Past exam of the mathematics course of the University of Cambridge / 2001 / iii

Identify the Lie algebra $\mathfrak{g}$ with the tangent space $T_{e}G$. For $X\in \mathfrak{g}$, its left-invariant vector field is $X^L(g)=(d{L}_g{)}_{e}X$. Let ${\gamma }_X$ be its integral curve with ${\gamma }_X(0)=e$. Uniqueness of integral curves and left invariance give ${\gamma }_X(s+t)={\gamma }_X(s){\gamma }_X(t)$ wherever initially defined. Repetition of a local curve extends it to all real times. Thus ${\gamma }_X$ is the unique one-parameter subgroup with derivative $X$ at zero, and the Exponential map of a Lie group is defined bySmooth dependence of solutions of differential equations on their initial data and parameters makes this map smooth.

The derivative at zero is particularly simple:It is the identity linear map from $\mathfrak{g}$ to $T_{e}G$. The inverse function theorem therefore proves that the Exponential map of a Lie group is a local diffeomorphism at zero, producing a local exponential chart. This proves the requested local assertion.

There is a necessary qualification to the other assertion: the image of every one-parameter subgroup lies in the identity component of a Lie group $G^{\circ }$. The correct statement without a connectedness hypothesis isIndeed, if $\exp$ is a group homomorphism, its image is an abelian subgroup. It contains an identity neighborhood by the local result. An open subgroup of a connected group is the entire group: every coset is open, so its complement is also open. Thus $\exp (\mathfrak{g})=G^{\circ }$, proving that this component is abelian.

Conversely, if $G^{\circ }$ is abelian, then $\exp (tX)\exp (tY)$ is a one-parameter subgroup: its group law follows by commuting the factors. Its derivative at zero is $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$ and, at $t=1$, additivity. This proves the additivity of the Lie exponential map criterion. For connected $G$ it is exactly the stated equivalence with an abelian group. For disconnected $G$ the unrestricted claim is false: $S_3\times \mathbb{R}$ is a nonabelian Lie group, but its exponential is the homomorphism $x\mapsto (e,x)$. The original PDF does not impose connectedness in that sentence, so that qualification cannot be omitted.

Now suppose $G$ is connected and abelian, of dimension $n$. The exponential is a surjective group homomorphism, and its kernel $\Lambda$ is discrete by local injectivity at zero. The induced mapis a bijective Lie group homomorphism and a local diffeomorphism, hence a Lie group isomorphism. To identify the quotient, take linearly independent elements ${\lambda }_1,\dots ,{\lambda }_b\in \Lambda$ spanning the real linear span of $\Lambda$, and let ${\Lambda }_0={\sum }_i\mathbb{Z}{\lambda }_i$. Every coset of ${\Lambda }_0$ in $\Lambda$ has a representative in the bounded fundamental parallelepiped of these vectors. A discrete additive subgroup has finite intersection with a compact set: otherwise differences of arbitrarily close elements would approach zero, contradicting discreteness at zero. Thus $\Lambda /{\Lambda }_0$ is finite.

It follows that $\Lambda$ is a finitely generated torsion-free abelian group of rank $b$, hence has a $\mathbb{Z}$-basis of $b$ elements. These basis elements also span its $b$-dimensional real span and are linearly independent over $\mathbb{R}$. Extending them to a real basis identifies $\Lambda$ with ${\mathbb{Z}}^b\times \{0{\}}^{n-b}$. Consequently the classification of connected abelian Lie groups isThe compact circle factors record the periods of the one-parameter subgroups.

The Lie algebra of SL2R is the space of traceless real two-by-two matrices. For $A$ in this space, the Cayley-Hamilton theorem gives $A^2=cI$, where $c=-\det A$. Expanding the matrix exponential, if $c=s^2>0$ thenIf $c=0$, then $\exp A=I+A$ and its trace is $2$. If $c=-s^2<0$, thenIn every case $\mathrm{tr}(\exp A)\ge -2$. Buthas determinant one and trace $-5/2$. Thus $g\in S{L}_2(\mathbb{R})$ is not an exponential, and the exponential map is not surjective.

Let $E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, so $A(t)=I+t{E}_{12}$ and $E_{12}^2=0$. Multiplication givesThe nilpotence also gives $(I+t{E}_{12}{)}^k=I+kt{E}_{12}$ for every nonnegative integer $k$, since all higher binomial terms vanish. Hence, for each positive integer $m$,These are unipotent matrices in SL2R. The integer is positive, since the displayed diagonal matrix is undefined at $m=0$.

