Past exam of the representation theory of semisimple Lie algebras course of the University of Oxford / 2010 (C2.2b, DMAC 2471)

Source: Mathematics release, bundle Part C 2007 11.pdf, pages 313–316. The Mathematics Part C paper was sat on 31 May 2010, 2.30–4.00pm.

Bounded exact-paper availability checks found no matching advertised answers. The general even-year release statement establishes coverage back to at least 2016 and does not itself establish this 2010 paper. All fourteen printed parts are addressed. Several literal source indexing and sign errors are retained in the private transcription and explained at their corresponding solutions.

Put $e=E_{+}$ and $f=E_{-}$. The supplied identities giveIf no $f^{k}v$ vanishes, this is the first alternative. Otherwise choose the least $N\ge 1$ with $f^{N}v=0$. Applying $e$ gives $N(\chi -N+1)f^{N-1}v=0$. Since $f^{N-1}v\ne 0$ and the field has characteristic zero, $\chi =N-1$. ThusThe endpoint must include zero: the trivial one-dimensional module has $\chi =0$ and $fv=0$, so the printed claim of strict positivity has that counterexample. All nonzero vectors in the displayed chain have distinct $H$-weights, and hence are linearly independent.

For the intended module, index a basis by $v_k=a_{l-2k}$ for $0\le k\le l$, and usewith $v_{-1}=v_{l+1}=0$. This makes the basis endpoints distinct, identifies the lowering operator, and includes its missing raising action at $k=l$. These are the necessary indexing corrections to the printed display. The same definition for $l=0$ supplies the trivial module needed for classification.

Directly on $v_k$,These hold at the endpoints too, proving the three defining ${\mathfrak{sl}}_2$ relations. FinallyThus the Casimir element acts by that scalar on the whole module.

Let $V\ne 0$ be finite-dimensional. Choose an $H$-eigenvector and apply $e$ repeatedly until the last nonzero vector $v$ in that chain. Such a last vector exists because $e$ raises the $H$-eigenvalue by two and $H$ has only finitely many eigenvalues. Then $ev=0$ and $Hv=lv$ for some $l$.

The nonzero $f$-chain cannot continue indefinitely, since its weights are distinct. Part (a), with its zero endpoint included, gives $l\in {\mathbb{Z}}_{\ge 0}$ and the basis $v,fv,\dots ,f^{l}v$. The raising formula in part (a) shows its span is a submodule isomorphic to $A_l$.

To prove $A_l$ irreducible, a nonzero submodule stable under diagonal $H$ contains a weight vector: polynomial interpolation in the distinct eigenvalues projects onto individual representation weight lines. Repeated raising reaches $v_0$, and repeated lowering then reaches every $v_k$. Thus the submodule is all of $A_l$. Distinct $l$ give different dimensions $l+1$, so the modules are nonisomorphic. Every finite-dimensional irreducible module contains one of these nonzero simple submodules and hence equals it. ThereforeThe zero module has no nonzero simple submodule and is dealt with as the empty direct sum in part (d).

We first prove that every extension of two simple modules splits. ConsiderIf $n\ne m$, the central Casimir element has distinct scalars $n(n+2)$ and $m(m+2)$. On $V$ its minimal polynomial divides the product of the corresponding two distinct linear factors. Its eigenspaces are submodules, and the eigenspace for $m(m+2)$ maps isomorphically onto the quotient. It is the required complement.

Now suppose $n=m=l$. Use generalized weight spaces for $H$; their possible weights are $l,l-2,\dots ,-l$, because the characteristic polynomial on an extension is the product of those on its two factors. Lift the quotient's highest vector to $w$ in the generalized weight space of representation weight $l$. Then $ew=0$ and $f^{l+1}w=0$, since these would lie in generalized weight spaces of weights $l+2$ and $-l-2$, which are absent. Also $(H-l)w=a{v}_0$ for a highest vector $v_0$ in the submodule $A_l$. The commutator identity givesThus $a=0$. The vector $w$ is an actual highest-weight vector. Its lowering chain maps to the full basis of the quotient, so it generates an $A_l$ complementary to the original submodule. This proves splitting also when the two simple types coincide, including $l=0$.

