4
$\begingroup$

In his book, Lie Groups, Lie Algebras, and Representations, Hall asks us to consider the nine-dimensional representation of $\mathfrak{sl}(3,\mathbb{C})$ obtained by taking the tensor product of the representations with highest weights (1,0) and (0,1) (that is, the standard and dual representations). We want to decompose this representation as a direct sum of irreducibles. I am hoping that there is a method for doing this similar to the Clebsch-Gordan procedure in the case of $\mathfrak{sl}(2,\mathbb{C})$. Is this the case? If not, how does one usually proceed here?

I compute the possible weights and the corresponding weight vectors for the tensor product representation to be:

  • (1,1) --- $e_1\otimes f_1$ (highest weight)
  • (-1,2) --- $e_2\otimes f_1$
  • (2,-1) --- $e_1\otimes f_2$
  • (1,-2) --- $e_3\otimes f_2$
  • (-1,-1) --- $e_3\otimes f_3$
  • (-2,1) --- $e_2\otimes f_3$
  • (0,0) --- $e_1\otimes f_3, e_2\otimes f_2, e_3\otimes f_1$

where $\{e_i\}$ is the standard basis for $\mathbb{C}^3$ and as in Hall's book, we define $f_1=e_3$, $f_2=-e_2$, and $f_3=e_1$ to be the basis for the second copy of $\mathbb{C}^3$ in our tensor product (recall the dual space is isomorphic to $\mathbb{C}^3$).

Note: There is an answer to a similar question here. If this is the procedure I'm looking for, my question becomes what does one mean by "Remove a copy of this irreducible representation from $\mathbb{C}^3\otimes V_{1,1}$" in step (iv)? I suspect that it has something to do with the decomposition of the tensor product representation into a direct sum of its weight spaces.

$\endgroup$

1 Answer 1

5
$\begingroup$

The statement "Remove a copy of this irreducible representation from $\mathbb{C}^3\otimes V_{1,1}$" just means remove the weights occurring in the irreducible representation from those appearing in your representation, and see what's left. In your case, the irreducible representation with highest weight (1,1) is 8-dimensional (and is actually isomorphic to the adjoint representation), and its weights are listed in Section 6.5 of Hall: they are (1,1), (-1,2), (2,-1), (0,0) (twice), (1,-2), (-2,1), and (-1,-1). Removing them from your list of weights leaves only (0,0), which is the weight of the 1-dimensional trivial representation (with highest weight (0,0)). So your tensor product representation decomposes into the 8-dimensional adjoint representation plus a 1-dimensional trivial representation. This is sometimes written as $3\otimes\overline{3} = 8\oplus 1$.

$\endgroup$
5
  • $\begingroup$ Extremely helpful, I was making a silly mistake involving the (0,0) representation. Thank you. $\endgroup$
    – zbrads2
    Dec 12, 2020 at 1:20
  • $\begingroup$ No problem, glad it was helpful. $\endgroup$
    – user17945
    Dec 12, 2020 at 1:28
  • $\begingroup$ What would happen if we run out of weights (by discarding them one at a time), but there are still dimensions left to be counted for a decomposition? That is, suppose we consider the representation formed by two copies with highest weight $(1, 0)$. Then, there are six weights each with multiplicity one. If we apply this approach, all the weights are removed but $3$ dimensions remain to be taken into account. How will we decompose $\mathbb{C}^3 \otimes \mathbb{C}^3$ in this case? $\endgroup$ May 26, 2021 at 6:22
  • $\begingroup$ This is really a separate question, but briefly: it's true there are six weights in $\mathbb{C}^3\otimes\mathbb{C}^3$, but they don't all have multiplicity one. Three of them $(0,1), (1,-1)$, and $(-1,0)$, have multiplicity two. For example, the $(0,1)$ weight space is spanned by $e_1\otimes e_2$ and $e_2\otimes e_1$. Now if you try to construct the irrep with highest weight $(2,0)$, you get a six-dimensional irrep, and the weight vector in this irrep corresponding to $(0,1)$ is $e_1\otimes e_2+e_2\otimes e_1$. $\endgroup$
    – user17945
    May 27, 2021 at 0:00
  • $\begingroup$ To get the rest of $\mathbb{C}^3\otimes \mathbb{C}^3$, start with $e_1\otimes e_2-e_2\otimes e_1$ instead. You can check this is a maximal weight vector (i.e. satisfies $\gamma(X_i)v=0$ for $i=1,2,3$). Constructing the irrep corresponding to this gives a three dimensional irrep that is isomorphic to the dual fundamental rep. This is sometimes written as $3\otimes 3 = 6\oplus\overline{3}$. $\endgroup$
    – user17945
    May 27, 2021 at 0:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .