Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 102 / 1

For an affine algebraic group $G$, its Lie algebra is the tangent space at the identity,Equivalently, it is the space of left-invariant derivations of the coordinate ring $\mathbb{C}[G]$. The Lie bracket is the commutator of derivations,It is antisymmetric because $[Y,X]=-[X,Y]$. Associativity of composition givesafter all six triple products cancel in pairs, proving the Jacobi identity. This construction is the Lie algebra of an affine algebraic group.

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.

The finite-dimensional irreducible rational representations of $S{L}_2(\mathbb{C})$ areof dimension $n+1$. The central element $-I$ acts on $V_n$ by $(-1{)}^n$. Therefore precisely the even-indexed representationsdescend to irreducible representations of $PG{L}_2(\mathbb{C})$. This is the Descent of an irreducible SL2 representation to PGL2.

A bilinear form $B$ on a Lie algebra is invariant whenequivalently $B([x,y],z)+B(y,[x,z])=0$.

Choose a basis $(x_i)$ of $\mathfrak{g}$ and its $B$-dual basis $(x^i)$. The Casimir element iswhere $U(\mathfrak{g})$ is the universal enveloping algebra. The tensor ${\sum }_i{x}_i\otimes x^i$ corresponds under $B:\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$ to the identity endomorphism, so invariance of $B$ makes it fixed by the diagonal adjoint action. Applying multiplication $U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$ givesfor every $x\in \mathfrak{g}$. Thus $\Omega$ lies in the center of an associative algebra of $U(\mathfrak{g})$.

Use the standard basisof ${\mathfrak{sl}}_2$. For the invariant trace form $B(x,y)=\mathrm{tr}(xy)$, the dual basis is $(f,e,h/2)$, soOn a highest-weight vector $v$ in the $(n+1)$-dimensional irreducible module, $ev=0$ and $hv=nv$. Since $efv=[e,f]v=nv$ and $fev=0$,Centrality and Schur lemma make this the eigenvalue on the whole module. Thus $\lambda =n(n+2)/2$, the Casimir eigenvalue for sl2. If the form is instead the Killing form, which is four times the trace form on ${\mathfrak{sl}}_2$, the corresponding Casimir and eigenvalue are divided by four.

No. Let $\mathfrak{g}={\mathfrak{sl}}_2$, let $\mathfrak{s}=\mathbb{C}e$ be the one-dimensional subalgebra generated by the standard raising operator, and let $L={\mathbb{C}}^2$ be the irreducible defining representation. On restriction to $\mathfrak{s}$, the element $e$ acts by a nonzero Nilpotent Jordan block. A direct sum of irreducible representations of the one-dimensional abelian Lie algebra would make $e$ diagonalizable, so this restriction is not completely reducible. The Complete reducibility of semisimple Lie algebra representations applies when the restricting algebra is semisimple, which $\mathfrak{s}$ is not.