Page 1 of 2 12 LastLast
Results 1 to 15 of 16
Like Tree9Thanks

Math Help - Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    I am seeking to understand Dummit and Foote's chapter on Representation Theory and Character Theory (see pages 840-843 attached).

    On Page 840 D&F write: (see attachment)

    Suppose we are given an FG-Module V.

    We obtain an associated vector space over F and a representation of G as follows:

    Since V is an FG-Module, it is an F-Module i.e. it is a vector space over F.

    Also for each g  \in G we obtain a map from V to V, denoted by  \phi (g)  , defined by

     \phi (g) (v) = g \cdot v for all  v \in V

    where g \cdot v is the given action of the ring element g on the element of V.

    Since the elements of F commute with each  g \in G it follows by the axioms of a module that for all v, w \in V and all  \alpha , \beta \in F we have:

     \phi (g) ( \alpha v + \beta w ) = g \cdot ( \alpha v + \beta w )

     = g \cdot ( \alpha v ) + g \cdot ( \beta w )

      = \alpha ( g \cdot v ) + \beta ( g \cdot w )

      = \alpha \phi (g) (v) + \beta \phi (g) (w)

    that is, for each  g \in G , \phi (g) is a linear transformation.

    ================================================

    My problem with the above text is

    1. I do not see why we can assume that " the elements of F commute with each  g \in G  "

    2 Hence I do not follow the assertion above that

    = g \cdot ( \alpha v ) + g \cdot ( \beta w ) = \alpha ( g \cdot v ) + \beta ( g \cdot w )

    Can someone please show me explicitly how this is the case?
    Would appreciate some help?

    Peter
    Last edited by Bernhard; March 12th 2012 at 03:30 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    an element of F in F[G] is in the form \alpha 1_G, \alpha \in F and an element of G in F[G] is in the form 1_Fg, g \in G. now, if you look at the definition at the first page of your link, you see that the multiplication in F[G] is defined by (\alpha g)(\beta h) = (\alpha \beta)gh, for all \alpha, \beta \in F and g, h \in G. if you put g = 1_G, \beta = 1_F, you'll get (\alpha 1_G)(1_F h) =  \alpha h = (1_F h)(\alpha 1_G). so every element of F commutes with every element of G. this also implies that every element of F commutes with every element of F[G] because F is commutative.
    Thanks from Bernhard and mash
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    Thanks

    Just working through your post

    Peter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    Thanks fo your help.

    I can see that  ( \alpha 1_G ) ( 1_F h ) = \alpha h

    But then you claim that

     \alpha h  = ( 1_F h ) ( \alpha 1_G )

    it appears that you are assuming   \alpha h = h \alpha

    Is that correct?

    If so why is   \alpha h = h \alpha ?

    Can you help? I am obviously missing something.

    Peter
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    Quote Originally Posted by Bernhard View Post
    Thanks fo your help.

    I can see that  ( \alpha 1_G ) ( 1_F h ) = \alpha h

    But then you claim that

     \alpha h  = ( 1_F h ) ( \alpha 1_G )

    it appears that you are assuming   \alpha h = h \alpha

    Is that correct?

    If so why is   \alpha h = h \alpha ?

    Can you help? I am obviously missing something.

    Peter
    ( 1_F h ) ( \alpha 1_G ) =(1_F \alpha)(h1_G)=\alpha h.
    Thanks from mash and Bernhard
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    a different way to look at it has to do with how F[G] is created (I will assume G finite for this post):

    what we do is start with a group G = \{e,g_1,g_2,.....g_n\}.

    to make this set into a vector space, we have two approaches, both of which are "the same" (but look different):

    1) consider all "formal linear combinations" of elements of G over F.

