Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 1 / 5

For a finite-dimensional Lie algebra representation $V$, a weight of a representation relative to $\mathfrak{h}$ is a linear functional $\mu \in {\mathfrak{h}}^{\ast }$ for which the simultaneous eigenspaceis nonzero. Its elements are weight vectors, and $\dim V_{\mu }$ is the weight multiplicity. The set of weights is $\boxed{{\Lambda }_V=\{\mu :V_{\mu }\ne 0\}}$. For a complex semisimple Lie algebra, the Cartan action on a finite-dimensional module is simultaneously diagonalizable, giving $V={⨁}_{\mu \in {\Lambda }_V}{V}_{\mu }$. The set of weights and their multiplicities are distinct information: several basis vectors can occupy the same point of a weight diagram.

Let $E$ be the real span of the roots in ${\mathfrak{h}}^{\ast }$. The weight lattice iswhere $h_{\alpha }$ is the coroot in the root's sl2 subalgebra associated with a root, normalized by $\alpha (h_{\alpha })=2$. It suffices to test the simple coroots. Equivalently, if ${\omega }_i(h_{{\alpha }_j})={\delta }_{ij}$ defines the fundamental weights, then $P={⨁}_i\mathbb{Z}{\omega }_i$.

To prove integrality of representation weights on coroots, restrict a finite-dimensional module $V$ to the root's sl2 Lie algebra. By the classification of finite-dimensional sl2 representations, the eigenvalues of its Cartan element $h_{\alpha }$ are integers. Thus every $\mu \in {\Lambda }_V$ has $\mu (h_{\alpha })\in \mathbb{Z}$. The simple coroots form a basis of the real Cartan space, so these integral values also put $\mu$ in $E$. Therefore $\boxed{{\Lambda }_V\subseteq {\Lambda }_W}$ for every finite-dimensional representation in this setting. The weight lattice need not coincide with the smaller root lattice.

Choose a real linear functional $\ell :E\to \mathbb{R}$ nonzero on every root. Its positive system is $\boxed{{\Phi }^{+}=\{\alpha \in \Phi :\ell (\alpha )>0\}}$. The other roots form ${\Phi }^{-}=-{\Phi }^{+}$. Thus exactly one of $\alpha ,-\alpha$ is positive, and a sum of positive roots that is a root is again positive because $\ell$ is additive.

The simple roots are the positive roots not expressible as a sum of two positive roots. They form a fundamental system of a root system; every positive root is a nonnegative integral combination of them. Positivity is a choice of ordering, rather than a property of a root independent of that choice.

For the chosen simple roots $\Delta$, the open fundamental chamber of a root system isEquivalently $\lambda (h_{\alpha })>0$ for every simple coroot. Since all positive roots are nonnegative combinations of simple roots, requiring the inequality for all positive roots gives the same chamber. It is a connected component of the complement of the root reflection hyperplanes; the Weyl group permutes those chambers simply transitively.

The Closed dominant Weyl chamber is obtained by replacing the inequalities by $\ge 0$. It is the cone generated by the fundamental weights. When the fundamental Weyl chamber is taken to include its walls, this closed convention is intended; both conventions determine the same positive system and boundary rays.

For the usual diagonal Cartan subalgebra, write ${ϵ}_i(H)=H_{ii}$. The roots are ${ϵ}_i-{ϵ}_j$ for $i\ne j$. The three diagonal entries of the specified $H_0$ have orderingTherefore the positive roots selected by $\ell (\mu )=\mu (H_0)$ areTheir positive evaluations are respectively $2-\sqrt{2}$, $2\sqrt{2}-1$ and $1+\sqrt{2}$. For this nonstandard ordering the simple roots are ${\alpha }_1={ϵ}_2-{ϵ}_3$ and ${\alpha }_2={ϵ}_3-{ϵ}_1$; the third positive root is their sum. These simple roots differ from the usual choice used in Question 1.

Identify the real weight plane with triples $(u_1,u_2,u_3)$ of sum zero. Pairing with these simple coroots gives the differences $u_2-u_3$ and $u_3-u_1$. Consequently the fundamental chamber of a root system isIts closure replaces the inequalities by weak inequalities. The fundamental weights for this ordering are ${\omega }_1={ϵ}_2$ and ${\omega }_2=-{ϵ}_1$, represented by $(-1/3,2/3,-1/3)$ and $(-2/3,1/3,1/3)$. Their simple-coroot pairings are $(1,0)$ and $(0,1)$, so the two boundary rays areThe sketch uses ${ϵ}_1=(1,0)$, ${ϵ}_2=(-1/2,\sqrt{3}/2)$ and ${ϵ}_3=(-1/2,-\sqrt{3}/2)$ in a Euclidean plane. The shaded sector is the sixty-degree closed chamber; its interior is the open chamber. The ordering vector $H_0$, identified with its dual direction, lies inside it.