18 Jun A Short Note on Certain Vector Spaces Associated with Finite Groups
Throughout this article is a finite group.
- Free vector space generated by : Given a field we take the set of all formal linear combinations of form where . We define addition on the set as
and scalar multiplication as
for all . We take to be the zero in the set. This set now forms a vector space called the free vector space generated by over . Clearly, the dimension of this vector space is equal to the order of the .
- Permutation Representation : Let be a set with a finite number of elements. Let act on . To this set we can associate a vector space spanned by a basis, where each basis element corresponds to a unique element in . Hence we can index the basis elements with the elements of , i.e. let the basis be . We can now define a -action on this basis and extend this action by -linearity to the entire vector space. The action is
We call this vector space a permutation representation of .
- Class Function : A function , where is a field said to be a class function if
for all .
The set of all class functions for the given group forms a vector space over . What is the dimension of this vector space? We observe that every class function is invariant on the conjugacy classes of . Hence an easy guess would be that the dimension of this vector space is equal to the number of conjugacy classes of . After one makes this guess it is very easy to see why it is so by producing a basis for .
- One can notice that we haven’t used the group structure in 1. Hence, we can extend this definition to any set with a finite cardinality. Infact we can further extend this definition to any set Y where we define formal linear combinations to be where atmost finitely many of the ‘s can be non-zero.
- In 1. we can generalize the construction to make a module over a ring in a similar way. We can infact make a ring by defining the multiplication in a way similar to the multiplication of polynomials. We call the group so formed as the group ring or the group algebra.
- In 3. if the underlying field is algebraically closed and its characteristic doesn’t divide the order of (this is not the strongest condition though) then the characters of all irreducible representations of (all of which are class functions) form an orthogonal basis for the vector space with respect to the following inner product :
where and are class functions.
I am a PhD student at University College London. I am also an editor of the English section.