    2) consider all mappings f:G-->F. in this view of things, we can use the addition and multiplication in F to define a vector sum and scalar multiplication on the function space.

    either way, the end result is that we simply declare the elements of G to be a basis of a vector space over F (in view #2, the basis element g_i is the map that sends g_i to 1, and all other elements of G to 0).

    but this construction takes no advantage of the group structure of G (indeed, we could just have started with a set X, and still obtained a vector space). well, if we want a ring structure consistent with our vector space (that is, we desire an algebra) then we want the distributive laws to hold. and for these distributive laws to hold, we need "scalars" (the elements of F) to commute with "basis vectors" (the elements of G). this is what motivates the definition of the product of terms:

    (\alpha g)(\beta h) = (\alpha\beta)(gh),

    which we then extend by linearity to linear combinations ("collecting like terms" as we go).

    now, these same considerations hold if R is merely a ring, in which case we get a (left) R-module, instead of a vector space. in particular, one can take R = Z, the ring of integers, and one can interpret a linear combination like:

    3g + 4h + 5k (where g,h,k are elements of G)

    to mean: we have 3 g's, 4h's and 5 k's. this is not that different than the way we think of polynomials: each power of x is regarded as "having it's own place"....if one allows negative powers of x to be included as "allowable" powers (i think that the technical term is Laurent polynomial), then the Laurent polynomials with integer coefficients are just the group ring \mathbb{Z}[\mathbb{Z}].
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    I am still struggling with the proof that  \alpha h  = h \alpha (see above posts by NonCommAlg)

    You advise me that  ( 1_F h ) ( \alpha 1_G )  = ( 1_F \alpha ) ( h 1_G ) = \alpha h ............. (1)

    I can follow (1) as it stands.

    But how to explicitly show that  \alpha h  = h \alpha ???

    I can see that if we assume that

    1_Fh = h = h and  \alpha 1_G = \alpha that we get  ( 1_F h ) ( \alpha 1_G ) = h \alpha

    ie the extreme LHS of (1) above is  h \alpha and so we get the result we want i.e.  \alpha h = h \alpha

    But here I need some advice ... is it legitimate to assume that 1_Fh = h given that this is the identity element of a field multiplied by a group element equalling a group element - so why is this legitimate?

    Similarly is it legitimate to assume that  \alpha 1_G = \alpha given that this is a field element multiplied by the identity element from a group equalling a field element.

    Can someone clarify this for me please.

    Peter
    Last edited by Bernhard; March 14th 2012 at 01:55 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    perhaps you should think of it this way:

    F[G] doesn't really contain F OR G, instead we have the two isomorphisms:

    \alpha \to \alpha 1_G
    g \to 1_Fg

    the former is a field isomorphism, the latter is a group isomorphism. these are embeddings of F and G into F[G]. as is customary with embeddings, the identity elements are often omitted in the notation: for example, we do not write the rational number 2 as 2/1, nor do we write the negative integer -k as 0-k. identity elements are "transparent".

    to give another example, in matrices, say Mat(n,F) we have the isomorphism:

    \lambda \to \lambda I

    and what we MEAN when we write Aλ (note the scalar is on the right) is actually A(λI), where λI is the diagonal matrix with all λ's on the diagonal. but λI is in the center of the algebra Mat(n,F), so Aλ = A(λI) = (λI)A = λ(IA) = λA.

    in this same way, in F[G], what we mean by an element h\alpha is the group ring element (1_Fh)(\alpha 1_G). but the group identity commutes with every element of the group (including h), and all field elements commute with each other. so after we "clump" the field elements together, and the group elements together (which is how we define the multiplication in F[G]), we can switch the order of h and 1_G and 1_F and \alpha.

    in other words, we build "linearity" into our definition of the multiplication of F[G].

    perhaps we should back up a bit. how do we define g.v? we have a homomorphism φ from G to Aut(V). this is in direct analogy with a group acting on a set, where we have a homomorphism from G to Aut(X) = Sym(X) (automorphisms of a SET are called bijections, or permutations). it's just here, V has "extra structure" that of a vector space over F. but the autmorphisms of a vector space are invertible linear mappings (that is Aut(V) = GL(V)). so g.v isn't really "g times v" it's the homomorphic image of g (which can be thought of as a matrix) mapping v to φ(g)(v). and matrices (or linear mappings) are linear: T(av) = aT(v) (now just substitute φ(g) for "T").
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    Thanks for that - that clarifies things a lot!

