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Representations of Groups - S3

  1. Feb 5, 2012 #1
    I am reading Dummit and Foote Ch 18, trying to understand the basics of Representation Theory.

    I need help with clarifying Example 3 on page 844 in the particular case of [itex] S_3 [/itex].

    (see the attahment and see page 844 - example 3)

    Giving the case for [itex]S_3 [/itex] in the example we have the following situation:

    Let [itex]G = S_3 [/itex] and let V be an 3-dimensional vector space over F with basis [itex] e_1 , e_2, e_3 [/itex].

    Let [itex] S_3 [/itex] act on V by defining for each [itex] \sigma \in S_3 [/itex] [tex] \sigma \circ e_i = e_{ \sigma (i)} , \ \ \ 1 \leq i \leq n [/tex]

    i.e [itex] \sigma [/itex] acts by permuting the subscripts of the basis elements


    My problem:

    I followed Example 3 page 844 - JUST!!!

    I am now trying without success to derive (or write it down anyway) the representation [itex] \phi : G \rightarrow GL(V) [/itex] for the above, and then write down the matrices [itex] \phi (g) [/itex] for every element [itex] \sigma \in S_3 [/itex]

    I would be very appreciative of some help.

    Peter
     

    Attached Files:

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  3. Feb 5, 2012 #2

    Deveno

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    i will explicitly describe φ by listing the matrices φ(g) for every g in S3:

    φ(e) = I, the 3x3 identity matrix.
    φ((1 2)) =

    [0 1 0]
    [1 0 0]
    [0 0 1].

    φ((1 3)) =

    [0 0 1]
    [0 1 0]
    [1 0 0].

    φ((2 3)) =

    [1 0 0]
    [0 0 1]
    [0 1 0].

    φ((1 2 3)) =

    [0 1 0]
    [0 0 1]
    [1 0 0].

    φ((1 3 2)) =

    [0 0 1]
    [1 0 0]
    [0 1 0].

    these matrices, unsurprisingly enough, are called permutation matrices.
     
  4. Feb 5, 2012 #3
    Thanks for the help

    Peter
     
  5. Feb 5, 2012 #4
    By the way - is there a neat 'formula' for the representation?

    Peter
     
  6. Feb 5, 2012 #5

    Deveno

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    yes, but you already posted it:

    [tex]\varphi(\sigma):e_i \to e_{\sigma(i)}[/tex]

    to expand on this, a linear map (including an element of GL(V)), is completely determined by its action upon a basis of V. so by specifying what φ(g) does to a basis, you've given the matrix φ(g), just write down the column φ(g)(ei) for each i in {1,2,...,n}, where {e1,e2,...,en} is your basis of V, and dim(V) = n.

    in this particular case, the basis given is the "standard" basis for F3, which is particularly nice, since it is orthonormal (with respect to the usual inner product), which makes φ into a unitary representation (provided F is a subfield of the complex numbers, which is usually the case).

    you'll see more of this: often matrices are easier to deal with than abstract representations, and there is an isomorphism between Mat(n,F) (nxn matrices in F) and linear mappings V→V, when dimF(V) = n. however, to establish this isomorphism, you need to pick a basis, and there are many ways to do this (that is, different bases for V yield different matrices for φ(g)). the matrices you're interested in, are the invertible matrices in Mat(n,F), which (in the same way as above) are in 1-1 correspondence with the elements of GL(V).
     
  7. Feb 6, 2012 #6
    Thanks - your expansion on the formula was most helpful

    I was intently focussed on finding [itex] \phi : G \rightarrow GL(V) [/itex] that I was looking for a function that took a group element to a linear transformation.

    However I can now see that the action of [itex] S_3 [/itex] on V is actually such a relationship if you realise that specifying what the action does to the basis then implicitly specifes what the action does to all elements of V.

    Further that the action actually specifies a function [itex] \phi [/itex] of the form [itex] \phi ( \sigma ) [/itex] [itex] : e_i \rightarrow e_{ \sigma (i)} [/itex] - this I did not see before! [so specifying an action also specifies a function?]

    By the way I was a bit spooked in reading D&F by the fact that the image of [itex] \phi [/itex] , namely, [itex] \phi ( \sigma ) [/itex] is used as a function - but then I gues that is because [itex] \phi [/itex] is a map from G to GL(V) where GL(V) is a group of linear transformations (functions) and so the image of [itex] \phi [/itex] is a function

    Yet another issue/question I have is the following:

    On page 842 of D&F - see attachement - we are told to make V into an FG-module by considering the action of a group ring element on an element of V.

    However when we come to Example (3) on page 844 (see attachement) we find D&F talking about the action of [itex] S_n [/itex] on V rather than [itex] F S_n [/itex] on V ... until we come to the specific example of 3-dimensional space where D&F resume talking about the group ring [itex] F S_3 [/itex]

    Can anyone clarify these issues for me - be grateful for some help!


