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Tangent space, derivation defifinition

  1. Nov 13, 2006 #1
    I'm reading "tensor analysis on manifolds" by Bishop and Goldberg. I have taking a course in differential geometry i R^3. The course was held on Do Carmos book. Do carmo deffined the tangent at a point on a surface as all tangents to all curves on the surface going through that point (or something like that). This diffenition is pretty clear and easy to imagine.

    In the other book it is defined as all darivations on all ****tions from the manifold at the point to R. Then they define the tangent of a curve in the manifold to be and operator. And then they define the tangent to the ith coordinate curve of gamma, and proof a theorem about that these coordinate tangents at a point m on the manifold is a basis for all tangents.

    The proof are not that hard, but my problem is that i can't see that the two definitions are the same when i choose one of the "manifolds" we worked on in do carmo, fx. the unit sphere in R^3. I now the one is a operator and the other are vectors, but there most be a clear connection?

    I've seen an example where they use R^2 as there manifolds, and then state that the directional derivative corospones to the operators. And then it is clear that given and directional derivative in the direction v lets call it Dv, then i have the operator and the if i define a function f(Dv)=v then i get all tangent vectors to all points in R^2.

    But somehow i can't find the connection when my surface gets more complicated, like the unit sphere. I can see that the operator is connected to the tangent of curves through m, but it is not so clear to me, in my mind there most be a function (possible a bijection) on the "tangent operators" such that i get all tanget vectors, when my manifold is so simple like the unit sphere that a "normal" tangent vector/plane makes sence. My point is that if it makes sence to define the tangents like operaors, I most have just as much information as if I had defined it the "usual" way (of cause the new defintion makes more sence when i have obscure manifolds where tangentplanes are not clear, but in R^3 there most be a clear connection in my head).

    I hope someone can understand what my problem is, anbd have the commitment to read the whole post (it got a bit long, sorry). And by the way, i'm sorry for the bad englsih.
  2. jcsd
  3. Nov 13, 2006 #2
    This is a sticking point for many students of diffl. geometry. Keep in mind that, if p is a point on the manifold living in a coordinate patch U with coordinates (x1,...,xn), then any differential function f defined on U can be seen as a function of the variables (x1,...,xn), so that the directional derivatives of the manifold at p are given as the linear span of {d/dx1,..., d/dxn} (yes, the d's should be partials but I don't feel like TeXing right now)

    Now, suppose the manifold M is a surface in R^3, whose coordinates we will call (x,y,z). Locally, M can be thought of as a function of two of those coordinates, i.e. z=f(x,y) for some function (of maximal rank) f. If p is a point in such a neighborhood, then the local coordinate system of p can be given as (x,y). Furthermore, the directional derivatives on that coordinate patch will be the linear span of {d/dx,d/dy}.

    Now, how is the linear span of {d/dx,d/dy} related to the space of all lines tangent to M at p? They are related by the linear transformationn given by: d/dx --> (1,0,df/dx) and d/dy-->(0,1,df/dy). (again those should all be partials, not d's)

    For points on M where M is given by, say, x=g(y,z), we do something similar.
  4. Nov 13, 2006 #3


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    to elaborate on doodle bobs comments, a tangent vector is something that gives a directional derivative.

    thus there is a pairing between functions and tangent vectors. thus if tangent vectors have not yet been defined they can be defined as objects which operate on functions in the way that directional derivatives do.

    here are some equivalent definitions for algebraic manifolds, that also make sense for differential ones:

    Tangent Spaces

    There are several possible definitions of the tangent space Tp(X) to a variety X at a point p, all equivalent, and each having various good points.

    Definition: Let X in A^n be a closed affine set with ideal of defining equations I(X) = (f1,....,fm), and assume that 0 is a point of X. We say that the line L = ta, is tangent to X at 0 iff L intersects every hypersurface V(fj) at 0 with multiplicity > 1, i.e. iff t^2 divides every polynomial fj(ta) = the restriction of fj to L.

