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Covariant vs contravariant

  1. Aug 30, 2009 #1
    I'm having some trouble breaking into tensors. What is specifically bewildering me is contravariant vs. covariant indices. Could someone please explain this to me (or link me)? I barely understand what each one means in its own right, let alone the differences between the two.
     
  2. jcsd
  3. Aug 30, 2009 #2
    http://www.mathpages.com/rr/s5-02/5-02.htm" [Broken] is a great description from a classical level.

    The idea of contravariant vectors is that they transform with the Jacobian of a map and the classical example of this is the tangent vector of a curve. The classical example of a covariant vector is the gradient of a scalar field. This will transform with the inverse of the Jacobian. The geometric interpretation of the gradient at a point is as a normal vector to the level set that the point belongs to.

    Given that a vector maps into the dual space, we can interpret a vector as being a normal to the kernel of its image under that map. Then projecting a second vector onto that vector is just determining which coset of that kernel the second vector lies in. At this point, we are concerned not with the vector so much as with the factor space that it induces when mapped into the dual space and this is determined by what (n-1)-dimensional subspace it is normal to. So when we transform the vector space, the subspace comes along for the ride and we need the inverse Jacobian to transform the normal vector if we want it to still be normal to the image of the same subspace after the transformation. So it's covariant.

    Essentially, covariant vectors live in the dual space and contravariant vectors live in the regular vector space. When we transform the vector space, we implicitly transform the dual space with the inverse Jacobian of the transform.

    I'm just figuring all of this stuff out myself though so I'm sure that I said something wrong. At least my mistake might induce someone that actually knows what they're talking about to correct it and provide a better explanation ;)
     
    Last edited by a moderator: May 4, 2017
  4. Aug 31, 2009 #3

    dx

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    Given a differentiable manifold M, there are two natural structures that we are led to consider, functions R → M and functions M → R, called 'parametrized curves' and 'functions' respectively. If you think about it, you will see that these objects have natural derivatives, i.e. local linear behaviors near a point. The space of derivatives of a parametrize curve is called the tangent space, and the space of derivatives of functions is called the cotangent space, whose elements are called vectors and covectors respectively.

    To understand which of these things behave covariantly and which of them behave contravariantly, we must first understand what these terms mean. A smooth map f: M → N between two smooth manifolds induces two operations [tex] f^* [/tex] and [tex] f_{*} [/tex], called the push-forward and the pull-back. What do these operations act on? The push-forward acts on the space of covariant structures on M and the pull-back acts on the space of contravariant structures on N. We have just given names to these operations, and the spaces on which they act. But what are they supposed to represent? What they represent is the idea that a map between two manifolds is able to take structures on them back and forth. For example, if you have a covector field ω on N, then f is able to induce a naturally determined covector field (by f and ω) on M, which corresponds in our language to [tex] f^{*} \omega [/tex], i.e. the pull-back of ω by f. For this reason covector fields are contravariant (remember that the pull-back acts on the space of contravariant structures). Let us consider an example of this. If you have a real function g on N, then the exterior derivative of g on N is a covector field called dg. What is the pull-back of dg by f? This is the answer:

    [tex] f^{*}dg = d(g \circ f) [/tex]

    Please don't be confused by the unfamiliar notation or terminology. These are simple ideas if you think about them geometrically. If you've been visualizing all the things I just said, you will see easily that covectors are cotravariant structures and vectors are covariant structures.

    A (k, l)-tensor at a point p on the manifold is a multilinear map

    [tex] T : V \times V \times ... \times V \times V^{*} \times V^{*} \times ... \times V^{*} \rightarrow \mathbb{R}[/tex]

    (k copies of V and l copies of V*), where V is TpM and V* is T*pM. Any coordinate system {xα} induces a basis for the space of such tensors (which form a vector space):

    {[tex] \partial_{\alpha_1} \otimes \partial_{\alpha_2} \otimes ... \otimes \partial_{\alpha_k} \otimes dx^{\beta_1} \otimes dx^{\beta_2} \otimes ... \otimes dx^{\beta_l} [/tex]}.

    In this basis, the tensor T can be written as

    [tex] T = T^{\alpha_1 \alpha_2 ... \alpha_k}_{\beta_1 \beta_2 ... \beta_l} \partial_{\alpha_1} \otimes \partial_{\alpha_2} \otimes ... \otimes \partial_{\alpha_k} \otimes dx^{\beta_1} \otimes dx^{\beta_2} \otimes ... \otimes dx^{\beta_l} [/tex]

    where [tex] T^{\alpha_1 \alpha_2 ... \alpha_k}_{\beta_1 \beta_2 ... \beta_l} [/tex] are real numbers. The indices are classified as covariant and contravariant according to whether they are connected to the objects dxβ or the objects ∂α.

    This is all quite abstract, and I don't expect people who are seeing this for the first time to get it fully, but I have tried to give an honest answer to your question. If you want me to explain any of the points in more detail, please ask.
     
    Last edited: Aug 31, 2009
  5. Aug 31, 2009 #4

    Fredrik

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    I would prefer to start with a simpler version of what dx said. Forget about manifolds for a moment and let V be an arbitrary finite-dimensional vector space over the real numbers. Now define V* as the set of all linear functions from V into [itex]\mathbb R[/itex]. Then define the sum of two members of V*, and the product of a member of [itex]\mathbb R[/itex] and a member of V* by

    [tex](f+g)(v)=f(v)+g(v)[/tex]

    [tex](kf)(v)=k(f(v))[/tex]

    These definitions give V* the structure of a vector space. It's called the dual space of V. Since V* is a vector space, the members of V* are vectors. However, when V is the tangent space of a manifold, the members of V are called "tangent vectors" and the members of V* are called "cotangent vectors". This is often shortened to the misleading "vectors" and "covectors", or worse (much worse actually) "covariant vectors" and "contravariant vectors".

    From now on, I'll put an arrow on top of the members of V, and a tilde on top of members of V*, so that it will be perfectly clear what each symbol represents. I won't write out any summation symbols, because it's easy to remember that we must sum over every index that appears twice (once upstairs and once downstairs) from 0 to 3. (Assuming that we're talking about relativity. In a more general setting, the sum would be from 1 to dim V).

    Now consider two bases of V, [tex]\{e_\alpha\}[/tex] and [tex]\{\vec e_\alpha{}'\}[/tex].

    [tex]v=v^\alpha \vec e_\alpha=v'^\alpha\vec e_\alpha{}'[/tex]

    A member of a basis can of course be expressed as a linear combination of members of another basis:

    [tex]\vec e_\alpha{}'=M^\beta{}_\alpha \vec e_\beta[/tex]

    So we have

    [tex]v^\alpha \vec e_\alpha=v'^\alpha\vec e_\alpha{}'=v'^\alpha M^\beta{}_\alpha \vec e_\beta=v'^\beta M^\alpha{}_\beta \vec e_\alpha[/tex]

    All I did in the last step was to change the labels of the summation indices, which is obviously OK. Since a basis by definition is a linearly independent set, the above implies that

    [tex]v^\alpha=M^\alpha{}_\beta v'^\beta[/tex]

    This is clearly a matrix equation in component form. I'll write the matrix equation as

    [tex][v]=M[v]'[/tex]

    Here [v] is the 4x1 matrix that consists of the components of v in the unprimed basis, [v]' is the 4x1 matrix that consists of the components of v in the primed basis and M is the 4x4 matrix that has [itex]M^\alpha{}_\beta[/itex] on row alpha, column beta. We can solve for [v]':

    [tex][v]'=M^{-1}[v]=\Lambda [v][/tex]

    where [itex]\Lambda=M^{-1}[/itex]. We can of course express this in component form too:

    [tex]v'^\alpha=\Lambda^\alpha{}_\beta v^\beta[/tex]

    What this equation tells you is how the components of a vector in a particular basis changes when you switch to another basis.

