# The Unapologetic Mathematician

## Reducibility

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 it always contains a copy of the trivial representation as a nontrivial submodule. Notice that we have to be careful about what we mean by each use of “trivial” here.

If the $n$-dimensional representation $V$ has a nontrivial $m$-dimensional submodule $W\subseteq V$$m\neq0$ and $m\neq n$ — then we can pick a basis $\{w^1,\dots,w^m\}$ of $W$. And then we know that we can extend this to a basis for all of $V$: $\{w^1,\dots,w^m,v^{m+1},\dots,v^n\}$.

Now since $W$ is a $G$-invariant subspace of $V$, we find that for any vector $w\in W$ and $g\in G$ the image $\left[\rho(g)\right](w)$ is again a vector in $W$, and can be written out in terms of the $w^i$ basis vectors. In particular, we find $\left[\rho(g)\right](w^i)=\rho_j^iw^j$, and all the coefficients of $v^{m+1}$ through $v^n$ are zero. That is, the matrix of $\rho(g)$ has the following form:

$\displaystyle\left(\begin{array}{c|c}\alpha(g)&\beta(g)\\\hline{0}&\gamma(g)\end{array}\right)$

where $\alpha(g)$ is an $m\times m$ matrix, $\beta(g)$ is an $m\times(n-m)$ matrix, and $\gamma(g)$ is an $(n-m)\times(n-m)$ matrix. And, in fact, this same form holds for all $g$. In fact, we can use the rule for block-multiplying matrices to find:

\displaystyle\begin{aligned}\left(\begin{array}{c|c}\alpha(gh)&\beta(gh)\\\hline{0}&\gamma(gh)\end{array}\right)&=\rho(gh)\\&=\rho(g)\rho(h)\\&=\left(\begin{array}{c|c}\alpha(g)&\beta(g)\\\hline{0}&\gamma(g)\end{array}\right)\left(\begin{array}{c|c}\alpha(h)&\beta(h)\\\hline{0}&\gamma(h)\end{array}\right)\\&=\left(\begin{array}{c|c}\alpha(g)\alpha(h)&\alpha(g)\beta(h)+\beta(g)\gamma(h)\\\hline{0}&\gamma(g)\gamma(h)\end{array}\right)\end{aligned}

and we see that $\alpha(g)$ actually provides us with the matrix for the representation we get when restricting $\rho$ to the submodule $W$. This shows us that the converse is also true: if we can find a basis for $V$ so that the matrix $\rho(g)$ has the above form for every $g\in G$, then the subspace spanned by the first $m$ basis vectors is $G$-invariant, and so it gives us a subrepresentation.

As an example, consider the defining representation $V$ of $S_3$, which is a permutation representation arising from the action of $S_3$ on the set $\{1,2,3\}$. This representation comes with the standard basis $\{\mathbf{1},\mathbf{2},\mathbf{3}\}$, and it’s easy to see that every permutation leaves the vector $\mathbf{1}+\mathbf{2}+\mathbf{3}$ — along with the subspace $W$ that it spans — fixed. Thus $W$ carries a copy of the trivial representation as a submodule of $V$. We can take the given vector as a basis and throw in two others to get a new basis for $V$: $\{\mathbf{1}+\mathbf{2}+\mathbf{3},\mathbf{2},\mathbf{3}\}$.

Now we can take a permutation — say $(1\,2)$ — and calculate its action in terms of the new basis:

\displaystyle\begin{aligned}\left[\rho((1\,2))\right](\mathbf{1}+\mathbf{2}+\mathbf{3})&=\mathbf{1}+\mathbf{2}+\mathbf{3}\\\left[\rho((1\,2))\right](\mathbf{2})&=\mathbf{1}=(\mathbf{1}+\mathbf{2}+\mathbf{3})-\mathbf{2}-\mathbf{3}\\\left[\rho((1\,2))\right](\mathbf{3})&=\mathbf{3}\end{aligned}

The others all work similarly. Then we can write these out as matrices:

\displaystyle\begin{aligned}\rho(e)&=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\\\rho((1\,2))&=\begin{pmatrix}1&1&0\\{0}&-1&0\\{0}&-1&1\end{pmatrix}\\\rho((1\,3))&=\begin{pmatrix}1&0&1\\{0}&1&-1\\{0}&0&-1\end{pmatrix}\\\rho((2\,3))&=\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&1&0\end{pmatrix}\\\rho((1\,2\,3))&=\begin{pmatrix}1&0&1\\{0}&0&-1\\{0}&1&-1\end{pmatrix}\\\rho((1\,3\,2))&=\begin{pmatrix}1&1&0\\{0}&-1&1\\{0}&-1&0\end{pmatrix}\end{aligned}

Notice that these all have the required form:

$\displaystyle\left(\begin{array}{c|cc}1&\ast&\ast\\\hline{0}&\ast&\ast\\{0}&\ast&\ast\end{array}\right)$

Representations that are not reducible — those modules that have no nontrivial submodules — are called “irreducible representations”, or sometimes “irreps” for short. They’re also called “simple” modules, using the general term from category theory for an object with no nontrivial subobjects.

September 23, 2010 -

## 9 Comments »

1. [...] Today I’d like to cover a stronger condition than reducibility: decomposability. We say that a module is “decomposable” if we can write it as the [...]

Pingback by Decomposability « The Unapologetic Mathematician | September 24, 2010 | Reply

2. [...] than any particular invariant form is this: if we have an invariant form on our space , then any reducible representation is decomposable. That is, if is a submodule, we can find another submodule so that [...]

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

3. [...] saw last time that in the presence of an invariant form, any reducible representation is decomposable, and so any representation with an invariant form is completely [...]

Pingback by Maschke’s Theorem « The Unapologetic Mathematician | September 28, 2010 | Reply

4. [...] that we call a -module irreducible or “simple” if it has no nontrivial submodules. In general, an object in any category [...]

Pingback by Schur’s Lemma « The Unapologetic Mathematician | September 30, 2010 | Reply

5. [...] start our considerations by letting by any matrix irrep, and let’s calculate its commutant algebra. By definition for any we have for all . We can [...]

Pingback by Endomorphism and Commutant Algebras « The Unapologetic Mathematician | October 1, 2010 | Reply

6. [...] We want to calculate commutant algebras of matrix representations. We already know that if is an irrep, then , and we’ll move on from [...]

Pingback by Some Commutant Algebras « The Unapologetic Mathematician | October 4, 2010 | Reply

7. [...] More Commutant Algebras We continue yesterday’s discussion of commutant algebras. But today, let’s consider the direct sum of a bunch of copies of the same irrep. [...]

Pingback by More Commutant Algebras « The Unapologetic Mathematician | October 5, 2010 | Reply

8. [...] one copy of the trivial representation and no copies of the signum representation. In fact, we already knew about the copy of the trivial representation, but it’s nice to see it confirmed again. [...]

Pingback by One Complete Character Table (part 1) « The Unapologetic Mathematician | October 26, 2010 | Reply

9. [...] The secret is to look at the block diagonal form from when we defined reducibility: [...]

Pingback by An Alternative Path « The Unapologetic Mathematician | October 28, 2010 | Reply