Representation Theory of Finite Groups - CH 18 Dummit and Foote

[Math Amateur]

I am reading Dummit and Foote on Representation Theory CH 18

I am struggling with the following text on page 843 - see attachment and need some help.

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

 

[lavinia]

The group acts on the vector space by linear transformations. So the formulas just express linearity.

This just says that the vector space is a module over the group ring.

BTW:I do not see why these ideas require a representation of the group. Why can't the linear transformations be singular?

For instance take any linear transformation of a vector space. Then the vector space is a module of the group ring using exactly the same rules. In fact, in this case the vector space becomes a module over the ring of polynomials in one indeterminate.

 

[Math Amateur]

Are you saying that if I go to the axioms of a vector space over a group ring (not sure where these are actually explicitly specified [I like to be very sure what I am doing is legitimate :-) ] then I will find justification in the axions

By the way it would help me a lot if you could explicitly justify g \cdot ( \alpha v ) = \alpha ( g \cdot v ) - perhaps explictly referring to the appropriate axioms of a module over a group ring.

Thanks for your help so far by the way!

Peter

 

[lavinia]

Are you saying that if I go to the axioms of a vector space over a group ring (not sure where these are actually explicitly specified [I like to be very sure what I am doing is legitimate :-) ] then I will find justification in the axions

By the way it would help me a lot if you could explicitly justify g \cdot ( \alpha v ) = \alpha ( g \cdot v ) - perhaps explictly referring to the appropriate axioms of a module over a group ring.

Thanks for your help so far by the way!

Peter

Certainly the relations you are asking about follow from linearity of the action of G on the vector space.

But the scalars can be thought of as elements of the group ring. Just multiply the scalar times the identity element of the group. Then linearity follows from the associativity of the action.

 

[Math Amateur]

Thanks

Will reflect on your advice ... in particular will re-read D&F on group rings and modules

Peter

 

[lavinia]

Thanks

Will reflect on your advice ... in particular will re-read D&F on group rings and modules

Peter

BTW: The idea of a group ring does not require a field. One can use any commuataive ring as scalars. A field is used in your case because the subject is representations of the group on a vector space.

Suppose though that you wanted to study representations of a group on a lattice such as the points in Euclidean space that have all integer coordinates. Then your scalars would have to be restricted to the integers. This integer group ring is also used to define the cohomology of a group.

 

[Math Amateur]

Hi Lavinia

You write "The idea of a group ring does not require a field." ... yes understand ..

Yes, follow the next point ... appreciate your help

Peter

 

[Math Amateur]

Just to be sure ...

Are you saying that g \cdot ( \alpha v ) = \alpha ( g \cdot v ) follows from the axioms of a module over a group ring?

Peter

 

[lavinia]

Just to be sure ...

Are you saying that g \cdot ( \alpha v ) = \alpha ( g \cdot v ) follows from the axioms of a module over a group ring?

Peter

I think so because \alpha = \alpha. [identity]. So the associative axiom of the group ring allows you to factor \alpha out. Then use that the identity symbol in the group ring commutes with the other symbols.

In terms of the representations, you can think of scalar multiplication as multiplication by the scalar times the identity transformation.