2
$\begingroup$

I am struggling to understand the precise definition of "direct sum of irreducible representations", when speaking of representations of Lie algebras.

Given a representation $\phi:g \rightarrow End(V)$, we say that this is "completely reducible" if it is decomposable into a direct sum of irreducible representations. I am not sure what a direct sum of representations is.

I understand that a representation is irreducible if $V$ has no invariant subspaces under $\phi(x)$ for all $x \in g$.

This leads me to think that a direct sum of irreducible representations means that we can write $\phi$ as some sort of sum of irreducible representations, such that subspaces $V_i$ are invariant under those subspaces, but I don't really know what it would mean to write $\phi$ as a sum of irreducible representations.

$\endgroup$
  • 3
    $\begingroup$ Surely any text on representation theory defines a direct sum of representations near the very beginning, and it's quite easy to google "direct sum of representations." My guess is your confusion stems from only being familiar with calling the homomorphism $\phi:{\frak g}\to{\frak gl}(V)$ itself a representation, when actually we even more often just call $V$ itself the representation (with notation for the action of $\frak g$ on $V$ suppressed). So, yes, $V$ is a direct sum of irreps means it can be written as a direct sum of subspaces each of which is closed under $\frak g$'s action. $\endgroup$ – anon Oct 25 '17 at 0:04
  • $\begingroup$ The text I'm using unfortunately does not. Googling as you suggest gives me the definition in terms of representations of groups, and I'm not sure whether that's the same. Nevertheless, it appears that you've answered my question with your comment. Does this have any implication for $\phi$? I understand that $V$ is a direct sum of subspaces (each invariant), and I know what that means, but what I'm asking is what this means for the map $\phi$? Is it expressable as the sum of maps on the invariant subspaces? $\endgroup$ – Matt Oct 25 '17 at 0:23
  • $\begingroup$ Uh, sure, if you want to talk about $\phi$. If you use matrices it would mean $\phi$ output block diagonal matrices (each block corresponding to one of the irreps). (I notice you haven't said which text you're using.) $\endgroup$ – anon Oct 25 '17 at 0:24
  • $\begingroup$ @anon $\phi$ is completely reducible does it mean $\phi(x) = \sum_{j=1}^r P_j^* \rho_j(x) P_j$ where $P_j P_j^* P_j = P_j, P_j P_i^* = 0$ and $\rho_j([x,y]) = \rho_j(x)\rho_j(y)-\rho_j(y)\rho_j(x)$ and the $\rho_j$ have no $\frak g$ invariant subspaces ? (the $P_j$ are projections $V \to W_j$ and $P_j^*$ the adjoint) $\endgroup$ – reuns Oct 25 '17 at 0:24
  • $\begingroup$ @anon J-P. Serre $\endgroup$ – Matt Oct 25 '17 at 1:11
2
$\begingroup$

Suppose that $V$ is a vector space with a direct sum decomposition $V = U \oplus W$. Suppose that we have a linear operator $f: V \to V$ respecting this decomposition, so $f(U) \subseteq U$ and $f(W) \subseteq W$. Then we get a pair of operators $(U \to U, W \to W)$ by restriction: $(f|_U, f|_W)$.

Conversely, suppose that we have a pair of operators $g: U \to U$ and $h: W \to W$, then there is a uniquely defined operator $g \oplus h: V \to V$ by setting $(g \oplus h)(v) = g(u) + h(w)$, where $v = u + w$ is the direct sum decomposition.

You can show that, having fixed the decomposition $V = U \oplus W$, the above gives a bijection between operators $V \to V$ respecting the direct sum decomposition, and pairs of operators $(U \to U, W \to W)$.

So, having a direct sum of representations is the same as having a sum of smaller operators.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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