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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.

For an oriented three-manifold M with prescribed framed boundary endpoints P, the relative Kauffman bracket skein module is the complex vector space generated by properly embedded framed tangles, modulo framed isotopy and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted circle multiplies by $\delta =-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a circle has value $\delta$. For a nonempty uncolored link this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the quantum dimensions. Gluing compatible boundary pieces composes skeins; disjoint union gives tensor products. With no endpoints this is the Kauffman bracket skein module. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed isotopy.

For $A\ne 0$, the Temperley-Lieb algebra ${\mathrm{TL}}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed circle removed with factor $\delta$. There are $\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands givesEvery crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.

Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum ${\sum }_{s=0}^{m-1}{A}^{2(m-1-2s)}$ if its denominator vanishes. Set ${\Delta }_j=(-1{)}^j[j+1]$. These quantum integers satisfy ${\Delta }_0=1$, ${\Delta }_1=\delta$, and ${\Delta }_j=\delta {\Delta }_{j-1}-{\Delta }_{j-2}$. The Jones-Wenzl idempotent $f_n$ is characterized by identity-matching coefficient one andStarting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes 1$ and $C=F{e}_{n-1}F$. Whenever the denominators are nonzero the recursion isHere is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $({\Delta }_{n-1}/{\Delta }_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu ={\Delta }_{n-1}/{\Delta }_{n-2}$. Also $FC=CF=C$. Expanding $(F-{\mu }^{-1}C{)}^2$ proves idempotence. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $C{e}_{n-1}=\mu F{e}_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.

For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta -{\Delta }_{n-2}/{\Delta }_{n-1})f_{n-1}=({\Delta }_n/{\Delta }_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\mathrm{tr}(f_n)={\Delta }_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.

A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}{e}_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted Jones-Wenzl twist eigenvalue identity, including its exponent:A negative framing twist acts by the inverse scalar.

Choose $A=e^{\pi i/(2r)}$, with an integer $r\ge 3$. Then $[r]=0$ but $[1],\dots ,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the reduced Temperley-Lieb skein theory by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\dots ,r-2$, with dimensions and twistsTheir dimensions are nonzero. The recurrence decomposes $f_j\otimes 1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup idempotent is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all matrix blocks split. Applying the recurrence to the decompositions proves the fusion of Jones-Wenzl colors ruleOne can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.

We next prove the Kirby color handle-slide identity, rather than assume it. In the annular skein space let ${\hat{f}}_b$ be the closure of color b and define the Kirby color $\Omega ={\sum }_b{d}_b{\hat{f}}_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_c{i}_c=1_c$ and ${\sum }_c{i}_c{p}_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtainingClosing the resulting diagrams and straightening the bent strand gives $\mathrm{tr}(V_c^b{U}_c^b)=\mathrm{tr}(p_c{i}_c)=d_c$. Since b is simple, $V_c^b{U}_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. ThereforeTo slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution ${\sum }_c{i}_c{p}_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed handle slide.

We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed Hopf link with colors i,j. On a fusion channel k, its double braiding has eigenvalue ${\theta }_k/({\theta }_i{\theta }_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identityFor $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey ${\hat{f}}_j=X{\hat{f}}_{j-1}-{\hat{f}}_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the Hopf pairing of Jones-Wenzl colorsIndeed the recurrence is the sine identity $\sin ((j+1)x)=2\cos x\sin (jx)-\sin ((j-1)x)$ with $x=(i+1)\pi /r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now givesFor example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is ${\sum }_{k=1}^{r-1}\sin (ik\pi /r)\sin (jk\pi /r)=\frac{r}{2}{\delta }_{ij}$. Thus no degeneracy remains in this pairing.

The Gauss sums of Jones-Wenzl colors are $p_{\pm }={\sum }_j{d}_j^2{\theta }_j^{\pm 1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion proveIn the last step ${\sum }_i{d}_i{N}_{ij}^k=d_j{d}_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_{+}{T}^{-1}d$, with $T=\mathrm{diag}({\theta }_j)$. The matrix $H$ and vector $d$ are real, and each twist has modulus one, so conjugation gives $H{T}^{-1}d=p_{-}Td$. Apply H again and use its orthogonality to obtain $\boxed{p_{+}{p}_{-}={\mathcal{D}}^2}$. Hence neither factor vanishes.

We may use the surgery existence theorem and Kirby calculus: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by handle slides and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery link L let $\sigma (L)$ be the signature of its symmetric surgery linking matrix, put $\kappa =p_{+}/\mathcal{D}$, and define the Jones polynomial surgery invariantA handle slide preserves the colored bracket by the proved identity and changes the linking matrix by congruence, preserving its signature. Adding a disjoint plus-one-framed unknot multiplies the bracket by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so the net factor is $p_{+}/(\mathcal{D}\kappa )=1$. For a minus-one-framed unknot it is $\kappa p_{-}/\mathcal{D}=p_{+}{p}_{-}/{\mathcal{D}}^2=1$. Kirby calculus therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has ${\tau }_r(S^3)=1$; reversing orientation conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the Jones polynomial.

For an oriented three-manifold M with prescribed framed boundary endpoints P, the relative Kauffman bracket skein module is the complex vector space generated by properly embedded framed tangles, modulo framed isotopy and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted circle multiplies by $\delta =-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a circle has value $\delta$. For a nonempty uncolored link this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the quantum dimensions. Gluing compatible boundary pieces composes skeins; disjoint union gives tensor products. With no endpoints this is the Kauffman bracket skein module. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed isotopy.

