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Proof that grad V is a (1,1) tensor

  1. Oct 16, 2004 #1
    i am trying to prove that
    [tex]\nabla V[/tex]
    which has components
    [tex](\nabla V)^\alpha{}_{\beta} = V^\alpha{}_{;\beta}[/tex]
    is a (1,1) tensor using the equation
    [tex]\frac{\partial V}{\partial x^\beta} = V^\alpha{}_{;\beta}e_{\alpha}[/tex]

    this is pretty much chinese to me, but after reading a few things on this site and trying to decipher my lecture notes this is the progress i made (not much at all). i dont understand any of this tensor and one-form business but i found out that a (1,1) tensor is a 3x3 matrix which sounds easy enough to deal with
    i said
    [tex]V^\alpha{}_{;\beta} = \frac{\partial V^\alpha}{\partial x^\beta}+V^\mu \Gamma^\alpha_{\mu\beta}[/tex]
    the RHS is the components of the vector
    [tex]\frac{\partial V}{\partial x^\beta}[/tex]
    as my understanding of it goes, the superscripts and subscripts can take values of 0,1,2,3 since they are greek characters so that would make me guess that [tex]\nabla V[/tex] is a 4x4 matrix and god knows what that is as a tensor.

    any one have any ideas on how to do this
  2. jcsd
  3. Oct 18, 2004 #2
    I was studying for electromagnetism and I wanted to proof the same, but in christian, that is, without tensor notation.

    We define gradient of f(x,y,z) as:

    [tex]\vec{\nabla}f = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k}[/tex]

    We will want to generalize this for an escalar [tex]x_i[/tex] which depends on n coordinates from [tex]x_0[/tex] to [tex]x_n[/tex].

    [tex]\vec{\nabla} x_i = \frac{\partial x_i}{\partial x_1} \vec{u_1} + \frac{\partial x_i}{\partial x_2} \vec{u_2} + ... [/tex]

    So, in the general case we get this definition of gradient:

    [tex]\vec{\nabla} x_i = \frac{\partial x_i}{\partial x_j} \vec{u_j}[/tex]

    So, you have that we got the gradient as the product of a contravariant transform (0,1) (as the partial derivative is) with a vector, as a covariant tensor (1,0), so the product must be a 2nd rank mixed tensor.

    Is this right? :confused: :confused:
  4. Oct 18, 2004 #3


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    Возьмите отличную книгу Дубровина Новикова и Фоменко Современная геометрия (есть на английском) и почитайте главы 1,3 про ковариантные и контравариантные тензоры.
  5. Oct 18, 2004 #4


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    Please do not post in languages other than English here.

    - Warren
  6. Oct 18, 2004 #5
    When I say that tensors are a bit difficult to understand, I mean really difficult to understand :biggrin:

    Google translates it as:

    Take the outstanding book of Dubrovin Novikov and Fomenko the contemporary geometry (there is in the English) and read a little chapters 1,3 about the covariant and contravariant tensors.
  7. Oct 18, 2004 #6
    I said Google when I wanted to say Babelfish™ Altavista (babel.altavista.com)
  8. Oct 18, 2004 #7
    good question, i wouldnt have a clue

    thnx for the replies
  9. Oct 19, 2004 #8
    There is a mistake, a vector is a (0,1) contravariant tensor, not covariant as I said!
  10. Oct 19, 2004 #9


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    It is so frustrating to see people still struggling with just manipulating tensor notation instead of trying to understand what tensors are. First learn what you are trying to describe, then afterwards learn how to describe them.

    advice : start by understanding what a tangent space is and what a cotangent space is.

    clue: the velocity vector of a curve is a tangent vector,

    and the gradient vector of a function is a cotangent vector.
  11. Oct 20, 2004 #10
    Then, lit a clandle in our darkness...
  12. Oct 20, 2004 #11


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    If you will search a little on these pages you will find dozens of pages of text and explanation i have written on this topic.

    i already gave you the first candle above. a tangent vector is a velocity vector to a curve, and a cotangent vector is (locally) the gradient of a function.

    a tensor is merely a formal product of several tangent vectors and several cotangent vectors. If there are r tangent vectors and s cotangent vectors then the tensor is called of type (r,s) or maybe (s,r), i don't know about this, as it is not an idea but a convention. I can look it up though. Thus a tensor is an element of the space:


    where T denotes tangent space and T* denotes cotangent space.

    the point is to learn what they are for, what kinds of things they measure.

