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according to wikipedia the product metric between 2 metrics is the metric given by: $d(x,y)=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}$

Now if $(M,g_m)$ and $(N,g_n)$ are 2 Riemannian manifolds we can construct the product $M\times N$ equipped with the riemannian metric $g_m+g_n$.

Is there a link between the "product metric" and the natural metric on $M\times N$ or is it two different things ?

Thanks

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Yes. If you endowed $M = \mathbb{R}^1$ as a manifold with the usual metric $g = g_{ij}dx^i \ dx^j = 1.dx\ dx = dx^2$, then $M \times M$ has induced metric

$$ds^2 = dx^2 + dy^2$$ which gives Euclidean distance.


Relatedly all Riemannian metrics induced by Euclidean metrics.

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  • $\begingroup$ I feel like there might be a confusion, when i wrote $g=g_m+g_n$ i meant $g(x,y)=g_m(x_m,y_m)+g_n(x_n,y_n)$, not $d(x,y)=d_m(x_m,y_m)+d_n(x_n,y_n)$. In the euclidian case, $g$ would correspond to the product metric, but does it also in the general case? $\endgroup$ – Chevallier Nov 6 '14 at 14:21
  • $\begingroup$ I see. Then yes. Sorry for the confusion. Let me edit my post. $\endgroup$ – Simon S Nov 6 '14 at 14:31

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