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Differentiating a function on a manifold

  1. Oct 7, 2011 #1
    Hey everybody!

    Physicists have no problem differentiating a function of many variables - in flat space R^n.

    But I dont like how many books dont give examples of how this done in a manifold- even if it may be easy when one finally understands it.

    For example, how do I differentiate a function f on a circle if that function only lives on the circle and not on an ambient space? For example, if given a prescription of a differentiation (contained in some vector V at a point p), how does V differentiate f explicitly? Can someone explain + give a non-trivial example?

    Also, I would very much like to see an example on a sphere too - where many directions may be chosen in which to differentiate a function.
     
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  3. Oct 7, 2011 #2

    Deveno

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    it's been so long that i may have forgotten how to say this correctly, but here goes:

    a manifold M comes with a local diffeomorphism to R^m at every point x in M. so what you do is you use the diffeomorphism to pull your function back to a copy of R^m (the tangent space at x). now you're in R^m, there's no problem.

    if your function f:M-->N where N is another manifold, the derivative of f is a linear map between the tangent space of x in M to the tangent space of f(x) in N (which is basically just a linear map from R^m-->R^n). the matrix for this derivative will depend on what diffeomorphisms you use (the "charts" or coordinate systems for your manifolds).

    a circle is "locally" a line (the circle is locally diffeomorphic to R, you can use an exponential map e^(it) as the diffeomorphism (you'll have to restrict the parameter t so you don't go "too far around")). suppose we mapped f:(cos(t),sin(t)) ---> cos(t).

    the graph of f looks like a "wavy ring" in 3 dimensions (it would be sooo much easier to draw this). for any particular value of t, the derivative of f give a line lying in the tangent plane (at the point (cos(t), sin(t)) to the infinite tube going through our circle. if you were to plot this as t changed, you'd get another wavy ring "out of phase" with f (by pi/2). as we might expect, f' can be parameterized as g(cos(t),sin(t)) = -sin(t).

    on a sphere, you have the same choice of directions for differentiation that you do on the plane. at any point on the surface of the sphere, you have a tangent plane, to compute a derivative of f (defined on the sphere) in some direction (which direction? well, to specify a direction you need a coordinate system for the sphere), you pull the coordinate system for the sphere back to a coordinate system for the tangent plane, and calculate the directional derivative in that direction in the tangent plane.

    alternately, you can define a curve on the sphere, say c(t), and define the derivative of f in the direction of c at c(t) to be: d/dt(f(c(t)).

    anyway, i believe that's how it goes. if i am mistaken, i'm sure someone else will correct me.
     
  4. Oct 7, 2011 #3

    lavinia

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    differentiate along a curve that fits the tangent vector.
     
  5. Oct 7, 2011 #4

    mathwonk

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    the thing to remember is that on a manifold the only thing you get from a derivative is the rank, not an explicit number or matrix. so for a real valued function all you get is that the derivative is zero or not zero.
     
  6. Nov 26, 2011 #5

    Bacle2

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    You may want to look into the concept of connections in Riemannian geometry, which was designed just for that: the idea is that (maybe someone else can correct/comment) while in R^n the tangent spaces at different points are naturally isomorphic to each other, (so that you can translate tangent vectors from one vector space to another ), this is not the case for other manifolds , so that you need to have a choice of isomorphism between tangent spaces at different points. The connection gives you such a choice using , in part, curves connecting different points, since curves are contractible, so that their bundles are trivial (the connection lives on the tangent bundle to the manifold ).
     
  7. Nov 26, 2011 #6
    If you have a vector field, then you can get a derivative of a real-valued function on a manifold. It just measures "the change of the function in the direction of the vector field" and returns another real valued function on the manifold. I think this is called the directional derivative. You could view regular differentiation on the real line in this way- you differentiate along the vector field which is just the value "1" at each point of the tangent space of the real line.
     
  8. Nov 26, 2011 #7
    (note that the generalisation of this to arbitrary tensor fields etc. is the Lie derivative)
     
  9. Nov 27, 2011 #8
    You have to first remember how it is done in multivariable calculus. Differentiation is done by chosing a direction for the derivative i.e a vector.

    The answer to your question is not however that trivial. It requires some manifold theory. Definition of vectors on real manifolds are generally done by using vector fields. To define vector fields on a manifold you first need to define local coordinates. What changes at this point is, if you want to get a value for the derivative then you need to work in the local coordinates.

