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Could someone critique this? (Lie Group with the Lie Algebra)

  1. May 31, 2015 #1
    Is the following correct?

    We begin with a set.

    Then, we specify a certain collection of subsets and thereby create a topology. This endows the set with certain properties, one of which is “nearness” and “boundedness.”

    Then we specify that the topology be smooth. In so doing, our topology become a manifold.

    Finally, we specify a metric. The metric enables measurements: angles, distances, etc.

    We now have a set on which we can perform calculus. And by perform “calculus,” I mean study how things change, ON THE SET. And, more specifically: tangent vectors, tangent bundles, etc.

    Now, this is different from how we performed calculus in R3. In R3, we have all the rules of calculus. But a manifold restricts R3, except in small local regions where it is still like it. (Not to mention that the coordinates of a manifold could be temperature, pressure, etc.: all different.. whatever that means.)

    They could be, say, rotation matrices...

    Now we move onto Lie Groups. A Lie Group is a manifold because it is a continuous group. So once we map, say, the nine components of a matrix, and realize that if the matrix is orthogonal, then we have 3 independent coordinates. We can use the Euler angles.

    And now, in the tangent space to the manifold we have a linear space. And now we can use Lie Algebra.

    AND THAT is where I have the last problem. What does it mean to replace the Lie Group with the Lie Algebra?

    Finally, as an aside, in the PROCESS of learning manifolds, I now understand exterior algebra. And from that, forms. And from forms, Stokes theorem. But while that is beautiful in itself (seeing the nature of the divergence, gradient and curl, along with the Green’s theorem and calculus of variations as a special case of the generalized Stokes), I do NOT need exterior forms if I am to understand dynamics from the perspective of manifolds and lie derivatives.

    (This does not mean I understand... rather it means I finally know what I do not know...and where I can now begin learning.)

    And if you can add to this statement in ANY WAY.. . PLEASE...
     
  2. jcsd
  3. May 31, 2015 #2

    MarcusAgrippa

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    There are many errors in what you say. Here is one. "Topology endows a set with boundedness" - this is not true for several reasons: 1. Boundedness is a concept that belongs only to ordered sets, like the real numbers. Perhaps you mean "compactness", but topology does not make a set compact. Compactness is a special property of sets with a topology, and so is not a necessary property. Topology does not mean that a set is compact.

    It would be helpful if you could respond to this comment before we proceed to the next point.
     
  4. May 31, 2015 #3

    mathwonk

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    A lie group is a smooth manifold with a lot of symmetry. In particular it is possible to translate every point to the orign, and hence also to translate every tangent space to the tangent space at the origin. Hence it suffices to understand the tangent space of a lie group at the origin. that tangent space is precisely the lie algebra. i repeat: the lie algebra of a lie group is exactly its tangent space at the origin. thus replacing the lie group by the lie algebra is the process of linearizing the study of the lie group, or "taking a derivative". i.e. differential calculus is the process of approximating a smooth map locally by a linear map, namely its derivative. analogously, we replace, i.e. approximate, a lie group by its lie algebra, its tangent space at the origin.


    Read the wiki article on this. the section on lie algebra of a lie group is a little less than 1/2 way down the page:

    http://en.wikipedia.org/wiki/Lie_group
     
    Last edited: Jun 1, 2015
  5. May 31, 2015 #4

    micromass

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    Boundedness makes sense outside ordered sets too, like in metric spaces.
     
  6. Jun 1, 2015 #5

    WWGD

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    But I think he refers then, to distance used to bound set, to be a Real number and then ordered in the usual way, i.e., (X,d) is bounded iff (this is the def. I have
    seen, Sup {## d(x,y): (x,y) \in X \times X =M < \infty##}

    which makes use of the (ordered) Real numbers, since ## d(x,y) \in \mathbb R ##.
     
  7. Jun 1, 2015 #6
    Yes, I have been lazy (and thereby disrespectful...) I am sorry.

    How about this...

    Could you rewrite what I wrote... IN A SHORT a statement... to put all of this in context.
    I think I understand the individual pieces (or, more like it: I know what I do not know about the individual pieces and what I must still learn), but I in this question ,I am looking for the glue between them.... a kind of "motivating" glue that makes me see, stepwise, why I must take this path.

    Essentially, I am hoping for a statement that can minimmize the details of what each item is ( can forego that) and specify how a specific detail of the defintion BECKONS (<- deliberate word choice) the next level up the ladder.
     
    Last edited: Jun 1, 2015
  8. Jun 1, 2015 #7
    OK.. but I have read, time and time again now:
    A lie group is a smooth manifold with a lot of symmetry
    But you just wrote
    A lie algebra is a smooth manifold with a lot of symmetry

    I can see a lie group being smooth: SO(3) depends on continuous functions of rotations.

    Now, I think I am hitting the trifecta---

    You said
    the lie algebra of a lie grouop is exactly its tangent sopace at the origin.

    WHOA! I really sense that the missing piece for me lies in an elaboration of that sentence.

