2
$\begingroup$

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?

$\endgroup$

1 Answer 1

5
$\begingroup$

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 $v=A_1v=0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.