    I am still struggling with the formal and explicit proof of the following:

    Given what you have said how would you prove (explicitly and formally) that  (1_F h ) ( \alpha 1_G ) = ( \alpha 1_G ) (1_F h )   ?

    {I know I am being a bit slow with this :-) }

    Peter
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    by definition (of the multiplication in F[G]):

    (αg)(βh) = (αβ)(gh) = (βα)(gh).

    if g and h commute, then this equals:

    (βα)(hg).

    and the identity of G commutes with ANY h.
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    Thanks to Deveno and NonCommAlg for the helpful posts above, but although much clearer on the above I still have a problem

    I now am comfortable that the elements of F commute with each  g \in G .

    But when I read the following text in Dummit and Foote page 843 I still have some concerns.

    The text I am referring to reads as follows - see attachment page 843 for details

      \phi ( g ) ( \alpha v + \beta w ) = g \cdot ( \alpha v + \beta w )

     = g \cdot ( \alpha v ) + g \cdot ( \beta w )

      = \alpha ( g \cdot v ) + \beta ( g \cdot w )

     = \alpha \phi ( g ) ( v ) + \beta \phi ( g ) (w)


    Now my problem with the above concerns   g \cdot ( \alpha v ) + g \cdot ( \beta w )   = \alpha ( g \cdot v ) + \beta ( g \cdot w )

    This looks like it is just using the fact that elements of F commute with elements of g as in  g \cdot ( \alpha v ) = \alpha ( g \cdot v )

    BUT ... this is not just an element of F commuting with an element of G as in  (1_F h ) ( \alpha 1_G) = ( \alpha  1_G ) ( 1_F h ) ......

    the statement above involves the  \cdot operation which is (to quote D&F) " the given action of the ring element g on the element v of V" { why "ring" element? }

    Doesn't the fact that we are dealing with an action mixed with terms like   \alpha v involving a field element multiplied by a vector complicate things ...

    how do we formally and explicitly justify  g \cdot  ( \alpha v )  =  \alpha ( g \cdot v )?

    How do we justify taking  \alpha  out through the the action  \cdot ? Why are we justified in doing this?


    Peter
    Last edited by Bernhard; March 16th 2012 at 02:50 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    because g.v is defined to be the image of the (invertible) linear map φ(g) on v:

    so g.(av) = φ(g)(av) = a(φ(g)(v)) = a(g.v)

    recall, that in a vector space, av is just another vector, since scalar multiplication is a map: F \times V \to V.

    perhaps a worked out example will help you see how this works.

    let G = {e,a,b,ab}, where a^2 = b^2 = e, ab = ba, the klein 4-group.

    let V = \mathbb{R}^2, the euclidean plane.

    we will define the homomorphism (which in this case is an isomorphism) φ as follows:

    \varphi(a) = A = \begin{bmatrix}1&0\\0&-1 \end{bmatrix}
    \varphi(b) = B = \begin{bmatrix}-1&0\\0&1 \end{bmatrix}

    it should be clear that A^2 = B^2  = I, AB = BA = -I.

    let's look at what each element of φ(G) does to a typical vector (x,y):

    I(x,y) = (x,y)
    A(x,y) = (x,-y)
    B(x,y) = (-x,y)
    AB(x,y) = (-x,-y).

    now, we define g.v to be φ(g)(v), so:

    e.(x,y) = I(x,y) = (x,y)
    a.(x,y) = A(x,y) = (x,-y)
    b.(x,y) = B(x,y) = (-x,y)
    ab.(x,y) = AB(x,y) = (-x,-y)


    the R-linearity of these maps is clear. for example:

    a.((x,y)+(x',y')) = a.(x+x',y+y') = (x+x',-(y+y')) = (x,-y) + (x',-y') = a.(x,y) + a.(x',y'), and for a real number r:

    a.(r(x,y)) = a.(rx,ry) = (rx,-ry) = r(x,-y) = r(a.(x,y))
    Last edited by Deveno; March 16th 2012 at 11:36 AM.
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    Thanks so much. Very clear ... most helpful to have an example

    Thanks to your help I can now go on in representation theory with more confidence!


