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No difference between covectors and functions?

  1. May 6, 2010 #1
    I'm reading into an introductory book on manifolds (Tu) and they start out by showing vectors are isomorphic to derivations at a point. They go on to introduce covectors, k-tensors, and then I've just gotten to the point where they introduce the "d" operator which constructs a 1-form from a continuous function.

    It seems like vectors in R^n can be interpreted isomorphically as points in R^n (though I haven't tried to prove it.) This suggests that there is really no difference between functions (R^n --> R) and covectors, and so I feel a little bit confused or unsure about things. Is this just a fluke for this specific type of k-tensor (i.e., the covector) or am I interpreting things incorrectly?
     
  2. jcsd
  3. May 6, 2010 #2
    You're correct in a certain sense, but I'm not quite sure I see the benefit of the identification. If you're working on n-dimensional Euclidean space, the base space, the tangent space at a point, and the cotangent space at a point are all diffeomorphic (i.e. we can treat them all as points in R^n).

    But it's certainly not the same as functions from R^n to R - this is infinite dimensional no matter how you look at it.
     
  4. May 7, 2010 #3

    dx

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    Covectors are linear functions, and therefore must be defined on a vector space. If you have a function on a manifold M, then applying the d operator to that function will give you a covector in each of the tangent spaces TpM.

    In your example you have a space Rn. A function on Rn is arbitrary, and does not need to respect the linear structure of Rn, whereas a covector on Rn has to be a linear function on the space of tangent vectors, which in this case is isomorphic to Rn as you said. (a vector space and its dual space are isomorphic as vector spaces.)
     
  5. May 7, 2010 #4

    Ben Niehoff

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    Any finite-dimensional vector space is isomorphic to its dual. However, in the infinite-dimensional case, the dual space might be strictly larger than the original vector space.
     
  6. May 7, 2010 #5
    Ahh ok, I think I am understanding, especially the post from dx gets to the heart of my confusion. I forgot that regular functions don't have to be linear, whereas the covectors (or generally k-tensors) are supposed to be multilinear by definition. Thanks a lot.
     
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