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What's the deal on infinitesimal operators?

  1. Jul 1, 2013 #1

    Stephen Tashi

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    Is there a treatment of "infinitesimal operators" that is rigorous from the epsilon-delta point of view?

    In looking for material on the infinitesimal transformations of Lie groups, I find many things online about infinitesimal operators. Most seem to be by people who take the idea of infinitesimals seriously and I don't think they are talking about the rigorous approach to infinitesimals [a la Abraham Robinson and "nonstandard analysis".

    I suppose people who work on mainfolds and various morphisms also can deal with infinitesimal operators via some abstraction. However, I'd like to know if there is an approach to infinitesimal operators ( in general, not simply Lie group operators) that is essentially from the "advanced calculus" point of view.

    (If not, I suppose I'll have to think about them like a physicist.)
     
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  3. Jul 1, 2013 #2

    lurflurf

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    http://en.wikipedia.org/wiki/Infinitesimal_transformation

    Any book on differential geometry or Lie algebras should be rigorous. You can just go through and add epsilons and deltas.

    Maybe you are talking about things like

    $$e^{t \, D} \mathop{f}(x)=\mathop{f}(x+t) \\
    e^{t \, D} \, \sim \, 1+t \, D $$

    That is entirely rigorous, it is just that the equivalence is to first order.
     
  4. Jul 2, 2013 #3

    Stephen Tashi

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    That's the correct subject, but it isn't the correct approach. (I find it amusing that calculus students are discouraged from proving things in terms of infinitesimals, but when we get to some advanced topics, reasoning by way of infinitesimals suddenly becomes respectable.)

    I have yet to see such a book that was compatible with an rigorous advanced calculus point of view. The abstract approaches are rigorous, but that's not my interest. The down to earth books speak in terms of infinitesimals. Do you know of a book with a practical orientation that treats infinitesimal transformations without using infinitesimals?

    (I first saw that result in a volume of Oliver Heaviside's Electrical Papers.)


    As far as I can interpret the materials I've seen on applications of Lie groups, the arguments that begin with showing things are true "to the first order" proceed to claim things that are true with "=" meaing equal, not "equal to the first order". Perhaps there is a rigorous way to phrase these arguments, but I have not seen it done.





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  5. Jul 2, 2013 #4

    HallsofIvy

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    I'm not sure what you mean by "rigorous from the epsilon-delta point of view"- the whole point of "epsilon delta limit arguments" is to avoid "infinitesmals", which, in order to be introduced rigorously require an extended number system.
     
  6. Jul 2, 2013 #5
    I am positively surprised to find that I'm not the only one who's annoyed by this.

    Eventually I have not evolved to become an expert on this stuff, so I cannot answer the original question. But I am under impression that for example "Lie Groups Beyond an Introduction" by Anthony W. Knapp would be rigorous like real mathematics. What I do not know is how much it actually answers to those who are puzzled by theoretical physics.
     
  7. Jul 2, 2013 #6

    Stephen Tashi

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    I do mean that I would like to see a treatment of "infinitesimal operators" that did not make reference to "infinitesimals" but rather used the more familiar (to me) concepts of limits (in the epsilon-delta sense) of functions, sequences of functions etc. My web search on "infinitesimal operators" didn't even turn up a source that defined "infinitesimal operators" except in terms of infinitesimals.

    Perhaps I'm using terminology that only turns up things written by physicists. Or perhaps "infinitesimal operators" are one of these areas where physicists have gotten ahead of mathematicians and no non-infinitesimal treatment of them has been written.

    An example that I've given in another thread (on using Lie Groups to solve differential equations) is the treatment of the "infinitesimal operator" of a 1-parameter Lie group.

    Avoiding the traditional zoo of Greek letters, I prefer the terminology:

    Let [itex] T(x.y,\alpha) = ( \ T_x(x,y,\alpha), T_y(x,y,\alpha )\ ) [/itex] denote an element of a Lie group of 1-parameter transformations of the xy-plane onto itself.

    Let [itex] f(x,y) [/itex] be a real valued function whose domain is the xy-plane.

    Let [itex] u_x(x,y) = D_\alpha \ T_x(x,y,\alpha)_{\ |\alpha=0} [/itex]
    Let [itex] u_y(x,y) = D_\alpha \ T_y(x,y,\alpha)_{\ |\alpha =0} [/itex]

    ([itex] u_x [/itex] and [itex] u_y [/itex] are the " infinitesimal elements ".)

