Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 5

Choose a Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$. For $\lambda \in {\mathfrak{t}}^{\ast }$, let ${\mathbb{C}}_{\lambda }$ be the one-dimensional $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isIts universal property of a Verma module says that any vector of weight $\lambda$ annihilated by ${\mathfrak{n}}^{+}$ receives the canonical highest-weight vector under one unique module homomorphism from $M(\lambda )$.

The sum of the proper submodules of $M(\lambda )$ is its unique maximal proper submodule, because no proper submodule contains the highest-weight vector. Its quotient $V(\lambda )$ is therefore the unique irreducible quotient of a Verma module, and hence the unique irreducible highest-weight module of weight $\lambda$.

The module $V(\lambda )$ is finite-dimensional exactly when $\lambda$ is a dominant integral weight:for every simple root ${\alpha }_i$.

For ${\mathfrak{sl}}_2$, the Verma module has basisby the Poincare-Birkhoff-Witt theorem. If $hv=\lambda v$, thenFor $\lambda =-d$ with $d>0$, the coefficient isfor every $r\ge 1$. Thus no $f^{r}v$ with $r>0$ is a singular vector.

Any nonzero submodule contains a weight vector $f^{r}v$ because the $h$-weights are distinct. Applying $e^r$ gives a nonzero multiple of $v$, after which applying powers of $f$ generates all of $M(-d)$. HenceThis is also the negative-highest-weight case of Reducibility of an sl2 Verma module.

For a dominant integral weight $\lambda$, the Weyl character formula iswhere $W$ is the Weyl group, $\ell$ is Coxeter length, and $\rho$ is the half-sum of positive roots. The Weyl denominator formula is

Set $\lambda =k\rho$. Apply the denominator identity after replacing every formal exponential $e^{\mu }$ by $e^{(k+1)\mu }$:Dividing this by the ordinary denominator gives

Realize the B2 root system in ${\mathbb{R}}^2$ asChoose the short simple root and long simple rootThenas follows from $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. The roots form a square from the long roots with the four short roots on the coordinate axes; the double edge in the Dynkin diagram points toward ${\alpha }_1$.

The positive roots areFor $\lambda =a{\omega }_1+b{\omega }_2$, substituting their coroot pairings in the Weyl dimension formula givesThis is the Weyl dimension formula for B2.

The weights of the defining five-dimensional representation areIts highest weight is ${\epsilon }_1={\omega }_2$, so irreducibility identifies it as

If $v_{+}$ is a highest-weight vector, then $v_{+}\otimes v_{+}$ is a highest-weight vector of weight $2{\omega }_2$ in $V\otimes V$. The subrepresentation it generates is therefore $V(2{\omega }_2)$. Equivalently, it is the Traceless symmetric square of the defining so5 representation; the invariant quadratic form supplies the complementary trivial line in $S^{2}V$.

Putting $a=0$ and $b=2$ into the formula from part i gives