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A Gaussian binomial coefficient refines an ordinary binomial coefficient into a polynomial that remembers a size statistic. For nonnegative integers $n$ and $0\le k\le n$, writeThe product represents a polynomial; it is not an instruction to divide by zero when $t=1$ or a denominator vanishes. The Gaussian binomial coefficient recurrence gives a way to compute it with only addition and multiplication. The empty product for $k=0$ is one, and $G_{n,n}(t)=1$ as well. Setting $t=1$ after obtaining the polynomial recovers $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$.

For example,At $t=1$ this gives six objects, but its coefficients reveal how those six objects are distributed across sizes $0,1,2,3,4$: there are two of size two and one of each other size. The lattice-path interpretation of Gaussian binomial coefficients makes that information visible. Take unit steps north or east from $(0,0)$ to $(2,2)$. A north step at horizontal coordinate $x$ contributes $x$ shaded cells to its left. The six step words have total shaded areas $0,1,2,2,3,4$.

This is useful because an aggregate count can hide the feature one actually wants. A generating function retains the distribution of a statistic, so coefficient extraction answers questions such as “how many allowed diagrams contain exactly two cells?” The partition interpretation of Gaussian binomial coefficients makes every coefficient a nonnegative integer without relying on cancellation in the product formula.

An unexpected second use comes from linear algebra. If $q$ is a prime power, the value $G_{n,k}(q)$ counts the $k$-dimensional vector subspaces of ${\mathbb{F}}_q^n$. Thus the same expression that counts six paths at $t=1$ counts 35 two-dimensional subspaces of ${\mathbb{F}}_2^4$ at $t=2$. The Gaussian binomial coefficient counts finite-field subspaces article explains why selecting a vector subspace differs from selecting a subset of coordinates. ^[ref]

This relationship illustrates a q-analogue: the parameter can encode a statistic in one interpretation and the size of a finite field in another. Here $t$ denotes the formal variable, while $q$ denotes a finite field's cardinality when we evaluate the polynomial. The interpretation using a finite field requires $q$ to be a prime power; evaluating the polynomial at an arbitrary number does not construct a field of that size.

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.

One form of the Jacobi triple product is, for $0<|q|<1$ and $z\ne 0$,Both sides converge locally uniformly on the punctured $z$ plane. Here is a finite-product proof which also determines the multiplicative constant.

Put $Q=q^2$ and $(Q;Q{)}_m={\prod }_{j=1}^m(1-Q^j)$, with value one at $m=0$. Define the Gaussian binomial coefficient ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q=(Q;Q{)}_m/[(Q;Q{)}_k(Q;Q{)}_{m-k}]$. The recurrence ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q={\left[\genfrac{}{}{0ex}{}{m-1}{k}\right]}_Q+Q^{m-k}{\left[\genfrac{}{}{0ex}{}{m-1}{k-1}\right]}_Q$ proves by induction thatRewrite the two finite factors asThe finite identity, with $n=k-N$, makes this ${\sum }_{n=-N}^N{z}^n{q}^{n^2}{\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q$. Multiply by $(Q;Q{)}_N$. For each fixed integer $n$, the resulting coefficient tends to one, because $(Q;Q{)}_m$ tends to a nonzero limit and ${\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q\to 1/(Q;Q{)}_{\infty }$.

All finite $(Q;Q{)}_m$ are uniformly bounded above and away from zero. Thus these coefficients have one uniform bound, and ${\sum }_{n\in \mathbb{Z}}|q{|}^{n^2}|z{|}^n$ is uniformly summable on compact annuli. Dominated convergence passes the limit through the finite sums and proves the boxed formula. At $q=0$ the formula follows by its limiting value one. The frequently used alternative form isIt follows by the substitution $q^2=t$, $z=u/q$, and is the version used for the charge-counting identity in Question 8.

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.

A parameter-dependent generalization which recovers a familiar object at a specified value or limit of the parameter. In enumerative combinatorics, a generating function can refine an ordinary count by retaining a statistic; Gaussian binomial coefficients are one example.

A row echelon form is in reduced row echelon form when every pivot element is one and is the only nonzero entry in its column.^[ref] This extra reduction displays each pivot unknown directly in terms of the free variables in a linear system. The zero matrix also satisfies the definition: it has no pivot elements.

Every matrix has exactly one reduced row echelon form under elementary row operations, with its size and column order held fixed.^[ref] Ordinary row echelon form is not unique: multiplying a nonzero row by two over the rational numbers keeps the staircase but changes the matrix. The reduced conditions remove that freedom.

For a fixed number of columns, each row space has a unique reduced matrix after zero rows are removed; at a fixed full size, the number of appended zero rows is also determined. Thus reduced row echelon form represents a vector subspace independently of the starting basis. This is the uniqueness used by Gaussian binomial coefficients to count vector subspaces over a finite field.

The fixed column order matters. Elementary column operations can change the row space, the kernel and the resulting reduced row echelon form. For instance, swapping the columns of $(12)$ over the rational numbers gives $(21)$, whose reduced row echelon form is $(11/2)$ instead of $(12)$. When solving a linear system, a column interchange also requires tracking the corresponding interchange of unknowns.

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.

For $0\le k\le n$, let $V$ be the $n$-dimensional irreducible sl2 Lie algebra representation. Its weights are $n-1,n-3,\dots ,1-n$. A wedge basis shows $\mathrm{ch}{\Lambda }^{k}V=q^{-k(n-k)}{G}_{n,k}(q^2)$, where $G_{n,k}(t)$ is the Gaussian binomial coefficient. By the Weyl complete reducibility theorem, this character is a nonnegative sum of irreducible strings $q^m+q^{m-2}+\cdots +q^{-m}$. All its weights have parity $k(n-k)$, so their coefficients increase toward zero and decrease afterward. Consequently the coefficients of $G_{n,k}(t)$ are symmetric and unimodal. The symmetric quantum binomial coefficient is this character, a Laurent polynomial; its coefficients are unimodal on the occupied parity. The full integer-exponent sequence need not be unimodal, as $q+q^{-1}$ demonstrates.