Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 (Lie Algebras and Their Representations)

The Adjoint representation of a Lie algebra isThe Killing form is the symmetric invariant bilinear form

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A vector subspace $I\subseteq \mathfrak{g}$ is an ideal of a Lie algebra when $[\mathfrak{g},I]\subseteq I$. A Nilpotent Lie algebra is one whose lower central serieseventually vanishes.

By the Engel theorem, the operators ${\mathrm{ad}}_x$ for a nilpotent complex Lie algebra can be represented simultaneously by strictly upper triangular matrices. Their products are strictly upper triangular and have zero trace. Hence

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A Solvable Lie algebra is one whose derived serieseventually vanishes. By the Lie theorem, the adjoint operators of a solvable complex Lie algebra are simultaneously upper triangular. If $x\in [\mathfrak{g},\mathfrak{g}]$, then ${\mathrm{ad}}_x$ is a sum of commutators of upper triangular matrices and is therefore strictly upper triangular. For every $y\in \mathfrak{g}$, the product ${\mathrm{ad}}_x{\mathrm{ad}}_y$ is strictly upper triangular, soThus

For a nonzero example, let $\mathfrak{g}$ have basis $h,e$ with $[h,e]=e$. It is solvable because $[\mathfrak{g},\mathfrak{g}]=\mathbb{C}e$ is abelian, but in the ordered basis $(h,e)$,

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Invariance of the Killing form givesIf $x\in {\mathfrak{g}}^{\perp }$, the right-hand side vanishes for every $y$, so $[a,x]\in {\mathfrak{g}}^{\perp }$. Thus ${\mathfrak{g}}^{\perp }$ is an ideal.

Let $\mathfrak{h}=\mathrm{ad}({\mathfrak{g}}^{\perp })\subseteq \mathfrak{gl}(\mathfrak{g})$. For $x\in [{\mathfrak{g}}^{\perp },{\mathfrak{g}}^{\perp }]$ and $y\in {\mathfrak{g}}^{\perp }$,The Cartan solvability criterion makes $\mathfrak{h}$ solvable. The kernel of the adjoint map on ${\mathfrak{g}}^{\perp }$ lies in its center and is abelian, so ${\mathfrak{g}}^{\perp }$ is itself solvable. This proves the Solvability of the radical of the Killing form.

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The Killing form of a complex Simple Lie algebra is nondegenerate. Since $B$ is also nondegenerate, there is a unique endomorphism $T$ of $\mathfrak{g}$ satisfyingInvariance of both forms givesso $T$ intertwines the adjoint representation. That representation is irreducible because its invariant subspaces are ideals. The Schur lemma therefore gives $T=cI$. Since $\kappa$ is nondegenerate, $c\ne 0$, andThis is the uniqueness of an invariant bilinear form on a simple Lie algebra.

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For ${\mathfrak{sl}}_3(\mathbb{C})$, the Killing form of the special linear Lie algebra isFor $H_1=E_{11}-E_{22}$ and $H_2=E_{22}-E_{33}$, its matrix on the Cartan part iswhose determinant is $108$. Each $E_{ij}$ pairs only with $E_{ji}$, with value $6$. In the stated ordering, the remaining block iswhose determinant is $-6^6$. Therefore

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A Weyl chamber is a connected component ofA root basis $\Delta$ is a basis of $E$ made of roots such that every root is an integer combination of $\Delta$ whose nonzero coefficients all have one sign.

Choose a regular vector $\gamma$, meaning $(\gamma ,\alpha )\ne 0$ for every root. DeclareThe indecomposable roots in ${\Phi }_{\gamma }^{+}$ form a root basis ${\Delta }_{\gamma }$, and every root basis arises in this way. Its chamber is the component containing $\gamma$.

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Root bases correspond bijectively to Weyl chambers: the walls of a chamber determine its inward simple roots. The Weyl group acts transitively on the chambers. One proof joins interior points of two chambers by a generic line segment. Each time the segment crosses one reflecting hyperplane, reflect the remaining segment across that wall; the resulting product of root reflections sends the first chamber to the second. It consequently sends the first root basis to the second. Thus $W$ acts transitively on root bases.

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We induct on the Coxeter length $\ell (w)$. There is nothing to prove when $\ell (w)=0$. Otherwise choose a simple root $\alpha$ such thatequivalently, $w\alpha$ is negative. Since $\lambda$ and $\mu =w\lambda$ lie in the Closed dominant Weyl chamber,Therefore $⟨\lambda ,{\alpha }^{\vee }⟩=0$, and the simple reflection $s_{\alpha }$ fixes $\lambda$. Moreover,The induction hypothesis writes $w{s}_{\alpha }$ as a product of simple reflections that fix $\lambda$. Multiplying on the right by $s_{\alpha }$ gives the required expression for $w$. This is the Weyl stabilizer of a dominant point lemma.

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For existence, choose $w\in W$ maximizing $(w\lambda ,\rho )$, where $\rho$ lies in the interior of the dominant chamber. If $⟨w\lambda ,{\alpha }^{\vee }⟩<0$ for a simple root $\alpha$, thencontradicting maximality. Hence $w\lambda$ is dominant.

