The Unapologetic Mathematician

Mathematics for the interested outsider


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


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”.

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


  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

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: