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Attempting to learn tensor calculus

  1. Feb 1, 2014 #1
    Hello all,

    After a brief break from attempting to learn tensor calculus, I'm once again back at it. Today, I started reading this: http://web.mit.edu/edbert/GR/gr1.pdf. I got to about page 4 before things stopped making sense, right under equation 3. Question 1: apparently a "one-form" is a scalar function of a vector; thus, if P is a one-form and V is a vector, then P(V) makes sense. However, V(P) obviously doesn't; a vector isn't a function. Can someone please explain this?

    Regardless of my confusion, I continued on. Immediately under the introduction to tensors on page 5, it says that "a tensor of rank (m,n) is defined to be a scalar function of m one-forms and n vectors. Thus, a vector is a tensor of rank (1,0)..." Question 2. How can this be so? Shouldn't a vector be a tensor of rank (0,1), based on the above definition?

    I ran into my biggest confusion on page 6 under the metric tensor. At the top, it says that the metric tensor is a "symmetric bilinear scalar function of two vectors" (just like before- the tensor was said to be a scalar function). Lower on the same page, it says, "thus, the metric g is a mapping from the space of vectors to the space of one-forms." Question 3. What???? Didn't it just say that tensors map vectors and one-forms to scalars?

    I'm really sorry if these questions are trivial. I would greatly appreciate if anyone could help me out here, because I'm getting a little frustrated.
     
  2. jcsd
  3. Feb 1, 2014 #2

    WannabeNewton

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    It's just bad phrasing on the part of the notes. A vector ##V^{\mu}## can map a one-form ##\omega_{\mu}## to a real (scalar) number simply by ##V^{\mu}\omega_{\mu}##. Obviously this is the same thing as the one-form mapping the vector to a real number since all we're doing is contracting indices.

    No. A tensor ##T^{\mu_1...\mu_m}{}{}_{\nu_1...\nu_n}## maps m one-forms and n-vectors to the reals by index contraction just as above i.e. ##T^{\mu_1...\mu_m}{}{}_{\nu_1...\nu_n}\omega_{\mu_1}...\omega_{\mu_m}V^{\nu_1}...V^{\nu_n}##. A vector maps 1 one-form to the reals so it is a rank (1,0) tensor.

    Bad phrasing on their part again. The metric tensor ##g_{\mu\nu}## maps two vectors to the reals again by index contraction i.e. ##g_{\mu\nu}V^{\mu}W^{\nu}##; this is of course just the inner product of ##V^{\mu}## and ##W^{\mu}##. However at the same time, we can also use ##g_{\mu\nu}## to map between vectors and one-forms by ##V_{\mu} = g_{\mu\nu}V^{\nu}## and ##\omega^{\mu} = g^{\mu\nu}\omega_{\nu}##. This is informally called "raising and lowering of indices" and formally called a musical isomorphism. Note then that the inner product between ##V^{\mu}## and ##W^{\mu}## is simply given by ##V_{\mu}W^{\mu}##. Of course the operation of raising and lowering indices in this way only makes sense because the musical isomorphism is bijective.

    See here for more: http://en.wikipedia.org/wiki/Musical_isomorphism

    P.S. I would recommend getting an actual book on the subject of tensor calculus.
     
  4. Feb 1, 2014 #3

    jcsd

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    For a vector space V over K the one-forms are the set of all linear functionals p:V->K. The space of all the one-forms P is called the (algebraic) dual vector space of V

    We can define vectors in V as linear functionals v:P->K using v(p) = p(v), for real finite-dimensional vector spaces.

    A vector is a scalar function of 1 one-form, so it is a tensor of rank (1,0)

    If v and w are in V, then we can use g(v,w) = p(w) to map v to p.
     
  5. Feb 1, 2014 #4

    stevendaryl

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    If [itex]V[/itex] is a vector space, then you can form a second vector space [itex]V^*[/itex] which is the set of linear functions that take an element of [itex]V[/itex] and return a scalar. If you do that operation twice, you get [itex]V^{**}[/itex]. Those are linear functions that take an element of [itex]V^*[/itex] and return a scalar. For finite-dimensional vector spaces, [itex]V[/itex] and [itex]V^{**}[/itex] are isomorphic. That is, if [itex]v[/itex] is an element of [itex]V[/itex], then you can define a corresponding element of [itex]V^{**}[/itex] as follows:

    [itex]v' = [/itex] that function such that for any element [itex]\phi[/itex] of [itex]V^*[/itex],
    [itex]v'(\phi) = \phi(v)[/itex]

    So the distinction between [itex]V[/itex] and [itex]V^{**}[/itex] is usually ignored in talking about tensors, since you can go back and forth between them.

    So in this way, a vector [itex]v[/itex], which is an element of [itex]V[/itex] can also be regarded as an element of [itex]V^{**}[/itex]. As an element of [itex]V^{**}[/itex], it is a function that takes 1 one-form and returns a scalar. So it is a [itex](1,0)[/itex] tensor. A one-form takes a 1 vector and returns a scalar, so it is a [itex](0,1)[/itex] tensor.
     
