Metric tensor of total space in trivial bundles

Hi everyone, sorry if my question may be wrong (i'm an engineer not a mathematician!). I was wondering, given a total space E of a trivial bundle, such that E=MxF, where $gm_{ij}$, and $gf_{ij}$ are the metric tensors on the base space M and the fiber F resp., what is the metric tensor of E?

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That would be the direct sum of the two, corresponding to a block matrix with zero blocks off the diagonal. –  Gunnar Magnusson Dec 13 '11 at 18:03
Thanks for the answer Gunnar. If i understand correctly, in a general fiber bundle (not a trivial one), if M⊂R${^n}$ and F⊂R${^m}$ it would be the case that $g_{ij}=\mathbf{gm_i}\mathbf{gf_j}$ for i=1...n, j=n+1...n+m (the upper block), but in the case of a product space, $\mathbf{gm_i}\mathbf{gf_j}=0$? –  Jorge Dec 13 '11 at 19:10
A metric tensor on a manifold is a piece of extra data. One can always put a metric (if the manifold admits partitions of unity, which it will in all practical applications) but there are uncountably many possible metrics. Gunnar's comment is true in the sense that this does give a metric on $M \times F$, and this is certainly the most natural, given that you have metrics on $M$ and $F$ separately, but it's certainly not the only possibility. It really depends on what you are studying. –  Spiro Karigiannis Dec 13 '11 at 19:23
@Jorge: I don't understand your comment. For a nontrivial fibre bundle, one cannot give it the product metric, because it's not a product. One way to get a metric on the total space of a fibre bundle, given metrics on the base and the fibre, is by using a connection (called an Ehresmann connection in this context.) You can probably find this in Kobayashi-Nomizu. –  Spiro Karigiannis Dec 13 '11 at 19:25