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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.

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

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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 $W$).

Completely reducible is a synonym for semisimple, but I think it can be a confusing term and try to avoid it. The reason I think it is confusing is because the term "reducible" does not require the existence of a complementary subrepresentation, but "completely reducible" does.

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).

Note that an irreducible representation is semisimple, but not reducible.

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.

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