    There's a natural way to associate a basis of V* with each basis of V. We define [tex]\tilde e^\alpha[/tex] by

    [tex]\tilde e^\alpha(\vec e_\beta)=\delta^\alpha_\beta[/tex]

    [tex]\{\tilde e^\alpha\}[/tex] is said to be the dual basis of [tex]\{\vec e_\alpha\}[/tex]. We can of course write an arbitrary member of V* as

    [tex]\omega=\omega_\alpha \tilde e^\alpha = \omega'_\alpha \tilde e^\alpha{}'[/tex]

    We would like to find the relationship between the components of [itex]\omega[/itex] in the primed basis and its components in the unprimed basis, just as we did for v. This isn't difficult to do.

    [tex]\tilde\omega(\vec e_\alpha)
    = \omega_\beta\tilde e^\beta(\vec e_\alpha)
    = \omega_\beta\delta^\beta_\alpha
    = \omega_\alpha[/tex]

    [tex]\omega_\alpha{}'
    =\tilde\omega(\vec e_\alpha{}')
    =\tilde\omega((\Lambda^{-1})^\beta{}_\alpha\vec e_\beta)
    =(\Lambda^{-1})^\beta{}_\alpha\omega(\vec e_\beta)
    =(\Lambda^{-1})^\beta{}_\alpha\omega_\beta[/tex]

    If we're talking about special relativity, we can write [itex]\Lambda_\alpha{}^\beta[/itex] instead of [itex](\Lambda^{-1})^\beta{}_\alpha[/itex]. This follows from [itex]\Lambda^T\eta\Lambda=\eta[/itex] and the convention to use the metric to raise and lower indices. (Note that the formula, which can be taken as the definition of a Lorentz transformation, implies that [itex]\Lambda^{-1}=\eta\Lambda^T\eta[/itex]).

    So what does any of this have to do with coordinate systems? There's a natural way to associate a basis of the tangent space at a point p of a manifold with each coordinate system, so a change of coordinates can also be thought of as a change of basis for all of the tangent spaces at the points where both coordinate systems are defined. See my posts in this thread.
     
    Last edited by a moderator: Dec 31, 2011
  6. Aug 31, 2009 #5
    This is really a totally simple idea.

    A vector space always has a dual space, the space of linear maps of it into the base field.
    A linear map of vector spaces,V and W, maps vectors in V into vectors in W but does not map duals in V to into duals in W. For duals the reverse happens. A dual on W get mapped into a dual on V.

    This is the difference between covariant and contravariant. That is it.

    For the special case where the linear map is a change of coordinates the same thing applies - no difference.

    At a point of the tangents space of a manifold one has tangent vectors and dual tangent vectors just as in any other vector space. The Jacobians of coordinate transformations are just change of basis linear maps.
     
  7. Aug 31, 2009 #6

    dx

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    When I hear the word 'tensor', my mind immediately thinks of manifolds. Wofsy and Fredrik just explained the ideas of covariance and contravariance in the context of vector spaces rather than manifolds. A specific context is actually not necessary and the ideas can be understood with respect to general mathematical structures in category theory using the idea of a functor.
     
  8. Dec 1, 2010 #7
    Just curious:
    I am a math student with unfortunately primitive understanding of physics.

    I am curious to know a couple of things, please:

    1)Is there a physical interpretation for the dual of a tangent space.?

    Also, if we have a linear map L:V-->W, both V,W fin.dim V.Spaces

    L the matrix with resp. to some basis. Is there some relation between L^t

    (matrix transpose of L ) and V*.

    2) Why are we interested in functions defined on, say, VxVXV*xV*;

    what kinds of objects are these, and what do they tell us.?. I understand the

    Riemannian metric is defined on VxV, but I am having trouble conceiving of

    other functions of interest in spaces of the sort VxVxV*xV* , etc.

    Are these k-linear maps linear approximations to functions, or do we

    have other types of functions.

    Thanks.
     
  9. Dec 1, 2010 #8

    haushofer

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    1) The "physical interpretation" would be of the objects living in that space; it turns out that gauge potentials, essential for describing interactions in particle physics, live in these dual spaces.

    2) For instance, in particle physics one is concerned about tensors transforming under the Lorentz group. These tensors describe a particle with a certain spin. By looking at tensor products, one can construct objects with higher spin. Also, in string theory one encounters naturally so-called "n-forms" in the spectrum, objects living in the tensor product of n dual tangent spaces.
     
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