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The distinct affine algebraic groupsboth have the special linear Lie algebra ${\mathfrak{sl}}_2(\mathbb{C})$. Quotienting a Lie group by a discrete central subgroup does not change its tangent Lie algebra.

Let $V={\mathbb{C}}^4$. The special linear Lie algebra $\mathfrak{sl}(V)$ acts on ${\Lambda }^{2}V$. The wedge productis a nondegenerate symmetric bilinear form, and the action preserves it because $\mathfrak{sl}(V)$ acts trivially on ${\Lambda }^{4}V$. This gives an injective homomorphismBoth Lie algebras have dimension $15$, so the map is an isomorphism. This realizes the Isomorphism between so6 and sl4.

The Lie algebra ${\mathfrak{sl}}_2$ has generators $e,h,f$ with $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$.

This is the $n=2$ instance of the special linear Lie algebra. Classifying its modules is a separate representation-theory problem.

Over a field $k$, the trace-zero $2\times 2$ matrices form the special linear Lie algebra ${\mathfrak{sl}}_2(k)$. The matrix trace of a commutator is zero, so this is a Lie subalgebra of the general linear Lie algebra. A basis isIn fact $\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)=aH+bE+cF$, uniquely. The individual products areSubtracting in the stated order givesFor characteristic different from two, this is the standard sl2 triple; see Etingof, Section 11.4. The products remain valid in characteristic two, but there $H=I$ is central and $2E=2F=0$, so the familiar characteristic zero structure changes.

The following table gives all nine basis brackets. Each entry is $[\text{row},\text{column}]$; the diagonal vanishes by alternation, and exchanging row and column reverses the sign.

Bilinearity now computes any commutator, for exampleThe three structure constants of a Lie algebra relations encode all of this multiplication. They also show why the sl2 triple is useful in Lie algebra representations: the actions of $E$ and $F$ move between eigenvalues of the action of $H$ in steps of two.

For the defining three-dimensional special linear Lie algebra representation $V$, contraction $(uv)\otimes f\mapsto f(u)v+f(v)u$ maps $(S^{2}V)\otimes V^{\ast }$ onto $V$. The kernel contains the highest-weight vector $e_1^2\otimes e_3^{\ast }$ of Dynkin labels $(2,1)$. The Weyl dimension formula gives its irreducible module dimension $15$, the entire kernel dimension; the Weyl complete reducibility theorem supplies the displayed splitting.

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 $24+15+6=45$.

For the special linear Lie algebra ${\mathfrak{sl}}_3$, the irreducible highest-weight representation with Dynkin labels $(a,b)$ has the displayed formal character on diagonal matrices with $x_1{x}_2{x}_3=1$. The interlacing rows $(a+b,b,0)$, $(p,q)$ and $(r)$ enumerate a Gelfand–Tsetlin basis; each triple contributes the exponent differences $(r-(p+q-r),(p+q-r)-(a+2b-p-q))$ as its weight. This gives exact integer weight multiplicities, including repeated weights, without floating-point character evaluation.

For the special linear Lie algebra $\mathfrak{sl}(3)$, with Dynkin labels indexing irreducible highest-weight representations,The summands have dimensions $60,24,15,15,6$, summing to $120=15\cdot 16/2$. The identity follows from the formal character of a weight module formula ${\chi }_{S^{2}W}(z)=({\chi }_W(z{)}^2+{\chi }_W(z^2))/2$ and the Weyl character formula.

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.