# The Unapologetic Mathematician

## Decomposability

Today I’d like to cover a stronger condition than reducibility: decomposability. We say that a module $V$ is “decomposable” if we can write it as the direct sum of two nontrivial submodules $U$ and $W$. The direct sum gives us inclusion morphisms from $U$ and $W$ into $V$, and so any decomposable module is reducible.

What does this look like in terms of matrices? Well, saying that $V=U\oplus W$ means that we can write any vector $v\in V$ uniquely as a sum $v=u+w$ with $u\in U$ and $w\in W$. Then if we have a basis $\{u_i\}_{i=1}^m$ of $U$ and a basis $\{w_j\}_{j=1}^n$ of $W$, then we can write $u$ and $w$ uniquely in terms of these basis vectors. Thus we can write any vector $v\in V$ uniquely in terms of the $\{u_1,\dots,u_m,v_1,\dots,v_n\}$, and so these constitute a basis of $V$.

If we write the matrices $\rho(g)$ in terms of this basis, we find that the image of any $u_i$ can be written in terms of the others because $U$ is $G$-invariant. Similarly, the $G$-invariance of $W$ tells us that the image of each $w_j$ can be written in terms of the others. The same reasoning as last time now allows us to conclude that the matrices of the $\rho(g)$ all have the form $\displaystyle\left(\begin{array}{c|c}\alpha(g)&0\\\hline0&\gamma(g)\end{array}\right)$

Conversely, if we can write each of the $\rho(g)$ in this form, then this gives us a decomposition of $V$ as the direct sum of two $G$-invariant subspaces, and the representation is decomposable.

Now, I said above that decomposability is stronger than reducibility. Indeed, in general there do exist modules which are reducible, but not decomposable. Indeed, in categorical terms this is the statement that for some groups $G$ there are short exact sequences which do not split. To chase this down a little further, our work yesterday showed that even in the reducible case we have the equation $\gamma(g)\gamma(h)=\gamma(gh)$. This $\gamma$ is the representation of $G$ on the quotient space, which gives our short exact sequence $\displaystyle\mathbf{0}\to W\to V\to V/W\to\mathbf{0}$

But in general this sequence may not split; we may not be able to write $V\cong W\oplus V/W$ as $G$-modules. Indeed, we’ve seen that the representation of the group of integers $\displaystyle n\mapsto\begin{pmatrix}1&n\\{0}&1\end{pmatrix}$

is indecomposable.

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September 24, 2010 -

## 5 Comments »

1. sorry, just a question: what software do you use to prepare mathematical formulas in your blog? Comment by Ng Foo Keong | September 26, 2010 | Reply

• WordPress.com-hosted weblogs have native support for $\LaTeX$. Before the formula, you put $latex, and after it you close with another $. Comment by John Armstrong | September 26, 2010 | Reply

2. […] 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 as […]

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 reducible. Maschke’s theorem […]

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

4. […] into a direct sum of irreducible submodules, we say that is “completely reducible”, or “decomposable”, as we did when we were working with groups. Any module where any nontrivial proper submodule has a […]

Pingback by Reducible Modules « The Unapologetic Mathematician | September 16, 2012 | Reply