    In geometry the two most important ones are: dot product, an element of T*tensT*, which eats two tangent vectors and produces a number, and curvature, which i don't understand as well, seems to be a field of elements of T*tensT*tensT*tensT, since it eats three vector fields and produces another one.

    again please refer to spivak, differential geometry volume 2.
  13. Oct 20, 2004 #12


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    see the 6th post in the thread "Math "Newb" Wants to know what a Tensor is", posted by chroot. he gives notation and meanings.

    you might ALSO be amused by reading some of the thread "what is a tensor"
    Last edited: Oct 20, 2004
  14. Oct 23, 2004 #13


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    I have decided to try taking my own advice above: "first learn what you are trying to describe then learn how to describe it."

    This seems to imply that it is more important and more useful to understand first some quantities that require tensors for their description, than to understand the mathematical definition of a tensor.

    From this perspective I have been quite wrong to pretend to some mathematical superiority in understanding what "tensors" are, simply from having a familiarity with the language of bundles and "tensors". I.e. any physicist who understands some physical quantity which is measured mathematically by a "tensor" knows what a tensor is better than I.

    That is to say, if we first understand something that should be measured by a tensor we can analyze it to see what the tensor measuring that quantity must be.

    Since apparently tensors arose in riemann's study of differential geometry, and in particular to measure curvature, perhaps that is the place to begin.

    Now we all know that Gauss first defined curvature for surfaces, and Riemann only generalized his definition to higher dimensional manifolds, so let us begin with curvature of surfaces and even curvature of curves.

    The curvature of a plane curve is constant if that curve is a circle, and the curvature of another curve is measured by comparing that curve with a circle. So first we want a map from our curve to a circle.

    So we parametrize our curve by arc length, i.e. we run along our curve at unit speed, and consider at each point the velocity vector, which then will be a unit tangent vector. now after translation to the origin, a unit tangent vector will have its head on the unit circle, hence this gives natural map from our curve to the unit circle.

    now if our curve is a line, this map will be constant, mapping every point of our line to the same point of the unit circle, and if our curve is itself a circle, this map will have image a moving point on the unit circle, moving indeed around the circle at a constant rate.

    If our curve is not a circle, then the rate at which the image vector moves around the circle will vary. so perhaps it is the rate at which our image velocity vector moves around the circle that should be the curvature. I.e. we should probably define the curvature as the derivative of the unit velocity vector, as a vector, with respect to arc length. Or rather the magnitude of that derivative. Indeed i believe this is the definition in my calculus book.

    Now Gauss gave a generalization of this measure to the case of a surface in three space, i.e. he found a natrual map from a surface to the unit sphere such that in some sense, the derivative of that map measured curvature of the surface. The natural choice is still called the Gauss map of the surface, and takes each point of an oriented surface, to the unit normal vector at that point. Then we translate the unit normal vector to the origin, and then it too has its head on the unit sphere.

    Now just as the angle cut out by a curve is measured by the arc length it cuts out on the unit circle, the "solid" angle of a piece of surface should be measured by the size of the patch of surface it cuts out on the unit sphere. So we want to measure whether the Gauss map is area - preserving as a map from our surface to the unit sphere or not, or more precisely, we want to measure how the Gauss map distorts area, in odrer to measure how the curvature of our surface differs from that of a sphere.

    As everyone knows, change in area under a map is measured by the change of variables formula, i.e. by the variation of the jacobian matrix of the transformation. Thus the determinant of the 2by2 matrix derivative of the Gauss map, is the curvature of a surface. [As I recall, this is a symmetric matrix, and its eigenvalues are the curvatures of the two curves passing through the given point, and having respectively the greatest and least curvatures of all curves through that point. These curves are also perpendicular, I believe.]

