that $T_{(p,q)}M\times N \cong T_pM \oplus T_qN$? Jul 20, 2012 at 7:20

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 $M \times N$.

• 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. Jul 20, 2012 at 7:43
• Nice, that's a concise way to put it. Jul 20, 2012 at 8:41
• Is there a nice book which discusses this?
– ABIM
Jul 3, 2018 at 16:56