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Connections in differential geometry

  1. Nov 10, 2007 #1
    Hi, I have been reading some stuff about differential geometry (with the ultimate goal of trying to understand loop quantum gravity...) and something I have been having trouble with for awhile is the idea of a connection. Basically, I have yet to find a definition that actually enables me to understand what a "connection" is.

    In general, I think I understand that the motivation of a connection is that when spaces are or can be curved, then the notion (normally taken utterly for granted) that a vector at one coordinate can be compared to a vector at another coordinate becomes false. In order to compare two vectors under these circumstances, we must effectively "transport" the first vector to the location of the second vector. However, since the space may be curved, then the value of the first vector on arrival at the position of the second vector will depend on the path the first vector followed when getting there! Therefore, we must define a notion of "parallel" transport, which will allow us to link up paths between all coordinates in such a way that if we have arbitrarily-positioned vectors A, B, and C, then transporting C to A by our parallel-transport method will have the same effect as moving C first to B and then to A by the same parallel-transport method. Connections provide such a method of "parallel" transport.

    What I don't get though is: Exactly what is the connection, itself? What is the definition of a connection as a mathematical object? For example if someone makes a statement like "the holonomy of the connection can be identified with a Lie group, the holonomy group" then exactly what is it that this holonomy group is being identified with?

    When we talk about a particular "connection". Do we mean that the connection is a system of parallel-transport paths, which can be used to pick a parallel-transport path given any two points? Or do we mean simply the connection is the the parallel-transport path between the two points itself?

    Along similar lines: I recently read a book called "Geometrical Methods of Mathematical Physics" which cleared up a lot of terminology issues for me, but left me still confused about connections. For one thing, they actually did not address connections at all, but only explained something called "affine connections" in an appendix. (Is a connection and an affine connection the same thing?) I ultimately was unable to get anything out of this appendix.

    Something they did do however in the main part of the book was introduce the idea of a "congruence". They did this because they wanted to introduce the concept of "lie dragging". As I understood it, the point of this was that they needed some unambiguous notion of "direction" on a manifold, in order that that notion of direction could be used to introduce coordinates; the congruence would provide an unambiguous path between any two points on the manifold, and "lie dragging" along the congruence would map the manifold to itself in such a way that each point would be "moved" along its path a certain amount. After defining a sort of basis set of congruences for some manifold, you can then introduce coordinates for that manifold because each of the different ways of "lie dragging" along the basis set of congruences your a manifold would produce one infinitesimal generator of that manifold's lie group. (Does that sound right?) This by itself all made sense, but in some ways it confused me even more, because lie dragging sounded to me exactly like parallel transport! When they first started explaining this, I at first thought that the congruence was going to either be related to, or even the same as, a connection. But apparently they are different.

    So: What is the difference between a congruence and a connection? What is the difference between lie-dragging along a congruence, and parallel transport along a connection?

    One last thing I do not understand is why connections are identified with fiber bundles. It seems that this idea of parallel transport makes sense on any manifold with a vector space attached. Or is the idea that if a vector space is "attached" to the manifold in the first place, then this by definition means that we have a fiber bundle?

    Any help would be appreciated, thanks.
     
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  3. Nov 10, 2007 #2
    I would start by examining an introductory book on Riemannian Geometry. One possibility is Riemannian Geometry written by Gallot, Hulin, and Lafontaine.

    I recommend doing some calculations and understanding some examples. The field of geometry is motivated by visual "intuitions", but not everything is comprehensible in terms of slick geometric visuals.

    However, I'm looking forward to some of the other responses.

    I believe the Levi-Civita connection has the following interpretation. I didn't think through this carefully; so, please check it.

    Let (M,g) be a Riemannian Manifold and assume it is isometrically embedded in R^N for some N. Now, let X and Y be two vector fields on M. Extend X and Y to vector fields on R^N and let DX denote its derivative in R^N. Now, (DX)Y is a vector field in R^N in a neigborhood of M. Let proj be the projection from a neigborhood of M in R^N to M. Then,

    \del_X Y= (D proj)[(DX)(Y)],

    where \del_X Y is the Levi_Civita connection.
     
    Last edited: Nov 11, 2007
  4. Nov 12, 2007 #3

    Ben Niehoff

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    Personally, I think the way that the "connection" is usually explained can be rather confusing. Misner, Thorne and Wheeler describe the connection as a sort of "correction factor" which provides a covariant derivative on curved manifolds when added to the flat-space partial derivatives. I see the same sort of explanations on Wikipedia and other online sources, so I'm guessing this is pretty standard.