    Let [itex] U [/itex] be the differential operator defined by the operation on the function [itex]g(x,y)[/itex] by:

    [itex] U g(x,y) = u_x(x,y) \frac{\partial}{\partial x} g(x,y) + u_y(x,y)\frac{\partial}{\partial y} g(x,y) [/itex]

    (The operator [itex] U [/itex] is "the symbol of the infinitesimal transformation" .)

    Every book that takes a concrete approach to Lie Groups proves a result that says

    [tex]f(x_1,y_1) = f(x,y) + U f\ \alpha + \frac{1}{2!}U^2 f \ \alpha^2 + \frac{1}{3!}U^3 f \ \alpha^3 + ..[/tex]

    by using Taylor series.

    However, the function they are expanding is (to me) unclear.

    If I try to expand [itex] f(x_1,y_1) = f(T_x(x,y,\alpha),T_y(x,y,\alpha)) [/itex] in Taylor series, only the first two terms of that result work.

    If I expand [itex] f(x_1,y_1) = f( x + \alpha \ u_x(x,y), y + \alpha \ u_y(x,y) ) [/itex] then I get the desired result. So I think this is equivalent to what the books do because they do not give an elaborate proof of the result. They present it as being "just calculus" and expanding [itex] f(x_1,y_1) = f( x + \alpha \ u_x(x,y), y + \alpha \ u_y(x,y) ) [/itex] is indeed just calculus.

    The books then proceed to give examples where the above result is applied to expand [itex] f(x_1,y_1) = f(T_x(x,y,\alpha),T_y(x,y,\alpha)) [/itex] I haven't found any source that justifies this expansion except by using the concept of infinitesimals.
     
    Last edited: Jul 2, 2013
  8. Jul 2, 2013 #7

    lavinia

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    I am not sure I understand the question. But it seems to me that epsilon delta proof are avoided for simplicity of exposition.
     
  9. Jul 2, 2013 #8

    Stephen Tashi

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    I'd be happy to have answers to either of the following questions.

    1. If you are famililar with books about Lie groups that do a Taylor expansion in terms of the operator [itex]U [/itex] what is the definition of the function [[itex] f(x_1,y_1) [/itex] that they are expanding?

    2. What justifies using that expansion for the function [itex] f(T_x(x,y,\alpha),T_y(x,y,\alpha) ) [/itex]?



    I suppose lack of clarity is a type of simplicity.
     
  10. Jul 3, 2013 #9

    WannabeNewton

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    The only context in which I've ever seen the phrases "infinitesimal operators" and "infinitesimal transformations" are in QM and GR texts respectively with the latter being related to one-parameter Lie groups (specifically one-parameter diffeomorpism groups of space-times). I've never seen a math book that uses the terminology. In the case of "infinitesimal transformations", one starts with a one-parameter group of diffeomorphisms ##\varphi_t## on a smooth manifold ##M## generated by a vector field ##v^{a}## and defines the Lie derivative of a tensor field ##T^{a_1...a_k}{}{}_{b_1...b_l}## on ##M## as [tex]\mathcal{L}_{v}T^{a_1...a_k}{}{}_{b_1...b_l} = \lim_{t\rightarrow 0}\{\frac{\varphi^{*}_{-t}T^{a_1...a_k}{}{}_{b_1...b_l} - T^{a_1...a_k}{}{}_{b_1...b_l}}{t}\}[/tex] I guess in that sense the "infinitesimal transformations" are codified by regular ##\epsilon-\delta## limits.
     
    Last edited: Jul 3, 2013
  11. Jul 3, 2013 #10

    Stephen Tashi

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    My simplistic view of the above is that the vector field in my example is [itex] (u_x,u_y) [/itex] and one can define a directional derivative of a function at each point (x,y) that is taken with respect to the field vector at that point. Those concepts are defined using the ordinary notion of limit. However, I don't understand how Taylor expansions of functions in terms of the operator [itex] U [/itex] are proven without resorting to arguments using infinitesimals.


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  12. Jul 3, 2013 #11

    WannabeNewton

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    Ah yes. The book "Modern Quantum Mechanics"-Sakurai does that over and over and the justifications are by mathematical standards abysmal. All he does is ignore the terms of second order and higher in the infinitesimals. To give an example: http://postimg.org/image/fxhr3sgmf/ [Broken]

    This is one of the reasons I absolutely hate that book. Apparently the author had some kind of grudge against mathematical rigor. I don't get how anyone can settle for such hand-waviness. The book by Ballentine is supposed to be more rigorous mathematically but: http://postimg.org/image/h29z9wlzb/ [Broken]

    You might be interested in Stone's theorem: http://en.wikipedia.org/wiki/Stone's_theorem_on_one-parameter_unitary_groups and this thread: http://physics.stackexchange.com/questions/62876/translator-operator

    I'm sure there is a QM for mathematicians book out there that treats these "infinitesimal operators" with more care. I agree with you that the notion of infinitesimal generators in the context of vector fields is much simpler to make mathematically rigorous than in the context of operators.
     
