Isomorphisms in tangent bundles

Question

Given the tangent bundle $T$, is the following isomorphism $$\Lambda^{d-1}T\simeq T^*\otimes\Lambda^dT^* $$ true on a $d$-dimensional manifold? i.e. are $(d-1)$-vectors equivalent to objects which are tensor products of a 1 form and a $d$ form? And if so, why?

Answer 1

Let $V$ be a finite-dimensional vector space. The wedge product $$ \wedge \colon \Lambda^{d - 1}(V) \times V \rightarrow \Lambda^n(V) $$ gives you a non-degenerate pairing as one can easily verify by choosing a basis for $V$. This means that we have a canonical isomorphism $$ \Lambda^{d-1}(V) \xrightarrow{\varphi} \operatorname{Hom}_k(V, \Lambda^d(V)) $$ given by $\varphi(\alpha)(v) = \alpha \wedge v$. Since $\operatorname{Hom}_k(V, \Lambda^d(V))$ is isomorphic to $V^{*} \otimes \Lambda^d(V)$ you get an isomorphism $\Lambda^{d-1}(V) \cong V^{*} \otimes \Lambda^d(V)$. If $V$ is equipped with a volume form (a non-zero element of $\Lambda^d(V^{*})$) then the volume form gives you an identification between $\Lambda^d(V)$ and $\Lambda^d(V^{*})$ so you also get an isomorphism $\Lambda^{d-1}(V) \cong V^{*} \otimes \Lambda^d(V^{*})$.

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This generalizes fiberwise to manifolds. Without any assumptions on $M$ you have a canonical isomorphism of vector bundles $\Lambda^{d-1}(TM) \cong T^{*}M \otimes \Lambda^d(TM)$. If $M$ is oriented by a volume-form $\omega$ then you get an ($\omega$-dependent) isomorphism $\Lambda^{d-1}(TM) \cong T^{*}M \otimes \Lambda^d(T^{*}M)$. If you follow all the identifications involved, you can write this isomorphism succintly as $$ \alpha \mapsto \iota_{\alpha}(\omega) \otimes \omega $$ where $\iota_{\alpha}(\omega)$ is the one-form obtained from $\omega$ by contracting it with the multivector $\alpha$.