    For instance take the circle as you said. Let f be the height function of the unit circle i.e f(x,y) = y (restricted to circle) or f(θ)=sinθ (where θ are the local coordinates given by (x,y) = (cosθ, sinθ) but I dont delve into intracies of defining the local coordinates for different patches etc since our example is trivial). We want to differentiate lets sat at the point (x',y')= (θ') along the direction of the unit vector pointing in clockwise direction.

    I will first do this in terms of circle coordinates. In terms of the circle coordinates this vector is given as ∂θ. Then the derivative of the function along this vector is simply ∂θsinθ (at θ') = cosθ' = x'. However I know that what you really ask is a computation in the coordinates (x,y). So you write ∂θ in these coordinates as (∂x/∂θ)∂x + (∂y/∂θ)∂y = -y∂x + x∂y (by chain rule basically). Now apply this to f(x,y)=y at the point (x',y') we get after applying the vector field to the function and evaluating again x (at x') = x'.

    So what you really need to is to find the functions defining those surfaces you mention. Then fix a tangent vector to take derivative with. Find how that tangent vector looks in your ambient space coordinates (in our case for instances vectors tangent to the circle look like (∂x/∂θ)∂x + (∂y/∂θ)∂y) then apply this vector (infact vector field) to the function. You can easily extend this to a sphere after writing its coordinates. But on the sphere now any vector has the general form a∂θ + b∂ where θ ,∅ are the coordinates of the sphere.

    Because you are embedding this manifold in R^n, you can in a sort of way worry less about being rigorous on the matter of locality. But if you want to make a global statement about the derivative of a function f along a vector field X, all you can say is that its derivative is Xf :p. Finding values for this derivatives means choosing local coordinates. And in physics usually things boils down to calculating things in local coordinates after you formulate them abstractly.
     
    Last edited: Nov 27, 2011
  10. Nov 27, 2011 #9
    Could you please explain this a bit more? Are you meaning finding some numerical value which is global?

    In any way, the concept of Lie derivative, covariant derivative etc is used (as an extension of the concept of derivation to manifolds) to define "derivatives" globally so locality does not seem like a concern. Since vector fields are defined globally (on non-analytic manifolds however), making calculations in coordinates should not be a problem I suppose no?
     
  11. Nov 27, 2011 #10
    I don't think that local coordinates matter- using a different coordinate system should give the same derivative (it would be pretty rubbish otherwise).

    In fact, you can actually consider vector fields simply as being the derivations on the ring of continuous functions on a manifold.
     
  12. Nov 27, 2011 #11
    Well any such equation is invariant under coordinate transformation. Indeed if you define a coordinate transformation θ then θ*X((θ-1)*f) = (θ*-1)*X)(f) = X(f). Covariance is a general property of equations written in terms of vector fields and form fields this is also why mathematical physics revolves writing equations in terms of tensors, vector fields and form fields etc.
     
    Last edited: Nov 27, 2011
  13. Nov 28, 2011 #12

    Bacle2

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    Sina wrote, in part:

    "I will first do this in terms of circle coordinates. In terms of the circle coordinates this vector is given as ∂θ. Then the derivative of the function along this vector is simply ∂θsinθ (at θ') = cosθ' = x'. However I know that what you really ask is a computation in the coordinates (x,y). So you write ∂θ in these coordinates as (∂x/∂θ)∂x + (∂y/∂θ)∂y = -y∂x + x∂y (by chain rule basically). Now apply this to f(x,y)=y at the point (x',y') we get after applying the vector field to the function and evaluating again x (at x') = x'.

    So what you really need to is to find the functions defining those surfaces you mention. Then fix a tangent vector to take derivative with. Find how that tangent vector looks in your ambient space coordinates (in our case for instances vectors tangent to the circle look like (∂x/∂θ)∂x + (∂y/∂θ)∂y) then apply this vector (infact vector field) to the function. You can easily extend this to a sphere after writing its coordinates. But on the sphere now any vector has the general form a∂θ + b∂∅ where θ ,∅ are the coordinates of the sphere. "


    I agree overall, but I am having some dimensional trouble re the form:

    (∂x/∂θ)dx+(∂y/∂θ)dy

    for vectors in the tangent space of S1 ; isn't the tangent space of S1 1-dimensional, so that tangent vectors should be spanned by a single
    basis element; either dx or dy?
     
