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Given two Riemannian manifolds $(M,g^M)$ and $(N,g^N)$, we can construct a product Riemannian manifold $(M\times N,g^{M \times N})$ as described in Product of Riemannian manifolds? .

Is there a simple description of the volume form of the product manifold in terms of the volume forms of $M$ and $N$?

Dimensional reasons make me think the volume form of the product $\omega^{M \times N}$ should be something along the lines of $\omega^{M \times N} = \omega^M + \omega^N$, where $\omega^M$ and $\omega^N$ are the volume forms of $M$ and $N$ respectively, but I have no idea how to show this.

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Mark B
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1 Answers1

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If $\pi_{M}:M \times N \to M$ and $\pi_{N}:M \times N \to N$ are the projections, the formula you're looking for is $$ \omega^{M \times N} = \pi_{M}^{*}(\omega^{M}) \, \pi_{N}^{*}(\omega^{N}). $$ In local coordinates, the product metric is block diagonal, and the volume form is the determinant, a.k.a., the product of the determinants of the blocks. (This should jibe with your intuition about $\mathbf{R}^{2}$, the Riemannian product of two flat lines. :)

Andrew D. Hwang
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