Results 1 to 4 of 4

Thread: Modules over group algebra

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    11

    Modules over group algebra

    I'm having a little trouble understanding (intuitively or otherwise) what is meant by a module over a group algebra FG (F a field, G a finite group) and how it relates to transformations.

    For example, how do I go about thinking about this question:

    Identify $\displaystyle S_3$ with the subgroup of $\displaystyle S_4$ which fixes 4 ($\displaystyle S_n$ is of course the symmetric group on $\displaystyle n$ letters). Then $\displaystyle S_3$ acts on $\displaystyle S_4$ by conjugation (???). Find the character of the linearization $\displaystyle \mathbb{C} S_4$ as a $\displaystyle \mathbb{C} S_3$-module, and determine the multiplicities of each simple $\displaystyle \mathbb{C} S_3$-module in $\displaystyle \mathbb{C} S_4$.

    I would be very very grateful for any light anyone could shed on this question; especially any general comments about FG-modules and representations.

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by AtticusRyan View Post
    I'm having a little trouble understanding (intuitively or otherwise) what is meant by a module over a group algebra FG (F a field, G a finite group) and how it relates to transformations.

    For example, how do I go about thinking about this question:

    Identify $\displaystyle S_3$ with the subgroup of $\displaystyle S_4$ which fixes 4 ($\displaystyle S_n$ is of course the symmetric group on $\displaystyle n$ letters). Then $\displaystyle S_3$ acts on $\displaystyle S_4$ by conjugation (???). Find the character of the linearization $\displaystyle \mathbb{C} S_4$ as a $\displaystyle \mathbb{C} S_3$-module, and determine the multiplicities of each simple $\displaystyle \mathbb{C} S_3$-module in $\displaystyle \mathbb{C} S_4$.

    I would be very very grateful for any light anyone could shed on this question; especially any general comments about FG-modules and representations.

    Thanks in advance!
    to get you started:

    well, a (finite dimensional) vector space $\displaystyle V$ over a field $\displaystyle F$ is called an $\displaystyle FG$ module ($\displaystyle G$ is a group usually finite) or a $\displaystyle G$ module if there exists a homomorphism $\displaystyle \rho: G \longrightarrow \text{GL}(V).$

    in your problem: $\displaystyle F=\mathbb{C}, \ G=S_3, \ V=\mathbb{C}S_4.$ the homomorphism $\displaystyle \rho$ is defined by $\displaystyle \rho(\sigma)(\alpha)=\sigma \alpha \sigma^{-1},$ for all $\displaystyle \sigma \in S_3, \ \alpha \in \mathbb{C}S_4.$ since the value of $\displaystyle \chi,$ the character corresponding

    to the representation $\displaystyle \rho,$ over a conjugacy class is constant, you only need to find $\displaystyle \chi(\sigma_j)=\text{Tr}(\rho(\sigma_j)), \ j=1,2,3,$ where $\displaystyle \sigma_1=1_{S_3}, \ \sigma_2=(1 \ 2), \ \sigma_3=(1 \ 2 \ 3).$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    11
    I think I understand; could you perhaps help me find the matrix $\displaystyle \rho(1_{S_3})$? In particular what is our basis of $\displaystyle V$? This is the kind of stuff I don't fully understand how to do.

    Obviously $\displaystyle \chi(1_{S_3})=\dim(V)$ but what is the dimension of $\displaystyle V$ as a $\displaystyle \mathbb{C}S_3$-module?
    Last edited by AtticusRyan; Oct 11th 2009 at 01:36 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by AtticusRyan View Post
    Identify $\displaystyle S_3$ with the subgroup of $\displaystyle S_4$ which fixes 4 ($\displaystyle S_n$ is of course the symmetric group on $\displaystyle n$ letters). Then $\displaystyle S_3$ acts on $\displaystyle S_4$ by conjugation (???). Find the character of the linearization $\displaystyle \mathbb{C} S_4$ as a $\displaystyle \mathbb{C} S_3$-module, and determine the multiplicities of each simple $\displaystyle \mathbb{C} S_3$-module in $\displaystyle \mathbb{C} S_4$.
    This is my attempt. Hope it helps.

