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Interpretation of the Gradient Vector?

  1. Jul 24, 2013 #1
    I've always thought of the gradient of a scalar function (id est, ##\nabla\varphi##) as being a vector field. However, I started thinking about it just now in terms of transformation with respect to coordinate changes, and I noticed that the gradient transforms covariantly. Thus, shouldn't the gradient be represented with a row vector?

    I don't know why this is confusing for me. After looking through a couple websites, I saw that there were some who said "yes" and some who said "no," so I don't know what to think.
     
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  3. Jul 24, 2013 #2

    WannabeNewton

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    It's technically not a vector field but rather a 1-form. The components ##\partial_{i}f## transform as those of a 1-form so that's a simple reason for why i.e. ##\partial_{i'}f = \frac{\partial x^{i}}{\partial x^{i'}}\partial_{i}f##. A slightly more general explanation is as follows: https://www.physicsforums.com/showpost.php?p=4374094&postcount=18
     
    Last edited: Jul 24, 2013
  4. Jul 24, 2013 #3
    I thought that ##\nabla\varphi=(d\varphi)^\sharp##. Wouldn't that imply that the gradient is a vector field? Or, am I just confused as to the effects of raising the index? Or, is the equation not valid?
     
  5. Jul 24, 2013 #4

    WannabeNewton

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    The musical isomorphism is trivial for ##\mathbb{R}^{n}## so it really doesn't matter but if you look at the gradient component wise as ##\partial_{i}f## then it's a 1-form.
     
  6. Jul 24, 2013 #5

    micromass

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    The gradient of a scalar field is actually a section of the cotangent bundle of the manifold. However, if we are given a metric tensor, then we have a natural morphism between the cotangent bundle and the tangent bundle. As such, in Riemannian manifolds, the gradient can be seen as a section of the tangent bundle. When no natural metric is given, it can not be seen as such.

    In algebraic geometry, we can see the gradient as an ##A##-derivation map ##d:B\rightarrow \Omega_{B\A}##, where ##A##, ##B## is an ##A##-algebra and where ##\Omega_{B\setminus A}## is the module of relative differential forms. We can further generalize this by defining ##\Omega_{X\setminus Y}## as a quasicoherent module of sheafs, when ##X## and ##Y## are schemes.
     
    Last edited: Jul 24, 2013
  7. Jul 24, 2013 #6

    WannabeNewton

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    Let me just add on a bit to what I said before. We can think of ##\partial_{a}## as a derivative operator on ##\mathbb{R}^{n}## that sends (n,m) tensor fields to (n,m+1) tensor fields (I'm sure you've seen this before in a more general setting!) so in this sense it sends a scalar field (i.e. a (0,0) tensor field) to a 1-form (i.e. a (0,1) tensor field); now let's say we have a metric tensor ##g_{ab}## then we can define the associated vector field ##\partial^{a}\varphi = g^{ab}\partial_{b}\varphi## which is just the musical isomorphism. In ##\mathbb{R}^{n}## we usually take ##g_{ab} = \delta_{ab}## so if ##\partial_{i}\varphi## are the components then ##\partial^{i}\varphi = \delta^{ij}\partial_{j}\varphi = \delta^{i1}\partial_{1}\varphi + \delta^{i2}\partial_{2}\varphi + \delta^{i3}\partial_{3}\varphi## so ##\partial^{1}\varphi = \partial_{1}\varphi## etc.
     
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