# Sections of a bundle

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.

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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: $$\Gamma(E\otimes F)\simeq \Gamma(E)\otimes_{C^{\infty}}\Gamma(F)$$
Thank you very much. Can you please say me with more details, why do we have, in général : $$\Gamma(E\otimes F)\simeq \Gamma(E)\otimes_{C^{\infty}}\Gamma(F)$$ I don't understund this point. Thank you. – Bryan Dec 3 '12 at 18:35
It's the same argument: there are finite open covers $(U_i)$ and $(V_i)$ by $n+1$ open sets that trivialise $E$ and $F$ respectively, so you get an finite open cover $(U_i\cap V_j)$ that trivialises both $E$ and $F$, so also $E\otimes F$, and you take a partition of unity subordinate to this cover, and argue similarly. – Olivier Bégassat Dec 3 '12 at 18:39