    Lemma 1. Let N be a normal subgroup of G and let U be a $\displaystyle \mathbb{C}(G/N)$-module. Then U admits a canonical $\displaystyle \mathbb{C}G$-module structure, with a subspace of U being a $\displaystyle \mathbb{C}G$-submodule iff it is a $\displaystyle \mathbb{C}(G/N)$-submodule. If $\displaystyle \psi$ is the character of the $\displaystyle \mathbb{C}(G/N)$-module U, then the character of the $\displaystyle \mathbb{C}G$-module U is $\displaystyle \psi \cdot \mu$, where $\displaystyle \mu:G \rightarrow G/N$ is the natural map.

    Lemma 1 says that given $\displaystyle g \in G$, $\displaystyle u \in U$ and $\displaystyle N \triangleleft G$, we can define gu = (gN)u such that the linear transformation of U induced by g under the action of G on U is the same as that induced by gN under the action of G/N on U. This gives U a $\displaystyle \mathbb{C}G$-module structure where $\displaystyle \mathbb{C}G$-submodule of U are exactly the $\displaystyle \mathbb{C}(G/N)$-submodule of U.

    Let N={1, (1 2)(3 4), (13)(24), (14)(23)}. Then $\displaystyle S_4/N \cong S_3$, where $\displaystyle N \triangleleft S_4$. Now we lift the characters of $\displaystyle S_3$ to the characters of $\displaystyle S_4$.

    The character table of $\displaystyle S_3 \cong S_4/N$ is

    $\displaystyle \begin{matrix}
    & N & (12)N & (123)N\\
    \phi^1 (trivial) & 1 & 1 & 1\\
    \phi^2 (sign) & 1 & -1 & 1\\
    \phi^3 (permutation) & 2 &0 & -1
    \end{matrix}$

    Since $\displaystyle S_3$ have three conjugacy classes, there are three irreducible characters in $\displaystyle S_3$. Note that characters are constant in conjugacy classes.

    Now we use the lifting property and find the character table of $\displaystyle S_4$. The character table of $\displaystyle S_4$ is

    $\displaystyle \begin{matrix}
    & 1 & (12) & (123) & (1234) & (12)(34)\\
    \chi^1 (trivial) & 1 & 1 & 1 & 1 & 1\\
    \chi^2 (sign) & 1 & -1 & 1 & -1 & 1\\
    \chi^3 (permutation) & 2 &0 & -1 & 0 & 2\\
    \cdots & \cdots& \cdots& \cdots& \cdots& \cdots\\
    \end{matrix}$

    The fourth and fifth column is simply acquired by using the lifting property that $\displaystyle \chi^i(1234) = \phi^i((12)N)$ and $\displaystyle \chi^i((12)(34)) = \phi^i(N)$.

    While $\displaystyle S_3$ has three conjugacy classes, $\displaystyle S_4$ has five conjugacy classes. This implies that $\displaystyle S_4$ has five irreducible characters and two of them cannot be lifted from $\displaystyle S_3$. To get $\displaystyle \chi^4$, suppose $\displaystyle S_4$ act transitively on the set X ={1, 2, 3, 4}; let $\displaystyle \pi$ be the character of the $\displaystyle \mathbb{C}S_4$-module $\displaystyle \mathbb{C}X$. $\displaystyle \pi(g)$ equals the number of fixed points under the action of g on X, and we have $\displaystyle \chi^4 = \pi - \chi^1$ (I'll leave it to you to verify that $\displaystyle \chi^4$ is irreducible). Using the fact that $\displaystyle \chi^2\chi^4$ is irreducible, the last irreducible character is $\displaystyle \chi^5 = \chi^2\chi^4$.

    Anyhow, I was not able to figure out the multiplicity of $\displaystyle \mathbb{C}S_3$-module in $\displaystyle \mathbb{C}S_4$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. why is Lie algebra better to work with than with Lie group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 4th 2011, 12:37 AM
  2. Replies: 2
    Last Post: Feb 16th 2011, 06:41 PM
  3. abstract algebra (group)
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Feb 10th 2010, 10:07 AM
  4. Abstract Algebra- order of a group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jul 1st 2009, 06:23 PM
  5. Algebra, group action
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 2nd 2006, 03:19 PM

Search Tags


/mathhelpforum @mathhelpforum