Put $U=\varphi (A(t))$. Applying the group homomorphism $\varphi$ to the identity just proved shows that $U$ is conjugate to $U^{m^2}$ for every positive integer $m$. Thus the multiset of its eigenvalues is unchanged by taking their $m^2$th powers.

Take $m=2$. The fourth-power operation permutes this finite multiset. For each eigenvalue $z$, iterating its permutation cycle gives $z^{4^r}=z$ for some positive integer $r$. Since $U$ is unitary, $z\ne 0$, so $z^{4^r-1}=1$: every eigenvalue is a root of unity. Choose a positive integer $d$ divisible by all their orders. Every eigenvalue of $U^{d^2}$ is then one. Because its eigenvalue multiset is also the original one, we obtainThis is square-power rigidity of a finite nonzero spectrum. Finiteness of the dimension is essential to the permutation-cycle argument.

Let $N$ be the normal closure of the upper unipotents $U(t)=A(t)$. Conjugation by $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ givesThus $N$ also contains every lower unipotent $L(s)=\left(\begin{array}{cc}1& 0\\ s& 1\end{array}\right)$. We prove explicitly that these elementary unipotent generators of SL2R generate the full group.

For $a\ne 0$, multiplication givesSo all determinant-one diagonal matrices lie in $N$. If $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ has $a\ne 0$ and $ad-bc=1$, thenbecause its lower-right entry is $(1+bc)/a=d$. If $a=0$, then $c\ne 0$ and left multiplication by $U(1)$ makes the upper-left entry equal to $c$. The resulting matrix lies in $N$ by the preceding factorization, and $U(1{)}^{-1}\in N$ restores $g$. Therefore

Every unitary matrix is unitarily diagonalizable. Part (ii) therefore proves the stronger matrix statement $\varphi (A(t))=I$, not merely a statement about its eigenvalues. The kernel of a group homomorphism is a normal subgroup, and it contains every upper unipotent. By part (iii) their normal closure is the entire group. ThusThis proves the triviality of finite-dimensional unitary representations of SL2R. In fact the argument used only the algebraic homomorphism property and finite-dimensional unitarity; it did not need continuity.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

We prove uniform approximation, not just approximation in $L^2$. The analytic facts used are: a continuous square-integrable kernel gives a compact integral operator on $L^2$; the spectral theorem for compact self-adjoint operators decomposes the closure of its range into its finite-dimensional nonzero eigenspaces; and continuous functions are dense in $L^2$ for normalized Haar measure on a compact Lie group. Normalized Haar measure on a compact group is both left- and right-invariant and is preserved by inversion.

Choose a continuous nonnegative function $k$ of integral one, supported in a sufficiently small symmetric neighborhood of the identity, with $k(g^{-1})=k(g)$. Such functions are obtained from a local bump and its inverted bump followed by normalization. Define the convolution operatorIts kernel is continuous on $G\times G$, so $T$ is compact. The symmetry of $k$ makes it self-adjoint. It commutes with the left translations $(L_{g}h)(x)=h(g^{-1}x)$ by a change of variable $y=gz$.

By Cauchy-Schwarz inequality and invariance of Haar measure,The same integral formula and continuity of the kernel show $Th$ is continuous for every $h\in L^2(G)$. Moreover, rewriting $y=x{z}^{-1}$ gives $(Tf)(x)=\int f(x{z}^{-1})k(z)dz$. Uniform continuity on the compact group therefore makes $\Vert Tf-f{\Vert }_{\infty }$ arbitrarily small when the support of $k$ is small. Also $\Vert T{\Vert }_{\infty \to \infty }\le 1$ because $k\ge 0$ and has integral one.

Fix $\epsilon >0$ and take $k$ so $\Vert Tf-f{\Vert }_{\infty }<\epsilon /3$. ThenLet $P_N$ be the orthogonal projection onto increasing finite sums of the nonzero eigenspaces of $T$. Since $Tf$ lies in the closure of the range, the spectral theorem gives $P_{N}Tf\to Tf$ in $L^2$. Applying the smoothing estimate givesThus some $h=T{P}_{N}Tf$ satisfies $\Vert h-f{\Vert }_{\infty }<\epsilon$.

Each nonzero eigenspace is finite dimensional, invariant under $L_g$, and consists of continuous functions: if $Tv=\lambda v$ with $\lambda \ne 0$, then $v=Tv/\lambda$. The finite sum $M$ containing $h$ therefore carries the finite-dimensional unitary representation $\rho (g)=L_g{|}_M$. It is continuous, since translating each of its finitely many continuous basis functions depends continuously on $g$ in the uniform and hence the $L^2$ norm.

Let $\ell$ be evaluation at the identity on $M$, and let the dual representation on $M^{\ast }$ be $\theta (g)\ell =\ell \circ \rho (g^{-1})$. For the linear functional $L_h$ on $M^{\ast }$ defined by $L_h(u)=u(h)$,The rank-one endomorphism $\alpha =\ell \otimes L_h$ of $M^{\ast }$ satisfies $\mathrm{tr}(\alpha \theta (g))=L_h(\theta (g)\ell )$. ThereforeThis is the requested Peter-Weyl theorem, proved via the convolution proof of uniform Peter-Weyl approximation. The extra convolution after the spectral truncation is what upgrades the $L^2$ estimate to a uniform one.

To deduce a faithful finite-dimensional representation, first note that these representations separate points. For $g\ne e$, choose a continuous function taking different values at $g$ and $e$, and approximate it closely enough by a trace coefficient to retain that difference. The representation appearing in that trace must satisfy $\theta (g)\ne I$.

There is an identity neighborhood $U$ containing no nontrivial subgroup. Here is the Lie groups have no small subgroups argument: choose an injective exponential chart on a Lie-algebra ball of radius $r$, and take $U=\exp (B_{\delta })$ with $0<\delta <r/2$. If a subgroup contained a nonidentity $\exp X$ in $U$, choose the first integer $m$ with $m\Vert X\Vert \ge \delta$. Then $\delta \le m\Vert X\Vert <2\delta <r$, so its power $\exp (mX)$ lies outside $U$, a contradiction. For a zero-dimensional group, take $U=\{e\}$.

For every $g\notin U$, choose a representation nontrivial at $g$. The sets on which these representations are not the identity are open and cover the compact set $G\setminus U$. A finite subcover gives finitely many representations, whose direct sum has kernel contained in $U$. A subgroup contained in $U$ is trivial, so this direct sum is faithful. Averaging its Hermitian form makes it unitary. Thus the finite faithful representation from point-separating representations yieldsUsing the no-small-subgroups neighborhood is essential; $G\setminus \{e\}$ alone need not be compact, so a finite-subcover argument on that punctured set would not be valid.

A maximal torus in a compact connected Lie group is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak{t}$ for its Lie algebra.

Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak{g}$ over the adjoint action to obtain an $\mathrm{Ad}(G)$-invariant positive definite inner product. It defines a bi-invariant Riemannian metric. Compactness makes this metric complete, and the Hopf-Rinow theorem joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is ${\nabla }_{X}Y=[X,Y]/2$, so the geodesic starting with $X$ is $\exp (tX)$. Consequently every $g$ has the form $\exp X$.

The Lie-algebra centralizer of $\mathfrak{t}$ is exactly $\mathfrak{t}$. If $X$ commutes with $\mathfrak{t}$, its one-parameter subgroup commutes with $T$. The closure of the subgroup generated by $T$ and $\exp (\mathbb{R}X)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in \mathfrak{t}$. The torus action on ${\mathfrak{g}}_{\mathbb{C}}$ decomposes into finitely many character weight spaces. The zero-weight space is ${\mathfrak{t}}_{\mathbb{C}}$. Choose $H\in \mathfrak{t}$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak{t}$.

For an arbitrary $X\in \mathfrak{g}$, the continuous function $g\mapsto ⟨{\mathrm{Ad}}_{g}X,H⟩$ attains a maximum. Put $Y={\mathrm{Ad}}_{g_0}X$ at a maximizer. Differentiating along $\exp (sZ)g_0$ givesHence $[Y,H]=0$, so $Y\in \mathfrak{t}$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This adjoint-orbit criterion for maximal-torus conjugacy proves

Restriction of representations is a ring homomorphism $\mathrm{Res}:R(G)\to R(T)$. If a virtual representation restricts to zero, its character vanishes on $T$. Since characters are class functions and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the character map proved in Question 3 then givesThe Weyl group is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho (n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto nt{n}^{-1}$. Therefore every restricted representation, and every virtual difference of them, is fixed by $W$:

For the special orthogonal group $G=SO(2m+1)$, takewhere $R_{\theta }$ is a real planar rotation. It is a maximal torus: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its character lattice is ${\mathbb{Z}}^m$, so, writing $z_i=e^{i{\theta }_i}$,Indeed commuting unitary torus matrices diagonalize simultaneously into one-dimensional characters, and the characters of the circle are exactly the integral powers.

The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus character pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving

It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant Laurent polynomials. In this case we can prove equality. Invariance under individual inversions first givesFor one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x{s}_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the Fundamental theorem of symmetric polynomials then givewhere $e_k$ are the elementary symmetric polynomials.

Let $V$ be the complexification of the defining $(2m+1)$-dimensional real representation. Its torus weights are $1,z_i,z_i^{-1}$. The characters of its exterior powers obeyFor $k\le m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k={\chi }_{{\Lambda }^{k}V}$ isIts leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the character of a genuine representation and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the representation ring of an odd-dimensional special orthogonal group descriptionThe triangular change of generators is integral, so $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb{Z}[V_3]$, with the three-dimensional defining representation as generator. These are genuine representations of the orthogonal group, not the half-integral spin weights of its simply connected double cover.

Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime }+a{c}^{\prime },c+c^{\prime })$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^c{z}^b{x}^a$. ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.

Here is the needed skew Hilbert basis theorem, including its proof. If $A$ is a left Noetherian ring and $\sigma$ an automorphism, let $J$ be a left ideal of $A[t;\sigma ]$, where $ta=\sigma (a)t$. DefineThese are left ideals of $A$: adding lifts adds the coefficient, and left multiplication by ${\sigma }^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient ${\sigma }^n(u_{nj})$.

For $f\in J$ of degree $m\ge N$, write its normalized leading coefficient as ${\sigma }^{-m}(\mathrm{lc}f)={\sum }_j{r}_j{u}_{Nj}$. Thenhas smaller degree: the displayed summands have leading coefficients ${\sigma }^m(r_j{u}_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma ]$ is a left Noetherian ring. Passing to the opposite ring replaces $\sigma$ by ${\sigma }^{-1}$ and proves the corresponding right-sided result.

This proof includes the ordinary Hilbert basis theorem when $\sigma$ is the identity. Starting with the field $k$, it makes $k[y,z]$ Noetherian; localization at powers of $y,z$ makes $B$ Noetherian as well. Finally, for a left ideal $L$ in $B[x,x^{-1};\sigma ]$, its intersection $J$ with $B[x;\sigma ]$ has finitely many generators. For each $f\in L$, some $x^{q}f$ lies in $J$, so $f=x^{-q}(x^{q}f)$ is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore $kG$ is a left Noetherian ring and a right Noetherian ring.

The commutative Laurent polynomial ring $B$ is an integral domain and $\sigma$ is invertible. Write nonzero elements asThe largest exponent in their product is $m+n$, with coefficientThere is only one pair of exponents contributing at this degree, and the coefficient cannot vanish in the integral domain $B$. Hence $rs\ne 0$. Equivalently, $rs=0$ forces $r=0$ or $s=0$. This is skew Laurent extensions of domains are domains, and the argument works over the original field in any characteristic.

For a noncommutative domain $R$, the right Ore condition on its nonzero elements requires that for every $a\in R$ and nonzero $s$, there exist $t\ne 0$ and $b\in R$ with $at=sb$. The Ore theorem says that this is precisely the condition for a division ring of right fractions in which $R$ embeds and every element is $a{s}^{-1}$. The left-handed condition similarly gives left fractions.

For completeness, a right Noetherian domain satisfies the condition. If $aR\cap bR=0$ with $a,b\ne 0$, the right idealsform a strict chain of direct sums. Indeed, in a relation ${\sum }_{i=0}^n{a}^{i}b{r}_i=0$, the first term lies in $bR$ and the rest in $aR$. Their zero intersection makes $b{r}_0=0$, hence $r_0=0$; cancel $a$ and repeat. Every new summand is nonzero in a noncommutative domain, contradicting the ascending chain condition. Thus $au=bv\ne 0$ exists, with $u,v\ne 0$, giving right common multiples. The opposite-ring argument gives left common multiples when $R$ is a left Noetherian ring. This proves right Noetherian domains satisfy the Ore condition.

A sketch of the fraction construction also explains the theorem. Pairs $(a,s)$ represent $a{s}^{-1}$. They represent the same fraction when a common right denominator $su=tv\ne 0$ also satisfies $au=bv$. Addition uses $(au+bv)(su{)}^{-1}$. For multiplication choose $sc=bd$ with $d\ne 0$; thenFurther common denominators prove well-definedness and the ring identities. The original map is injective because all nonzero denominators are regular; a zero fraction numerator would satisfy $au=0$ for $u\ne 0$. Every nonzero fraction has inverse $s{a}^{-1}$, so the ring is a division ring. This construction has the universal property of inverting all nonzero elements.

