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Math Help - Modules over group algebra

  1. #1
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    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 S_3 with the subgroup of S_4 which fixes 4 ( S_n is of course the symmetric group on n letters). Then S_3 acts on S_4 by conjugation (???). Find the character of the linearization \mathbb{C} S_4 as a \mathbb{C} S_3-module, and determine the multiplicities of each simple \mathbb{C} S_3-module in \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!
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  2. #2
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    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 S_3 with the subgroup of S_4 which fixes 4 ( S_n is of course the symmetric group on n letters). Then S_3 acts on S_4 by conjugation (???). Find the character of the linearization \mathbb{C} S_4 as a \mathbb{C} S_3-module, and determine the multiplicities of each simple \mathbb{C} S_3-module in \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 V over a field F is called an FG module ( G is a group usually finite) or a G module if there exists a homomorphism \rho: G \longrightarrow \text{GL}(V).

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

    to the representation \rho, over a conjugacy class is constant, you only need to find \chi(\sigma_j)=\text{Tr}(\rho(\sigma_j)), \ j=1,2,3, where \sigma_1=1_{S_3}, \ \sigma_2=(1 \ 2), \ \sigma_3=(1 \ 2 \ 3).
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  3. #3
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    I think I understand; could you perhaps help me find the matrix \rho(1_{S_3})? In particular what is our basis of V? This is the kind of stuff I don't fully understand how to do.

    Obviously \chi(1_{S_3})=\dim(V) but what is the dimension of V as a \mathbb{C}S_3-module?
    Last edited by AtticusRyan; October 11th 2009 at 01:36 PM.
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  4. #4
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    Quote Originally Posted by AtticusRyan View Post
    Identify S_3 with the subgroup of S_4 which fixes 4 ( S_n is of course the symmetric group on n letters). Then S_3 acts on S_4 by conjugation (???). Find the character of the linearization \mathbb{C} S_4 as a \mathbb{C} S_3-module, and determine the multiplicities of each simple \mathbb{C} S_3-module in \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 \mathbb{C}(G/N)-module. Then U admits a canonical \mathbb{C}G-module structure, with a subspace of U being a \mathbb{C}G-submodule iff it is a \mathbb{C}(G/N)-submodule. If \psi is the character of the \mathbb{C}(G/N)-module U, then the character of the \mathbb{C}G-module U is \psi \cdot \mu, where \mu:G \rightarrow G/N is the natural map.

    Lemma 1 says that given g \in G, u \in U and 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 \mathbb{C}G-module structure where \mathbb{C}G-submodule of U are exactly the \mathbb{C}(G/N)-submodule of U.

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

    The character table of S_3 \cong S_4/N is

    \begin{matrix}<br />
& N & (12)N & (123)N\\<br />
\phi^1 (trivial) & 1 & 1 & 1\\<br />
\phi^2 (sign) & 1 & -1 & 1\\<br />
\phi^3 (permutation) & 2 &0 & -1<br />
\end{matrix}

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

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

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

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

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

    Anyhow, I was not able to figure out the multiplicity of \mathbb{C}S_3-module in \mathbb{C}S_4.
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