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Sorry if the question is dumb. I am trying to learn representation theory of finite groups from J.P.Serre's book by myself. In section 2.6 on canonical decomposition, he says that let V be a representation of a finite group G, $W_1,...,W_h$ be the distinct irreducible representations of G, and let V = $U_1 \oplus ... \oplus U_m$ be some decomposition of V into irreducible subrepresentations. Then we can write V = $V_1\oplus ...\oplus V_h$, where $V_i$ is the direct sum of irreducible subrepresentations among $U_i$'s which are $isomorphic$ to $W_i$. This much is clear. But then he says that :

Next, if needed, one chooses a decomposition of $V_i$ into a direct sum of irreducible representations each isomorphic to $W_i$: $$V_i = W_i \oplus ...\oplus W_i$$ The last decomposition can be done in infinitely many ways; it is just as arbitrary as the choice of a basis in a vector space.

I am confused with this part. I understand $external$ direct sums of same spaces, but how is the $internal$ direct sum of same spaces $W_i$ defined in general? I think I might be facing some notational difficulty. Thanks in advance.

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2 Answers 2

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In the internal decomposition, the spaces aren't all equal to $W_i$ in a literal sense, just isomorphic to it as representations of $G$ - what you really have is $V_i=V_{i,1}\oplus\cdots\oplus V_{i,n}$ with $V_{i,j}\cong W_i$ for all $j$.

My preference would be to write $V_i\cong W_i\oplus\cdots\oplus W_i$ instead of $V_i=W_i\oplus\cdots\oplus W_i$, which I thinks makes it a bit clearer (and emphasises the fact that the decomposition is non-canonical).

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  • $\begingroup$ Thank you! So, whenever I come across any notation where a space is written as an internal direct sum of same spaces, can I always safely assume that actually it means direct sum of spaces isomorphic to a common space? $\endgroup$
    – Hajime_Saito
    Commented Jun 12, 2013 at 19:02
  • $\begingroup$ If the space you're direct summing copies of is non-zero, then it doesn't make sense to write an internal direct sum anyway, so it must be an external direct sum in some sense. Note that in this example, we can say $V_i\cong W_i\oplus\cdots\oplus W_i$ where we mean an external direct sum of $W_i$s, and then the choice of isomorphism allows us to translate that into a decomposition of $V_i$, each component of which is isomorphic to $W_i$. (I guess all I'm really saying is that internal and external direct sums are isomorphic when both defined). $\endgroup$
    – mdp
    Commented Jun 12, 2013 at 20:13
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Notice the language "each isomorphic to $W_i$." You're right to say that $V_i=W_i\oplus\ldots\oplus W_i$ doesn't make much sense as an internal direct sum, but we write it that way just to make notation convenient. Perhaps it would be more clear to write:

$$V_i=W_{i_1}\oplus\ldots\oplus W_{i_s}$$

with $W_{i_j}\cong W_i$ for all $j=1,\ldots, s$.

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  • $\begingroup$ thank you for the clarification! both answers were helpful, I wish I could upvote both the answers. $\endgroup$
    – Hajime_Saito
    Commented Jun 12, 2013 at 18:50
  • $\begingroup$ @usersujo You can upvote both answers, although I'm guessing you meant to write "accept both answers." $\endgroup$
    – rschwieb
    Commented Jun 12, 2013 at 19:33

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