Tensor product of Lie algebra representations ($V\otimes W$)

The tensor product of Lie algebra representations has action $x(v\otimes w)=xv\otimes w+v\otimes xw$. The Lie algebra representation identity follows because operators on the two different tensor factors commute. This action descends to exterior powers and symmetric powers.

The summands have Dynkin labels $(2,1)$, $(0,2)$ and $(1,0)$. Convolving the triplet and octet weight diagrams and applying highest-weight character subtraction removes these three characters in that order. Their dimensions $15+6+3=24$ match the product.

// Target: quantum-field-theory.bigb

For the defining two-dimensional sl2 Lie algebra module with coordinates $x,y$, put $u_{ij}=x^{3-i}{y}^i\otimes x^{2-j}{y}^j$. Its three highest-weight vectors are $u_{00}$, $u_{10}-u_{01}$ and $u_{20}-2u_{11}+u_{02}$, of weights five, three and one. Lowering the last once gives $u_{30}-2u_{21}+u_{12}$, an explicit second vector of its two-dimensional irreducible summand.

Give $\mathfrak{g}$ its Adjoint representation. The representation identity gives $a(y\cdot (x\otimes v))=[y,x]v+xyv=y(xv)$, so the action map is a Lie algebra representation homomorphism. For a nontrivial simple module it is surjective. If the acting algebra is semisimple and the module is finite dimensional, Weyl complete reducibility theorem splits the surjection and makes $V$ a direct summand.

Take the Verma module for the sl2 Lie algebra of highest weight $-2$, with basis $v_m$ and $f{v}_m=v_{m+1}$, $h{v}_m=(-2-2m)v_m$, $e{v}_m=-m(m+1)v_{m-1}$. It is simple. In ${\mathfrak{sl}}_2\otimes V$, the highest vectors of weight $-2$ are multiples of $w=e\otimes v_1-h\otimes v_0=f(e\otimes v_0)$. Every homomorphism to $V$ kills $e\otimes v_0$, whose weight is zero, and hence kills $w$. Every embedded copy of $V$ would have its highest vector proportional to $w$, so no projection onto such a copy can exist. Thus semisimplicity of the Lie algebra alone does not provide an action summand in infinite dimension.

For $V={\mathbb{C}}^n$, $n\ge 3$, the tensor product of Lie algebra representations for the special linear Lie algebra splits as displayed. The exterior product map $v\otimes (u\wedge w)\mapsto v\wedge u\wedge w$ has an equivariant right inverse obtained by dividing its three-term alternating insertion by three. Its kernel is the Schur functor of shape $(2,1)$, with highest weight ${\omega }_1+{\omega }_2$ and dimension $n(n^2-1)/3$. The other summand has highest weight ${\omega }_3$ for $n\ge 4$ and is trivial for $n=3$.

The weight space of $2{\epsilon }_i+{\epsilon }_j$ has dimension one when $i\ne j$. The weight space of ${\epsilon }_i+{\epsilon }_j+{\epsilon }_k$ has dimension three for distinct indices. Restricting these functionals to the traceless diagonal Cartan subalgebra creates no additional coincidences; for $n=3$ the latter weight is zero.

For the complex special linear Lie algebra ${\mathfrak{sl}}_3$, tensoring an irreducible highest-weight representation by the defining representation gives the displayed sum, omitting terms with negative Dynkin labels. It follows by multiplying its Weyl character formula by $x_1+x_2+x_3$. In particular ${\Gamma }_{2,1}\otimes {\Gamma }_{1,0}={\Gamma }_{3,1}\oplus {\Gamma }_{1,2}\oplus {\Gamma }_{2,0}$, with dimensions $24+15+6=45$.