Apply this to the two-sided Noetherian noncommutative domain $kG$. Its left Ore property expresses any right fraction as a left one: choose $u\ne 0$ with $ua=bs$, yielding $a{s}^{-1}=u^{-1}b$. Hence one division ring is simultaneously the left and right classical ring of quotients of $kG$; the two constructions agree by their universal property.

Work over $\mathbb{C}$, or another field of characteristic zero. The Weyl algebra $A_n$ is generated by $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ with commuting $x$'s, commuting $\partial$'s and $[{\partial }_i,x_j]={\delta }_{ij}$. Its ordered monomial basis of a Weyl algebra consists of the monomials $x^{\alpha }{\partial }^{\beta }$. One way to verify independence is to apply a finite operator relation to the formal exponential $e^{\mathbf{t}\cdot \mathbf{x}}$: after canceling that exponential the relation becomes a polynomial identity $\sum c_{\alpha \beta }{\mathbf{x}}^{\alpha }{\mathbf{t}}^{\beta }=0$, so every coefficient vanishes. The relations already reorder every word into that basis.

Give both generators degree one. The Bernstein filtration isThus $\dim F_m{A}_n=\left(\genfrac{}{}{0ex}{}{m+2n}{2n}\right)$. Commutator terms have lower degree, accounting for the commutative symbol algebra. Since that algebra is Noetherian, leading-symbol generation followed by finite downward degree cancellation proves that $A_n$ is Noetherian on both sides.

A left finitely generated module $M$ has a good filtration, for example $M_m={\sum }_j{F}_m{A}_n{m}_j$ for a finite generating set of the module. More generally one allows fixed degree shifts of the generators. The associated graded module is finite over the symbol algebra. DefineThis is the characteristic variety of a Weyl algebra module for the Bernstein algebra filtration. It records where the module's leading symbols remain nonzero and makes noncommutative module growth accessible to commutative algebra.

For a fixed algebra filtration it is independent of the good filtration. Any two such filtrations differ by bounded shifts $M_i\subseteq {\tilde{M}}_{i+c}$ and conversely, since each finite set of generators occurs in bounded degree in the other filtration. If a degree-$d$ operator has symbol annihilating $\mathrm{gr}M$, its $r$th power maps $M_i$ into $M_{i+rd-r}$. For $r>2c$, bounded-shift comparison forces the power of that symbol to annihilate $\mathrm{gr}\tilde{M}$. Reversing the argument gives equal radicals of ideals of annihilators. This is independence of characteristic support from a good module filtration. Changing the algebra filtration itself, for example to the order filtration of a Weyl algebra with the $x_i$ in degree zero, need not preserve the same variety; the convention matters.

The Hilbert-Serre theorem gives an eventual cumulative polynomialHere $d$ is also the Gelfand–Kirillov dimension of a module, and $e(M)>0$ is an integer multiplicity of a filtered module for nonzero $M$. Finite generation bounds $d\le 2n$.

Here is a proof sketch of the essential lower bound, rather than an appeal to its name. For a nonzero $f\in F_m{A}_n$, choose a nonzero monomial of greatest total degree in its ordered expansion, say $x^{\alpha }{\partial }^{\beta }$. The operations $\mathrm{ad}{\partial }_i$ and $\mathrm{ad}(-x_i)$ differentiate this expansion with respect to $x_i$ and ${\partial }_i$. Apply them ${\alpha }_i$ and ${\beta }_i$ times. Lower-degree terms die, and every other monomial of the same degree also dies unless it has exactly those exponents. The result is a nonzero scalar multiple of the chosen coefficient. Characteristic zero is essential for the factorials. Expanding the repeated commutators therefore givesChoose $0\ne v\in M_0$. If $f$ acted as zero on all $M_m$, every $f{v}_{j}v$ would vanish, contrary to $v={\sum }_j{u}_{j}f{v}_{j}v$. Hence the linear action mapis injective. It follows that $\dim F_m{A}_n\le (\dim M_m)(\dim M_{2m})$. Comparing polynomial degrees gives $2n\le 2d$. ThusThis proves Bernstein inequality for Weyl algebra modules through the faithful finite-step action of a Weyl algebra.