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

Need help understanding the pushforward

  1. Feb 5, 2012 #1
    In my notes, the following two functions are defined:

    Suppose [itex]M^m[/itex] and [itex]N^n[/itex] are smooth manifolds, [itex]F:M \to N[/itex] is smooth and [itex]p \in M[/itex]. We define:
    [tex]F^*:C^\infty (F(p)) \to C^\infty (p)\ ,\ F^*(f) = f \circ F[/tex]
    [tex]F_{*p}: T_pM \to T_{F(p)}N\ ,\ [F_{*p}(X)](f) = X(F^*f) = X(f \circ F)[/tex]

    I understand the first function, [itex]F^*[/itex]; it maps [itex]f[/itex], a function on [itex]C^\infty(F(p))[/itex], to [itex]f \circ F[/itex], a function on [itex]C^\infty(p)[/itex].

    However, I don't understand the second one, [itex]F_{*p}[/itex]. Since [itex]X(f) \in T_pM[/itex], it follows that [itex]f \in C^\infty (p)[/itex]. But then how is
    [tex][F_{*p}(X)](f) = X(F^*f)[/tex]
    defined? After all, in the definition of [itex]F_{*p}(X)[/itex], [itex]f[/itex] is a function on [itex] C^\infty (p)[/itex], not [itex]C^\infty(F(p))[/itex], so how can we evaluate [itex]F^*f[/itex]?
     
  2. jcsd
  3. Feb 5, 2012 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    X(f) isn't in TpM -- X is.

    [itex]F_\ast[/itex] takes X to [itex]F_\ast(X)[/itex]. The question now is, what is [itex]F_\ast(X)[/itex]? We want it to be an element of TF(p)N, i.e., it should be a point derivation at F(p) on N, i.e., you need to know how to evaluate Fp(X) at smooth germ f at F(p).
     
  4. Feb 5, 2012 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So F*p takes a vector X at p (derivation on the germs of smooth function at p) and sends it to a vector F*pX at F(p) (derivation on the germs of smooth function at F(p)). So what is this vector F*pX then? How does it act on a germ f at F(p)? This is what the formula is telling you: it says (F*pX)(f) is just X(F*f). And this makes sense, since F*f =f o F is indeed a germ of functions at p.
     
  5. Feb 5, 2012 #4
    Sorry, I've never heard of the term 'germ' before, can you explain please?
     
  6. Feb 5, 2012 #5
    A germ is essentially just a local topological structure.
     
  7. Feb 5, 2012 #6

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    Analytically, a tangent vector at a point,p, on a manifold is a linear operator that acts on differentiable functions defined in an open neighborhood of p. A function,g, on N composed with F is a function on M. So a tangent vector at p now acts on the composition of g with F. But this may also be viewed at an action on g at F(p).
     
  8. Feb 5, 2012 #7
    Thanks everyone :)
     
  9. Feb 5, 2012 #8

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I assumed that by [itex]C^{\infty}(p)[/itex] you mean the set of real-valued functions f that are defined and smooth on some neighborhood U of p, modulo the equivalence relations according to which f~g iff f and g coincide on some small nbhd of p.

    If so, then the elements of [itex]C^{\infty}(p)[/itex] are called germs of smooth functions.
     
  10. Feb 6, 2012 #9
    Just to make sure I've got it, in

    [tex][F_{*p}(X)](f) = X(f\circ F)[/tex]
    [itex]f[/itex] is kind of placeholder, in the sense that the [itex]f[/itex] on the LHS is an arbitrary function in [itex]C^\infty(p)[/itex] and the [itex]f[/itex] on the RHS is an arbitrary function in [itex]C^\infty (F(p))[/itex]


    So on the left and right sides of the equation, [itex]f[/itex] does not represent functions with the same germs. I think this is where I got confused.
     
  11. Feb 6, 2012 #10

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The f in the LHS is the same as the f in the RHS, and in both case, it is a function in [itex]C^\infty (F(p))[/itex]. Indeed, it better be so that f o F is in [itex]C^\infty(p)[/itex] so that X(f o F) makes sense!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Need help understanding the pushforward
  1. Product of pushforward (Replies: 3)

Loading...