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Maps of Ringed Spaces

  1. Jun 18, 2014 #1
    I recently bought a copy of S. Ramanan's Global Calculus. I skimmed around a bit. Naturally, I was confused when it defined a differentiable function ##f:M\to N## between differentiable manifolds as a continuous map such that, for each ##x\in M## and for each ##\phi\in\mathcal{O}_N(V)##, where ##V\ni f(x)## is some neighborhood of ##f(x)## in ##N## and ##\mathcal{O}_N## is the structure sheaf of ##N##, the composition ##\phi\circ f## is in ##f_*\mathcal{O}_M(V)##.

    I'd think the more appropriate definition would be that ##f:M\to N## is differentiable if (and only if) it is a homomorphism of locally ringed spaces. I believe these two definitions are equivalent, but I haven't checked yet, due to a more general question to muse over: if two locally ringed spaces have structure sheaves that would allow for the composition in the first definition to make sense, would an analogue of the first definition define homomorphisms of locally ringed spaces between two such structures?

    I do not think this holds for ringed spaces of the same nature, since definition #1 is essentially a "local" condition, but I don't know how to show this. Could someone please explain where this goes wrong for ringed spaces? Thank you.
     
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  3. Jun 18, 2014 #2

    micromass

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    A morphism of (locally) ringed spaces ##(X,\mathcal{O}_X)## and ##(Y,\mathcal{O}_Y)## is given by two data, a continuous function ##f:X\rightarrow Y## and a morphism of sheafs ##f^\sharp: \mathcal{O}_Y\rightarrow f_*\mathcal{O}_X##. You have not specified what this morphism of sheafs would be.
     
  4. Jun 18, 2014 #3
    Indeed, being cocky with notation has the added effect of not being clear.

    I implicitly assumed that the morphism of sheaves is given by ##f^\sharp:\mathcal{O}_N\to f_*\mathcal{O}_M##, where on each open ##U\subseteq N##, ##f^\sharp_U:\mathcal{O}_N(U)\to f_*\mathcal{O}_M(U)## is defined by ##f^\sharp_U(\varphi)=\varphi\circ f##. Now that I've been motivated to check it, this is clearly a homomorphism of rings, and in fact induces a local homomorphism between stalks (!) since ##\mathfrak{m}_{f(x)}=\{\phi\in\mathcal{O}_{N,f(x)}\vert \phi(f(x))=0\}## and ##\mathfrak{m}_x=\{\psi\in\mathcal{O}_{M,x}\vert \psi(x)=0\}##.

    In this light, I'd say that sheaves of functions would satisfy this, correct?

    Edit: Now that I'm thinking about it, I have yet another thought.

    Conjecture:
    Because of the necessary correspondence of maximal ideals, I don't even have to define what my morphism of sheaves is!

    Any homomorphism of locally ringed spaces in this case will necessarily be "pullback-like." Is this correct?
     
    Last edited: Jun 18, 2014
  5. Jun 18, 2014 #4

    micromass

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    Yes, in differential geometry context this follows quite easily from usual function theoretic arguments. In algebraic geometry it doesn't follow so easily.
     
  6. Jun 18, 2014 #5
    Could you please explain why?
     
  7. Jun 18, 2014 #6

    micromass

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    Because the structure sheafs in algebraic geometry are abstract rings and might not have anything to do with functions. Of course, we can still view elements of the structure sheafs as functions and this gives us good intuiton, but pure formally they can be anything.

    The good situation here was that a map between the topological spaces actually implied a map between the structure sheafs by just composition. This fails dramatically in the algebraic geometry setting and we need to actually write down the two maps there. Furthermore, we need to demand axiomatically that the maps satisfy the local property.

    Of course, in the case of affine schemes, things behave nicely.
     
  8. Jun 18, 2014 #7
    Oh. I thought I was clear when I said "sheaves of functions." By this, I meant sheaves [strike]that have[/strike] whose sections [strike]that[/strike] are functions. Sorry for being confusing.

    Once again, thank you!
     
  9. Jun 20, 2014 #8
    I think I misinterpreted what was being said. Is there an example where the structure sheaf of a (real) manifold is not naturally isomorphic to a sheaf of real-valued functions?
     
  10. Jun 20, 2014 #9

    micromass

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    I was talking of algebraic geometry, not differential geometry.
     
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