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A Abstract tensor notation

  1. May 5, 2016 #1
    In Wald's "General Relativity", in his section on abstract tensor notation, he lets [itex]g_{ab}[/itex] denote the metric tensor. When applied to a vector [itex]v^a[/itex], we get a dual vector, because [itex]g_{ab}(v^a, \cdot)[/itex] is just a dual vector. Okay cool. But then he says that this dual vector is actually [itex]g_{ab}v^b[/itex], which is a contraction. But don't we have [itex]g_{ab}v^b = \sum\limits_{i=1}^n g(\cdot, v^i) = \sum\limits_{i=1}^n g_{ab}(v^i, \cdot)[/itex], which in general is not going to be equal to [itex]g_{ab}(v^a, \cdot)[/itex]? Where am I messing up here?

    EDIT: [itex] \{v^i\}[/itex] is a basis for the tangent space.
     
    Last edited: May 5, 2016
  2. jcsd
  3. May 5, 2016 #2
    Well, the metric tensor is a symmetric tensor, so the order of the arguments doesn't matter.
     
  4. May 5, 2016 #3
    I know, but suppose for example [itex] v^a = \sum\limits_{i=1}^n c_i v^i[/itex] for scalars [itex]c_i > 1[/itex]. Then [itex] g_{ab}(v^a, \cdot) \neq \sum\limits_{i=1}^n g_{ab}(v^i, \cdot) [/itex]

    EDIT: My issue is, when you apply the metric tensor to the vector [itex] v^a [/itex] then you get the dual vector [itex] g_{ab}(v^a, \cdot) [/itex]. But Wald writes that this is also the tensor [itex] g_{ab}v^b [/itex]. Now, I don't see how [itex] g_{ab}(v^a, \cdot) = g_{ab}v^b [/itex]. In fact, according to my calculations above, they are not, in general, equal.

    EDIT 2: Hold on, is [itex] g_{ab}v^b [/itex] really a contraction? I thought contractions were applied to a single tensor? I mean, you can contract [itex] T^{abc}_{de} [/itex], but I don't see how [itex] g_{ab}v^b [/itex] is a contraction.
     
    Last edited: May 5, 2016
  5. May 5, 2016 #4
    Ah okay I get what you are trying to say now; the notation used in that book is really really terrible. Let me switch to friendlier notation: using an overbar for vectors and a tilde for dual vectors (also known as one-forms).
    So, we have
    [tex]\tilde{V} = \mathbf{g} (\bar{V}, \cdot)[/tex]
    To find out the components of this dual vector, we evaluate
    [tex]V_{i} = \tilde{V}(\bar{e}_{i}) = \mathbf{g} (\bar{V}, \bar{e}_{i}) = \mathbf{g} (V^{j} \bar{e}_{j}, \bar{e}_{i}) = V^{j} g_{ij}[/tex]

    As for the strange notation of ##g_{ab}(v^{a}, \cdot)##, I have absolutely no idea what it means...when you put superscripts or subscripts you are referring to the components already evaluated in some basis.
     
  6. May 5, 2016 #5
    If ##v^a## is a general vector, why does it have a superscript - and why does it share the same index as that of the metric tensor? I think the source of your confusion stems from the notation.
    Yes, in standard notation, it is a contraction. You can always define [itex]T_{ab}^{c} \equiv g_{ab}v^c[/itex], no?
     
  7. May 5, 2016 #6
    I will type exactly what the section says in my textbook. It's only a few lines:

    Additional notational rules apply to the metric tensor, both in the index and component notations. Since a metric ##g## is a tensor of type ##(0,2)##, it is denoted ##g_{ab}##. If we apply the metric to a vector, ##v^a##, we get the dual vector ##g_{ab}v^b##.
     
  8. May 5, 2016 #7
    I would like to go over some things in regards to this abstract tensor notation..Suppose we have a tensor of type ##(3,2)##. We can denote it by ##T^{abc}_{de}##. This means that ##T: V^* \times V^* \times V^* \times V \times V##. So, let ##\{\phi^*_i\}## denote the elementary contravariant tensors on ##V## and let ##\{\phi_i\}## denote the elementary covariant tensors on ##V##. Then we can write ##T^{abc}_{de} = \sum\limits_{i,j,k,u,v = 1}^n \phi^*_i \otimes \phi^*_j \otimes \phi^*_k \otimes \phi_u \otimes \phi_v ##. Do we write ##T## is equal to this summation? Or do we write ##T^{abc}_{de}## is equal to this summation? Because at first I thought ##T^{abc}_{de}## denoted the tensor, but now I am starting to think that it just denotes the general component of the tensor?

    EDIT: Sorry, it should be ## T^{abc}_{de} = \sum\limits_{i,j,k,u,v = 1}^n \phi_u \otimes \phi_v \otimes \phi^*_i \otimes \phi^*_j \otimes \phi^*_k ## right?
     
  9. May 5, 2016 #8
    After racking my brain, I honestly have no idea what Wald is saying, although I guess that whatever he is trying to say is essentially equivalent to whatever I posted in #4. More accurately, ##g_{ab}V^b = V_{a}## is the a-th component of the dual vector ##\tilde{V} = \mathbf{g}(\bar{V}, \cdot)##.

    Well, it can sometimes mean the former when used loosely, i.e. the tensor ##T^{abc}_{de}## blahblahblah, when when used mathematically, it refers to the component - after all, when you perform manipulations and summations, you are working with numbers (i.e. components).

    I'm not sure that this makes any sense. We usually write
    [tex]\mathbf{T} = T^{abc}_{de} \bar{e}_{a} \otimes \bar{e}_{b} \otimes \bar{e}_{c} \otimes \tilde{\omega}^{d} \otimes \tilde{\omega}^{e} [/tex]
    where ##\bar{e}## and ##\tilde{\omega}## are the basis vectors and dual vectors (one-forms).
     
  10. May 5, 2016 #9

    George Jones

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    Consider tensor products of the vector space (over the reals) ##V## and its dual space

    In Penrose's abstract index notation, which Wald uses, and which I dislike, an index that uses the latin alphabet labels the space in which the object lies, so ##v^a## is an element of ##V##, not a real number (component with respect to a basis), ##g_{ab}## is an element of ##V^* \otimes V^*##, etc. Latin indices unambiguously give the space in which the object lives.

    Indices that uses the greek alphabet specify components with respect to a particular basis, i.e., are real numbers.

    ##g_{ab} v^b## lives in ##V^*## (not in ##\mathbb{R}##), and in more traditional notation might be written as a contraction of ##\bf{g}\otimes\bf{v}##
     
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