# A question about (completely) reducible representations and isoytpic components

Let's consider a group $$G$$, a finite-dimensional vector space $$V$$ and a representation:

$$\rho : G \rightarrow GL(V)$$

Now I have read a lot about reducible and completely reducible representations, but what is the concrete difference?

As far as I'm concerned, if $$V$$ is finite-dimensional and the group $$G$$ is finite then we can write

$$\rho = \rho_1 \oplus \dots \oplus\rho_k$$

where $$\rho_i$$ are irreducible representations (the only invariant subspaces are trivials). Does this mean that $$\rho$$ is (completely) reducible? Considering these subrepresentations

\begin{align} \rho_1 : G &\rightarrow V_1 \\ \rho_2 : G &\rightarrow V_2 \\ &\vdots\\ \rho_k : G & \rightarrow V_k \end{align}

can we write then $$V = V_1 \oplus \dots \oplus V_k$$ where $$V_i$$ are called isotypic components?

I'm trying to better frame the situation here.

• Reducible just means not irreducible. For "completely reducible" see the duplicate. Commented Jan 4, 2022 at 11:51
• Does this answer your question? Definition completely reducible group representation Commented Jan 4, 2022 at 11:52
• The $V_i$ may not be isotypic components: an isotypic component is the sum of all subrepresentations isomorphic to some specific irreducible representation. Commented Jan 11, 2022 at 4:00

A representation $$V$$ is reducible if it contains a proper subrepresentation $$U \subseteq V$$, proper meaning that $$0 \neq U$$ and $$U \neq V$$. In other words, reducible means precisely not irreducible.
A representation $$V$$ is semisimple if for every subrepresentation $$U \subseteq V$$ there exists a complementary subrepresentation $$W \subseteq V$$, meaning $$U + W = V$$ and $$U \cap W = 0$$ (i.e. $$V$$ is a direct sum of $$U$$ and $$W$$).
You are correct that if a representation $$V$$ breaks into a direct sum $$V = V_1 \oplus \cdots \oplus V_k$$ of irreducible subrepresentations $$V_i$$, then $$V$$ is semisimple. (There is something to be proved here though, since the decomposition of $$V$$ may not be unique).
For an example of a reducible representation which is not semisimple, we need to look for slightly more exotic representation theory (not just finite groups over $$\mathbb{R}$$ or $$\mathbb{C}$$), for example where the order of the group divides the underlying field. Consider the group $$G$$ with two elements, acting on the vector space $$\mathbb{F}_2^2$$ over the field $$\mathbb{F}_2$$ of two elements, where the nontrivial element of $$G$$ acts by swapping coordinate vectors $$(1, 0) \leftrightarrow (0, 1)$$. This representation is reducible because $$U = \{(0, 0), (1, 1)\}$$ is a proper (and indeed irreducible) subrepresentation, but it has no complementary subrepresentations, since the only subrepresentations of $$\mathbb{F}_2^2$$ are $$0$$, $$U$$, and $$\mathbb{F}_2^2$$ itself.