    Now the challenge is to unmderstand how this is related to the "curvature tensor" which is apparently a section of the bundle T*tensT*tensT*tensT, in fact of T*wedgeT*tensT*tensT, hence takes a pair of vectors to a map from T to T, and does so in an alternating way.

    Now a normal vector is somehow derived from to a pair of tangent vectors via cross product, which is also alternating. So how does the Gauss map yield such a tensor? I.e given a pair of tangent vectors, suppose we wedge them to get a normal vector, then map that to the unit sphere by translation to the origin and scaling. Hmmm....

    But we want to go from a pair of tangent vectors to a map from T to T, right? so from a pair of tangent vectors after we get a single normal vector, lets make a map out of it from T to T. I.e. given a tangent vector how do we get another tangent vector? by crossing it with the normal vector? sure! that would do it!

    So maybe that is how Gauss's method of measuring curvature is linked up with a "tensor"!

    My point is just that we should not quibble over what the literal definition of a tensor is, but should have faith in our understanding of what it is meant to do, to try to understand it. Thus we should not go back to the dictionary definition so much, i.e. we should argue that "if this is not what a tensor is, then this is what it should be".

    Of course ultimately with such a well traveled concept, we should hope to regain the usual definition at last. What is your reaction MiGui?"
    Last edited: Oct 23, 2004
  15. Oct 24, 2004 #14


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    it seems my guess as to how to render gauss' notion of curvature into tensor form is much too simple minded. it seems there should be some derivatives in there. but i still suggest as a useful exercise, to try to understand how riemann's definition for the curvature tensor captures gauss' intuitive measure of a surface curving, by comparing areas of small patches of surface with their images on the unit sphere.

    i.e. this seems more important than arguing about whether the curvature tensor lives in T*tensT*tensT*tensT, or in some other space, or whether the indices are up or down.
  16. Oct 24, 2004 #15


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    by theway, if f:M-->N is a map from one manifold to another, then the derivative of f, is a family of linear maps. i.e. at each point p of the domain manifold, f'(p), or df(p), is a linear map from the tangent space of M at p, to the tangent space of N at f(p).

    To define it, given a tangent vector v to M at p, just take any curve passing through p with velocity vector equal to v. Then compose this curve with f getting a curve through f(p) in N. Then the velocity vector of this composed curve at f(p), is the image df(p)(v).

    Thus at each point p of M df gives a linear map from T(M)(p) to T(N)(f(p)). Hence df is a family or "field" of linear maps. If the atrget manifold N is the real numbers R, i.e. if f is a function, then f' or df is a field of linear maps from the tangent spaces T(X)(p), to the real numbers R = T(R)(f(p)). I.e. df is a field of covectors on M. This is the "gradient of f". Thus for a function f, the gradient df is a covector field, i.e. a section of the dual tangent bundle T*(M).

    Whether you call this a (1,1) or (0,1) tensor or whatever, is a matter of terminology, less important than knowing what it means. Presumably since it is a section of T*, it is either a (0,1) or (1,0) tensor depending on which index counts the number of copies of T*, and which counts the number of copies of T.
  17. Oct 24, 2004 #16


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    I realize now that I am taking advantage, in gauss definition, of the curvature of the ambient space. riemanns definition used only a metric on the manifold and not an embedding. so perhaps one could translate back to the unit sphere in the tangent space at the given point to imitate gauss definition. or perhaps it is more subtle.
  18. Oct 27, 2004 #17
    proof by definition

    vladimir, are you required to prove this by the route you are asking? because a proof follows simply by definition.

    recall the definition of the gradient of f:

    [tex] df(\vec{v}) = <\vec{\nabla}f, \vec{v}> [/tex]

    so this says that given some vector [tex]\vec{v}[/tex], there exists a vector [tex]\vec{\nabla}f[/tex] such that the inner product of the two is equal to the covector [tex]df[/tex] acting on [tex]\vec{v}[/tex]

    so expanding this a bit:

    [tex] <\vec{\nabla}f, \vec{v}> = \nabla f^i v^j <\partial _i, \partial _j> = \nabla f^i v^j g_{ij} = \nabla f_j v^j = \nabla f^i v_i [/tex]

    and for the left-hand side:

    [tex] df(\vec{v}) = \vec{v}(f) = v^j \frac {\partial f} {\partial x^j} [/tex]

    this all implies that:

    [tex] \nabla f^i = \frac {\partial f} {\partial x^j} g^{ij} = dx^i(\nabla f) [/tex]

    (Notice that the components of [tex] \nabla f [/tex] are NOT the same as the components of [tex] df [/tex], but rather depend on the metric. In Euclidean space, the metric is just the identity matrix, so many folks mistakenly just refer to [tex] df [/tex] and [tex] \nabla f [/tex] as if they were the same.)