    Last edited by a moderator: May 6, 2017
  13. Jul 4, 2013 #12

    lavinia

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    Since I don't have your book let's start with an example to see if this is the right track.

    SO(2) parameterized as

    G(x) =

    cos(x) -sin(x)
    sin(x) cos(x)


    G can be expressed as a Taylor series around the identity matrix. This series can be computed directly.

    On the other hand since G(x) is a homomorphism from the real line with addition, to the group of rotations under matrix multiplication, one can derive the equation

    dG/dx = VG where VG is the matrix product of V and G and V is the matrix,

    0 -1
    1 0

    V is the derivative of G at zero and the iterated derivatives of G at zero are just the iterated powers of V.

    So in this case G(x) = exp(xV).

    V is called the infinitesimal generator of G because every element of G is the exponential of V times the parameter,X. This simplifies G since now it can be derived from a single matrix.

    Another way to look at infinitesimal generators of SO(2) is to look at the infinitesimal effect of a rotation of the plane upon a differentiable function.

    The function, F(x,y) under rotation becomes a function of θ. That is F(x,y) = F(x(θ),y(θ))

    where x(θ) = cosθx[itex]_{0}[/itex] - sinθy[itex]_{0}[/itex] and y'(θ) = sinθx[itex]_{0}[/itex] + cosθy[itex]_{0}[/itex] and (x[itex]_{0}[/itex],y[itex]_{0}[/itex]) is the point at which F is being differentiated and θ is the angle of rotation.

    so dF/dθ = (∂F/∂x)(∂x/∂θ) + (∂F/∂y)(∂y/∂θ) = (∂F/∂x)(-x[itex]_{0}[/itex]sinθ - y[itex]_{0}[/itex]cosθ) + (∂F/∂y)(x[itex]_{0}[/itex]cosθ -y[itex]_{0}[/itex]sinθ)

    evaluated at θ = 0. This is -y[itex]_{0}[/itex]∂F/∂x + x[itex]_{0}[/itex]∂F/∂y.

    F(x,y) = F(x[itex]_{0}[/itex],y[itex]_{0}[/itex]) + (-y[itex]_{0}[/itex]∂F/∂x + x[itex]_{0}[/itex]∂F/∂y)dθ up to first order.

    The infinitesimal generator of the rotation is defined as the operator

    -y∂/∂x + x∂/∂y
     
  14. Jul 4, 2013 #13

    Stephen Tashi

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    The question that I've asked (I think) is why, in general, does such a derivative turn out to be iterated powers of a single operator? Matrix groups are special case and ordinary calculus may suffice to show this. Perhaps that gives insight into the general situation.


    The original post asks how to look at it without using arguments that assume existence of "infinitesimals".
     
  15. Jul 4, 2013 #14

    lavinia

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    The generalization is the 1 parameter subgroup whose derivative at the identity is a particular element of the Lie algebra. In the case of SO(2), the rotation matrix of sines and cosines is a 1 parameter group.

    All of the properties of the exponential follow because it is a homomorphism of the real numbers under addition into the Lie group.



    Not sure about your problem here. You can think of dθ as the differential of the function,θ. For this first order approximation it just means that for small enough increments in theta the approximation is close and higher order terms become insignificant. This is just your epsilon delta proof. But the important observation is that (the Higher Order Terms/Δθ) all go to zero. So evetually the first order approximation is arbitrarily accurate.
     
    Last edited: Jul 4, 2013
  16. Jul 4, 2013 #15

    Stephen Tashi

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    I think that's the problem I have described, if we assume [itex] T_x , T_y [/itex] are infinitely differntiable functions.


    To justify the exponential in the first place, you need establish that the expansion can be done using powers of the same operator, which, as I said, is what I want to see demonstrated.



    I don't have a problem with differentials provided a "differential" can be defined without resorting to the terminology of "infinitesimals". The idea that that a first order approximation can be defined is not a problem. The problem is how that implies that the Taylor expansion can be written with an "=" (instead of a [itex]\approx [/itex]) using operators that are a first order approximations in the variable [itex] \alpha [/itex] that is used in the expansions.
     