    Peter
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    Just been reflecting on your post and re-reading D&F - page 843 in particular (see attached) and now have a further problem

    My problem refers to your post and D&F s approach to representations through defining an FG-module - an approach in which D&F define  \phi as a "map from V to V" [not a linear map mind - they take 4 lines of a proof on page 843 to show that  \phi is a LINEAR map or transformation.] So my question is how do we show  g \cdot ( \alpha v) = \alpha ( g \cdot v ) if we cannot assume  \phi  is linear? [Note that D&F seem to assume that  \phi is linear in their proof of same on page 843.

    Can you please help clarify this? I must say I cannot see how to do this!



    To help I will now include some of the relevant text from D&F papge 843 - (see attachment) - as follows:

    ================================================== ========================

    Suppose now that conversely we are given an FG-Module V. We obtain an associated vector space over F and representation of G as follows. Since V is an FG-module, it is an F-Module, i.e. it is a vector space over F. Also for each  g \in G we obtain a map from V to V, denoted by  \phi (g)  , defined by

     \phi (g) (v) = g \cdot v for all  v \in V

    where   g \cdot v is the given action of the ring element g on the element v of V.

    Since the elements of F commute with each  g \in G it follows by the axioms for a module that for all  v,w \in V and all  \alpha , \beta \in F we have

     \phi (g) ( \alpha v + \beta w ) = g \cdot ( \alpha v + \beta w )

     = g \cdot ( \alpha v ) + g \cdot ( \beta w )

     = \alpha ( g \cdot v ) + \beta ( g \cdot w )

      = \alpha \phi (g) (v) + \beta \phi (g) (w)


    That is for each  g \in G  \ ,  \ \phi (g)  is a linear transformation.

    ================================================== ==================

    My question, thus, relates to justifying the following (without assuming linearity of  \phi )


     g \cdot ( \alpha v ) + g \cdot ( \beta w )  = \alpha ( g \cdot v ) + \beta ( g \cdot w )

    I suspect that D&F are using the fact that "elements of F commute with each  g \in G " but how, explicitly and formally do you justify this when we are not talking about an operation between elements of F and G, but are dealing with an action of a ring element on an element of a vector space.


    Peter
    Attached Files Attached Files
    Last edited by Bernhard; March 16th 2012 at 06:01 PM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: Representation Theory - Ch18 - Dummit and Foote - FG-Modules and Representations

    ok, let's approach it from the OTHER direction...we are given that we have an F[G]-module. first, let's show that this F[G]-module is also an F-module, that is: a vector space.

    since we have a module, let's call it V, we know automatically that (V,+) is an abelian group (since any module has this property). so vector addition is taken care of, we can just use the module addition. so all we need to do is show that we can define a "scalar multiplication" that is: define αv for v in V, and α in F. the way we will do this is to define:

    \alpha v = (\alpha 1_G) \cdot v,

    where \alpha 1_G is an element of F[G], and \cdot is the given F[G]-action (our ring being F[G]) on the abelian group V.

    the "distributive laws" and "associative laws" for a vector space's scalar multiplication, namely:

    (\alpha + \beta)v = \alpha v + \beta v
    \alpha(u + v) = \alpha u + \alpha v
    \alpha(\beta v) = (\alpha\beta)v

    are straight-forward consequences of the same laws that hold (for modules) when considering the elements of F as identified with their isomorphic images in F[G].

    so V is an F-vector space. again, identifying g in G with the element 1_Fg in F[G], we let g.v be the element of V, we get when multiplying v by the "ring scalar" g. this determines a mapping \varphi_g : V \to V given by \varphi_g(v) = g\cdot v.

    now, let's show that this mapping is F-linear. we can do this in one step or two, i'll do it in two steps, for clarity's sake (it might help to think of this mapping as g.__).

    first, we need to show that \varphi_g(\alpha v) = \alpha(\varphi_g(v)).

