In mathematics, specifically in representation theory, a semisimple representation (also called a completely reducible representation) is a linear representation of a group or an algebra that is a direct sum of simple representations (also called irreducible representations). It is an example of the general mathematical notion of semisimplicity.

Many representations that appear in applications of representation theory are semisimple or can be approximated by semisimple representations. A semisimple module over an algebra over a field is an example of a semisimple representation. Conversely, a semisimple representation of a group G over a field k is a semisimple module over the group algebra k[G].

Let V be a representation of a group G; or more generally, let V be a vector space with a set of linear endomorphisms acting on it. In general, a vector space acted on by a set of linear endomorphisms is said to be simple (or irreducible) if the only invariant subspaces for those operators are zero and the vector space itself; a semisimple representation then is a direct sum of simple representations in that sense.

The following are equivalent:

The equivalence of the above conditions can be proved based on the following lemma, which is of independent interest:

Lemma—Let p:V → W be a surjective equivariant map between representations. If V is semisimple, then p splits; i.e., it admits a section.

Proof of the lemma: Write ${\displaystyle V=\bigoplus _{i\in I}V_{i}}$ where ${\displaystyle V_{i}}$ are simple representations. Without loss of generality, we can assume ${\displaystyle V_{i}}$ are subrepresentations; i.e., we can assume the direct sum is internal. Now, consider the family of all possible direct sums ${\displaystyle V_{J}:=\bigoplus _{i\in J}V_{i}\subset V}$ with various subsets ${\displaystyle J\subset I}$. Put the partial ordering on it by saying the direct sum over K is less than the direct sum over J if ${\displaystyle K\subset J}$. By Zorn's lemma, we can find a maximal ${\displaystyle J\subset I}$ such that ${\displaystyle \operatorname {ker} p\cap V_{J}=0}$. We claim that ${\displaystyle V=\operatorname {ker} p\oplus V_{J}}$. By definition, ${\displaystyle \operatorname {ker} p\cap V_{J}=0}$ so we only need to show that ${\displaystyle V=\operatorname {ker} p+V_{J}}$. If ${\displaystyle \operatorname {ker} p+V_{J}}$ is a proper subrepresentatiom of ${\displaystyle V}$ then there exists ${\displaystyle k\in I-J}$ such that ${\displaystyle V_{k}\not \subset \operatorname {ker} p+V_{J}}$. Since ${\displaystyle V_{k}}$ is simple (irreducible), ${\displaystyle V_{k}\cap (\operatorname {ker} p+V_{J})=0}$. This contradicts the maximality of ${\displaystyle J}$, so ${\displaystyle V=\operatorname {ker} p\oplus V_{J}}$ as claimed. Hence, ${\displaystyle W\simeq V/\operatorname {ker} p\simeq V_{J}\to V}$ is a section of p. ${\displaystyle \square }$

Note that we cannot take ${\displaystyle J}$ to the set of ${\displaystyle i}$ such that ${\displaystyle \ker(p)\cap V_{i}=0}$. The reason is that it can happen, and frequently does, that ${\displaystyle X}$ is a subspace of ${\displaystyle Y\oplus Z}$ and yet ${\displaystyle X\cap Y=0=X\cap Z}$. For example, take ${\displaystyle X}$, ${\displaystyle Y}$ and ${\displaystyle Z}$ to be three distinct lines through the origin in ${\displaystyle \mathbb {R} ^{2}}$. For an explicit counterexample, let ${\displaystyle A=\operatorname {Mat} _{2}F}$ be the algebra of 2-by-2 matrices and set ${\displaystyle V=A}$, the regular representation of ${\displaystyle A}$. Set ${\displaystyle V_{1}={\Bigl \{}{\begin{pmatrix}a&0\\b&0\end{pmatrix}}{\Bigr \}}}$ and ${\displaystyle V_{2}={\Bigl \{}{\begin{pmatrix}0&c\\0&d\end{pmatrix}}{\Bigr \}}}$ and set ${\displaystyle W={\Bigl \{}{\begin{pmatrix}c&c\\d&d\end{pmatrix}}{\Bigr \}}}$. Then ${\displaystyle V_{1}}$, ${\displaystyle V_{2}}$ and ${\displaystyle W}$ are all irreducible ${\displaystyle A}$-modules and ${\displaystyle V=V_{1}\oplus V_{2}}$. Let ${\displaystyle p:V\to V/W}$ be the natural surjection. Then ${\displaystyle \operatorname {ker} p=W\neq 0}$ and ${\displaystyle V_{1}\cap \operatorname {ker} p=0=V_{2}\cap \operatorname {ker} p}$. In this case, ${\displaystyle W\simeq V_{1}\simeq V_{2}}$ but ${\displaystyle V\neq \operatorname {ker} p\oplus V_{1}\oplus V_{2}}$ because this sum is not direct.