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Let $D_n(a)$ denote the constant term, retaining the dimension as a subscript when it changes. We use the Good recurrence for the Dyson constant term. For algebraically independent variables, Lagrange interpolation polynomial applied to the constant polynomial one givesAt $t=0$, this becomes the rational identityWhen every $a_i>0$, multiplying by the given Laurent polynomial cancels one factor in row $i$ of the product and givesThis is a polynomial identity after cancellation; no choice of an expansion of a rational function is involved.

The boundary case is essential. If $a_i=0$, the factors in row $i$ are absent. The only factors involving $X_i$ are then $(1-X_j/X_i{)}^{a_j}$ for $j\ne i$, and all their powers of $X_i$ are nonpositive. To obtain power zero in $X_i$, one must select the constant term one from each of them. Taking the constant term in $X_i$ therefore deletes that variable and exponent:For $n=1$, the empty product is one, so $D_1(a_1)=1$. The all-zero exponent vector also gives one.

Now put $M_n(a)=m!/{\prod }_i{a}_i!$, where $m={\sum }_i{a}_i$. This multinomial coefficient obeys the same deletion rule for a zero exponent. For positive exponents,Induction on $n$, and within each dimension on $m$, consequently determines $D_n$ uniquely and identifies it with $M_n$. This proves the Dyson constant-term identity:

Apply the given skein relation to a positive and a negative kink in a diagram of the unknot. Both crossed link diagrams represent the unknot, whereas the oriented smoothing represents the two-component unlink. ThusFactor the left-hand side and cancel the nonzero factor in the Laurent polynomial ring:The positive sign follows from the skein relation as printed in this paper. It differs from the frequently used Jones polynomial convention whose two-component unlink has value $-(q^{1/2}+q^{-1/2})$. With the printed convention, an $r$-component link has invariant $(-1{)}^{r-1}$ times that convention's Jones polynomial.

Insert a kink on any component of an oriented link diagram for $L$. Switching the kink's crossing gives another link diagram of $L$; its oriented smoothing separates off a small unknot in a ball disjoint from the rest of $L$. The skein relation therefore givesCancel the nonzero factor $q^{1/2}-q^{-1/2}$ in the Laurent polynomial ring to obtain

Let $H_{+}$ be the positively oriented Hopf link obtained by closing a two-strand braid with two positive crossings. Switching one crossing gives the two-component unlink; the oriented smoothing gives the unknot. The printed skein relation givesRepresent the right-handed trefoil knot by the closure of the two-strand braid with three positive crossings. Switching one crossing gives the unknot, while its oriented smoothing gives $H_{+}$. Consequentlyso the two Jones polynomials areThe second expression follows from the mirror of a link identity. They are unequal Laurent polynomials, so invariance under link isotopy proves that the right-handed and left-handed trefoils are not isotopic.

Consider the symmetric Laurent polynomialIt satisfies $f(1)=1$, so the Alexander polynomial realization theorem gives a knot $K_3$ with ${\Delta }_{K_3}(t)\doteq f(t)$. The polynomial is not a unit of $\mathbb{Z}[t^{\pm 1}]$, whereas

Here functional calculus convergence means that for every $f\in C(\mathbb{T})$,

Embed ${\mathcal{H}}_n$ in $\mathcal{H}$ and write ${\tilde{A}}_n=P_n^{\ast }{A}_n{P}_n$. The hypothesis gives ${\tilde{A}}_n\rightharpoonup A$ in the weak operator topology. Since $A_n$ and $A$ are unitary operators,Thus ${\tilde{A}}_n\to A$ in the strong operator topology. Applying the same argument to the adjoints gives ${\tilde{A}}_n^{\ast }\to A^{\ast }$ strongly.