  6. Feb 1, 2014 #5

    Nugatory

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    Physicists tend to be sloppy about the formalism, especially after you've used the machinery enough to get comfortable with it. Half-jokingly, I've suggested that one way of distinguishing physicists from mathematicians is that:
    - A physicist will tell you that all vectors are tensors, because a vector is obviously just a rank-one tensor with one [STRIKE]contravariant[/STRIKE] upper index.
    - A mathematician will tell you that all tensors are vectors, because tensors meet the mathematical definition of a vector space.

    OK, so with these preliminaries out of the way (and at this point the mathematicians may wish to avert their eyes)....

    ##g(A,B)=g_{ij}A^iB^j## is clearly a function that maps two vectors to a real scalar.

    But I can also use the metric tensor as a function of one vector: ##g(A)=g_{ij}A^j=A_i## and that's mapping the vector ##A## to a one-form. ##A_i## has a [STRIKE]covariant[/STRIKE] lower index and can be used to map a vector onto a real via ##A_iB^i = g(A,B)## which makes it a perfectly reasonable one-form.

    OK, the formalists can come back now :smile:
     
  7. Feb 1, 2014 #6
    I don't understand this. V* is a scalar, and the * operator as you have defined it takes a vector input, not a scalar, so V** should be undefined. What is wrong with that logic?

    Edit: never mind, I see what you're saying

    Also, thanks to everyone for the replies. Can someone give me a specific example of a vector, one-form, and tensor relation (instead of all the indices)? That would be really helpful. Also, is the original definition of a one-form correct: a scalar valued function of a vector?
     
    Last edited: Feb 1, 2014
  8. Feb 1, 2014 #7

    jcsd

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    If V is a finite-dimensional real vector space than any linear function from V to R is a one-form.

    An example of a one-form would be v*(w) = <v,w> and an example of a (higher rank) tensor would be the metric tensor g(v,w) = <v,w>.

    The definition as you state it is not quite sufficient t because the function must be linear.
     
    Last edited: Feb 1, 2014
  9. Feb 1, 2014 #8

    stevendaryl

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    No, [itex]V[/itex] is a SET, a set of vectors. [itex]V^*[/itex] is another set, the set of all linear functions [itex]f[/itex] such that if [itex]v[/itex] is a vector, then [itex]f(v)[/itex] is a scalar. The operator [itex]^*[/itex] takes a set as an input, and returns a set as an output.

    In some places, the notation [itex]A \rightarrow B[/itex] is used to mean the set of all functions from set A to set B. So in terms of that notation,

    [itex]V^* = V \rightarrow \mathcal{R}[/itex]
    [itex]V^{**} = V^{*} \rightarrow \mathcal{R} = (V \rightarrow \mathcal{R}) \rightarrow \mathcal{R}[/itex]
     
  10. Feb 1, 2014 #9
    Also, can someone explain what's up with raising and lowering indices?
     
  11. Feb 1, 2014 #10

    pervect

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    This just says that a vector is a linear map from a one-form to a scalar. Which is correct as writtten, but you need the "duality" idea to understand it. Look in your linear algebra textbook for "duals of vector spaces".

    The duality idea starts with the fact that if V is a vector space, linear maps from V to scalars form another vector space (this is generally proved in most linear algebra textbooks). This vector space is called the "dual" of the original vector space, and denoted by V*.

    If we go on, we can define a map from V* to scalars, and call this map V**. The duality idea is that V** === V. This is also proved in most linear algebra textbooks. You need to understand this before you start.


    If A and B are vectors, we have the fact that the metric maps (A,B) into a scalar. For any given value of B, though, this means that the metric defines a map from A to a scalar.

    [edit]
    Since a map from A to a scalar is a one form, for any specified value of B, we have a one form, Thus we have a map from a vector, B, to a one form, the one form being the map from A to a scalar given by g_ab A B.

    [edit2]
    More generally, we can regard a tensor of type (m,n) as something that takes
    a one-forms and b-vectors and returns a tensor of type (m-a, n-b). This is just a partitioning operation. You fill a of the one-form slots with the one-forms you specified, and b of the vector slots with the vectors you specified. You then have remaining a map of from (m-a) one forms and (n-b) vectors to a scalar, which is a tensor of rank (m-a, n-b).
     
    Last edited: Feb 2, 2014
  12. Feb 1, 2014 #11

    robphy

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  13. Feb 3, 2014 #12
    Last edited: Feb 3, 2014
  14. Feb 3, 2014 #13

    Nugatory

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    You can organize the components of any rank-two tensor, including the metric tensor, as a matrix. That doesn't make it a matrix though - what really makes a tensor a tensor is that its components transform from one coordinate system to another in a way that maintains the validity of the tensor equation.

    For example, the dot product of two vectors is given by the coordinate-independent equation ##A\cdot{B} = g(A,B)##. If I'm working in two-dimensional Cartesian coordinates, the matrix representing the metric tensor is ##diag(1,1)##; if I'm working in polar coordinates the matrix would be ##diag(1,r^2)## and of course the components of the vectors would be represented differently in that coordinate system so that the dot-product comes out the same; and if I use some coordinate system in which the axes are not orthogonal, the matrix will pick up some off-diagonal elements.
     
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