# The Unapologetic Mathematician

## 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”.

September 22, 2010 -

## 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 […]

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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 […]

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