I've come across a conceptual issue working through Fulton and Harris' representation theory text, stemming from understanding the relation between the map $\rho : \mathfrak g \to \mathfrak{gl}(V)$ and $V$, which we often call the representation, even though it's $\rho$.

For example, the standard representation of $\mathfrak{sl}_2 \mathbb C$ is $V \cong \mathbb C^2$, and we have, using the notion of breaking up any representation into eigenspaces, that,

$$\mathrm{Sym}^2 V \otimes \mathrm{Sym}^3 V \cong \mathrm{Sym}^5 V \oplus \mathrm{Sym}^3 V \oplus V.$$

I have two questions:

  • Is the motivation of this decomposition to express it as a direct sum of spaces which behave 'nicely' under the action of the Lie algebra? In the same way one can decompose a tensor?
  • How does this relation read in terms of how $\rho$ acts on elements on the left hand side and the right hand side of this decomposition? I can imagine I could choose some arbitrarily complicated way to define a $\rho$ to act on a space, and it is not immediately obvious that such a decomposition would always be possible.

PS: Should it be relevant, I am a physicist trying to more rigorously grasp representation theory.

  • $\begingroup$ The motivation is that $\operatorname{Sym}^k V$ for all $k$ is the set of finite-dimensional irreducible representations of $\mathfrak{sl}_2\mathbb{C}$. Decomposing the tensor product tells you how much of each irreducible is present. For spin networks, this is useful for understanding what can be at vertices, since the dimension of the space of such vertex forms is the amount of the trivial representation in a tensor product of three representations. (Also, the $\cong$ doesn't mean you know where the irreducibles are. You still have to compute bases for that.) $\endgroup$ – Kyle Miller Nov 4 '17 at 23:12
  • $\begingroup$ @KyleMiller Can you make more precise the "doesn't mean you know where the irreducibles are"? $\endgroup$ – JamalS Nov 4 '17 at 23:19
  • $\begingroup$ I just mean that you can do the analysis to see that there has to be such a decomposition, but you have to do more work to compute the actual linear isomorphism between the two sides to see how to read off how $\rho$ acts on corresponding elements. $\endgroup$ – Kyle Miller Nov 4 '17 at 23:27
  • $\begingroup$ @KyleMiller Where could I read about how to do that? That seems to answer my second question precisely. (Unless it's covered in Fulton and Harris' and I just have to keep reading.) $\endgroup$ – JamalS Nov 4 '17 at 23:32
  • $\begingroup$ For $\mathfrak{sl}_2\mathbb{C}$, you look for the highest-weight vectors, from there it's easy to compute bases. It's in Fulton and Harris, so keep reading! $\endgroup$ – Kyle Miller Nov 4 '17 at 23:45

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