Let $M$ and $N$ be two Riemannian manifolds with Riemannian metrics $g$, $h$ respectively. We consider the product $M \times N$ with metric $g \oplus h$. By the metric we get an isomorphism of bundles $T(M \times N) = TM \times TN$.
My question is: is this also true for the cotangent bundle?
I.e. is $T^{*}(M \times N) = T^{*}M \times T^{*}N$?
I hope for a lot of answers and want to thank in advance.
It doesn't require much to define a tangent space and then a cotangent space. Relatively speaking a Riemannian metric is a very strong instance of a metric, which gives raise to a natural isomorphism of the bundle and here follows the illustration of the construction:
If you think of $V=V^i\partial_i$ as your vectors and $g(V,W)=g_{ij}V^iW^j$, then
$$g(V,W) =g_{ij}V^iW^j =g_{ij}(g^{ik}\alpha_k)(g^{jl}\beta_l) =g^{kl}\alpha^V_k\beta^W_l =:G(\alpha^V,\beta^W),$$
where I used $g_{ij}g^{ik}=\delta^k_j$. So for each $V=V^i\partial_i$ you have a $\alpha^V=\alpha^V_i\text dx^i$ and for them a metric $G$. More abstractly speaking, you pull back the metric to $T^{*}(M \times N)$, thus making it a tangent space with a Riemannian metric too (with $T^{*}(M \times N)$ its cotangent space). Hence all such Riemann geometry statements for tangent bundles translate.
Btw. it's $\ TM \oplus TN$.