The Unapologetic Mathematician

Mathematics for the interested outsider

Submodules

Fancy words: a submodule is a subobject in the category of group representations. What this means is that if (V,\rho_V) and (W,\rho_W) are G-modules, and if we have an injective morphism of G modules \iota:W\to V, then we say that W is a “submodule” of V. And, just to be clear, a G-morphism is injective if and only if it’s injective as a linear map; its kernel is zero. We call \iota the “inclusion map” of the submodule.

In practice, we often identify a G-submodule with the image of its inclusion map. We know from general principles that since \iota is injective, then W is isomorphic to its image, so this isn’t really a big difference. What we can tell, though, is that the action of g sends the image back into itself.

That is, let’s say that \iota(w) is the image of some vector w\in W. I say that for any group element g, acting by g on \iota(w) gives us some other vector that’s also in the image of \iota. Indeed, we check that

\displaystyle\left[\rho_V(g)\right](\iota(w))=\iota\left(\left[\rho_W(g)\right](w)\right)

which is again in the image of \iota, as asserted. We say that the image of \iota is “G-invariant”.

The flip side of this is that any time we find such a G-invariant subspace of V, it gives us a submodule. That is, if (V,\rho_V) is a G-module, and W\subseteq V is a G-invariant subspace, then we can define a new representation on W by restriction: \rho_W(g)=\rho_V(g)\vert_W. The inclusion map that takes any vector w\in W\subseteq V and considers it as a vector in V clearly intertwines the original action \rho_V and the restricted action \rho_W, and its kernel is trivial. Thus W constitutes a G-submodule.

As an example, let G be any finite group, and let \mathbb{C}[G] be its group algebra, which carries the left regular representation \rho. Now, consider the subspace V spanned by the vector

\displaystyle v=\sum\limits_{g\in G}\mathbf{g}

That is, V consists of all vectors for which all the coefficients c_g are equal. I say that this subspace V\subseteq\mathbb{C}[G] is G-invariant. Indeed, we calculate

\displaystyle\left[\rho(g)\right](cv)=c\left[\rho(g)\right]\left(\sum\limits_{g'\in G}\mathbf{g'}\right)=c\sum\limits_{g'\in G}\left[\rho(g)\right](\mathbf{g'})=c\sum\limits_{g'\in G}\mathbf{gg'}

But this last sum runs through all the elements of G, just in a different order. That is, \displaystyle\left[\rho(g)\right](cv)=cv, and so V carries the one-dimensional trivial representation of G. That is, we’ve found a copy of the trivial representation of G as a submodule of the left regular representation.

As another example, let G=S_n be one of the symmetric groups. Again, let \mathbb{C}[G] carry the left regular representation, but now let W be the one-dimensional space spanned by

\displaystyle w=\sum\limits_{g\in G}\mathrm{sgn}(g)\mathbf{g}

It’s a straightforward exercise to show that W is a one-dimensional submodule carrying a copy of the signum representation.

Every G-module V contains two obvious submodules: the zero subspace \{0\} and the entire space V itself are both clearly G-invariant. We call these submodules “trivial”, and all others “nontrivial”.

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September 22, 2010 - Posted by | Algebra, Group theory, Representation Theory

4 Comments »

  1. […] 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 […]

    Pingback by Reducibility « The Unapologetic Mathematician | September 23, 2010 | Reply

  2. […] invariant form on our space , then any reducible representation is decomposable. That is, if is a submodule, we can find another submodule so that as […]

    Pingback by Invariant Forms « The Unapologetic Mathematician | September 27, 2010 | Reply

  3. […] know that there’s a copy of the the trivial representation as a submodule of the defining representation. If we use the standard basis of , this submodule is the line […]

    Pingback by One Complete Character Table (part 2) « The Unapologetic Mathematician | October 27, 2010 | Reply

  4. […] in sight has been a vector space over that field. But remember that a representation of is a module over the group algebra , and we can take tensor products over this algebra as […]

    Pingback by Tensor Products over Group Algebras « The Unapologetic Mathematician | November 9, 2010 | Reply


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