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A tensor-product top weight does not determine all constituents

Statement

False: the highest weight of a tensor product $V \otimes W$ of finite-dimensional irreducible modules over a complex semisimple Lie algebra determines the complete irreducible decomposition of $V \otimes W$ .

Facts & Assumptions

Given: The Lie algebra ${s l}_{2} (C)$ with basis $e, f, h$ and its Cartan subalgebra $h = C h$ (The special linear Lie algebra sl_2), the standard two-dimensional module $V = C^{2}$ with basis $u_{1}, u_{2}$ on which $h \cdot u_{1} = u_{1}$ , $h \cdot u_{2} = - u_{2}$ , $e \cdot u_{1} = 0$ , $e \cdot u_{2} = u_{1}$ , $f \cdot u_{1} = u_{2}$ , $f \cdot u_{2} = 0$ , and the tensor product $V \otimes V$ with the usual action $x \cdot (u \otimes w) = x \cdot u \otimes w + u \otimes x \cdot w$ (Weight and weight space).

[L1]

A finite-dimensional ${s l}_{2}$ -module is a direct sum of irreducible submodules, and an irreducible submodule with highest weight $m \geq 0$ (that is, with $h$ acting with top eigenvalue $m$ ) has dimension $m + 1$ with $h$ -eigenvalues $m, m - 2, \dots, - m$ (Finite-dimensional representations of sl_2).

[L2]

The module $V$ is irreducible of top weight $1$ : its weight vectors are multiples of $u_{1}$ and $u_{2}$ , and the polynomials $(h + 1) / 2$ and $(1 - h) / 2$ project any nonzero submodule onto at least one of those lines; the submodule generated by $u_{2}$ contains $u_{1} = e \cdot u_{2}$ and hence is $V$ , and the submodule generated by $u_{1}$ contains $u_{2} = f \cdot u_{1}$ and hence is $V$ ; the vector $u_{1}$ is killed by $e$ and has $h$ -eigenvalue $1$ , so it is a nonzero vector killed by the positive root vector $e$ and its weight $1$ is the maximum of the weights of $V$ (Irreducible, completely reducible, and faithful representations).

[L3]

In $V \otimes V$ , put $S = span {u_{1} \otimes u_{1}, u_{1} \otimes u_{2} + u_{2} \otimes u_{1}, u_{2} \otimes u_{2}}, A = C (u_{1} \otimes u_{2} - u_{2} \otimes u_{1}) .$ Direct application of the displayed action in the Given line shows that $S$ and $A$ are submodules, that $A$ is trivial, and that on the displayed basis of $S$ the operators $e$ and $f$ join the three $h$ -weight spaces of weights $2, 0, - 2$ by nonzero arrows: writing the basis as $(s_{+}, s_{0}, s_{-})$ , one has $e s_{+} = 0$ , $e s_{0} = 2 s_{+}$ , $e s_{-} = s_{0}$ , $f s_{+} = s_{0}$ , $f s_{0} = 2 s_{-}$ , $f s_{-} = 0$ . [given]

Refutation

technique · direct

1.1

The ambient complex algebra is simple, hence semisimple: an ideal is invariant under $ad h$ , whose three distinct eigenspaces are $C e$ , $C h$ , $C f$ . Polynomial spectral projections put a nonzero member of one of these lines in any nonzero ideal; the displayed brackets then generate all three lines. Also the algebra is nonabelian and its derived algebra is itself. The factors $V$ are irreducible by [L2]. The four displayed symmetric and alternating tensors in [L3] form a basis of $V \otimes V$ , so $V \otimes V = S \oplus A$ and has top weight $2$ .

L2L3

2.1

The submodule $S$ is irreducible. Indeed, for a nonzero submodule $U \leq S$ , the three distinct $h$ -eigenvalues allow a polynomial in $h$ to project a nonzero vector of $U$ onto a nonzero weight vector. The nonzero $e$ - and $f$ -arrows in [L3] then put all three displayed basis vectors in $U$ , so $U = S$ . Thus [L1] identifies $S ≅ L (2)$ , while $A ≅ L (0)$ by [L3], and $V \otimes V ≅ L (2) \oplus L (0)$ .

L1L3step 1.1

3.1

Set $a = u_{1} \otimes u_{2} - u_{2} \otimes u_{1} \neq 0$ , so $A = C a$ is the one-dimensional trivial irreducible module. The second tensor product $T = S \otimes A$ has two irreducible factors by step 2.1. The linear map $T \to S$ , $s \otimes c a \mapsto c s$ , is bijective with inverse $s \mapsto s \otimes a$ , and it intertwines the Lie-algebra action because $x a = 0$ . Consequently $T$ is irreducible of top weight $2$ and has a single constituent, whereas $V \otimes V$ has the two constituents $S$ and $A$ . The existence of $S$ was established by the explicit construction, not inferred from a theorem about an already irreducible module.

L3step 2.1algebra

4.1

Both $V \otimes V$ and $S \otimes A$ are tensor products of finite-dimensional irreducible modules over the same complex semisimple algebra. Both have highest weight $2$ , but their decompositions differ: $V \otimes V ≅ L (2) \oplus L (0)$ has dimension four and $S \otimes A ≅ L (2)$ has dimension three. Thus the top weight alone does not determine the constituent list even within the stated class of tensor products.

step 1.1step 2.1step 3.1algebra∎

Depends on

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Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)
Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)