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Preliminary knowledge on tensor analysis

  1. Apr 24, 2014 #1

    ChrisVer

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    I am not sure whether this needs to be transported in another topic (as academic guidance). I have some preliminary knowledge on tensor analysis, which helps me being more confident with indices notation etc... Also I'm accustomed to the definition of tensors, which tells us that a tensor is an object which has a given transformation law... Like a tensor of rank [itex](n,k)[/itex] transforms under a coordinate transformation (diffeomorphism) [itex]\phi^{a}(x^b)[/itex] as:
    [itex] (T')^{a_{1}...a_{n}}_{b_{1}...b_{k}}= \frac{\partial \phi^{a_{1}}}{\partial x^{w_{1}}}...\frac{\partial \phi^{a_{n}}}{\partial x^{w_{n}}} \frac{\partial x^{r_{1}}}{\partial \phi^{b_{1}}}...\frac{\partial x^{r_{k}}}{\partial \phi^{b_{k}}} (T)^{w_{1}...w_{n}}_{r_{1}...r_{k}}[/itex]

    However, a tensor is not only defined as such... I learned that a (n,k) tensor can also be a mapping... In particular it takes vectors from a vector space (n times) and its dual (k times) and maps them to the real numbers:
    [itex]T: V \otimes V \otimes ... \otimes V \otimes V^{*}\otimes ... \otimes V^{*} \rightarrow R [/itex]
    so in that case instead of indices one can use arguments of the corresponding vectors in V....
    eg the metric is a tensor [itex]g: V \otimes V \rightarrow R[/itex] which gets translated to [itex]g_{ab}x^{a}x^{b}= g(x^{a},x^{b})= ds^{2}[/itex]
    But wouldn't also the metric send me from [itex]V \rightarrow V^{*}[/itex] like an object which lowers or raises indices? In that case: [itex]g(x^{a})=x^{*b}[/itex]

    Is there any good book that can give me a good insight on this way of defining things? Thanks...
     
    Last edited: Apr 24, 2014
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  3. Apr 24, 2014 #2

    Fredrik

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    Introduction to smooth manifolds by John M. Lee is great. I learned these things from A comprehensive introduction to differential geometry, vol. 1, by Michael Spivak, which is also good, at least about the definitions of tensors.

    Regarding the metric...

    We have ##g:V\times V\to\mathbb R##. Let ##g(\cdot, v)## denote the map ##u\mapsto g(u,v)## from V into ##\mathbb R##. This map is actually an element of V*. So the map ##v\mapsto g(\cdot,v)## is from V into V*.

    The components of ##g(\cdot,v)## are the numbers ##g(e_i,v)=g(e_i,v^je_j)=v^jg(e_i,e_j)=g_{ij}v^j##.

    Edit: Note that it's the components of the tensors that "transform" from one coordinate system to another. The tensors themselves are coordinate independent. This would be apparent even in the old-fashioned approach, if the books that use it had bothered to write the definition in a way that resembles actual mathematics.
     
    Last edited: Apr 24, 2014
  4. Apr 24, 2014 #3

    Matterwave

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    I would like to add that any rank (0,2) tensor can be regarded as either a map from ##V\times V\rightarrow \mathbb{R}## or as a map ##V\rightarrow V^*##. This is not restricted to the metric.
     
  5. Apr 24, 2014 #4

    ChrisVer

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    Thanks Fredrik..
    Hmmm Matterwave, you are true I guess (by definition)... but what distinguishes for example the Ricci Tensor to the metric with such a definition? The vector spaces? (I know that both are tensors so they must satisfy this condition, but the way they act on/map these spaces must be different?)
     
  6. Apr 24, 2014 #5

    Matterwave

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    They are different tensors, so the maps are different (and may be very different!) of course. The special thing about the metric is that it defines the inner product on your manifold, turning your manifold into a Reimmanian manifold. This allows you to start talking about things like distances (indeed, this is perhaps the aspect of the metric tensor) between two points.

    So, let's look at the map that ##g## induces on your manifold. If I put a vector in it, I get out a one form from the map: ##\tilde{V}(\quad)\equiv g(V,\quad)##. Now, what is the inner product of this vector with itself? Well it's simply: ##V\cdot V=g(V,V)=\tilde{V}(V)##. This means that the contraction of ##\tilde{V}## with ##V## is equal to the inner product of ##V## with itself. Similarly, the contraction of any one form ##\tilde{W}## with any vector ##V## is equal to the inner product between the vector ##V## and the vector ##W## which was mapped into ##\tilde{W}## by our metric tensor.

    This is a very natural thing. It's a very natural definition to define the "one form dual to the vector V" as: ##\tilde{V}(\quad)\equiv g(V,\quad)##.

    It is NOT natural to define "the one form dual to the vector V" as, for example, the one form to which the Ricci tensor maps V. In other words, I would not like to say: ##\tilde{V}=R(V,\quad)##. Why is this? Well the Ricci tensor doesn't define inner products on our vector space, the metric tensor does. So if we made the above definition, then the inner product of V with itself is no longer the contraction of V with its dual! (i.e. ##\tilde{V}(V)=R(V,V)\neq V\cdot V=g(V,V)##). This is very unnatural. And in fact, since the Ricci tensor might be 0 on our manifold, which would be the case if our manifold is Ricci flat (e.g. a vacuum solution to the Einstein field equations), then we might not get a unique mapping in that case, since if ##R(V,W)=0## for all ##V,W##, then it maps all vectors ##V## into the 0 one-form (an infinite to 1 mapping!).

    So what's special about the metric, that lets us call it the metric, rather than any rank (0,2) tensor? Well, it must satisfy some very important properties, namely:

    1. It must be symmetric: ##g(V,W)=g(W,V)\quad \forall V,W##
    2. It is non-degenerate, that means if ##V\neq0## then there exists a ##W## such that: ##g(V,W)\neq 0##
    3. It must be positive definite everywhere (necessary if we want to call our manifolds Riemannian): ##g(V,W)>0,\quad V\neq 0\quad and\quad W\neq 0##

    Property 3 may be relaxed, and a non-positive definite metric is called "pseudo-Riemannian", the most important example of such is a Lorentzian manifold (the manifolds of General relativity).

    Hopefully this helps!
     
  7. Apr 24, 2014 #6

    WWGD

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    To say part of it in other words, any non-degenerate form on a vector space gives rise to a canonical isomorphism ( i.e. an isomorphism independent of the choice of basis) between a finite-dimensional vector space and its dual ( note that a f.d vector space and its dual are _always_ isomorphic, by dimension reasons.)

    Like you said, a tensor is a multilinear map, i.e., a map that is linear in each argument separately, and the inputs of this tensor/map are both elements of the vector space often called the covariant part ) and elements of its dual (often called covariant ); you may also have purely covariant -- as in the case of a metric-- or purely contravariant tensors/maps.
     
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