The same splitting holds for a semisimple kernel and simple quotient by induction on the number of kernel summands. Write the kernel as $U_1\oplus U_2$ with $U_1$ simple. Modulo $U_2$, split the extension by the simple quotient; the inverse image of that chosen quotient copy is an extension with kernel $U_2$. Split it inductively, obtaining a submodule in $V$ meeting the entire original kernel trivially.

Finally induct on $\dim V$. A maximal proper submodule has smaller dimension and hence is semisimple by induction; the quotient is simple by maximality. The just-proved splitting completes the induction. Hencewith only finitely many nonzero multiplicities. This proof does not assume the complete-reducibility theorem it is establishing.

Use a triangular decomposition of a Lie algebra $\mathfrak{g}={\mathfrak{n}}_{-}\oplus \mathfrak{h}\oplus {\mathfrak{n}}_{+}$. The highest weight space of $M^{\chi }$ has dimension one. If $z$ is central, $z{m}_{\chi }$ is still of representation weight $\chi -\rho$ and is killed by ${\mathfrak{n}}_{+}$, soSince the module is generated by $m_{\chi }$ and $z$ commutes with $U(\mathfrak{g})$, the same scalar acts on every vector. Addition, multiplication and the unit show that $z\mapsto {\theta }_{\chi }(z)$ is an algebra homomorphism to $\mathbb{C}$.

For the polynomial assertion, the relevant variable for an element of $S(\mathfrak{h})$ is $\chi$. The dependence on $z$ itself is linear, as already follows from the scalar action. Apply the Poincaré-Birkhoff-Witt theorem ordering $U({\mathfrak{n}}_{-})U(\mathfrak{h})U({\mathfrak{n}}_{+})$. Because $z$ has adjoint representation weight zero, its surviving action on the highest vector is precisely the pure $U(\mathfrak{h})$ component $p_z$; terms with a positive root factor kill that vector, and a term with only a nontrivial negative root factor could not have representation weight zero. Since $U(\mathfrak{h})=S(\mathfrak{h})$,This specifies the intended polynomial despite the printed variable mismatch. The unadorned module symbol in that sentence is understood as $M^{\chi }$.

Write $h_{\alpha }$ for the simple coroot and choose the corresponding normalized ${\mathfrak{sl}}_2$ triple. The order condition meansThe highest vector has $h_{\alpha }$-representation weight $⟨\chi -\rho ,{\alpha }^{\vee }⟩=k-1$. The raising-lowering identity givesOther simple positive root vectors commute with $E_{-\alpha }$ and also kill this vector. Thus it is a highest vector of representation weightThe universal property of the Verma module gives the required homomorphism. It is injective: under the Poincaré-Birkhoff-Witt theorem identification of each Verma module with the free rank-one $U({\mathfrak{n}}_{-})$ module, it is right multiplication by $E_{-\alpha }^k$. The algebra $U({\mathfrak{n}}_{-})$ has no zero divisors, because its associated graded algebra is the polynomial domain $S({\mathfrak{n}}_{-})$. HenceFor $k=0$ this is the identity map, so the equality case of the order is covered.

For every $\chi$ on a hyperplane $⟨\chi ,{\alpha }^{\vee }⟩=k$, with $k\in {\mathbb{Z}}_{\ge 0}$, part (b) embeds one Verma module into the other. A central element acts on the nonzero image by both scalars, so $j_z(\chi )=j_z(s_{\alpha }\chi )$ there.

The difference of these two expressions is a polynomial on ${\mathfrak{h}}^{\ast }$. Fix all coordinates parallel to those hyperplanes and vary the remaining coordinate. It vanishes at every nonnegative integer, so as a polynomial in that coordinate it is identically zero. Its coefficient polynomials therefore vanish as well. Consequently the equality holds for all $\chi$. Since the simple reflections generate the Weyl group,

In the shifted normalization of part (a), the Harish-Chandra isomorphism is the algebra isomorphismThus the centre is identified with the Weyl-invariant polynomials on ${\mathfrak{h}}^{\ast }$. Central characters satisfy ${\theta }_{\chi }={\theta }_{\xi }$ exactly when $\chi$ and $\xi$ belong to the same Weyl orbit. The $\rho$ shift is already incorporated in $M^{\chi }$ having highest weight $\chi -\rho$; without that shift the corresponding Weyl action is the dot action.

