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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.

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Why would you need the metric for the isomorphism of the tangent bundles? –  Willie Wong Oct 26 '12 at 14:40
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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 $\ TM \oplus TN$.

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Re: your last statement, that's why I wrote my comment above. There is a canonical isomorphism $T(M\times N) = TM\times TN$ as smooth manifolds. A reference is Section 1.12 of Kolar-Michor-Slovak's Natural Operations in Differential Geometry. However, I am not quite sure what the OP meant by "isomorphism of bundles" since I am not sure what he intends the base manifold and bundle projection to be on $TM\times TN$. –  Willie Wong Oct 26 '12 at 14:54
    
@WillieWong: You can take an external direct sum of bundles. If $E_j \to X_j, j=1,2$ are vector bundles and $p_i : X_1 \times X_2 \to X_i$ denotes the projection, then $E_1 \times E_2 \to X_1 \times X_2$ can be thought of as $p_1^* E_1 \oplus p_2^* E_2$. –  Eric O. Korman Oct 26 '12 at 17:33
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