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Push Forward on a Product Manifold.

  1. Sep 23, 2014 #1
    Some words before the question.
    For two smooth manifolds [itex]M[/itex] and [itex]P[/itex] It is true that
    [tex]T(M\times P)\simeq TM\times TP [/tex]
    If I have local coordinates [itex]\lambda[/itex] on [itex]M[/itex] and [itex]q[/itex] on [itex]P[/itex] then ([itex]\lambda[/itex], [itex]q[/itex]) are local coordinates on [itex]M\times P[/itex] (right?). This means that in these local coordinates the tanget vectors are of the form [itex]a^{i}\frac{\partial}{\partial\lambda^{i}}+b^{i}\frac{\partial}{\partial q^{i}}[/itex]

    Now, I can compute push forwards in local coordinates. For example, for a function
    [tex]f(\lambda, q)\rightarrow(\lambda,Q(q,\lambda))[/tex]
    [tex]f^{*}\left(\frac{\partial}{\partial\lambda}\right)=\frac{\partial}{\partial\lambda}+\frac{\partial Q}{\partial\lambda}\frac{\partial}{\partial q}[/tex]
    where I just had to do the matrix product of the Jacobian to the column vector [itex](1,0)^{T}[/itex].

    Actual Question.

    For a function [itex]f:\, TM\times TP\longrightarrow TM\times TP[/itex]
    and without using local coordinates what can be said about the Push forward [itex]f^{*}:\, TM\times TP\longrightarrow TM\times TP[/itex] ?.

    Particularly interested if the push forward can be descomposed into something in [itex]TM[/itex] product something in [itex]TP[/itex].
  2. jcsd
  3. Sep 23, 2014 #2


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    There is a result that for m in M , n in N, ## T_{(m,n)} (M \times N) = T_m M (+) T_n N ## , where ##(+)## is the direct sum of (tangent) vectors. Where by '=' I mean isomorphic.
  4. Sep 23, 2014 #3
    Thanks for replying.

    Yes, I'm aware of the result, I actually implicitly used it in the example above.
  5. Sep 23, 2014 #4


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    I am not sure of what you are looking for, but you can also use properties of duality, since ##TM:= \cup (T_p M)^{*}##
    Then ## T(M\times N) = (T_{(m,n)} M \times N )^{*}=T_m^{*}M (+) T_n^{*}N = TM \times TN ##.

    Then a finite direct sum is a direct product .

    You may want to play with these properties of duals , duals of products, etc. to look for the
    result you want.
    Last edited: Sep 23, 2014
  6. Sep 23, 2014 #5
    I'm not sure to understand what you mean. Either way, What I'm intersted is in the push forward.

    If I have local coordinates, like in the example in the OP, for a given a function of [itex]M\times P[/itex] to itself I can compute the push forward of that function. If I do not have coordinates then I have no idea what to do.

    For example, let [itex]G[/itex] be a lie group (I'm interested in SU(n)). The left translation is given by [itex]L_{a}b=ab[/itex]. The push forward of the left translation aplied to a tangent vector at the identity, [itex]E\in T_{e}G[/itex] , would give a tangent vector at the new group element [itex](L_{a})_{*}E\in T_{a}G[/itex].

    Now, for [itex] (g,g\text{´)}\in G\times G[/itex] I would like to define [itex] L_{(1,h)}(g,g\text{´)}=(g,hg)[/itex], and find the push forward of this function as [tex] L_{(1,h)*}(E\oplus0)= Something \oplus Something[/tex] .
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