  17. Jul 4, 2013 #16

    lavinia

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    For matrix groups this is just a Taylor series expansion. You are solving the differential equation dH/dt = XH or X = (dH/dt)H[itex]^{-1}[/itex] which shows that the tangent vector to H is right invariant.

    On an abstract Lie group the differential equation is

    dR[itex]_{g_{-1}}[/itex]dH/dt = X where X is an element of the tangent space at the identity.
     
    Last edited: Jul 4, 2013
  18. Jul 4, 2013 #17

    Stephen Tashi

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    In the books that take a concrete approach to "abstract" Lie groups (i.e. don't restrict themselves to matrix groups) the differential equation approach is mentioned. However, as an independent demonstration of the operator expansion, they purport to do the Taylor expansion directly. When I try this the 3rd term of the expansion

    [itex] f(T_x(x,y,\alpha),T_y(x,y,\alpha)) = f(x,y) + Uf + \frac{1}{2!} U^2 f\ \alpha + .... [/itex]

    does not come out to be [itex] \frac{1}{2!} U^2f\ \alpha^2 [/itex].

    So a particular instance of my original question is whether there is a way to argue that the 3rd term is [itex] \frac{1}{2!} U^2f\ \alpha^2 [/itex]. using the special properties of the transformation.

    Those special properties being

    [itex] T_x(x,y,0) = x,\ \ T_y(x,y,0) = y [/itex]

    and the property that allows the homomorphism you mentioned:

    [tex] T_x(x,y,\alpha + \beta)= T_x(T_x(x,y,\alpha),\beta), T_y(T_y(x,y,\alpha),\beta) [/tex]
    [tex] T_y(x,y,\alpha + \beta)= T_x(T_x(x,y,\alpha),\beta), T_y(T_y(x,y,\alpha),\beta) [/tex]

    Or is the claim that the result can be established by doing Taylor expansion directly false?
     
  19. Jul 4, 2013 #18
    Wow, make sure you link this thread to our one in your next post, good stuff!

    Emanuel derives it on page 13, explicitly showing how you get the third term, is there something wrong with what he does?

    I was reading Gelfand's proof of Noether's theorem yesterday (Page 168) & level to which terms of second order & higher are abused in that proof are stunning. I love the book but in that proof there's just too much of it going on, & not coincidentally it's intimately related to the topic of infinitesimal transformations.

    Another place I've seen infinitesimal transformations come up is in the derivation of the Euler angles in classical mechanics, you could now have three subjects you';; have seen this stuff in :tongue:
     
  20. Jul 4, 2013 #19

    Stephen Tashi

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    only if this thread enlightens me.

    He just claims its true. And he's correct if [itex] f_1(x,y) [/itex] is defined to be [itex] f(x + \alpha\ u_x(x,y), y+ \alpha\ u_y(x,y)) [/itex], which isn't the function I want to expand.

    Try to derrive it using [itex] f_1(x,y) = f(T_x(x,y,\alpha), T_y(x,y,\alpha)) [/itex] and using ordinary calculus - the chain rule, the product rule. It doesn't work because [itex] \frac{d^2 f_1}{d\alpha^2} [/itex] has terms involving factors like [itex] \frac{d^2T_x(x,y,\alpha)}{d\alpha^2} [/itex] When you set [itex] \alpha = 0 [/itex] that factor becomes [itex]\frac{d^2 T_x(x,y,\alpha)}{d\alpha^2}_{|\ \alpha = 0} [/itex], not [itex] (u_x(x,y))^2 = \big( \frac{dT_x(x,y,\alpha)}{d\alpha}_{|\ \alpha = 0} \big)^2 [/itex]. So the derrivation (if there is one) requires something besides straightforward differentiation.
     
  21. Jul 5, 2013 #20

    lavinia

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    From the differential equation

    dH/dt = XH

    one gets d[itex]^{2}[/itex]H/dt[itex]^{2}[/itex] = XdH/dt = X[itex]^{2}[/itex]H

    Inductively, d[itex]^{n}[/itex]H/dt[itex]^{n}[/itex] = X[itex]^{n}[/itex]H

    H(0) = Identity matrix

    so the Taylor series at the identity follows.



    The differential equation derives from the homomorphism equation.

    H(x+y) = H(x)H(y)

    for instance,

    dH(x+y)/dx = dH/dxH(y) At x = 0 the right hand side is just XH where X is the derivative of H at 0.

    The left hand side can be rewritten as

    [dH(x + y)/x+y]d(x+y)/dx using the Chain Rule. This is just the derivative of H.

    In sum dH/dt = XH
     
    Last edited: Jul 5, 2013
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