    Theorem: All the following definitions define isomorphic linear spaces.
    1) T(X) = the union of all lines in A^n which are tangent to X at 0.
    2) T(X) = intersection {Lj = 0}, the common zero locus of the linear terms at 0, Lj, of the Taylor expansions of the generators fj of the ideal of X.

    3) T(X) = Homk( M/M^2, k) where M in k[X] is the ideal in k[X] of equations defining the point p.
    4) T(X) = Homk( m/m^2, k) where m in O(X,p) is the ideal in O(X,p) of the point p.
    5) T(X) = Derk(O(X,p) , k) = all k linear mappings D: O(X,p)-->k, which satisfy the Leibniz rule, i.e. such that D(fg) = f(p)D(g) + g(p)D(f), for all f,g in O(X,p).

    1)Note that the first definition has the intuition of a tangent line in it, i.e. as a line that meets the variety to higher order than one.
    2)The second definition implies that the tangent space is linear, and gives equations for it in terms of the partial derivatives of the generators of the ideal of X.
    3) Definition 3 shows that the tangent space depends only on the isomorphism class of X since it is given in terms of the coordinate ring of X, and independent of the embedding.
    4) Definition 4 shows that the tangent space is dependent only on the neighborhood of p in X, since it depends only on the local ring of p in X.
    5) Definition 5 is the one used in differential geometry, where a tangent vector v is thought of in terms of its associated differential operator Dv on functions, differentiation in the direction v.

    O(X,p) here denotes the ring of "germs" of functions at the point 0 or p, i.e. two functions are equated if they agree on some open nbhd of the point.
    Last edited: Nov 14, 2006
  5. Nov 14, 2006 #4


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    now that's what i call an answer. how bout them tangent vectors?

    oops, there is one of doodle bobs definitions that does not work in the algebraic case without modification, the implicit function approach. i.e. algebraically it is not possible to solve for z in terms of polynomial functions of x and y. there is a however a projection mapping from a surface in 3 space to at least one of the coordinate planes that induces an isomorphism on tangent planes.

    but i hope the others are useful, to someone.
    Last edited: Nov 14, 2006
  6. Nov 14, 2006 #5


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    oh there is yet another definition, that of descartes, in terms of algebra maps from the ring of germs O(X,p) to k[e]/(e^2) that send functions vanishing at 0, into the ideal (e). the idea is that (e^2) has codimension one in the dieal (e), so determines a hyperplane of functions vanishing at 0, those that vanish twice, i.e. that have zero derivative in some particular tangent direction.
  7. Nov 14, 2006 #6


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    the whole point is that a geometric tangent direction at 0, such as in definitions 1 and 2, defines an equivalence relation on M = functions vanishing at 0, namely two functions in M are equivalent if their difference has derivative zero in the given direction.

    Then any way of recovering this equivalence relation on M, also recovers the tangent vector, and can be used to define the tangent vector.

    algebraically, derivations on functions in M, are equivalent to linear functionals on equivalence classes of functions in M/M^2. so one gets the definitions 3,4, and 5.

    the one in post #5 is also another version of this.
  8. Nov 14, 2006 #7
    Yeah, well, my approach is really only good for a thought experiment. I only wanted to point out an example of equating the two approaches of defining the tangent space with an isomorphism. Actually, trying to calculate df/dx and df/dy for a specific f and then figuring out TM from that can be a bear.

    I've always found it easier to start from the actually implicit function approach: locally M is given by f(x,y,z)=0, so that TM=ker(df). As you pointed out, more often than not, this does not guarantee that you can solve one variable in terms of the other.
  9. Nov 16, 2006 #8


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    That Is Def 2 In Post 3
  10. Nov 21, 2006 #9
    Thanks for taking your time to answer my post. It helped a great deal. I think I know how to convince myself that it is a good definition.

    I have also found a book called:

    Modern Differential Geometry for Physicists 2nd ed. - C. Isham

    He defines the tangent space with deriviations and the more geometrical intuative way, and then compare them, in fact show that the two spaces are isomorph when we are looking in R^n, so this book may help others too.

    some of the definitions in this post are a bit too difficult for me, but maybe they will help me later, when I get a better overview of the subject.

    thanks again, Anders Berthelsen.
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