    \varphi_g(\alpha v) = g \cdot (\alpha v) = (1_Fg) \cdot (\alpha v) = (1_Fg) \cdot ((\alpha 1_G) \cdot v)
     = ((1_Fg)(\alpha 1_G)) \cdot v = ((1_F\alpha)(g1_G)) \cdot v = (\alpha g) \cdot v = \alpha \cdot (g \cdot v)
     = \alpha(g \cdot v) = \alpha(\varphi_g(v))

    next, we need to show that \varphi_g is an additive homomorphism of V:

    \varphi_g(u+v) = g \cdot (u+v) = g \cdot u + g \cdot v = \varphi_g(u) + \varphi_g(v)

    in the above we have used the MODULE axioms concerning g \cdot v several places, " \cdot" is the module "scalar multiplication" (i.e., ring action).

    so \varphi_g is a linear map, for each g in G. now we'll come full circle: we will exhibit a homomorphism between G and GL(V). in what will seem an obvious ploy, the homomorphism is this:

    \varphi : G \to GL(V), \varphi(g) = \varphi_g

    so we need to show 2 things again:

    1. \varphi(gh) = \varphi(g) \circ \varphi(h)
    2. \varphi(g) \in GL(V)

    so first things first:

    \varphi(gh) = \varphi_{gh}, to prove this is equal to  \varphi(g) \circ \varphi(h), we will evaluate it at all (any) v in V:

    \varphi_{gh}(v) = (gh) \cdot v = g \cdot (h \cdot v) = g \cdot (\varphi_h(v)) = \varphi_g(\varphi_h(v))
     = (\varphi_g \circ \varphi_h)(v).

    therefore, we have:

    \varphi(gh) = \varphi_{gh} = \varphi_g \circ \varphi_h = \varphi(g) \circ \varphi(h)

    so \varphi is a group homomorphism. next we need to show that each \varphi(g) is invertible (we showed the linear part above). recall that a function (and a group homomorphism is also a lowly function) is invertible (bijective) if and only if it posseses a (two-sided) inverse. what is the logical candidate for an inverse to \varphi(g)? why not try \varphi(g^{-1})? so let's see:

    \varphi(g) \circ \varphi(g^{-1}) = \varphi(gg^{-1}) = \varphi(1_G) = \text{id}_V since homomorphisms take identities to identities.
    \varphi(g^{-1}) \circ \varphi(g) = \varphi(g^{-1}g) = \varphi(1_G) = \text{id}_V

    so indeed \varphi(g^{-1}) = (\varphi(g))^{-1}, as desired. thus we can look at a representation 2 ways:

    1. a 4-tuple (F,G,V,φ) where φ:G→GL(V) is a group homomorphism, and V is a vector space over F (here φ is given "up-front").
    2. an F[G]-module V, which then DEFINES a homomorphism φ:G→GL(V) (by considering how "G-elements of F[G]" act on V as an F-module).

    it's really two sides of the same coin.
    Last edited by Deveno; March 16th 2012 at 07:29 PM.
    Thanks from Bernhard
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Orbits - Dummit and Foote and Fraleigh
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: December 24th 2011, 01:20 PM
  2. General Linear Group - Problem from Dummit and Foote
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 11th 2011, 09:07 AM
  3. Representation Theory
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 7th 2011, 12:55 PM
  4. Representation Theory
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 27th 2009, 08:01 AM
  5. Representation Theory problems -->
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: May 4th 2009, 04:33 AM

Search Tags


/mathhelpforum @mathhelpforum