We say that a module is “reducible” if it contains a nontrivial submodule. Thus our examples last time show that the left regular representation is always reducible, since it always contains a copy of the trivial representation as a nontrivial submodule. Notice that we have to be careful about what we mean by each use of “trivial” here.
If the 
-dimensional representation 
has a nontrivial 
-dimensional submodule 
— 
and 
— then we can pick a basis 
of 
. And then we know that we can extend this to a basis for all of : 
.
Now since is a 
-invariant subspace of , we find that for any vector 
and 
the image ](https://s-ssl.wordpress.com/latex.php?latex=%5Cleft%5B%5Crho%28g%29%5Cright%5D%28w%29&bg=e6e6e6&fg=333333&s=0 "\left[\rho(g)\right](w)")
is again a vector in , and can be written out in terms of the 
basis vectors. In particular, we find =\rho_j^iw^j](https://s-ssl.wordpress.com/latex.php?latex=%5Cleft%5B%5Crho%28g%29%5Cright%5D%28w%5Ei%29%3D%5Crho_j%5Eiw%5Ej&bg=e6e6e6&fg=333333&s=0 "\left[\rho(g)\right](w^i)=\rho_j^iw^j")
, and all the coefficients of 
through 
are zero. That is, the matrix of 
has the following form:
where 
is an 
matrix, 
is an 
matrix, and 
is an 
matrix. And, in fact, this same form holds for all 
. In fact, we can use the rule for block-multiplying matrices to find:
and we see that actually provides us with the matrix for the representation we get when restricting 
to the submodule . This shows us that the converse is also true: if we can find a basis for so that the matrix has the above form for every , then the subspace spanned by the first basis vectors is -invariant, and so it gives us a subrepresentation.
As an example, consider the defining representation of 
, which is a permutation representation arising from the action of on the set 
. This representation comes with the standard basis 
, and it’s easy to see that every permutation leaves the vector 
— along with the subspace that it spans — fixed. Thus carries a copy of the trivial representation as a submodule of . We can take the given vector as a basis and throw in two others to get a new basis for : 
.
Now we can take a permutation — say 
— and calculate its action in terms of the new basis:
&=\mathbf{1}+\mathbf{2}+\mathbf{3}\\\left[\rho((1\,2))\right](\mathbf{2})&=\mathbf{1}=(\mathbf{1}+\mathbf{2}+\mathbf{3})-\mathbf{2}-\mathbf{3}\\\left[\rho((1\,2))\right](\mathbf{3})&=\mathbf{3}\end{aligned}](https://s-ssl.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5B%5Crho%28%281%5C%2C2%29%29%5Cright%5D%28%5Cmathbf%7B1%7D%2B%5Cmathbf%7B2%7D%2B%5Cmathbf%7B3%7D%29%26%3D%5Cmathbf%7B1%7D%2B%5Cmathbf%7B2%7D%2B%5Cmathbf%7B3%7D%5C%5C%5Cleft%5B%5Crho%28%281%5C%2C2%29%29%5Cright%5D%28%5Cmathbf%7B2%7D%29%26%3D%5Cmathbf%7B1%7D%3D%28%5Cmathbf%7B1%7D%2B%5Cmathbf%7B2%7D%2B%5Cmathbf%7B3%7D%29-%5Cmathbf%7B2%7D-%5Cmathbf%7B3%7D%5C%5C%5Cleft%5B%5Crho%28%281%5C%2C2%29%29%5Cright%5D%28%5Cmathbf%7B3%7D%29%26%3D%5Cmathbf%7B3%7D%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}\left[\rho((1\,2))\right](\mathbf{1}+\mathbf{2}+\mathbf{3})&=\mathbf{1}+\mathbf{2}+\mathbf{3}\\\left[\rho((1\,2))\right](\mathbf{2})&=\mathbf{1}=(\mathbf{1}+\mathbf{2}+\mathbf{3})-\mathbf{2}-\mathbf{3}\\\left[\rho((1\,2))\right](\mathbf{3})&=\mathbf{3}\end{aligned}")
The others all work similarly. Then we can write these out as matrices:
Notice that these all have the required form:
Representations that are not reducible — those modules that have no nontrivial submodules — are called “irreducible representations”, or sometimes “irreps” for short. They’re also called “simple” modules, using the general term from category theory for an object with no nontrivial subobjects.
September 23, 2010
Posted by John Armstrong |
Algebra, Group theory, Representation Theory |
9 Comments