    Peter
     
    Last edited: Feb 6, 2012
  8. Feb 6, 2012 #7

    Deveno

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    there's a parallel to groups acting on a set. you can think of it (the action) as a homomorphism from a group to "a general enveloping group", so the image of a group element is a permutation function (the GL(V) of the set world). but you can also think of it as a group element pushing the set elements around.

    with a representation, you can think of it as a homomorphism from a group to a "general enveloping group (of linear transformations)", so the image of a group element is an invertible linear mapping (aka matix under the right circumstances), or you can think of it as a group pushing vectors around: g.v = φ(g)(v).

    in both cases, you examine the behavior of the group, in terms of the behavior of the "target structure", which is often easier to work with. the same kinds of ideas can be used for associative algebras or lie algebras instead of groups, allowing for very complicated interactions to be turned into matrices (and, at some level, then plain old arithmetic).

    in physics (or so i'm told), the vector spaces used can be hilbert spaces, allowing analysis methods to be applied to algebraic problems (so apparently that calculus will get used for something after all).

    this is yet another way of looking at a representation: as an FG-module (some texts just use G-module, if the underlying field is understood).
    basically, you can turn G into a vector space by just declaring the elements of G to be a basis (and identifying the identity "e" in the group G with 1). but since G also has a multiplication defined on it, you can multiply the G-vectors, to make a ring (collecting like terms in much the same way as when you multiply polynomials). you can then (if you have a representation of G handy) use a representation of G in V to turn V into a G-module (FG-module), which is a structure like a vector space, but the scalar multiplication only uses ring elements, instead of field elements (F-modules are what vector spaces are).

    what you do is turn the product of (a linear combination of G-elements (over F) times a vector) into linear combinations of: the images of the matrices (or more generally linear transformations) that represent each G-element acting on the vector, v.

    but it goes both ways: if you start with a G-module, then since G is a subset of the ring FG, you already have a notion of what gv means, so you just use that to define your representation φ, φ is the representation that maps g to the linear map v-->gv.

    so we can use either module theory, or linear algebra, depending on what works for us. a G-module on V is essentially the same thing as a representation of G in V, in the sense that given one, we can form the other that corresponds to it.

    in example (3), D&F define the representation first (in action form), and then demonstrate what the FG-module looks like. let me re-write their equations like this:

    let r = 2(1 2) - 3(1 2 3) in FS3, and let v = (x,y,z)

    (which is actually xe1+ye2+ze3).

    then r.v = [2(1 2) - 3(1 2 3)].(x,y,z) = 2(1 2).(x,y,z) - 3(1 2 3).(x,y,z)

    = 2(y,x,z) - 3(z,x,y) = (2y-3z,-x,2z-3y)

    in other words twice the "coordinate scramble" induced by (1 2), minus thrice the "coordinate scramble" induced by (1 2 3).
     
    Last edited: Feb 6, 2012
  9. Feb 6, 2012 #8
    Thanks so much for that - going away to think on that!

    By the when I first looked at the matrices that you gave for [itex] \phi [/itex] I thought they were very intuitive

    For φ((1 2)) =

    [0 1 0]
    [1 0 0]
    [0 0 1].

    the rows for [itex] e_1 [/itex] and [itex] e_2 [/itex] in the identity matrix were swapped around - sounded what should happen (although I could not justify it.

    But then with

    φ((1 2 3)) =

    [0 1 0]
    [0 0 1]
    [1 0 0].

    I was expecting that the change from the identity matrix would be row 1 would go to row 2 and row 2 would go to row 3 and row 3 to row 1 but this does not appear to have happened - can you clarify?

    Peter
     
  10. Feb 6, 2012 #9

    Deveno

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    permuting basis vectors swaps columns, not rows.
     
  11. Feb 6, 2012 #10
    Oh .... OK that explains it!

    Must go back and check the example - I thought it checked out for what I thought was intuitive - D&F give a process for determining the matrices.

    Just another question I am still struggling with ...

    On page 842 of D&F - see attachement - we are told to make V into an FG-module by considering the action of a group ring element on an element of V.

    However when we come to Example (3) on page 844 (see attachement) we find D&F talking about the action of Sn on V rather than FSn on V ... until we come to the specific example of 3-dimensional space where D&F resume talking about the group ring FS3

    Are you able to clarify this

    Peter
     
  12. Feb 6, 2012 #11

    Deveno

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    i edited my post above to address some of this. but one thing you should understand is this: they are using the representation of S3 (the action on F3) in order to define the FS3-module. that is, unless you know what g.v is, for any g in G, how are you going to say what (a1g1+...+angn).v is?
     
  13. Feb 6, 2012 #12
    Thanks once again

    In this set of exchanges you have taken me on miles from where our conversation started

    Going to print out your comments and go through the whole conversation again

    Thanks again!

    Peter
     
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