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I am currently working on the topic of Lie - Algebras and I have stumbled a few times over the expression "direct summand in a tensor product".

The text says that $\ V(\lambda) $ as an indecomposable direct summand is expressible as a tensor product.

$V(\lambda)$ denotes a semisimple L-module of highest weight $\lambda$.

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No offense, but I have to ask: How do you define a semi-simple module without knowing what a direct summand is? –  t.b. Jul 8 '11 at 21:25
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I'd wager that there is a mis-statement/question: probably $V(\lambda)$ is an indecomposable direct summand in a tensor product. –  paul garrett Jul 8 '11 at 21:42
    
Please excuse my lack of English language skills. –  nikki Jul 8 '11 at 22:16
    
Please excuse my lack of English language skills, it is not my native language. The topic Lie - Algebras is new to me and I have just begun reading a book about the topic. I meant "the direct summand in a tensor product". –  nikki Jul 8 '11 at 22:22
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There are two distinct notions here. Do you understand what a tensor product is? Do you understand what a direct summand is? –  Qiaochu Yuan Jul 8 '11 at 22:45

1 Answer 1

up vote 1 down vote accepted

To give a simple example, rather than a definition: with Lie algebra $g=sl(2)$, the "standard" repn $V(1)$ is 2-dimensional, and has highest weight $\pmatrix{1 & 0 \cr 0 & -1}\rightarrow 1$, where a highest-weight vector is $\pmatrix{1 \cr 0}$ and is annihilated by the raising operator $\pmatrix{0 & 1 \cr 0 & 0}$. The other weight is $-1$.

The tensor product of this std repn with itself is a _direct_sum_: $$ V(1) \otimes V(1) \approx V(2) \oplus V(0) $$ More generally, for $g=sl(2)$ and $n\ge 1$, $$ V(1) \otimes V(n) \approx V(n+1) \oplus V(n-1) \hskip30pt\hbox{(this was wrong earlier!)} $$ [The analogue for $V(1)$ replaced by $V(m)$ previously addled my wits, leading to an erroneous version of the previous.] Also, $V(2)$ has weights $2,0-2$, and $$ V(2)\otimes V(n) \approx V(n+2) \oplus V(n) \oplus V(n-2) \hskip30pt\hbox{(for $n\ge 2$)} $$ Each $V(\lambda)$ is irreducible, in the sense that it has no proper subrepns/submodules.

In very crude terms, somewhat parallel to the appearance of these operations in quantum mechanics a long time ago: "irreducibles/indecomposables" are elementary particles, the "tensor product" amounts to "crashing together" two repns, as you'd collide two particles, and "decomposing as a direct sum of irreducibles/indecomposables" is telling what elementary particles come out of the collision. Of course, this dramatically ignores the mathematical content... but perhaps the general shape of the action is of interest, prior to concrete definitions?

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Thank you very much, this was very helpful to me! –  nikki Jul 9 '11 at 11:44
    
Oop, some typos... sorry... thx, will repair it! –  paul garrett Jul 11 '11 at 19:50

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