Lemma 5

There exists a unique involutive automorphism $X \rightarrow X^*$ of the ring $\mathscr{R}(\mathfrak{a})$ such that $[E]^*= [E^*]$ for every finite dimensional $\mathfrak{a}$-module E.

The uniqueness is clear. By Lemma 3, there exists a homomorphism $X \rightarrow X^*$ from the additive group $\mathscr{R}(\mathfrak{a})$ to itself such that $[E]^*= [E^*]$ for every finite dimensional $\mathfrak{a}$-module E. We have $(X^*)^*= X$, so this homomorphism is involutive. It is an automorphism of the ring $\mathscr{R}(\mathfrak{a})$ since $(E\otimes F)^*$ is isomorphic to $E^*\otimes F^*$ for all finite dimensional $\mathfrak{a}$-modules E and F. Q.E.D.

Let $U(\mathfrak{a})$ be the enveloping algebra of $\mathfrak{a}, U(\mathfrak{a})^*$ the vector space dual of $U(\mathfrak{a})$. Recall (Chap. II, §1, no. 5) that the coalgebra structure of $U(\mathfrak{a})$ defines on $U(\mathfrak{a})^*$ a commutative, associative algebra structure with unit element. For any finite dimensional $\mathfrak{a}$-module E, the map $u \rightarrow$ Tr($u_E$) from $U(\mathfrak{a})$ to $k$ is an element $\tau_E$ of $U(\mathfrak{a})^*$. If $0\longrightarrow E'\longrightarrow E\longrightarrow E''\longrightarrow 0$ is an exact sequence of finite dimensional $\mathfrak{a}$-modules, then $\tau_E=\tau_{E'}+\tau_{E''}$. Hence, by Lemma 3 there exists a unique homomorphism, which we denote by Tr, from the additive group $\mathscr{R}(\mathfrak{a})$ to the group $U(\mathfrak{a})^*$ such that Tr[E] $=\tau_E$ for every finite dimensional $\mathfrak{a}$-module E. If $k$ denotes the trivial $\mathfrak{a}$-module of dimension 1, it is easy to check that Tr[$k]$ is the unit element of $U(\mathfrak{a})^*$. Finally, let E and F be finite dimensional $\mathfrak{a}$-modules. Let $u\in U(\mathfrak{a})$ and let $c$ be the coproduct of $U(\mathfrak{a})$. By definition of the U-module $E\otimes F$ (Chap. I, §3, no. 2), if $c(u) =\sum_iu_i\otimes u'_i$,

Consequently

$\tau_{E\otimes F}(u) =\sum_i$ Tr($u_i$)$_E$Tr($u'_i$)$_F=\sum_i\tau_E(u_i)\tau_F(u'_i)$

This means that $\tau_E\tau_F=\tau_{E\otimes F}$. Thus, Tr$:\mathscr{R}(\mathfrak{a})\rightarrow U(\mathfrak{a})^*$ is a homomorphism of rings.

Let $\mathfrak{a}_1$ and $\mathfrak{a}_2$ be Lie algebras, $f$ a homomorphism from $\mathfrak{a}_1$ to $\mathfrak{a}_2$. Every finite dimensional $\mathfrak{a}_2$-module E defines by means of $f$ an $\mathfrak{a}_1$-module, hence elements of $\mathscr{R}(\mathfrak{a}_2)$ and $\mathscr{R}(\mathfrak{a}_1)$ that we denote provisionally by $[E]_2$ and $[E]_1$. By Lemma 3, there exists a unique homomorphism, denoted by $\mathscr{R}(f)$, from the group $\mathscr{R}(\mathfrak{a}_2)$ to the group $\mathscr{R}(\mathfrak{a}_1)$ such that $\mathscr{R}(f)[E]_2=[E]_1$ for every finite dimensional $\mathfrak{a}_2$-module E. Moreover, $\mathscr{R}(f)$ is a homomorphism of rings. If U(f) is the homomorphism from U($\mathfrak{a}_1$) to U($\mathfrak{a}_2$) extending $f$, the following diagram is commutative

In what follows we take for $\mathfrak{a}$ the splittable semi-simple Lie algebra $\mathfrak{g}$. The ring $\mathscr{R}(\mathfrak{g})$ is called the representation ring of $\mathfrak{g}$. For all $\lambda\in P_{++}$, we denote by $[\lambda]$ the class of the simple $\mathfrak{g}$-module E($\lambda$) of highest weight $\lambda$.