Thread: 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 G we obtain a map from V to V, denoted by , defined by

for all

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

Since the elements of F commute with each it follows by the axioms of a module that for all and all we have:









that is, for each 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 "

2 Hence I do not follow the assertion above that



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

Peter

an element of in is in the form and an element of in is in the form . now, if you look at the definition at the first page of your link, you see that the multiplication in is defined by for all and . if you put you'll get so every element of commutes with every element of this also implies that every element of commutes with every element of because is commutative.

Thanks

Just working through your post

Peter

Thanks fo your help.

I can see that

But then you claim that



it appears that you are assuming

Is that correct?

If so why is ?

Can you help? I am obviously missing something.

Peter

Originally Posted by Bernhard

Thanks fo your help.

I can see that

But then you claim that



it appears that you are assuming

Is that correct?

If so why is ?

Can you help? I am obviously missing something.

Peter

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 .

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 is the map that sends 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:

,

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 .

I am still struggling with the proof that (see above posts by NonCommAlg)

You advise me that ............. (1)

I can follow (1) as it stands.

But how to explicitly show that ???

I can see that if we assume that

and that we get

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

But here I need some advice ... is it legitimate to assume that 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 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

perhaps you should think of it this way:

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




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:



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 is the group ring element . 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 and and .

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 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 ?

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

Peter

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 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 .

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










Now my problem with the above concerns

This looks like it is just using the fact that elements of F commute with elements of g as in

BUT ... this is not just an element of F commuting with an element of G as in ......

the statement above involves the 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 involving a field element multiplied by a vector complicate things ...

how do we formally and explicitly justify ?

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


Peter

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: .

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

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

let , the euclidean plane.

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




it should be clear that .

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))

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

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 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 is a LINEAR map or transformation.] So my question is how do we show if we cannot assume is linear? [Note that D&F seem to assume that 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 we obtain a map from V to V, denoted by , defined by

for all

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

Since the elements of F commute with each it follows by the axioms for a module that for all and all we have










That is for each is a linear transformation.

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

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




I suspect that D&F are using the fact that "elements of F commute with each " 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

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:

,

where is an element of F[G], and 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:





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 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 given by .

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 .





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



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

so 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:



so we need to show 2 things again:

1.
2.

so first things first:

, to prove this is equal to , we will evaluate it at all (any) v in V:


.

therefore, we have:



so is a group homomorphism. next we need to show that each 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 ? why not try ? so let's see:

since homomorphisms take identities to identities.


so indeed , 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.