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Push forward and pull back

  1. May 21, 2015 #1
    Hello

    I am a mechanical engineer who is teaching himself the math of exterior algebra and differential forms. It is not easy for me and I have had many SIMPLE stumbling blocks due to my not respecting algebra.

    May I ask for help on some simple aspects? (Please be patient with me.)

    My question is less a mathematical question and more one of semantics (perhaps stemming from ignorance).

    (As a caveat: I am not entirely sure I am even asking what I want to ask properly... so there might be followups.)

    In some places I have read: one PULLS BACK a FORM and PUSHES FORWARD a VECTOR. And the authors make an issue of this.

    In other places I have read one can pull back both and push forward both.

    First: what gives?

    Can I assume the SECOND case is ONLY possible when the mappings themselves are readily reversible? Could it be that?

    Why then do most authors make an issue of this? What is the historical reason?

    Also, many of the mappings are from M to N to R. And they... I really do not know how to say this without really revealing my ignorance, so here goes.... they go from left to right or in one direction. Is THAT the ORIGIN of the phrase PULL BACK or push FORWARD

    Now, if is true, WHY do books and theories focus MORE on the pull back?

    If one is pulling something BACK, then one must assume you are in the first space and yanking it back to you. If you are pushing it forward, you are in that same space and pushing it to a place where you are not. What is it about the space were you are that you are not happy with things THERE and things HERE: God I know I must sound like an idiot, but WTF.

    And now comes the odder questions…

    What is the big deal? HOW does pulling back FORMS matter to an engineer? How does pushing forward vectors matter? Yes, I understand FORMS are dual to Vectors. But what IS it about forms and what IS it about vectors that we pull one BACK and push the other FORWARD. What are we pulling them back TO? How does pulling them back help me?

    Could you make a comment in SIMPLE language?

    Also, I am getting the feeling that this thing I have come to know as the Material derivative in fluid mechanics, is intimately connected to this idea of a pullback or pushforward, but I cannot see it through the fog right now.

    Could you comment on that?

    Note that I am putting in hours to teach myself topology, manifolds, etc and learning a lot about how my educatino failed me. But for now, I really could use WORD responses that get to the core of the issue, qualitatively.
     
    Last edited: May 21, 2015
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  3. May 21, 2015 #2

    Fredrik

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    Can you tell us a bit more about how much differential geometry you know? In particular, are you familiar with the concepts manifold, tangent space, cotangent space, tensor and tensor field?
     
  4. May 21, 2015 #3
    So, I have read and PARTALLY understood the "Informal Overview of Caratns Exeterior Differential Forms" in Ted Frankels book "Geometry of Physics."
    I have read and understood everything leading up to General Stokes Theorem.
    I understand that 0 forms are functions, 1 forms are isomorphic to gradients, 2 forms to curl, and so on. I can reduce the general Stokes in 1D to First Fundamental Theorem of the Calculus, in 2D to Greens Theorem.

    But I LACK the PASSIONATE understanding. And I cannot even say what I mean by that. I can follow the stuff. I have a FEELING I can use it to improve my understanding. But I cannot wrap my head and RISE ABOVE (there, that is the key: I cannot RISE ABOVE what I am reading)

    I have read a bit on topology and metric spaces... and am learning about the difference.
    I know a LITTLE about manifolds and vector spaces.

    I cannot yet USE this to comfortabley say I understand tensors.

    Let me also add this...

    I understand that the integrand in a line integral is a one form pulled back, not a vector.

    Well... SO WHAT? Why does that matter to me? (I am being facetious with deliberation... I want to know why it matters to say it that way. To me, right now, having been improperly taught, it is just change of variables.)

    ALSO THIS: I understand that Force (acceleration vector weighted with mass) is a one form. One forms eat vectors to get R. Um... so what? That and some cash gets me on the subway?

    Now I have a FEELING that something is happening... forms on one manifold, vectors are on the other, things get pushed or pulled to enable the operation... But I cannot rise above it and I cannot continue reading until I see this realized. I have been struggling with about a month now (hours each night) and have this feeling it is coming together so please forgive what is ostenibly dissmissive.

    And WHAT about all this is EXTERIOR? Why is it called EXTERIOR algebra? What is it exterior to?
     
    Last edited: May 21, 2015
  5. May 21, 2015 #4

    Fredrik

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    OK, so you're probably aware that there's a vector space ##T_pM## associated with each point p of a smooth manifold M. ##T_pM## is called the tangent space of M at p. Its dual space ##T_pM^*## is called the cotangent space at p.

    Let M,N be two smooth manifolds and consider a function ##\phi:M\to N##. If ##f:N\to\mathbb R##, then ##f\circ\phi:M\to\mathbb R##. In this sense, ##\phi## "pulls back" a function on N to a function on M.

    Now if v is a tangent vector at p, then we define ##\phi_*v## as the tangent vector at ##\phi(p)## defined by
    $$(\phi_*v)(f)=v(f\circ\phi)$$ for all smooth ##f:N\to\mathbb R##. In this sense, ##\phi## "pushes forward" a vector from the point ##p## to the point ##\phi(p)##.

    Similarly, if ##\omega## is a cotangent vector at ##\phi(p)##, then we define ##\phi^*\omega## as the cotangent vector at ##p## defined by
    $$(\phi^*\omega)(v)=\omega(\phi_*v)$$ for all ##v\in T_pM##. In this sense, ##\phi## "pulls back" a cotangent vector from the point ##\phi(p)## to the point ##p##.

    If you want to pull back a tangent vector from ##\phi(p)## to ##p##, then you just use the inverse map ##\phi^{-1}## to do a push-forward. The push-forward of a cotangent vector is defined similarly. So you guessed it right. When you want to go in the opposite of the normal direction, the map you're using needs to be invertible.

    Not sure. Maybe because the pullback is the natural way to use the metric tensor field on one manifold to define a metric tensor field on another. If ##g## is a metric tensor field on N, the pullback ##\phi^*g## is the tensor field on M defined by ##(\phi^*g)_p=\phi^*g_{\phi(p)}## for all ##p\in M##. The right-hand side here is defined by ##(\phi^*g_{\phi(p)})(u,v)= g_{\phi(p)}(\phi_*u,\phi_*v)## for all ##u,v\in T_pM##.

    I don't think I can answer your question about why an engineer should know these things, but don't your books answer it? There should be examples in them where these concepts are used.
     
    Last edited: May 21, 2015
  6. May 21, 2015 #5
    OK; how about this...

    ##\omega## is a one form on M2. It is a map on vectors that live in M2.
    V is a vector in M1.
    Now the problem is that ##\omega## and V live in different places.
    ##\omega## cannot act on V's that live in M1
    ##\phi^*\omega## is a one form in M1 which acts on vectors in M1.
    So ##\phi^*\omega## on V in M1 id defiend by how ##\omega## acts on teh vector pushed forward to M2

    Is that all it is?
    Is that what all this pushing and pulling is doing? Making the operations work?

    AND!!! If my phrasing is correct, I now must ask this:

    How has it come to pass, in geometrical physics, that vectors and their dual one forms just so happen to live in different manifolds that requres us to use the push forward or pull back?

    I would, in my ignorance, expect that the one forms and their associated vectors live happily in the same manifold if their operations on each other are producing meaningful results.
     
    Last edited: May 21, 2015
  7. May 21, 2015 #6

    WWGD

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    The reasons why forms pullback instead of pushing forward is that a linear map ##L: V \rightarrow W ## between f.d vector spaces gives rise to a map _in the opposite direction_ between their respective duals ## W^*, V^*: L^*: W^* \rightarrow V^* ##, and this is said to be "functorial".
     
  8. May 21, 2015 #7
    Oh now THAT is interesting: that is certainly a new way of thinking for me. Can I restate in my words and ask for your confirmation? (and again, please be patient, for I may just be engaging in mental masturbation until clarity descends -- if it does.)

