Math Amateur said:
Thanks - your expansion on the formula was most helpful
I was intently focussed on finding \phi : G \rightarrow GL(V) 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 S_3 on V is actually such a relationship if you realize 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 \phi of the form \phi ( \sigma ) : e_i \rightarrow e_{ \sigma (i)} - 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 \phi , namely, \phi ( \sigma ) is used as a function - but then I gues that is because \phi is a map from G to GL(V) where GL(V) is a group of linear transformations (functions) and so the image of \phi is a function
Peter
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).
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 S_n on V rather than F S_n on V ... until we come to the specific example of 3-dimensional space where D&F resume talking about the group ring F S_3
Can anyone clarify these issues for me - be grateful for some help!
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 FS
3, and let v = (x,y,z)
(which is actually xe
1+ye
2+ze
3).
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).