    However, I think it simplifies the picture very much to simply return to the definition of a vector. The vector [itex]{\mathbf u}[/itex] expressed in terms of the basis vectors [itex]\{ {\mathbf e}_i \}[/itex] is:

    [tex]{\mathbf u} = u^i {\mathbf e}_i[/tex]

    where [itex]u^i[/itex] are the components of [itex]{\mathbf u}[/itex] in this basis. Now, we simply take the partial derivative in the direction of the j-th coordinate axis:

    [tex]{\partial \over \partial x^j} {\mathbf u} = {\partial \over \partial x^j} (u^i {\mathbf e}_i) = \left({\partial \over \partial x^j} u^i \right) {\mathbf e}_i + u^i \left({\partial \over \partial x^j} {\mathbf e}_i \right)[/tex]

    In ordinary Euclidean space, the basis vectors are constant, and so [itex]{\partial \over \partial x^j} {\mathbf e}_i[/itex] vanishes. But in curved space, the basis vectors change from place to place, so the second term in the above equation is in general nonzero.

    We note that the second term must be a vector quantity (because the first term is obviously a vector, and we're adding them together), so we can find its components in our current basis by taking the dot product with the basis vectors:

    [tex]\left({\partial \over \partial x^j} {\mathbf e}_i \right)_k = \left({\partial \over \partial x^j} {\mathbf e}_i \right) \cdot {\mathbf e}_k = \Gamma^k_{ji}[/tex]

    And so these are just the Christoffel symbols. To actually compute their values, you can do some algebra with the derivatives of the metric tensor.

    The reason that it's called a "connection" is because it tells us how to take derivatives in the manifold in such a way that agrees with the curvature of the manifold -- i.e., is "covariant". Since derivatives are inherently non-local objects (a derivative is formed by comparing values at two different points and dividing by the distance between the points), the covariant derivative specifies how the tangent spaces at each point of the manifold should be connected to one another, in such a way that is consistent with the curvature. The connection defines what it means to compare vectors at different points (because those vectors lie in different tangent spaces).
     
  5. Nov 15, 2007 #4
    Thanks, that helps a lot!

    To be clear though, am I understanding correctly that the Christoffel symbols and the Levi-Civita connection comprise a notion of connection which applies only for metric spaces? Or can they be generalized to describe a connection on any space?

    Also, will any given space/manifold/whatever tend to have one and exactly one possible connection, or is it possible for there to be some sense in which one space might have several different possible connections and which one you choose to work with actually makes a difference in some way (that is, a difference beyond just ease of computation and such)?
     
  6. Nov 15, 2007 #5

    Ben Niehoff

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    Hmm...according to my intuition, I don't think a connection could mean anything sensible in a non-metric space. In fact, I'm pretty sure a connection would induce a metric, just as vice versa. In a holonomic basis, the equations

    [tex]\Gamma^i_{kl} = {1 \over 2}g^{im}\left({\partial g_{mk} \over \partial x^l} + {\partial g_{ml} \over \partial x^k} - {\partial g_{kl} \over \partial x^m}\right)[/tex]

    would provide a set of partial differential equations for the metric tensor coefficients. However, I'm not sure if a connection might possibly be defined in a more abstract way that allows connections on non-metric spaces; I don't know enough about differential geometry to say.


    Since the connection defines the geodesics, any two connections on a manifold must agree on what the geodesics are. But you can define any coordinate system you like, so long as you adjust your metric tensor accordingly; as a result, these geodesics can be parametrized in any imaginable way. Drawing comparisons between different coordinate systems can be challenging.
     
  7. Nov 15, 2007 #6
    I see, that makes a lot of sense.

    Hm, okay. So just to be clear, it sounds like I can assume that, given a specific coordinate system for a manifold there I can assume there's only really one possible connection, but given different coordinate systems for one manifold I should not act as if the connections for each of the different coordinate systems are the same thing?

    Thanks again!
     
  8. Nov 15, 2007 #7
    Of course you can have connections without having a metric! You simply specify what the values of the Chrsitoffel symbols are. This cannot in general be inverted to give a metric, as a Levi-Civita connection of a metric has nice properties that a general connection need not have.
     