For uniqueness, suppose $\lambda$ and $\mu$ are dominant and $\mu =w\lambda$. Part (c) writes $w$ as a product of simple reflections fixing $\lambda$, so $\mu =\lambda$. Every Weyl orbit therefore has exactly one representative in the closed dominant chamber.

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Every diagonal element of the stated Cartan subalgebra has the formLet ${\epsilon }_i$ extract $t_i$. The roots are the B2 root system

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ChooseHere ${\alpha }_1$ is long and ${\alpha }_2$ is short. The Dynkin diagram consists of two vertices joined by a double edge, with its arrow pointing from ${\alpha }_1$ toward the shorter root ${\alpha }_2$:

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The Weyl reflection $s_{{\alpha }_1}$ swaps ${\epsilon }_1$ and ${\epsilon }_2$, whereas $s_{{\alpha }_2}$ changes the sign of ${\epsilon }_2$. Hence

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The highest root isSince $\beta$ is long and the ${\epsilon }_i$ are orthonormal, ${\beta }^{\vee }=\beta$. The coroot pairing gives

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The displayed calculation identifies the root system of ${\mathfrak{so}}_5(\mathbb{C})$ as $B_2$. The symplectic Lie algebra ${\mathfrak{sp}}_4(\mathbb{C})$ has root system $C_2$. After exchanging the two simple-root labels, the $B_2$ and $C_2$ Cartan matrices agree, so their Dynkin diagrams define the same complex simple Lie algebra. The classification of finite-dimensional complex simple Lie algebras therefore givesThis is the Isomorphism between so5 and sp4.

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Choose positive roots ${\Phi }^{+}$ and letbe the Weyl vector. For a dominant integral weight $\lambda$, the Weyl dimension formula isHere ${\alpha }^{\vee }$ is the coroot and $⟨,⟩$ is the natural weight-coroot pairing.

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Write the nonzero highest weight of $V$ asChoose $i$ with $m_i>0$. Then $\lambda -{\omega }_i$ is dominant. Every positive coroot is a nonnegative combination of simple coroots, so every numerator in the Weyl dimension formula for $\lambda$ is at least the corresponding numerator for ${\omega }_i$. ThusMinimality of $\dim V$ forces equality. If $\lambda \ne {\omega }_i$, some simple-coroot factor is strictly larger, making the product strict. Hence $\lambda ={\omega }_i$ andso $V$ is a fundamental representation.

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For an irreducible representation $V(\lambda )$, writeFormThe tensor product of highest-weight vectors is killed by all positive-root spaces and has weight $\lambda$. It therefore generates a highest-weight constituent isomorphic to $V(\lambda )$. By Complete reducibility of semisimple Lie algebra representations, this constituent is a subrepresentation of $T$. This proves the generation by fundamental representations statement.

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For ${\mathfrak{sl}}_3(\mathbb{C})$ there are two fundamental weights. The corresponding Fundamental representations of sl3 are

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Let $V=V({\omega }_1)$, so $V^{\ast }=V({\omega }_2)$. First,The highest-weight tensor-product rule givesandThe dimensions $15+3+6+3=27$ check the decomposition. Therefore the Triple tensor decomposition for the defining sl3 representation isThus one may take

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The Classification of finite-dimensional sl2 representations says that for every $n\ge 0$ there is one irreducible module $V(n)$ of dimension $n+1$, with weightsof multiplicity one, and every finite-dimensional ${\mathfrak{sl}}_2(\mathbb{C})$-module is a direct sum of these irreducibles.

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The weight lattice isFor each root $\alpha$, restrict $V$ to the subalgebraIf $V_{\lambda }\ne 0$, the ${\mathfrak{sl}}_2$ classification shows that the $h_{\alpha }$-eigenvalue $\lambda (h_{\alpha })=⟨\lambda ,{\alpha }^{\vee }⟩$ is an integer. Hence $\lambda \in X$.

The same classification makes every $\alpha$-string symmetric underand preserves weight multiplicity. Since the Weyl group is generated by these simple reflections,

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Take $\mathfrak{g}={\mathfrak{sl}}_3(\mathbb{C})$ and its adjoint representation $V=\mathfrak{g}$. This representation is irreducible because $\mathfrak{g}$ is simple. Its zero-weight space is the Cartan subalgebra $\mathfrak{t}$, so

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The dominance order isSuppose $\lambda \ne \mu$. Put $\delta =\lambda -\mu$. Since $\mu$ is dominant,Writing $\delta ={\sum }_i{n}_i{\alpha }_i$, some index with $n_i>0$ therefore satisfies $(\lambda ,{\alpha }_i)>0$, equivalently $⟨\lambda ,{\alpha }_i^{\vee }⟩>0$. The ${\mathfrak{sl}}_2$ lowering operatoris injective by the Injectivity of sl2 lowering above weight zero. HenceThe new weight still dominates $\mu$ in the partial order. Iterating until reaching $\mu$ givesThis is the weight multiplicity decreases away from a dominant weight property.

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