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Covariant and Contravariant Tensors

  1. Aug 11, 2010 #1
    we have studied in Tensor's analysis that there are two kinds of tensors that usually used in transformation. one is Contravariant & covariant. what is the difference between them and and why they are same for Rectangular coordinates?
     
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  3. Aug 11, 2010 #2

    Fredrik

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    I'll just quote myself:
    See this post for comments about the definition of "tangent space".

    Tangent vectors and cotangent vectors are never the same, since they live in different spaces. A cotangent vector is a function that takes a tangent vector to a number. However, if the manifold is endowed with a metric tensor, there's natural way to associate a tangent vector with each cotangent vector and vice versa. For example, if [itex]v^\alpha[/itex] are the components of a tangent vector v, the corresponding cotangent vector has components [itex]v_\beta=g_{\beta\alpha}v^{\alpha}[/itex]. If the components of the metric tensor are =0 when [itex]\alpha\neq\beta[/itex] and =1 when [itex]\alpha=\beta[/itex], then the tangent vector v and the corresponding cotangent vector have the same components.
     
  4. Aug 12, 2010 #3

    quasar987

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    I believe that the words covariant and contravariant refer to the way the components of the vector change with respect to the coordinates system. Suppose [itex](x^1,\ldots,x^n)[/itex], [itex](\tilde{x}^1,\ldots,\tilde{x}^n)[/itex] are two intersecting coordinate systems on a manifold M. Suppose for each coordinate system around a point p of M, there is a rule that associates to the coordinates [itex](x^1,\ldots,x^n)[/itex] of p a set of n numbers (a vector in R^n, then) [itex](v^1,\ldots,v^n)[/itex].

    If the components [itex]v^i[/itex], [itex]\tilde{v}^i[/itex] of the vector corresponding to two coordinate systems [itex](x^1,\ldots,x^n)[/itex], [itex](\tilde{x}^1,\ldots,\tilde{x}^n)[/itex] around p are related like so:

    [tex]\tilde{v}^i=\sum_jv_j\frac{\partial x^j}{\partial \tilde{x}^i}[/tex]

    then the vector [itex]v=(v^1,\ldots,v^n)[/itex], which we consider the same as the vector [itex]\tilde{v}=(\tilde{v}^1,\ldots,\tilde{v}^n)[/itex], is called a covariant vector.

    If, on the other hand, the components are related like so:

    [tex]\tilde{v}^i=\sum_jv_j\frac{\partial \tilde{x}^j}{\partial x^i}[/tex]

    then the vector [itex]v=(v^1,\ldots,v^n)[/itex], which we consider the same as the vector [itex]\tilde{v}=(\tilde{v}^1,\ldots,\tilde{v}^n)[/itex], is called a contravariant vector.

    So why the names? Probably because the formula as you go from [itex]v[/itex] to [itex]\tilde{v}[/itex] in a covariant vector involves the rate at which [itex]x[/itex] changes with respect to [itex]\tilde{x}[/itex], while in a contravariant vector, it is the contrary: it involves the rate at which [itex]\tilde{x}[/itex] changes with respect to [itex]\tilde{x}[/itex].

    Examples:
    (1) Suppose we have a curve on an n-manifold M passing through the point p at the time t=0. Then for each coordinate system around p, there corresponds a curve in R^n, and we may differentiate this curve at t=0 to obtain a vector in R^n. If you carry out the computation, you will discover that this is an example of a contravariant vector.

    (2) For f a function of a manifold, given a coordinate system around p, you can compute the gradient of the coordinate representation of f at p. This is an example of a covector.

    (3) [If you know some classical mechanics] If M=Q is the manifold of physical states of a system and [itex]L:TQ\rightarrow \mathbb{R}[/itex] is a lagrangian function, then for each chart [itex](q^1,\ldots,q^n)[/itex] of Q (i.e. each set of generalized coordinates) the generalized momenta are defined by

    [tex]p^i:=\frac{\partial L(q^1,\ldots,q^n,v^1,\ldots,v^n)}{\partial v^i}[/tex]

    This too defines a covector.


    Now, you will often read things like "a contravariant vector is an element of the tangent space and a covariant vector is an element of the cotangent space". What is meant by that is the following. Given a point p on a manifold, we call tangent space at p the vector space [itex]T_pM[/itex] consisting of all linear maps [itex]D:C^{\infty}(M)\rightarrow\mathbb{R}[/itex] satisfying the Leibniz rule "at p" (i.e. D(fg)=D(f)g(p)+f(p)D(g)). It turns out that for a coordinate system [itex](x^1,\ldots,x^n)[/itex] around p, there is a natural basis for [itex]T_pM[/itex] which we denote (by no accident) [itex](\partial/\partial x^1|_p,\ldots, \partial/\partial x^n|_p)[/itex]. So a general element of [itex]T_pM[/itex] is of the form

    [tex]v=\sum_{i}v^i\left.\frac{\partial}{\partial x^i}\right|_p[/tex]

    and the vector [itex](v^1,\ldots,v^n)[/itex] is contravariant. Indeed, if [itex](\tilde{x}^1,\ldots,\tilde{x}^n)[/itex] is another coordinate system around p, then by the chain rule

    [tex]v=\sum_{i}v^i\left.\frac{\partial}{\partial x^i}\right|_p=\sum_{i}v^i\left(\sum_j \frac{\partial\tilde{x}^j}{\partial x^i}(p)\left.\frac{\partial}{\partial \tilde{x}^i}\right|_p\right)=\sum_j\left(\sum_iv^i\frac{\partial\tilde{x}^j}{\partial x^i}(p)\right)\frac{\partial}{\partial \tilde{x}^i}\right|_p\right)[/tex]

    So, given any contravariant vector [itex](v^1,\ldots,v^n)[/itex] at p associated to a coordinate system [itex](x^1,\ldots,x^n)[/itex], you can identify [itex](v^1,\ldots,v^n)[/itex] with the element

    [tex]\sum_{i}v^i\left.\frac{\partial}{\partial x^i}\right|_p[/tex]

    of [itex]T_pM[/itex]. Therefor, from the mathematical perspective of structures, the only contravariant vectors at p are the elements of [itex]T_pM[/itex], since any other can be naturally identified with one of these.


    Similarly, if you consider [itex]T^*_pM[/itex], the dual space of [itex]T_pM[/itex], and if you note (again, by no accident) [itex](dx^1_p,\ldots,dx^n_p)[/itex] the basis of [itex]T^*_pM[/itex] dual to the basis [itex](\partial/\partial x^1|_p,\ldots, \partial/\partial x^n|_p)[/itex] of [itex]T_pM[/itex], then any element of [itex]T^*_pM[/itex] is of the form

    [tex]v=\sum_{i}v_idx^i_p[/tex]

    and the vector [itex](v_1,\ldots,v_n)[/itex] is covariant. Indeed, if [itex](\tilde{x}^1,\ldots,\tilde{x}^n)[/itex] is another coordinate system around p, then by definition of the differential of a function

    [tex]v=\sum_{i}v_idx^i_p=\sum_{i}v_i\left(\sum_j\frac{\partial x^i}{\partial \tilde{x}^j}d\tilde{x}_p^j\right)=\sum_j\left(\sum_iv_i\frac{\partial x^i}{\partial \tilde{x}^j}\right)d\tilde{x}^j_p[/tex]

    So, given any covariant vector [itex](v_1,\ldots,v_n)[/itex] at p associated to a coordinate system [itex](x^1,\ldots,x^n)[/itex], you can identify [itex](v_1,\ldots,v_n)[/itex] with the element

    [tex]\sum_{i}v_idx^i_p[/tex]

    of [itex]T^*_pM[/itex]. Therefor, from the mathematical perspective of structures, the only covariant vectors at p are the elements of [itex]T^*_pM[/itex], since any other can be naturally identified with one of these.


    Tensors of rank (k l) (read "tensor of k contravariant indices and l covariant indices") are defined similarly as a rule associating an array of number [tex]T^{i_1,\ldots,i_k}_{j_1,\ldots,j_l}[/tex] to each chart [itex](x^1,\ldots,x^n)[/itex] around p (where each i and j takes any values between 1 and n) such that if [tex]\tilde{T}^{i_1,\ldots,i_k}_{j_1,\ldots,j_l}[/tex] is the array of numbers associated with another coordinate system [itex](\tilde{x}^1,\ldots,\tilde{x}^n)[/itex], then

    [tex]\tilde{T}^{i_1,\ldots,i_k}_{j_1,\ldots,j_l}=\sum_{i_1'}\ldots\sum_{j_l'}T^{i_1',\ldots,i_k'}_{j_1',\ldots,j_l'}\frac{\partial \tilde{x}^{i_1'}}{\partial x^{i_1}}\ldots\frac{\partial x^{j_l}}{\partial \tilde{x}^{j_l'}}[/tex]

    But each of those can be canonically identified with an element of

    [tex]\otimes_{r=1}^kT_pM\otimes \otimes_{s=1}^lT^*_pM[/tex]

    so we often say that a tensor of rank (k l) is just an element of the above.
     
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