    You can consider the grad to be a (1,1) tensor in the sense that for any fixed v, the tensor takes in a contravariant del f and covariant df such that the definition is true. However, as others have pointed out, since df and f are related by the metric you could arguably call this a (0,1) tensor since it then takes in some vector v. or it could be a (1,0) tensor for similar reasoning...what you call it doesn't really matter as long as you know what it does.

    in any case, i don't think that you need to be using covariant differentiation and all of that stuff in order to show that this is a (1,1,) tensor...it all follows by definition with the caveat that you can show it is invariant under coordinate change (which i have not done here).
  19. Oct 31, 2004 #18


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    I think I have figured out the connection between Gauss' intuitve definition of curvature and riemann's tensor version.

    Gauss measured the curvature of a surface in 3 space at a point by comparing the area of a small nbhd of the point with the area of a small nbhd of a point on a sphere by the "Gauss map". I.e. map a point p on the surface S, to the point on the unit sphere obtained from translating the unit normal vector to S at p, to the origin.

    Then one has a map and one can measure the change in area via the change of coordinates transformation for integrals, i.e. by the jacobian matrix of the Gauss map at the point. We know that the determinant of this matrix is the Guass curvature at the point p. Hence more fundamental is the matrix itself.

    Now for a surface in 3 space, every tangent plane to the surface can be trams;lated in a unique way to the origin. Hence we may regard the Jacobian matrix of the Gauss map at p, as an endomorphism of the tangent space of S at p. Hence we can think of curvature of S at p, as an assignment of a linear map from the tangent space of S at p to itself. Such a thing is a candidate for a tensor "of type (1,1)" since it associates to every point a linear map from the tangent space to itself, or equivalently an element of the tensor product of the tangent space with its dual.

    Now Riemann generalized this to higher dimensional amnifolds apparently as follows:

    to each 2-plane in the tangent space at p, there is ( on a manifold with a metric) a corresponding local piece of surface swept out by geodesics corresponding to directions in the 2 plane of tangent vectors. To this surface we can apply Guass definition, or Riemann version of it, of curvature of that surface.

    This gives an assignment of an endomorphism of the tangent space to each 2 plane in that tangent space, i.e. an element of the tensor product of two copies ot the tangent space, with one more copy of the tangent space and one copy of the dual, i.e. an element of TtensTtensTtensT*.

    This is the original example of a tensor by the man who invented tensors, Riemann.

    Einstein's main idea in cosmology seems to have been (from mjy reading of the funny papers and little golden books of physics) to study gravitation in terms of the curvature it imposes on space time. hence Einstein's need for tensor calculus was to understand and represent curvature.

    I rest my case, fully aware that no one may be interested in this intuitive discussion of a formal concept. But I humbly submit it for the consideration of those who would first rather understand what tensors mean, and only later compute with them.

    Last edited: Oct 31, 2004
  20. Oct 31, 2004 #19


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    further confirmation of this point of view: as everyone knows: if at a point p of a surface, we associate to each geodesci through p the curvature of that curve (the sectional curvature), then there exist two orthogonal directions in which the curvatrures are repsectively maximal and minimal, and the Gauss curvature of the surface is the product of those two. (e.g. on a cylinder the circle with maximal curvature has positive curvature but the oertogonal line has zero curvature so the surface has Gauss curvature zero there.)

    This suggests clearly that the matrix we discussed above is diagonalized by choosing these orthogonal directions, and has these two extremal sectional curvatures as eigenvalues. Hence the Gauss' curvature is again the determinant of this diagonal matrix.
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