  9. Nov 15, 2007 #8
    You can indeed have a connection without needing to specify a metric beforehand. To wit, if [itex]\{\partial_a\}\equiv\{e_a\}[/itex] is some coordinate basis for vectors at a point [itex]p\in M[/itex], we can define an connection as a map [itex]\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\to\mathfrak{X}(M)[/itex] which satisfies

    [tex]1)\quad \nabla_X(Y+Z) = \nabla_XY + \nabla_XZ,[/tex]
    [tex]2)\quad \nabla_{X+Y}Z = \nabla_XZ + \nabla_YZ,[/tex]
    [tex]3)\quad \nabla_{fX}Y = f\nabla_XY,[/tex]
    [tex]4)\quad \nabla_X(fY) = X[f]Y + f\nabla_XY,[/tex]

    for where [itex]f[/itex] is some sufficiently smooth function from [itex]M\to\mathbb{R}[/itex] (at least [itex]C^1[/itex] on whatever domain you're interested in) and [itex]X,Y,Z\in\mathfrak{X}(M)[/itex]. So a connection is just some particular type of map which satisfies the above conditions. Using the above basis, the connection coefficients are then defined as

    [tex]\nabla_{e_i}e_j = \Gamma^k_{\phantom{k}ij} e_k[/tex]

    If you have a metric on your manifold as well, it is customary to demand that the metric be covariantly constant with respect to the connection, i.e., that the inner product [itex]g(X,Y)[/itex] between two vectors be constant under parallel transport. It's not too difficult to show that this requirement results in the so-called metric connection and fixes the form of the connection coefficients to be

    [tex]\Gamma^i_{\phantom{i}jk} = \frac{1}{2}g^{il}(\partial_jg_{kl} + \partial_kg_{lj} - \partial_lg_{jk}).[/tex]
     
    Last edited: Nov 15, 2007
  10. Nov 15, 2007 #9

    Hurkyl

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    A connection is a gadget that does the following:

    For any path on the manifold, the connection provides you with an isomorphism of the tangent spaces at the endpoints.

    We further require it should be a nice sort of gadget; e.g. it should be continuous and differentiable in the appropriate sense, play nicely with the concatenation of two paths, et cetera.


    Another picture is that it should lift a smooth path on the manifold to a smooth path on the tangent bundle. (Which projects down to the original path)


    This sounds right. The concept easily generalizes to any sort of fiber bundle -- it doesn't even have to be a vector bundle; all the fiber needs is a differential structure! One important case is when the fiber is a Lie group.

    And it is usually fruitful to consider not the product MxV of a manifold with a vector space, but the trivial vector bundle MxV --> M. (Or, more generally, a nontrivial vector bundle)


    From the description you gave, it sounds like a congruence is just a smooth group action; it doesn't tell you anything about tangent spaces, it just provides a (parametrized) mapping from the manifold to itself.
     
  11. Nov 16, 2007 #10
    Thanks to all.

    Shoehorn, I think the definition you give is actually exactly what I was fishing for when I asked for a definition of a connection "as a mathematical object". So I want to make sure I understand it: when you say [tex]\mathfrak{X}(M)[/tex]. Am I reading this right that M is the base manifold of a fiber bundle and [tex]\mathfrak{X}(M)[/tex] is the fiber bundle itself?

    Also, when you say:

    [tex]\nabla_{e_i}e_j = \Gamma^k_{\phantom{k}ij} e_k[/tex]

    What is the meaning of the uppercase gamma? I eventually found a specification for a vector bundle connection on mathworld and they use the same symbol several times as well.
     
  12. Nov 16, 2007 #11

    quasar987

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    [tex]\mathfrak{X}(M)[/tex] denotes the set of vector fields on M.

    the uppercase gamma are the christoffel symbols.
     
  13. Nov 16, 2007 #12
    Ah, okay. So when we say [itex]X,Y,Z\in\mathfrak{X}(M)[/itex], we're saying X, Y, Z are attached vector spaces in the bundle? Hm, in this case then what does it mean to "add" two of these spaces? Like when we say "X+Y" what operation do we mean by X?
     
  14. Nov 16, 2007 #13

    quasar987

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    X Y Z are vector fields. Sections of the tangeant bundle.
     
  15. Nov 16, 2007 #14

    WWGD

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    I apologize, I don't know how to use Tex yet. Please bear with me
    while I learn.

    I think actually that

    [itex]X,Y,Z\in\mathfrak{X}(M)[/ are V.Fields that are at least C^1.
    Vector fields can be seen as derivations on C^k functions:

    (Xf)(p)=X_p(f) , acts like partial differentiation in the dir. of X_p

    and X(fg)(p)=X_p(f)g(p)+f(p)X_p(g).

    And we can add vectors only if they are based at the same point, i.e, if
    they are in the same V.Space (tho in some special cases like in IR^n,
    you can add vectors in different tangent spaces as if they were in the
    same, because of the natural isomorphism between T_p(IR^n) and
    T_q(IR^n). I think the fact that tangent spaces at different points in
    a general mfld M are not naturally isomorphic (they are isomorphic, as
    they are both of the same dimension--the dimension of the mfld.--
    and they are V.Spaces over the same field) helps motivate the need
    for a connection, which I believe is not necessary when the tangent
    spaces are naturally isomorphic. I wonder if having zero curvature
    is equivalent to this last
     
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