# Does trivial cotangent bundle implies trivial tangent bundle?

by quasar987
Tags: bundle, cotangent, implies, tangent, trivial
 Sci Advisor HW Helper PF Gold P: 4,768 If the tangent bundle is trivial, then the cotangent bundle is trivial. To see this, consider (X_i) a global frame for TM. Then define a global frame (\alpha^i) for T*M by setting $\alpha^i(X_j)=\delta_{ij}$ and extend by linearity. Does trivial cotangent bundle implies trivial tangent bundle? A similar argument based on global frames does not seem to work in this direction: given a global frame (\alpha^i) for T*M, how do you define a global frame for TM? It does not makse sense to say "Let X_i be the vector field such that $\alpha^i(X_j)=\delta_{ij}$" because such a vector field might not exist. And if locally, $\alpha^i=\sum_j\alpha^i_jdx^j$, then defining a (global) vector field by setting $X_i:=\sum_j\alpha^i_j\partial_j$ is inconsistent because the coefficients $\alpha^i_j$ do no transform correctly.
 Sci Advisor HW Helper PF Gold P: 4,768 I may have spoken too fast. It appears now that it does make sense to say "Let X_i be the vector field such that $\alpha^i(X_j)=\delta_{ij}$".
 Sci Advisor P: 1,563 Yeah, if a matrix is invertible, then so is its inverse. :D The subtlety only happens when things are infinite-dimensional.
 Sci Advisor P: 1,563 The dual space V* of an infinite-dimensional vector space V can be strictly larger than V, and so they can't be isomorphic. For example, on a space of functions, the dual space includes distributions, which are not in the original space. Also, on an infinite-dimensional vector space, a linear operator might have a right inverse, but no left inverse (or vice versa). For example, on an L^2 function space in an interval [a,b], define the derivative operator D as $$Df = \frac{df}{dx}$$ and an integral operator J as $$Jf = \int_a^x f(x') \; dx'$$ Then D has a right inverse, given by J: $$DJf = f$$ but no left inverse, because $$JDf \neq f$$ for all possible f (specifically, all the constant functions are mapped to 0 by D, and J maps 0 to 0).