Indecomposable but reducible representation and their consequences. Examples.

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I have a couple of questions. First, I hope that this is a correct proof (I skipped some details on purpose):

Let $\rho:\mathbb{R} \to GL(\mathbb{R},2)$ be a representation of $(\mathbb{R},+)$ given by
$$a\mapsto\begin{bmatrix} 1 & a\\
0 & 1\end{bmatrix}.$$ I want to show that $\rho$ is reducible but is not decomposable.

$(1)$ $\rho$ is reducible since we can find a subspace $W=span\{e_1\}$ which is $\mathbb{R}$-invariant i.e. $\rho(a)W\subset W$.

$(2)$ $\rho$ is not decomposable. Indeed, if it is decomposable that $\rho(a)$ will be similar to a diagonal matrix $\begin{bmatrix} \lambda_1 & 0\\
0 & \lambda_2\end{bmatrix}$ i.e. $\rho(a)$ is diagonazible for all $a\in\mathbb{R}$ which is not true since the minimal polynomial of $\rho(a)$ has repeated roots.

Therefore, $\rho$ is indecomposable but reducible.

Second, what are the other interesting examples of indecomposable but reducible representation?

Also,

  • Does irreducible follows decomposable for finite groups since we can construct a $G$-invariant Hermitian form? So, this is always true for finite groups.
  • What is the consequence that we can find indecomposable but reducible representations?

Best Answer

First: your proof is correct.

Also:
For a finite group $G$ and a field $k$, the following statements are equivalent:
(i) every indecomposable representation of $G$ on a finite-dimensional vector space over $k$ is irreducible;
(ii) the group algebra $kG$ is semi-simple;
(iii) the characteristic of $k$ divides does not divide the order of $G$.
The equivalence (i)$\Leftrightarrow$(ii) is more or less by definition, the implication (iii)$\Rightarrow$(i) is Maschke's theorem. See for example Alperin's book Local representation theory (Cambridge University Press, 1986).

If every indecomposable representation is irreducible, then every finite-dimensional (or, more generally, finite length) representation is a direct sum of irreducible ones. For finite groups, there are only finitely many irreducible representations, so here you can get a complete understanding of the category of finite-dimensional representations by classifying the irreducible representations and the morphisms between them. On the other hand, for groups where there exist indecomposable representations that are not irreducible, the representation theory is generally much more difficult to understand. There will almost always be infinitely many indecomposable finite-dimensional representations and, with a few exceptions, the group algebra $kG$ will be wild (meaning, informally, that the category of finite-dimensional representations is "impossibly" complicated).

Second: Well, this depends on what your interests are. An easy example is the following family of representations of Klein's four group $V_4 = C_2\times C_2 = \langle a,b \mid a^2,\,b^2,\, aba^{-1}b^{-1} \rangle$ over a field $k$ of characteristic two:
Let $B=\begin{pmatrix} 1& \lambda\\ 0&1\end{pmatrix}$, where $\lambda\in k\setminus\{0\}$ (note that $B^2=\mathbb{I}_2$, since $\mathrm{char}(k)=2$). Define a $V_4$-action on $U = k^2$ by $a\cdot u = u$ and $b\cdot u = Bu$.
By the same argument as you gave in the question, this representation is indecomposable, but not irreducible. Note that if $k$ is infinite, this gives an infinite family of indecomposable finite-dimensional representations of the group $V_4$.

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