Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A contradiction in Spivak's Calculus on manifold?

  1. Aug 24, 2007 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    1. The problem statement, all variables and given/known data
    I don't have great expectation that this will get a reply but here goes, cuz this is bugging me.

    I will assume that you are familiar with the notation used by Spivak.

    In the last section of chapter 4, he shows how to integrate a k-form on R^m over a singular k-cube in R^m, namely,

    [tex]\int_c \omega = \int_{[0,1]^k}c^{*}\omega[/tex]

    He then goes on to define the integral of a k-form over a k-chain [itex]c=\sum a_i c_i[/itex] in R^m by

    [tex]\int_c \omega = \sum a_i\int_{c_i} \omega=\sum a_i\int_{[0,1]^k}c_{i}^{*}\omega[/tex]

    But actually, a k-chain in R^m is a k-cube in R^m, so in principle, the integral of a k-chain is already defined, namely by

    [tex]\int_c \omega =\int_{[0,1]^k}(\sum a_i c_i)^{*}\omega[/tex]

    and this will be consistent with the above definition if [itex](\sum a_i c_i)^{*}=\sum a_ic_i^{*}[/itex]. But is this so? Taking the case where w is a 1-form for simplicity,

    [tex](\sum a_i c_i)^{*}\omega (p)(v_p)=\omega (\sum a_i c_i(p))((\sum a_i c_i)_{*}(v_p))=\omega (\sum a_i c_i(p))(\sum a_i (Dc_i(p)(v))_{c(p)})=\sum a_i \omega (\sum a_i c_i(p))((Dc_i(p)(v))_{c(p)})[/tex]

    On the other hand, supposing we define "+" and multiplication by a scalar in the natural way on the f* operators,

    [tex](\sum a_ic_i^{*})\omega(p)(v_p)=\sum a_ic_i^{*}\omega(p)(v_p) = \sum a_i \omega(c_i(p))(c_{i*}(v_p))=\sum a_i \omega(c_i(p))(Dc_{i}(v))_{c_i(p)})[/tex]

    They are not at all the same.
     
  2. jcsd
  3. Aug 24, 2007 #2

    olgranpappy

    User Avatar
    Homework Helper

    This will not do. Math is obviously far too easy to be left to the mathematicians.
     
  4. Aug 24, 2007 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Better than nothing. :p

    (Where have the smiley faces gone?)
     
  5. Aug 24, 2007 #4

    olgranpappy

    User Avatar
    Homework Helper

    up in the toolbar-thing (look for the smileyface with a down-arrow next to it).

    :biggrin:
     
  6. Aug 25, 2007 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Could it be that chains are not really functions, and when we write [itex]c=\sum a_i c_i[/itex], it is purely notational?

    Because I happen to be reading at the moment that this whole cube & chains business can be extended to the general setting of manifold. Namely, if M is a manifold, a singular k-cube in M is a map c:[0,1]^k --> M. But since there is a priori no algebra on manifolds, it does not make sense to add and multiply cubes by constants, such that [itex]c=\sum a_i c_i[/itex] is only a formal sum.
     
  7. Aug 25, 2007 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think you have it. A chain is a piece of geometry. It's not the coordinate map, since the integration is independent of the coordinates. So, yes, it's a formal sum. Probably best to think of it as a 'weight' on the chain.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook