Search results

  1. K

    I Differential forms and bases

    Yes, I get it now. See my reply to @fresh_42. Thank you too!
  2. K

    I Differential forms and bases

    @fresh_42 I see it now. Thank you so much for your very detailed post! The book I'm reading does define the pullback of maps on manifolds. I got confused because it doesn't give an explicit formula for the pullback of forms. Instead, it says that the pullback can be extended to differential...
  3. K

    I Differential forms and bases

    It seems to me fresh_42 gave the same exact definition I'm using: ##(\phi^* \nu)(p) = \nu(\phi(p)) = (\nu\circ\phi)(p)##. His expression for differential forms is just a property of the ##d## operator, according to my book. In ##(f^*(w))(X_p) := w(f_* X_p)## you do the pullback on ##w## by...
  4. K

    I Differential forms and bases

    Speaking of push-forward, one book says that it's also called differential, but another book defines the differential differently: ##df(X_p) = X_p(f)##. Which is it? I also noticed that your definition of pull-back is somewhat different from mine. Your definition is ##(f^\star w)(X_p) :=...
  5. K

    I Differential forms and bases

    Basically, you have defined the tangent space and the cotangent space by the push-forward and pull-back induced by a map ##F:M\to N.## One can also note that the matrix associated with ##F_*## is just the Jacobian matrix of ##F## (which is more or less equivalent to your remark about the Chain...
  6. K

    I Differential forms and bases

    I understand that a covector is just a vector, but can we say that a cotangent space is just a tangent space? They're both vector spaces but are they both tangent to the manifold at a point? To me "tangent" means that it has to do with derivations, whereas cotangent means it's related to...
  7. K

    I Differential forms and bases

    I assume you meant cotangent space.
  8. K

    I Differential forms and bases

    In the exercises on differential forms I often find expressions such as $$ \omega = 3xz\;dx - 7y^2z\;dy + 2x^2y\;dz $$ but this is only correct if we're in "flat" space, right? In general, a differential ##1##-form associates a covector with each point of ##M##. If we use some coordinates...
  9. K

    I Cordinates on a manifold

    In my head I was identifying ##x^i## with ##\phi^i## and so the ##x^i## were local coordinates directly on ##U##. In the meantime I'd like to thank you all for your patience!
  10. K

    I Cordinates on a manifold

    I'm still not completely sure. We assumed that ##(U, \phi)## is a coordinate chart from the start. Let's say ##\phi## goes from the manifold ##M## to ##\mathbb{R}^n## and we also have some coordinates ##x^i:\mathbb{R}^n\to\mathbb{R}##. Then let's define ##\phi^i = x^i\circ\phi##. Would ##(U...
  11. K

    I Cordinates on a manifold

    Maybe I'm starting to see the problem with my definition: the coordinates would just be a local parametrization of the curve but maybe I'd lack enough structure to do regular calculus over them. The cleanest way is to define ##\phi## from ##M## to ##\mathbb{R}^n## and then the coordinates on...
  12. K

    I Cordinates on a manifold

    It says: --- starts --- Let ##(U,\phi)## be a coordinate chart with ##p \in U## and suppose that ##\phi(p)=q##. If ##x^1,\ldots,x^n## are the standard coordinate functions on ##\mathbb{R}^n## then ##q## has coordinates ##(x^1(q),\ldots,x^n(q))##. Thus we can write $$ \phi(p) =...
  13. K

    I Cordinates on a manifold

    Whether something is an abuse or not depends on the definitions you choose. You decided to define the ##x_i## the way you did, which makes my notation an abuse. But the point is this: what do you gain by using your definition? Or in other words, what do I lose by not using it?
  14. K

    I Cordinates on a manifold

    We're not talking about tangent spaces here. The manifold could be non-smooth. We're talking about the charts ##(U,\phi)## where ##\phi:U\to\mathbb{R}^n## is an homeomorphism. The point is whether we need to introduce coordinates ##x_1,\ldots,x_n## explicitly or if ##x_i## is just ##\phi_i##...
  15. K

    I Cordinates on a manifold

    An element of ##\mathbb{R}^n## has the form ##(x_1,\ldots,x_n)## by definition, so we must have ##\phi(p) = (\phi_1(p),\ldots,\phi_n(p))##. Why should we define a coordinate system for ##\mathbb{R}^n## when ##\mathbb{R}^n## is already a cartesian product? I would agree with you if we were...
  16. K

    I Cordinates on a manifold

    Let ##M## be an ##n##-dimensional (smooth) manifold and ##(U,\phi)## a chart for it. Then ##\phi## is a function from an open of ##M## to an open of ##\mathbb{R}^n##. The book I'm reading claims that coordinates, say, ##x^1,\ldots,x^n## are not really functions from ##U## to ##\mathbb{R}##, but...
  17. K

    A Definition of Tensor and... Cotensor?

    Thank you both!
  18. K

    A Definition of Tensor and... Cotensor?

    Why are there (at least) two definitions of a tensor? For some people a tensor is a product of vectors and covectors, but for others it's a functional. While it's true that the two points of view are equivalent (there's an isomorphism) I find having to switch between them confusing, as a...
  19. K

    I Hodge dual

    I'm reading Renteln's book.
  20. K

    I Hodge dual

    I'm making the inner product negative because if ##g(\sigma,\sigma)=(-1)^d##, then ##a\wedge\star b = (-1)^d g(a,b)\sigma##, according to the book. Here's the full theorem: Let ##V## be ##n##-dimensional with inner product ##g##. Let ##\eta,\lambda \in \bigwedge\nolimits^p V##, and choose...
  21. K

    I Hodge dual

    I assumed ##1\wedge\sigma=\sigma## but it doesn't make much sense. Maybe ##\star 1=\sigma## is just a convention.
  22. K

    I Hodge dual

    <Moderator's note: Moved from another forum.> The book I'm reading says that ##\star \sigma = 1## and ##\star 1 = \sigma##, but I'm not sure about the last one. The space is ##V = \mathbb{M}^4## and we choose the canonical base ##e_0,e_1,e_2,e_3##. This means that ##g_{ij} =...
  23. K

    I Covariant Derivatives

    OK, I think I understand now. The components ##V^j## and ##V^a## are related by the Jacobian matrix only because they're themselves derivatives and we can apply the chain rule. Thank you!
  24. K

    I Covariant Derivatives

    ##R## is the position vector, so it's a vector by definition. To introduce ##R##, Grinfeld chose an arbitrary origin. The problem is that it can't be expressed through a covariant basis for obvious reasons so the covariant derivative is not needed and doesn't make sense. I thought that it was...
  25. K

    I Covariant Derivatives

    I'm not trying to prove that ##Z^j## is not a vector. I already know that. I was basically trying to prove that ##\delta \pmb{R} / \delta t = d\pmb{R} / dt## by using the formula for the covariant derivative. What are the components of the position ##\pmb{R}## vector, by the way? ##\pmb{R}##...
  26. K

    I Covariant Derivatives

    That makes more sense. Unfortunately, Grinfeld explicitly defined it on the components and that confused me a little. But what about general tensors? His full definition works for any tensor ##T^{ijk\cdots}_{rst\cdots}##. Basically, there's the ordinary derivative and then an additional...
  27. K

    I Covariant Derivatives

    That's what I'm trying to prove! I'm watching the videos by Pavel Grinfeld and he defines ##\delta/\delta t## as a way to rewrite the derivative of the velocity vector in a cleaner way. (One can then prove that this definition makes sense (e.g. the Leibniz rule applies)). When he introduced the...
  28. K

    I Covariant Derivatives

    I've just learned about the covariant derivatives (##\nabla_i## and ##\delta/\delta t##) and I have a doubt. We should be able to say that $$ J^i = \frac{\delta A^i}{\delta t} = \frac{\delta^2 V^i}{\delta^2 t} = \frac{\delta^3 Z^i}{\delta^3 t} $$ where ##J## is the jolt. This...
  29. K

    I Christoffel symbol ("undotting")

    Ah, yes. After all that's exactly what the Christoffel symbol means: it "contains" the components of all the n^2 vectors wrt the covariant basis. In hindsight it really was a dumb question! Thank you. META: how should I thank people on this forum? Should I just press the "like" button?
  30. K

    I Christoffel symbol ("undotting")

    I hope you can understand my notation. The Christoffel symbol can be defined through the relation$$ \frac{\partial \pmb{Z}_i} {\partial Z^k} = \Gamma_{ik}^j \pmb{Z}_j $$ I can solve for the Christoffel symbol this way: $$ \frac{\partial \pmb{Z}_i} {\partial Z^k} \cdot \pmb{Z}^m = \Gamma_{ik}^j...
Top