Products of uniformly bounded strongly convergent operators converge strongly, so for every integer $k$,strongly, with negative $k$ interpreted through adjoints. Therefore convergence holds for every Laurent polynomial. The Stone-Weierstrass theorem says that Laurent polynomials are uniformly dense in $C(\mathbb{T})$. Since the continuous functional calculus is contractive, uniform approximation finishes the proof for every $f\in C(\mathbb{T})$.

The symmetric quantum integer is the displayed Laurent polynomial, with $[0{]}_q=0$. It obeys $[n+1{]}_q=q^{-1}[n{]}_q+q^n$ and $[r{]}_q[s+1{]}_q-[s{]}_q[r+1{]}_q=[r-s{]}_q$. At a primitive odd $\ell$th root of unity, $[n{]}_q=0$ exactly when $\ell$ divides $n$.

For the block-rotation maximal torus in the special orthogonal group $SO(2m+1)$, the Weyl group acts by permutations and independent inversions of $z_i$. Its invariant Laurent polynomials are the symmetric polynomials in $x_i=z_i+z_i^{-1}$, hence the polynomial ring in their elementary symmetric polynomials $e_1,\dots ,e_m$. Restriction of the group representation ring of a compact group is injective because every group element is conjugate into the torus and characters distinguish virtual representations. For the defining complexified representation $V$, the exterior-power characters satisfy ${\sum }_k{\chi }_{{\Lambda }^{k}V}{t}^k=(1+t){\prod }_i(1+x_{i}t+t^2)$. For $k\le m$, its coefficient is $e_k$ plus an integral linear combination of $e_j$ for $j<k$. Induction puts every $e_k$ in the restriction image, proving equality with the invariant ring and giving polynomial generators $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$.

Given a finite set $F\subset {\mathbb{Z}}^r$ of exponent vectors, there is $w\in {\mathbb{Z}}^r$ for which the integers $w\cdot m$, $m\in F$, are pairwise distinct. For $r=2$, take $w=(1,N)$ with $N$ larger than every first-coordinate difference in $F$. In higher dimensions take $w=(1,N,\dots ,N^{r-1})$ with $N$ sufficiently large. This proves that a nonzero Laurent polynomial can be detected by a substitution $X_i=T^{w_i}$ without cancellation of its distinct terms.

Let $A$ have no zero divisors and $\sigma$ be an automorphism. For nonzero Laurent polynomials $f=\sum a_i{x}^i$ and $g=\sum b_j{x}^j$ with greatest exponents $m,n$, the coefficient of $x^{m+n}$ in $fg$ is $a_m{\sigma }^m(b_n)$. No other exponent pair contributes to that highest degree. Both factors are nonzero, so their product is nonzero, proving the noncommutative domain property in every characteristic.

For backward differences $\Delta P_i=P_i-P_{i-1}$, a scalar stationary arity-$a$ subdivision mask has symbol $M(z)={\sum }_i{M}_i{z}^i$. The identity $(1-z)M(z)=B(z)(1-z^a)$ gives the displayed intertwining whenever the quotient is a finite Laurent polynomial. Iterating supplies higher-difference masks. The coefficients $a^{r\ell }{\Delta }^r{P}_i^{\ell }$ are candidate $r$th-derivative controls. A convergent derivative scheme, proved by contraction of its first differences, establishes corresponding smoothness of the original limit under the standard consistent parameterization.

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.

For the unitary group, the diagonal maximal torus has coordinates $z_i=e^{i{\theta }_i}$ and Weyl group $S_n$. The Weyl integration formula weights independent uniform angles by $n{!}^{-1}{\prod }_{i<j}|z_i-z_j{|}^2$. The Vandermonde determinant $\det (z_i^{n-j})$ has squared torus integral $n!$, by Fourier orthogonality. Multiplying an irreducible character of highest weight $\lambda$ by this determinant gives an alternating Laurent polynomial. Its leading alternant has coefficient one; character orthogonality for compact groups forces every other alternant coefficient to vanish. This yields the quotient $\det (z_i^{{\lambda }_j+n-j})/\det (z_i^{n-j})$.