The centre commutes with $\mathfrak{h}$, so every nonzero representation weight component of the common eigenvector $m$ has the same central character $\theta$. Pick such a component $m_{\lambda }$. In Category O, its $U({\mathfrak{n}}_{+})$-span is finite-dimensional and has finitely many weights. Choose a maximal representation weight in that span and a nonzero vector $v$ of that representation weight, say $\nu$. Every positive root operator kills $v$, since it would raise the representation weight within the same finite span. Centrality ensures $zv=\theta (z)v$.

The universal property supplies a nonzero map $M^{\nu +\rho }\to M$ taking its highest vector to $v$. The centre acts on that image both by $\theta$ and by ${\theta }_{\nu +\rho }$. Hence

Use the standard defining properties of Category O: finite generation over $U(\mathfrak{g})$, a direct sum of finite-dimensional weight spaces, and locally finite positive root action. In particular the weights lie in a finite union of downward root cones. Submodules and quotients retain the representation weight bounds and finite weight spaces; finite generation of submodules follows from the Poincaré-Birkhoff-Witt theorem and the noetherian property of $U(\mathfrak{g})$.

Choose finitely many weight vector generators. The sum $T$ of their full weight spaces is finite-dimensional and preserved by the centre. The kernel of its central action has finite codimension in $Z(U(\mathfrak{g}))$ and kills the generators, hence, by centrality, all of $M$. Thus only finitely many central characters ${\theta }_1,\dots ,{\theta }_r$ can occur in any subquotient: they are the characters of a fixed finite-dimensional commutative quotient algebra of the centre.

By the Harish-Chandra isomorphism each ${\theta }_i$ determines one finite Weyl orbit of shifted weights. Let $S$ be the finite union of these orbits after subtracting $\rho$. Every nonzero subquotient has a nonzero highest-weight vector, using the bounded-above weights. Within its finite-dimensional highest weight space, the commuting central operators have a common eigenvector. Part (e) and the central-character restriction show that its highest weight belongs to $S$.

Define for any subquotient $N$It is finite, additive in short exact sequences, and positive whenever $N\ne 0$. Hence each strict step of a submodule chain consumes at least one unit of $d(M)$. Refining a nonsimple factor into a proper submodule and quotient cannot continue more than $d(M)$ times. It ends with simple factors, provingThe argument proves the assertion for every object of $\mathcal{O}$, rather than just for a cyclic highest weight object.

For $f,g\in \mathcal{E}$, define their convolution byThe support restrictions make each sum finite: for fixed upper-cone representatives, the simple root coefficients in the two downward displacements are bounded nonnegative integers. This also proves closure in $\mathcal{E}$ and associativity by finite reordering at each coefficient. The unit is ${\delta }_0$, and ${\delta }_{\mu }\ast {\delta }_{\nu }={\delta }_{\mu +\nu }$.

With the source's negative root partition function, interpret $Q_{+}={\sum }_{\alpha \in B}{\mathbb{Z}}_{\ge 0}\alpha$ and allow zero coefficients in the partitions. Write $e^{\lambda }$ for ${\delta }_{\lambda }$, with multiplication meaning convolution. ThenThe denominator needed for the asserted character identities isThe repeated identical subscripts in the printed denominator would instead make it zero, so those literal factors cannot satisfy an inverse identity. Also the required inverse uses the negative $\rho$ shift. Geometric-series cancellation givesWith a positive shift the left side would be ${\delta }_{2\rho }$, which displays the second printed sign problem even after the denominator is repaired.

Poincaré-Birkhoff-Witt theorem gives ${\pi }_{M^{\xi }}={\delta }_{\xi -\rho }\ast K$, since each ordered negative root monomial lowers the highest weight by its corresponding partition. ThereforeThe remaining parts use $D$ for the intended Weyl denominator, while the private TeX preserves the printed signs.

For a finite-support formal character, use the same pullback rule $(wu)(\xi )=u(w^{-1}\xi )$; equivalently $w(e^{\lambda })=e^{w\lambda }$. This specifies the action on the denominator, which is a finite delta sum rather than a polynomial function. The whole infinite-support completion need not be Weyl-stable, so no such claim is required.

