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Let $M$ be a smooth manifold. Its structure sheaf $\mathcal{O}_M$ is the sheaf of smooth real-valued functions. Together they form a ringed space $(M,\mathcal{O}_M)$. The tangent sheaf $\mathcal{T}_M$ is a sheaf of modules over the structure sheaf. It can be defined as the sheaf of derivations of the structure sheaf.

A smooth map of manifolds $f: M \rightarrow N$ induces a morphism $df: \mathcal{T}_M \rightarrow f^*(\mathcal{T}_N)$ of $\mathcal{O}_M$-modules, where $f^*(\mathcal{T}_N)$ is the inverse image of $\mathcal{T}_N$. It is called the differential or pushforward of the map $f$.

Does anyone have a reference for the definition of the differential? In which kind of textbook would this be explained? It seems to be somewhere between differential geometry and algebraic geometry but I could not find it in any textbook in neither of these areas.

(I am not looking for the differential between tangent bundles which is explained in detail in every basic book on differential geometry.)

Thanks in advance!

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I seem to remember Milchor's 'Topics in Differential geometry' pays a lot of attention to the derivation-based/ germ theoretic definition (rather than the kinematic one) throughout, and might prove enlightening (plus there's a free copy here: mat.univie.ac.at/~michor/dgbook.pdf which could be handy...). I think the bits I've read mainly deal with the push forward case (rather than the pullback you describe), but it's a big book so you never know... –  Tom Boardman Jun 28 '10 at 12:31
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Since you are in the $C^{\infty}$ rather than analytic case, why does the sheaf of ($\mathbb{R}$-linear!) derivations of $O_M$ coincide with $T_M$? That is, why must such a derivation which kills local coordinates equal 0? What you call the "differential" literally is the usual map between tangent bundles, since a map of bundles $E' \rightarrow E$ over a map of spaces $f:X' \rightarrow X$ is "the same" as an $O_ {X'}$-linear map $\underline{E}' \rightarrow f^{\ast}(\underline{E})$, using the associated sheaves of modules $\underline{E}'$ and $\underline{E}$. Do you still need a reference? –  Boyarsky Jun 28 '10 at 12:35
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The map on sheaves can be directly constructed from the map on bundles. Since you are in the smooth case, the sheaves you are working with are determined by their global sections (this can be seen by patching together with a partial of unity). A map on tangent bundles will determine a map on global sections, and so it will determine to map on sheaves. –  Greg Muller Jun 28 '10 at 15:41
    
1) Unfortunately, I could not find anything in Michor, although it gave me a reference for another fact some time ago. 2) Yes, everything seems to boil down to the identification of vector bundle morphisms and morphisms between locally free sheaves. This is luckily explained in the reference below. Thanks to you all! –  C. Jost Jun 30 '10 at 19:41
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2 Answers 2

up vote 3 down vote accepted

You might want to try Ramanan's Global Calculus which does a bit of introductory manifold theory, and uses the mechanism of sheaves throughout.

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Yes, Ramanan treats the connection between morphisms of vector bundles and morphisms of locally free sheaves detailed and even applies it to tangent bundles and tangent sheaves. That was exactly what I was looking for. (Now I only would like to know who took the copy in our library without loaning it properly. But there is always google book search.) Thanks a lot! –  C. Jost Jun 30 '10 at 19:32
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If you are interested in the real $C^{\infty}$ case there are also: Jet Nestruev, Smooth Manifolds and observables, Springer Graduate Texts as well as Gonzales, Salas, C-differentiable spaces, Lecture notes in Mathematics, Springer.

The algebraic definition of the differential consist in composing a derivation $X:\mathcal{O}_M \to \mathcal{O}_M$ with $f^*:\mathcal{O}_N \to \mathcal{O}_M$ which gives you a derivation from $\mathcal{O}_N$ to $\mathcal{O}_M$. It remains to show that this sheaf of $\mathcal{O}_M$ modules is isomorphic to $f^*(\mathcal{T}_N)$. But maybe you did not ask for that.

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There seems to be an explanation in both books, but I had difficulties understanding them without having read the whole book. But thanks anyway, and thanks for the algebraic definition of the differential! It gives another point of view, which is always good and it might be even necessary for me. –  C. Jost Jun 30 '10 at 19:38
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