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Assume $0\le k\le n$. Let $V=L_{n-1}$ be the $n$-dimensional irreducible sl2 Lie algebra representation. Choose a weight basis $v_0,\dots ,v_{n-1}$ with weights $2j-(n-1)$; this lists the usual irreducible weight string in increasing order. Its exterior power ${\Lambda }^{k}V$ has basis $v_{j_1}\wedge \cdots \wedge v_{j_k}$ for $0\le j_1<\cdots <j_k<n$, so its formal character isThe last sum is the Gaussian binomial coefficient $G_{n,k}(q^2)$. To verify the product identity, splitting the subsets according to whether they contain $n-1$ givesThe product ${\prod }_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence, after putting both terms over their common denominator. Thus induction identifies it with the subset sum, proving it is a polynomial with nonnegative integer coefficients.

The quantum integer convention in the PDF hasMultiplying these factors in the factorial quotient givesThis establishes the required representation-theoretic interpretation.

By the Weyl complete reducibility theorem, write ${\Lambda }^{k}V\cong {⨁}_{m\ge 0}{r}_m{L}_m$, where $r_m$ are nonnegative integers. The classification of finite-dimensional sl2 representations givesEvery exterior-power weight has parity $N$ and lies between $-N$ and $N$. Therefore only $m\le N$ with $m\equiv N(\mathrm{mod}2)$ occur. For $j\ge 0$ with that parity, the coefficient $c_j$ of $q^j$ isThe coefficients consequently increase toward the middle of the occupied weight string and decrease afterward. If $G_{n,k}(t)={\sum }_{s=0}^N{a}_s{t}^s$, then $a_s=c_{-N+2s}$. HenceThis is the unimodality of Gaussian binomial coefficients, proved through parity unimodality of an sl2 character. It includes $k=0,n$, when the character is one.

There is a normalization qualification in the printed wording. The symmetric quantum-binomial quotient is a Laurent polynomial in $q$, with exponents spaced by two; the unimodality just proved is for that occupied parity, or equivalently for the ordinary polynomial $G_{n,k}(t)$ after setting $t=q^2$ and removing the monomial shift. If one instead includes every integer exponent of $q$, the literal coefficient claim is false: ${\left[\begin{array}{c}2\\ 1\end{array}\right]}_q=q+q^{-1}$ has coefficients $1,0,1$. The representation-theoretic conclusion is unimodality on the parity string, and ordinary unimodality of the Gaussian polynomial.

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.

The quotient $[n{]}_q[n-1{]}_q\cdots [n-k+1{]}_q/([k{]}_q\cdots [1{]}_q)$ is the displayed shifted Gaussian binomial coefficient. It is the formal character of ${\Lambda }^k{L}_{n-1}$, where $L_{n-1}$ is the $n$-dimensional irreducible sl2 Lie algebra representation. Indeed its exterior-power weights are sums of $k$ distinct terms of $n-1,n-3,\dots ,1-n$. Parity unimodality of an sl2 character proves unimodality after removing the intervening zero coefficients; equivalently, the ordinary Gaussian polynomial has unimodal coefficients.