    Vectors BEGIN the process: we must map them between two manifolds: from M1 forward to M2.
    As a CONSEQUENCE of the mapping, we get a dual map in the reverse direction: the "structure" of the algebra gives rise to this and one must accept it.
    And this reverse mapping involves the dual forms.

    OK... so let me push this higher.

    Why can we not start this process in reverse?
    Forms BEGIN the process: we must map them between two manifolds: from M2 backward to M1
    As a CONSEQUENCE of the mapping, we get a dual map in the reverse direction: the "structure" of the algebra gives rise to this and one must accept it.
    And this reverse mapping involves the dual vectors.

    So now I must ask: what came first, the chicken or the egg?
    And HOW is this (and by this, I mean this current issue of what gets pushed or pulled) relevant to physics?
    (I know this really must sound idiotic -- I am sorry.)

    And I am STILL confused as to why force is a form and acceleration is a vector. Why don't BOTH of these exist on the same manifold. Why must I push one or pull the other to do any meaningful operations?
     
    Last edited: May 21, 2015
  9. May 21, 2015 #8

    WWGD

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    The same issue of the mapping in the opposite direction is true for multilinear maps. And a form is just a multilinear map. I will look at it in more detail later.
     
  10. May 21, 2015 #9

    Please, please, please don't forget to respond later, esp, with regard to forces being forms acting on vectors and in two different manifolds.
    I am getting SO CLOSE to seeing it.
     
  11. May 21, 2015 #10

    Fredrik

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    That's essentially it, yes.

    But we use two different manifolds M and N in the definition, not out of necessity, but mainly because we can. The pullback and push-forward concepts are no less interesting or useful when M=N. Even if ##\phi:M\to M##, the tangent spaces at ##p## and ##\phi(p)## are different vector spaces. ##\phi## gives us a way to associate a tangent vector at ##\phi(p)## with each tangent vector at ##p##, and a cotangent vector at ##p## with each cotangent vector at ##\phi(p)##. If ##\phi## is invertible, you can use it to associate a tensor of any type, at either of these two points, with a tensor of the same type at the other point.

    This isn't the only way to associate tensors at one point with tensors at another. If the manifold has a connection (if it has a metric, you can always define a connection), you can use the connection to "parallel transport" tensors from any point in M to any point in M.

    The first two examples that come to mind where pullbacks or pushforwards are useful are isometries and Lie derivatives. Suppose that M and N are manifolds with metrics g and h respectively. If ##\phi:M\to N## is a diffeomorphism such that ##\phi^*h=g##, then ##M## is said to be an isometry. This concept is (at least for me) especially interesting when M=N. For example, if M is Minkowski spacetime (i.e. if ##M=\mathbb R^4## and ##g## is the Minkowski metric), then the isometries ##\phi:M\to M## are precisely the Poincaré transformations. The set of isometries of a manifold M is a group called the isometry group of M.

    Lie derivatives are a funny thing too. Let X and Y be vector fields on M. (For simplicity I will assume that their domains are M, rather than some subset). The Lie derivative at p of Y, in the direction of X, is defined by
    $$(\mathcal L_XY)_p=\lim_{t\to 0}\frac{(\phi_{-t})_*Y_{\phi_t(p)}-Y_p}{t}.$$ To understand this definition, you must understand ##\phi_t(p)##. For each ##t## in some interval of ##\mathbb R## that includes 0, the map ##p\mapsto\phi_t(p)## is a diffeomorphism from M to M. This map is denoted by ##\phi_t##. ##\phi_0## is the identity map on M, so ##\phi_0(p)=p##. For each p in M, the map ##t\to\phi_t(p)## is a curve through p. The tangent vector to this curve, at the point p, is ##X_p##. Such a curve is called an integral curve of X. So this derivative evaluates Y at ##\phi_{-t}(p)## (a point on the integral curve through p) to get a tangent vector at ##\phi_{-t}(p)##. Then it pushes it forward to ##p##. So the difference in the numerator is between two vectors in the same vector space, ##T_pM##.
     
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