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Tangent spaces and derivations

  1. Jun 29, 2007 #1

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    As far as I know, the tangent space of a manifold at a point is often defined abstractly as follows. First, we define the germ of smooth real valued functions at a point a, which is the set of equivalence classes of smooth functions where two functions are considered equivalent if they are the same on some neighborhood of a. Pointwise addition, multiplication and scalar multiplication makes this into an algebra.

    Then we define a derivation on this algebra. This is where I'm a little unclear. This is something like a linear functional on the algebra which satisfies:

    [tex] \partial(fg) = \partial(f) g(a) + f(a) \partial(g) [/tex]

    My question is, in the more general abstract setting, what are f(a) and g(a)? It seems like you need another linear functional e given by evaluation at a, and then the above condition is:

    [tex] \partial(fg) = \partial(f) e(g) + e(f) \partial(g) [/tex]

    Is this the correct general definition of a derivation, so that you must also specify a linear functional on the algebra? According to wikipedia, a derivation on an algebra satisfies:

    [tex] \partial(fg) = \partial(f) g + f \partial(g) [/tex]

    So that it takes values in the algebra. This gets around needing a linear functional, but it seems like this space of derivations would be much bigger.
     
    Last edited: Jun 29, 2007
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  3. Jun 29, 2007 #2

    Hurkyl

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    Huh. There is a glaring omission there, and I was sure that there didn't used to be an omission. At least the relevant information does appear in the Kahler differential page.

    Basically, the source and target of the derivation don't have to be the same structure -- in general, all you need is that the source is a R-algebra, and the target is an R-module. (And you can probably generalize further)

    Here, your derivation is a module homomorphism from the algebra of germs of smooth functions at a to the tangent space at a.

    Over what ring of scalars? I think you can use the reals, or you can use the ring of smooth functions, or even the ring of germs of smooth functions as your ring of scalars.

    Notice that the ring of germs of smooth functions has a unique maximal ideal: the set of all germs that vanish at a. Taking the corresponding quotient ring is effectively the same as evaluating at a.
     
  4. Jun 29, 2007 #3

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    I don't understand that. It seems to me like the target is either real numbers or germs of real smooth functions, and I'm trying to define the tangent space at a as the set of derivations on the algebra of germs of functions at a. If the target is real numbers, we need to use the first formula I wrote, which is fine except it doesn't work for general algebras, which is a problem if you want to use general algebraic techniques. And if it's germs, it seems like the set of derivations wouldn't be the tangent space, but something like the set of germs of smooth vector fields at a. Am I missing something?
     
  5. Jun 29, 2007 #4

    Hurkyl

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    I made a slight omission -- the target is not merely an R-module, but also a module over your R-algebra.


    It might be worth looking at the problem, using the fact that you already know the answer. You have a ring R of smooth functions on your manifold. You have the ring S of germs of smooth functions at a. You have the R-module T which is the tangent space to your manifold at a.

    You can give T an S-module structure with the scalar product
    f v := f(a) v (for f in S and v in T).​
    (check that this is actually an S-module structure!)

    Then, the given operator really is an R-derivation from S to T.


    Wait -- that's wrong. In this approach (as I know it), T is the cotangent space at a, not the tangent space...
     
    Last edited: Jun 29, 2007
  6. Jun 29, 2007 #5

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    Yea, I was confused at first, but the book I'm reading is definitely using this to define the tangent space, not the cotangent space. In euclidiean space, the derivation in question is not the gradient operator, but the linear functional sending a function f to the the real number [itex]\vec v \cdot \nabla f [/itex] for some vector v. It can be shown that all derivations at a point defined in the abstract way I described above give rise to such a functional for some vector v, and so there is an isomorphism between the derivations and the tangent space in euclidean space. This correspondence is then used to construct the tangent space for a general differentiable manifold. So it seems like we were talking about different things, but now do you see where my problem is?
     
    Last edited: Jun 29, 2007
  7. Jun 29, 2007 #6

    Hurkyl

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    Oh, right, the "tangent space = space of differential operators" construction. You know, I don't think I've actually ever gone through all of the details of that one. :frown:
     
  8. Jun 29, 2007 #7

    Hurkyl

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    Oh, right, the "tangent space = space of differential operators" construction. I don't think I've actually ever gone through all of the details of that one. :frown:
     
  9. Jun 29, 2007 #8

    mathwonk

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    usually you specify the kernel of the derivation and the target.

    i.e.c ertain multipiers make the derivation linear and others act like leibniz rule.
     
  10. Jul 1, 2007 #9

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    Ok, but what is the target of this derivation?
     
  11. Jul 1, 2007 #10

    mathwonk

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    the trget of a derivation on the ring of germs of functions is usually either the ring of germs of funcions oir the scalars. just look at your definition and see. it should be clear. im listening to janice joplin right now and assume you can do thus. pardon me if i am not helpful. later i will check back in.
     
  12. Jul 1, 2007 #11

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    Ok, well, if and when you get the time, I explained my confusion with both of those as target spaces in the third post.
     
  13. Jul 9, 2007 #12

    mathwonk

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    Theorem: If X in k^n is a closed affine variety then all the following definitions define naturally isomorphic linear spaces.
    1) Tp(X) = the union of all lines in k^n which are tangent to X at p (i.e. lines with intersection multiplicity at least 2 with X at p).
    2) Tp(X) = intersection of {Lj = 0}, the common zero locus of the linear terms at p, Lj, of the Taylor expansions of the generators fj of the ideal of X in k^n.
    3) Tp(X) = Homk( M/M^2, k) where M in k[X] is the ideal in k[X] of the point p.
    4) Tp(X) = Homk( m/m^2, k) where m in ØX,p is the ideal in ØX,p of the point p.
    5) Tp(X) = Derp(k[X], k) = all k linear mappings D: k[X] →k, which satisfy the Leibniz rule, i.e. such that D(fg) = f(p)D(g) + g(p)D(f), for all f,g in k[X].
    6) Tp(X) = Derp(ØX,p , k).
     
    Last edited: Jul 10, 2007
  14. Jul 10, 2007 #13

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    Ok, thanks for that, but I'm talking about a differentiable manifold. Is there some way to translate any of these over to that setting?
     
  15. Jul 10, 2007 #14

    mathwonk

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    probably 1,2,4,6 make sense almost exactly as stated in that case.
     
  16. Jul 11, 2007 #15

    mathwonk

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    this is easily derived from the smooth taylor polynomials
     
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