Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 4 / 5

Let $h$ be the standard Cartan element of the sl2 Lie algebra, with $[h,e]=2e$ and $[h,f]=-2f$. For a finite-dimensional representation, its formal character records its weight multiplicities:The classification of finite-dimensional sl2 representations and the Weyl complete reducibility theorem give $V={⨁}_{m\ge 0}{a}_m{L}_m$, with finitely many nonzero $a_m$. Each $L_m$ has weights $m,m-2,\dots ,-m$, each of multiplicity one. ThereforeIn particular $c_j=c_{-j}$, and for $j\ge 0$ one has $c_j-c_{j+2}=a_j\ge 0$. This proves parity unimodality of an sl2 character: within each parity, the coefficients increase towards zero and decrease away from zero. For a representation whose weights have a single parity, deleting the intervening zero coefficients gives a symmetric unimodal sequence.

The parity convention is necessary for the printed assertion. If unimodality means the full coefficient sequence at every consecutive integer exponent, it is false: the standard two-dimensional representation has character $q+q^{-1}$, whose coefficients at exponents $-1,0,1$ are $1,0,1$. The statement established above is unimodality on cosets of the root lattice for ${\mathfrak{sl}}_2$, and is the version needed in part ii. Infinite-dimensional representations need not even have Laurent-polynomial characters.

Assume $0\le k\le n$ with $n,k$ integers, as required for the quantum binomial expression. Let $L_{n-1}$ be the $n$-dimensional irreducible sl2 Lie algebra representation, with one-dimensional weight spaces of weights $n-1-2j$, $0\le j<n$. The exterior-power Lie algebra representation on ${\Lambda }^k{L}_{n-1}$ has characterThis is already a Laurent polynomial with nonnegative integer coefficients.

To identify it with the requested quotient, introduce $t=q^{-2}$ and the Gaussian binomial coefficientSplitting the subsets according to whether $n-1$ is included provesThe product ${\prod }_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence: after factoring common numerator and denominator terms, the needed identity is $(1-t^{n-k})+t^{n-k}(1-t^k)=1-t^n$. Induction on $n$ therefore proves the product identity, including its polynomiality.

Since the quantum integer is $[a{]}_q=q^{a-1}(1-q^{-2a})/(1-q^{-2})$, taking the quotient of the products givesThus this is the symmetric quantum binomial coefficient. All its weights have parity $k(n-k)$, because the exponent in the exterior-power sum differs from $k(n-k)$ by an even integer. Applying part i to this finite-dimensional representation proves symmetric unimodality in steps of two. Equivalently, the ordinary polynomial $G_{n,k}(t)$ has a unimodal coefficient sequence at consecutive powers of $t$.

Again, the full integer-exponent coefficient sequence in $q$, including the absent parity, need not be unimodal: for $n=2,k=1$ the answer is $q+q^{-1}$. This counterexample makes precise the convention under which the claimed quantum unimodality holds. The cases $k=0$ or $k=n$ give the constant one.