TITLE: Sections of a bundle
QUESTION [6 upvotes]: I would like that someone explain to me why in general $ \Gamma ( T^*M \otimes TM ) = \Gamma ( T^*M ) \otimes_{\mathcal{C}^{\infty} ( M )} \Gamma ( TM ) $ with $ \Gamma ( T^*M ) $ is a set of sections of the bundle $ T^*M $ ?
Thanks a lot.

REPLY [5 votes]: For a finite dimensional vector space $V$, there is a canonical isomorphism $$V^*\otimes V\simeq\mathrm{End}(V)$$ that sends $\phi\otimes v$ to $\lbrace x\mapsto \phi(x)v \rbrace$. This allows one to define an obviously $C^{\infty}(M)$-bilinear map $\Gamma(T^*M)\times\Gamma(TM)\to\Gamma(\mathrm{End}(TM))$ so a map $$\Gamma(T^*M)\otimes_{C^{\infty}(M)}\Gamma(TM)\to\Gamma(\mathrm{End}(TM)).$$ The question is wether this is an isomorphism.
There is a straightforward argument in case the base manifold is compact. Consider a finite cover of $M$ by chart domains $(U_1,\dots,U_m)$. These provide you with an open cover that trivialises the tangent and cotangent bundles. Take a partition of unity $(\chi_1,\dots,\chi_m)$ subordinate to this cover. Now work in one of those chart domains, say $U=U_a$ with chart $\phi=\phi_a$. The bundle of endomorphisms is trivialised by the local basis of smooth sections $dx^i\otimes \frac{\partial}{\partial x^j}:U\to\mathrm{End}(TM)|_U$, and so every endomorphism of $TM$ can locally be expressed (uniquely) as a sum with coefficients in $C^{\infty}(U)$
$$A|_U=a_i^j~dx^i\otimes \frac{\partial}{\partial x^j}$$
You can now define the inverse map $$\Gamma(\mathrm{End}(TM))\to\Gamma(T^*M)\otimes_{C^{\infty}(M)}\Gamma(TM)$$ by sending $A$ to the sum of the above things multiplied by the $\chi_a$ to turn the local sections into global smooth sections.
In case $M$ is not compact, you can use a theorem of topology that asserts that every bundle over a manifold of dimension $n$ can be trivialised over a finite cover of $M$ with at most $n+1$ open sets, and the above construction carries through word for word, and still works for other bundles:
$$\Gamma(E\otimes F)\simeq \Gamma(E)\otimes_{C^{\infty}}\Gamma(F)$$