    For me, in my primitive state, a GROUP is a set with certain operations.
    An ALGEBRA is a structure imposed on the group.
    These, to me, are two different things.
    So I do not even understand the semantics of the sentence you wrote (bold above),
    Is there a way you can explain this to me in the context of calculus on Rn where I seem to find more comfort?
     
  9. Jun 1, 2015 #8

    MarcusAgrippa

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    A Lie group is a group which has a smooth manifold structure defined on it as well as its group structure. Alternatively, it is a smooth manifold that also has a group structure defined on it. The rest of your first line quoted above is good.

    The set of all vector fields on a general manifold admits several operations that give it structure: addition, scaling by real numbers (together these operations make it an infinite dimensional vector space), scaling by smooth functions, and the operation called the Lie bracket. This operation takes two vector fields and produces a new vector field. These operations provide a formidable structure on the vector fields.

    In the set of these vector fields, there are no special fields which play a distinguished role (actually, there are, but not a role of the kind that is of interest in answering your question). If you like, there is a democracy among the vector fields. No single field lays a role that is more special than any other vector field.

    The manifold of a Lie group is not just a manifold. It possesses a very special structure - the group structure. It is reasonable to expect the special nature of this structure to reflect itself in special properties of the superstructure constructed on the manifold. And it does. There is a very special class of vector field that can be defined on the group manifold: the left invariant vector fields (and the right invariant one too, but this information is a carbon copy of that contained in the left invariant ones, so we can safely ignore them - this assertion requires proof, of course). There is also a special class of 1-forms: the left invariant 1-forms (the the right invariant ones too).

    Concentrate on the left invariant vector fields. These have several fantastically simple properties:

    1. they can be completely constructed from the field vector at the origin by the group properties (left translations - a distinguished set of manifold homeomorphisms)

    2. their Lie bracket is closed: if X,Y are left invariant, so also is [X,Y], so these fields form a Lie algebra.

    3. They form a vector space, so we can choose a basis for them X_i, i=1,...,r where r is the dimension of the manifold.

    4. For any basis, [itex] [X_i, X_j] = {c^k}_{ij} X_k [/itex] where the [itex] {c^k}_{ij} [/itex] are constants, and they transform as a mixed tensor of rank 3 when the basis fields are charged.

    These are fantastically special properties which are directly the consequence of the group structure of the manifold. The Lie algebra is the set of left invariant vector fields. Because these fields can be uniquely reconstructed from their values at the identity, we can transfer all of the properties of the Lie algebra of left invariant vector fields to the subspace of vectors that define them in the tangent space at the origin. So this subspace at the origin is also called the Lie algebra of the group.

    The Lie algebra of a group is uniquely defined by the group. The converse is not true. The converse of any proposition that requires differentiation is in general not true, because the derivative only probes the neighbourhood of a point and does not probe the manifold globally. Accordingly, many groups have the same Lie algebra. The partial converse to the theorem that every group uniquely defines a Lie algebra is that a given Lie algebra uniquely defines a simply connected group that has the given Lie algebra as its Lie algebra. This unique group is called the Universal Covering Group of the given Lie algebra. It can then be shown that every Lie group with the given Lie algebra arises as a factor group of the Universal Covering Group by a finite subgroup (of special type) of the Universal Covering Group.

    In any work involving a Lie group that probes local structure by differentiation, you will use only the Lie algebra of the group in your work. In this sense, you have replaced the Lie group with its algebra. In the strictest sense of the word "replace", you cannot replace a group with an algebra - they are completely different animals and cannot be substituted one for the other.
     
  10. Jun 1, 2015 #9
    And you did it again... THANK YOU...

    Here are my thoughts...
    Something has gone very very wrong with math education for engineers.
    I feel we are on the cusp of a new math (or may be you people already know that) but engineering is stuck for two reasons
    1. The math they teach is old
    2. The nature of a math textbook simplifies the concepts just to get the machinery across so students can pass tests.

    I think this stuff -- exterior algebra, calculus on manifolds, Lie Algebra, differential geometry -- is SOOOO important.
    But engineers are being left in the dust. And it is not easy to hitch a ride on this new language.

    When I read certain things, I come to get this feeling that I missed something when I studied Calculus 30 years ago. And I did: I studied an old math and I studied from textbooks that just wanted to... get the **** over with...

    There is such a need for a manuscript that can explain all this stuff but in way that helps the engineer step out of the sinking lifeboat.

    Right now, I am finally aware of how much I do not know.

    But I have the pieces in place and can justify things.
    So now begins a long process of teaching myself algebra again and calculus again... but a different way.

    Thank you so much.
     
  11. Jun 1, 2015 #10

    micromass

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    If you want, you have the concept of a bornology which makes sense outside the real numbers, and which even makes sense in topological spaces.
     
  12. Jun 1, 2015 #11

    MarcusAgrippa

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    Interesting. Can you expand a little on this?
     