For $A\ne 0$, the Temperley-Lieb algebra ${\mathrm{TL}}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed circle removed with factor $\delta$. There are $\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands givesEvery crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.

Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum ${\sum }_{s=0}^{m-1}{A}^{2(m-1-2s)}$ if its denominator vanishes. Set ${\Delta }_j=(-1{)}^j[j+1]$. These quantum integers satisfy ${\Delta }_0=1$, ${\Delta }_1=\delta$, and ${\Delta }_j=\delta {\Delta }_{j-1}-{\Delta }_{j-2}$. The Jones-Wenzl idempotent $f_n$ is characterized by identity-matching coefficient one andStarting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes 1$ and $C=F{e}_{n-1}F$. Whenever the denominators are nonzero the recursion isHere is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $({\Delta }_{n-1}/{\Delta }_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu ={\Delta }_{n-1}/{\Delta }_{n-2}$. Also $FC=CF=C$. Expanding $(F-{\mu }^{-1}C{)}^2$ proves idempotence. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $C{e}_{n-1}=\mu F{e}_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.

For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta -{\Delta }_{n-2}/{\Delta }_{n-1})f_{n-1}=({\Delta }_n/{\Delta }_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\mathrm{tr}(f_n)={\Delta }_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.

A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}{e}_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted Jones-Wenzl twist eigenvalue identity, including its exponent:A negative framing twist acts by the inverse scalar.

Choose $A=e^{\pi i/(2r)}$, with an integer $r\ge 3$. Then $[r]=0$ but $[1],\dots ,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the reduced Temperley-Lieb skein theory by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\dots ,r-2$, with dimensions and twistsTheir dimensions are nonzero. The recurrence decomposes $f_j\otimes 1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup idempotent is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all matrix blocks split. Applying the recurrence to the decompositions proves the fusion of Jones-Wenzl colors ruleOne can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.

We next prove the Kirby color handle-slide identity, rather than assume it. In the annular skein space let ${\hat{f}}_b$ be the closure of color b and define the Kirby color $\Omega ={\sum }_b{d}_b{\hat{f}}_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_c{i}_c=1_c$ and ${\sum }_c{i}_c{p}_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtainingClosing the resulting diagrams and straightening the bent strand gives $\mathrm{tr}(V_c^b{U}_c^b)=\mathrm{tr}(p_c{i}_c)=d_c$. Since b is simple, $V_c^b{U}_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. ThereforeTo slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution ${\sum }_c{i}_c{p}_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed handle slide.

We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed Hopf link with colors i,j. On a fusion channel k, its double braiding has eigenvalue ${\theta }_k/({\theta }_i{\theta }_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identityFor $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey ${\hat{f}}_j=X{\hat{f}}_{j-1}-{\hat{f}}_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the Hopf pairing of Jones-Wenzl colorsIndeed the recurrence is the sine identity $\sin ((j+1)x)=2\cos x\sin (jx)-\sin ((j-1)x)$ with $x=(i+1)\pi /r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now givesFor example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is ${\sum }_{k=1}^{r-1}\sin (ik\pi /r)\sin (jk\pi /r)=\frac{r}{2}{\delta }_{ij}$. Thus no degeneracy remains in this pairing.

The Gauss sums of Jones-Wenzl colors are $p_{\pm }={\sum }_j{d}_j^2{\theta }_j^{\pm 1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion proveIn the last step ${\sum }_i{d}_i{N}_{ij}^k=d_j{d}_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_{+}{T}^{-1}d$, with $T=\mathrm{diag}({\theta }_j)$. The matrix H and vector d are real, and each twist has modulus one, so conjugation gives $H{T}^{-1}d=p_{-}Td$. Apply H again and use its orthogonality to obtain $\boxed{p_{+}{p}_{-}={\mathcal{D}}^2}$. Hence neither factor vanishes.

We may use the surgery existence theorem and Kirby calculus: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by handle slides and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery link L let $\sigma (L)$ be the signature of its symmetric surgery linking matrix, put $\kappa =p_{+}/\mathcal{D}$, and define the Jones polynomial surgery invariantA handle slide preserves the colored bracket by the proved identity and changes the linking matrix by congruence, preserving its signature. Adding a disjoint plus-one-framed unknot multiplies the bracket by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so the net factor is $p_{+}/(\mathcal{D}\kappa )=1$. For a minus-one-framed unknot it is $\kappa p_{-}/\mathcal{D}=p_{+}{p}_{-}/{\mathcal{D}}^2=1$. Kirby calculus therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has ${\tau }_r(S^3)=1$; reversing orientation conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the Jones polynomial.

For real projective 3-space the surgery diagram is a single unknot of framing two. Thus $m=1$, $\sigma =1$, and its colored evaluation is ${\sum }_j{d}_j^2{\theta }_j^2$. Since $\mathcal{D}\kappa =p_{+}$, the explicit answer isThe denominator is nonzero by the proved Gauss-sum identity. For instance at $r=3$ the twists are $1,-i$, the squared dimensions are both one, and the numerator vanishes, so this normalized invariant is zero for ${\mathbb{RP}}^3$ at that level.

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.

For $yx=qxy$, the quantum enveloping algebra of sl2 acts by $K(x^r{y}^s)=q^{r-s}{x}^r{y}^s$, $E(x^r{y}^s)=[s{]}_q{x}^{r+1}{y}^{s-1}$ and $F(x^r{y}^s)=[r{]}_q{x}^{r-1}{y}^{s+1}$. Negative exponents here have zero coefficients and are omitted. The quantum integer identity $[r][s+1]-[s][r+1]=[r-s]$ verifies the commutator. These operations preserve total degree and make the quantum plane a module algebra.