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Any given representation (some matrices of some algebra) will be reducible if there exists a singular, but nonzero matrix S that commutes with all elements of the representation; conversely, if no such S exists, then the representation is irreducible. This is Schur's Lemma.

Question 1: Does there exists some analogously simple criterion for deciding whether or not a given representation is decomposable, i.e., whether or not there exists some change of basis (i.e., a similarity transformation) that brings the representation on (sub)block-diagonal form?

Only very recently I learned that the two concepts are not equivalent; as a physicist I have always thought so. A concrete example is "... the two-dimensional representation of the real numbers acting by upper triangular unipotent matrices" [http://en.wikipedia.org/wiki/Irreducible_representation].

Question 2: When are the two concepts equivalent? I expect that they are so at least for the specific case of finite-dimensional, unitary representations (or am I mistaken?), but are there other cases?

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I will assume our algebra to have an identity.

Question 1. How about: a representation is decomposable if and only if there exist two non-zero idempotent matrices $A_1$ and $A_2$ such that

  • both commute with all elements of the representation,
  • $A_1A_2=A_2A_1=0$ and
  • $A_1+A_2=I$, the identity.

If a representation is decomposable, then clearly such matrices exist. Conversely, the first condition implies that the images of $A_1$ and of $A_2$ are subrepresentations, the second one implies that they intersect trivially$^1$, and the third one implies that their sum (and therefore their direct sum) is the whole representation.

Question 2. Let $A$ be an algebra. The following are equivalent:

  • a representation of $A$ is irreducible if and only if it is indecomposable;
  • every representation is a direct sum of irreducibles;
  • $A$ is Artinian and as a (left, say) module over itself is a direct sum of irreducibles;
  • $A$ is isomorphic to a direct sum of matrix algebras over division algebras.

The last equivalence is the Artin-Wedderburn theorem, and it completely classifies the situation you are asking about. An important example of such algebras is given by group algebras $K[G]$, where $G$ is a finite group, and $K$ is a field of characteristic not dividing $|G|$.

$^1$ If $v$ is in the image of $A_2$, then $A_2v=v$, since $A_2$ is idempotent, so $A_1v=0$ by the second condition; so if $v$ is also in the image of $A_1$, then by the same argument $v=A_1v=0$.

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  • $\begingroup$ Answer to question 1 is very useful, thanks. Answer to question 2 is difficult to read; there seems to be missing some separating commas. $\endgroup$ Commented Jul 30, 2013 at 7:55
  • $\begingroup$ @John: I have edited the second part. $\endgroup$
    – Alex B.
    Commented Jul 30, 2013 at 8:03

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