  13. Jun 1, 2015 #12
    Marcus, may I? Comment on this please...


    We have a Lie Group which is a continuous group (and thus capable of modeling rotations -- sin, cos, etc.). The restriction of det=1 enables us to take the independent coordinates, equipped with the smoothness condition, and envision them as a manifold.

    So a Lie Group is a manifold.

    (incidentally, a manifold is a topological space of things equipped with continuity… of maps and such.)

    Manifolds have a tangent space. And this tangent space is a linear space. So the Lie Group’s behavior can be modeled by its Lie Algebra of the linear space (a vector space)

    And this Algebra involves multiplication of rotation matrices, Lie Brackets and all the tools of the Lie Algebra.

    Thus, we can now use the Lie Algebra so(3) to model the Lie Group SO(3) of rotations.

    Finally, as an aside, the math of the Manifold also gives me vectors and covectors. And while this is not DIRECTLY related to my trying to understand rotations, it does give me insight into fluid mechanics: stokes theorem, curl, divergence, gradient, etc.
     
  14. Jun 1, 2015 #13

    WWGD

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  15. Jun 1, 2015 #14

    mathwonk

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    i made a mistyping error. I meant a lie group is a smooth manifold with a lot of symmetry.
     
  16. Jun 2, 2015 #15

    MarcusAgrippa

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    I am not sure on what aspect of the above you are asking for a comment.

    I don't like your persistent use of the word "model". If you are attempting to describe the real world by mathematics, the word "model" is appropriate. In that situation, one is constructing a mathematical model of some real system. However, one does not "model" mathematics except, perhaps, in representation theory, but even there the word "model" is probably inappropriate.

    I would not say that "the Lie Group’s behavior can be modeled by its Lie Algebra". The Lie group defines the Lie algebra. Modern math, unlike classical math, is math without "dead wood". It strips away all that is unnecessary and leaves only the essentials of the structure. That way, you can concentrate exclusively on the things of interest without the morass of irrelevant garbage that clutters classical math. The idea is this: identify the essential elements of the properties of interest - these are the axioms of the theory - then deduce as much as you can from these essentials. One way to make deductions is to identify new structures that can be constructed in a natural way from the underlying theory. In the case of Lie groups, one of the new structures that the Lie group admits is its Lie algebra. Since the superstructure was constructed naturally (i.e. without arbitrarily importing things that do not belong), it must in some way incorporate properties of the original structure. The hope is that the superstructure has captured fewer properties and is able to manifest them in stark relief in a way to give us a better understanding of the original structure. This does not always happen, but often it does. So the Lie algebra of a Lie group is not a "model" - it is a construct that incorporates some of the properties of the group. Nothing is being "modelled". Math does not model itself.

    The Lie algebra is essentially the set of infinitesimal transformations of a group. The reason it is introduced is because a large part, if not all, of the group can be reconstructed from the infinitesimal transformations by exponentiation. This point is too long to elaborate in this blog. You will need to consult a book on it. The idea is this: a rotation by angle theta can be constructed by stating twice by angle theta/2, or three rotations of angle theta/3, ... , or n rotations by angle theta/n. If n is sufficiently large, you can approximate the rotation matrix R(theta/n) using the small angle approximation. So
    [itex] R(\theta/n) \approx \left( \begin{array}{cc} 1 & -\theta \\ \theta & 1 \end{array} \right) = I + \theta J [/itex]
    so
    [itex] R(\theta) \approx \left( 1 + \frac{\theta}{n} J\right)^n [/itex]
    Since the matrices I and J commute (everything commutes with the identity), the behaviour of the RHS is precisely as it would be were I replaced by 1 and J replaced by x. The right hand side, in the limit as n goes to infinity, is thus the exponential function, and we can write
    [itex] R(\theta) =\lim_{n\rightarrow \infty} \left( 1 + \frac{\theta}{n} J\right)^n = e^{\theta J} [/itex]
    It is easy to show that J^2=-1, J^3 = -J, J^4 = +1, so writing the power series expansion for the exponential, using these values of the powers of J, collecting the coefficients of I in one bracket and those of J in another, we get
    [itex] R(\theta) = \cos \theta\ I + \sin\theta\ J [/itex]
    which is the original rotation matrix. We can therefore recover the entire group SO(2) by exponentiation of the matrix J. The entire group has been generated from a single matrix.

    The matrix J can clearly be obtained by differentiation of the element [itex] e^{\theta J} [/itex] with respect to theta. J is thus a in the Lie algebra of SO(2). In fact, J is the basis of the lie algebra (1-dim) in this simple case.

    These results generalise, mutatis mutandis. Exponentiation of the Lie algebra generates at least a portion of the group, if not all of it. In fact, the Lie algebra generates the Universal Covering Group. All other groups with the same Lie algebra can then be constructed from the UCG.

    All manifolds are equipped with vectors and covectors, and much more besides. They form an essential part of the machinery of manifolds. They are needed to analyse a manifold and its properties whenever a manifold appears in a theory, or in a model of the real world. They are to the manifold what vectors are to elementary mechanics. You cannot do without them. They are part and parcel of the structure.
     
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