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If a vector bundle $E\to M$ has rank $k$, form the exterior power of a vector bundle fibrewise: its fibre at $p$ is ${\Lambda }^r{E}_p$. A vector bundle trivialization over $U$ induces one with fibre ${\Lambda }^r{\mathbb{R}}^k$. Transition matrices $A_{\alpha \beta }(p)$ induce ${\Lambda }^r{A}_{\alpha \beta }(p)$, which are smooth and obey the cocycle law because exterior powers preserve composition. They glue to a smooth bundle of rank $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$; for $r>k$ it is the zero bundle. At $r=k$ it is the determinant line bundle.

A real line bundle is trivial precisely when it has a nowhere-zero smooth section: a section $s$ gives the bundle isomorphism $(p,t)\mapsto ts(p)$ from $M\times \mathbb{R}$. Apply this to ${\Lambda }^n{T}^{\ast }M$. Suppose first that an oriented atlas is given. Its local coordinate formsare positive multiples of one another on every overlap. Choose a subordinate smooth partition of unity ${\rho }_{\alpha }\ge 0$. Extend ${\rho }_{\alpha }{\nu }_{\alpha }$ by zero outside its chart and set $\nu ={\sum }_{\alpha }{\rho }_{\alpha }{\nu }_{\alpha }$. Local finiteness makes this smooth. At any point at least one weight is positive, and all summands are nonnegative multiples of the same positive top form. Therefore $\nu$ never vanishes, proving triviality of the line bundle.

Conversely let $\nu$ be a nowhere-zero section of ${\Lambda }^n{T}^{\ast }M$. In each sufficiently small connected chart write $\nu =a_{\alpha }d{x}_{\alpha }^1\wedge \cdots \wedge d{x}_{\alpha }^n$. Its coefficient has constant nonzero sign. If necessary reverse the first coordinate, making $a_{\alpha }>0$. On overlaps the equality of the two expressions givesso each transition determinant is positive. These charts cover $M$, provingFor dimension zero both assertions hold with the empty exterior product equal to one.

For the orientability of a vector bundle total space, write $\pi :E\to M$, with rank $r$ and orientable base. The vertical tangent vectors at $v\in E_p$ identify with $E_p$ by differentiating translation in that fibre. The differential of $\pi$ therefore gives the exact sequence of vector bundles on $E$Taking determinant line bundles givesTo see the determinant identification directly, wedge a vertical frame with any lifts of a base frame; altering a lift by a vertical vector does not change that top wedge. Since $M$ is orientable, $\det TM$ is trivial. Consequently $\det TE\cong {\pi }^{\ast }\det E$. If $\det E$ is trivial, so is its pullback, making the total space orientable. Conversely, if $E$ is orientable as a manifold, its tangent determinant is trivial; restricting this isomorphism to the zero section gives triviality of $\det E$. The tangent and cotangent determinant lines are dual, so either tests orientability. We obtain

For a cotangent bundle the base need not be orientable. If base coordinates change from $x$ to $y=y(x)$ with Jacobian $J$, fibre coordinates of the same covector change from $\xi$ to $\eta =J^{-T}\xi$. The full coordinate change on $T^{\ast }M$ has block JacobianThus every induced cotangent coordinate change preserves orientation, even when the base change reverses it. These charts form an oriented atlas on the total space, proving every cotangent bundle is orientable. Equivalently the canonical symplectic form of the cotangent bundle has a nowhere-zero top exterior power, as in cotangent bundle orientation.