A simple reflection $s_{\alpha }$ permutes the positive roots other than $\alpha$ and sends $\alpha$ to $-\alpha$. In the product defining $D$, all factors are therefore permuted except the $\alpha$ factor, whose sign reverses. Thus $s_{\alpha }D=-D$. Simple reflections generate $W$, and each has determinant $-1$, so

Use the intended finite-dimensional irreducible highest weight module $A^{\varphi }$, with dominant integral highest weight $\varphi$. Its highest weight multiplicity is one. An arbitrary direct sum having that same largest representation weight would not in general satisfy this formula. Set $f={\pi }_{A^{\varphi }}$ and use the corrected denominator $D$ of part (a).

Restriction to each simple root ${\mathfrak{sl}}_2$ decomposes into symmetric finite representation weight strings, so $f$ is Weyl-invariant. Consequently $Df$ is Weyl-antisymmetric. We prove that no lower Weyl alternants occur, rather than assuming the desired character formula.

First establish the denominator identity itself. The finite sum $D$ has support in $\rho -Q_{+}$ and coefficient one at $\rho$. Antisymmetry kills all singular representation weight coefficients, since a reflection fixing the representation weight reverses their sign. Every remaining Weyl orbit has a unique regular dominant representative $\eta$. If it occurs in $D$, then $\eta \le \rho$. Since $\eta$ is integral and regular dominant, $\eta -\rho$ is dominant integral; it also belongs to $-Q_{+}$. A dominant vector $v=-\sum n_i{\alpha }_i$ satisfies $(v,v)=-\sum n_i(v,{\alpha }_i)\le 0$, so it must be zero. Thus the only possible orbit is $W\rho$. The leading coefficient givesFor the formal Laplacian $\mathcal{D}(e^{\lambda })=\Vert \lambda {\Vert }^2{e}^{\lambda }$, this implies $\mathcal{D}D=\Vert \rho {\Vert }^{2}D$.

Now choose root vectors dual under an invariant form, so $[E_{\alpha },E_{-\alpha }]=H_{\alpha }$ with $H_{\alpha }$ corresponding to the vector $\alpha$, and take the quadratic Casimirfor an orthonormal Cartan basis. On a highest vector its scalar is $c_{\varphi }=\Vert \varphi +\rho {\Vert }^2-\Vert \rho {\Vert }^2$, hence it has this scalar throughout the cyclic highest weight module.

Here is the computation of the weighted trace of its radial identity. For a positive root, let $X$ be the formal weighted trace of $E_{-\alpha }{E}_{\alpha }$ and $Y$ that of $E_{\alpha }{E}_{-\alpha }$. Cyclicity of trace and representation weight shifting give $Y=e^{\alpha }X$, while their difference is the weighted trace of $H_{\alpha }$, namely $(\alpha ,\nabla f)$. ThusTaking the trace of $C$ yieldsThe product expression for $D$ givesUsing $2\rho ={\sum }_{\alpha >0}\alpha$ and the product rule for the formal Laplacian therefore givesHence every representation weight in the support of $Df$ has norm $\Vert \varphi +\rho \Vert$. Its support is also contained in $\varphi +\rho -Q_{+}$, and it is antisymmetric. Any occurring regular dominant weight $\eta$ thus satisfies $\eta \le \varphi +\rho$. If the inequality were strict, thenbecause the first factor is a nonzero nonnegative sum of simple roots and the second is strictly dominant. This contradicts the equal-norm condition. Only the orbit $W(\varphi +\rho )$ remains. Its leading coefficient is one, proving

Convolve the identity of part (c) with the inverse denominator $K\ast {\delta }_{-\rho }$ found in part (a). Associativity givesTranslation by a Kronecker delta obeys $(K\ast {\delta }_{\lambda })(\psi )=K(\psi -\lambda )$. Taking the coefficient at $\psi$ therefore yieldsThis is the Kostant multiplicity formula in the paper's convention that $K$ counts negative root partitions. With a partition function counting positive root sums, each argument would have the opposite sign.