  14. Nov 28, 2011 #13
    they are indeed spanned by a single element which is -yd/dx + xd/dy any tangent vector on the circle is some constant times this vector. just because it involves the two basis vectors does not mean it can not be spanned by one vector =)

    consider for instance all the vectors spanned by d/dx + d/dy that is 1-d line on R^n
     
  15. Nov 28, 2011 #14

    Bacle2

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    Maybe this will help make sense of this ( at least for me): not every vector field on a manifold is a tangent vector field, e.g., maybe the most common (counter)example is the normal vector field, which we can call, say dy, and we let dx be the basis for the tangent space of S1. Then, if we had a function on the manifold whose derivative Xf is not a tangent vector, we would then project Xf onto the space spanned by dx (using the orthogonal decomposition of the ambient space as the sum of dx and dy ), and call the image under this projection the derivative. And, BTW , my understanding is that the kernel of this projection is related to the geodesic; any curve whose tangent vector field is the kernel of the projection is the geodesic; in this case, the normal vector to the circle is not a tangent vector field, but the projection of the normal vector field is the kernel of the projection into the tangent space.

    This definitely needs improvement; anyone? In the meantime, all this talk about the kernel made me hungry for chicken.
     
  16. Nov 28, 2011 #15
    There are integrability conditions that leads to such theorems as frobenius theorem that tell when a group of vector fields (or distributions) span the tangent space of a manifold. basically given a set of vector fields {X_i} if they all commute with each other wrt to lie bracket then they span the tangent space of a manifold at each point

    The second thing you said is precisely one of the definitions of covariant derivative on manifolds.
     
  17. Nov 28, 2011 #16

    Bacle2

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    I see, so it is like the case of a curve in ℝn parametrized as (x1(t),...,xn(t)) , despite the curve being intrinsically 1-dimensional (it is definable by a single parameter), it lives in ℝn. Right; I am way rusty on this.
     
  18. Nov 28, 2011 #17
    I really don't like this approach- you can do everything without resorting to embedding your manifold into Euclidean space. Vector fields on a manifold are usually understood to be tangent vector fields (i.e. sections of the tangent bundle).

    Your derivative along the vector field is as you'd expect- simply differentiate at each point on the manifold with directional derivative by appealing to the local coordinates.

    And non of this matters on the coordinate system.
     
  19. Nov 28, 2011 #18

    Bacle2

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    Your right, Jamma, but unfortunately I missed that part when I took the class, because I was fighting for a whole year with my landlord and super (everyday life in the city !); getting to figure out that layout would take me time; maybe if I saw some worked examples, would help, so not much chance.

    Still, while I agree with becoming coordinate-free, the embedding approach gives a nice and quick away of finding the geodesics for S^n.
     
  20. Nov 29, 2011 #19

    lavinia

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    If a vector is not tangent then you need to be able to draw a curve that moves off of the manifold to differentiate a function with respect to it. This means that the vector is tangent in a larger manifold.

    So the moral is that you can not differentiate a function except with respect to a tangent vector.

    However you can differentiate vectors that are not tangent to the manifold. In fact in any vector bundle you can find a connection that allows you to differentiate vectors in the bundle with respect to tangent vectors. This differentiation depends on the choice of connection It is not canonical.
     
  21. Dec 8, 2011 #20

    Bacle2

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    I don't mean to beat this one to death, but I think there is a distinction to be made here that is not pedantic:

    What we have calculated here is the exterior derivative /differential, and not the derivative strictly speaking. In cases of maps , e.g., f: ℝn→ℝ ,(x1,x2,..,xn)→ x the derivative (when it exists; more accurately here, when all the partials exist) , is a vector field given by
    (x1'(t),x2'(t),....,xn'(t)). The differential is a 1-form --as we can see by the fact that the answer had a form with coefficients dx, dy , or dθ , while the derivative is a vector field that assigns to a point the vector consisting of the directional derivatives.

    The choice of using the exterior product/exterior derivative has to see, AFAIK, with the fact that it is independent of the choice of coordinates.

    Note that while we can do a standard derivative as we did without any setup, if we wanted to differentiate along a vector field, we would have needed to use a connection. I think this would have been necessary if we had defined a map , e.g., from S2
    to S2.

    Hope this point is meaningful to the OP, maybe with someone else's added comments, since I am a bit weak on this material.
     
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