Finite affine polarizations and their canonical determinants

An explicit finite twisting bimodule. Let y be an original affine Weyl element and $n_y={\mathfrak{n}}_{-}\cap y^{-1}({\mathfrak{n}}_{+})$. Its finite inversion set consists only of real roots, so $n_y$ is finite-dimensional and nilpotent. Its adjoint action on g is locally nilpotent, jointly in its generators. To see the latter without an infinite-series assumption, an iterated bracket of length m starting in $g_{\eta }$ has original principal height at most ht(eta)-m, and after y has height at least ht(y eta)+m. For any affine root alpha+n delta the difference between those two heights belongs to a finite bounded set depending on y and the finite root alpha, because y fixes delta. Hence m is bounded. The same holds on each finite enveloping word by Leibniz.

Let $N_y$ be the affine space $n_y$ with its finite BCH group law. Its polynomial coordinate algebra $O(N_y)$ is the principal-graded dual $U(n_y)\mathrm{\#}$ via Taylor pairing at the identity; ordered exponential coordinates and PBW give a triangular pairing with nonzero factorial diagonal. For a in $n_y$ let $R_{a}f(n)={\frac{d}{dt}f(\exp (ta)n)\mid }_0$. These right-invariant vector fields have bracket -R_[a,b] and define the right $U(n_y)$-module structure. Set S_y=O(N_y) tensor_(U(n_y)) U.

Right U acts on the last factor. There is also the following explicit commuting LEFT g action. Since Ad_n X is polynomial in n by the local-nilpotence bound, write it as ${\mathrm{\Sigma }}_j{p}_j(n)X_j$ and put X(f tensor u)=Σ_j f p_j tensor X_j u.

This descends to the balanced tensor product. Indeed $R_a({\mathrm{Ad}}_{n}X)=[a,{\mathrm{Ad}}_{n}X]$, so applying the formula to ($R_a$ f) tensor u and using the balancing relation changes it to f tensor (Ad_n X) a u, the result of applying it to f tensor a u. Pointwise Ad preserves brackets, hence the left formula satisfies the Lie relations. It commutes with right multiplication on u. Thus a genuine U-bimodule has been constructed, with no assumed twisting theorem.

Its left PBW model is $U{\otimes }_{U(n_y)}O(N_y)$. Use the left regular transpose action $L_{a}f(n)={\frac{d}{dt}f(n\exp (ta))\mid }_0$ on $O(N_y)$, and define Phi(u tensor f)= f(n) Ad_n(u).

Here the right side is interpreted in $O(N_y){\otimes }_{U(n_y)}U$. This map is balanced. Write $b(n)=A{d}_{n}a={\mathrm{\Sigma }}_j{b}_j{x}_j$ in a basis of $n_y$. Then $L_a={\mathrm{\Sigma }}_j{b}_j{R}_{xj}$, and ${\mathrm{\Sigma }}_j{R}_{xj}{b}_j=-Tr(adb)=0$ since $n_y$ is nilpotent. Applying the balancing relation to the derivative of f Ad_n u gives (L_a f) Ad_n u = f (Ad_n u)(Ad_n a),

which is exactly the image of u a tensor f. The inverse uses $A{d}_{n^{-1}}$ and the reversed left/right calculation. These are finite polynomial maps. Phi intertwines the natural left U action with the action above. Therefore both PBW models are established without a dimension argument or a completed Hom space.

In particular the left Cartan action on the augmentation has shift gamma_y(H)=-Tr_(n_y)(ad H) =sum_(alpha positive, y alpha negative) alpha(H) =(rhohat-y^-1 rhohat)(H).

This is the actual normal-order shift, not an assertion of ordinary integrability under a rescaled loop algebra.

For length-additive y,z, the inversion-root decomposition gives a polynomial multiplication isomorphism between the corresponding ordered unipotent root-coordinate spaces. PBW orders n_(yz) as the z-inversion coordinates followed by the transported y-inversion coordinates. Fubini for the two balanced tensor products then identifies S_(yz) with the appropriately twisted tensor of $S_y$ and $S_z$: on functions this is pullback along group multiplication, and on U it is multiplication. The left actions intertwine because Ad_(n1 n2)=Ad_n1 Ad_n2; the Cartan shifts add as $\widehat{\rho }-(yz{)}^{-1}\widehat{\rho }=(\widehat{\rho }-z^{-1}\widehat{\rho })+z^{-1}(\widehat{\rho }-y^{-1}\widehat{\rho })$. The remaining negative root coordinates are free, so PBW also shows that no higher tensor Tor appears in this length-additive factorization. Thus the same factorization holds for derived tensor. Fixed Tits lifts may introduce a toral sign automorphism when words are compared; on Cartan-weight modules it is implemented by the corresponding scalar character, and can be included in the comparison isomorphism. No degree or convergence issue is hidden in that scalar choice.

Define $T_y$ by this bimodule followed by the Tits root relabeling. Its right model gives L_j T_y(M)=Tits_y Tor_j^(U(n_y))(O(N_y),M).

A Verma is free over $A=U({\mathfrak{n}}_y)$ by ordered PBW, so the Tor groups in step 11.1 vanish in positive degree. Here is the precise infinite-rank restricted-Hom argument; local finiteness of only one side would not suffice. Put $D=A^{\mathrm{\#}}$, with its commuting transpose left/right $A$ actions, and put ${\gamma }_y=-{\mathrm{Tr}}_{{\mathfrak{n}}_y}\mathrm{ad}$ as in step 8.1. If $V={⨁}_{\mu }{V}_{\mu }$ is the graded multiplicity space of a free presentation $A\otimes V$, require both sums ${\sum }_{\mu }\dim A_{\nu -\mu }\dim V_{\mu }$ and ${\sum }_{\mu }\dim D_{\nu -{\gamma }_y-\mu }\dim V_{\mu }$ to be finite for every full Cartan weight $\nu$. Call such a presentation admissible. This allows infinite PBW rank; it imposes the actual finite-coordinate conditions on both sides, rather than a completion. The two PBW bimodule models of steps 2.1–7.1 give $F(M)=S_y{\otimes }_{U}M\cong D\otimes V[{\gamma }_y]$ on restriction to $A$. For $a\in {\mathfrak{n}}_y$, its left action is the regular transpose action on $D$: in the right model, ${\mathrm{Ad}}_{n}a$ lies in ${\mathfrak{n}}_y$, so balancing moves it to the polynomial coordinate factor, and its divergence term is zero by the nilpotent trace computation in step 6.1. Thus this identification is an identification of $A$ modules. Define $G(X)={\mathrm{Hom}}_U^{\mathrm{w}\mathrm{t}}(S_y,X)$, where a weight-$\nu$ map satisfies $\phi (sH)=\nu (H)\phi (s)$; the $U$ action is $(u\phi )(s)=\phi (su)$. The left model identifies it with ${\mathrm{Hom}}_A^{\mathrm{w}\mathrm{t}}(D,X)$: the value on $D_{\delta }$ has weight $\nu +{\gamma }_y+\delta$. Both constructions are algebraic and $U$ stable; a root vector of weight $\alpha$ sends a weight-$\nu$ map to weight $\nu +\alpha$. The unit is ${\eta }_M(m)(s)=s\otimes m$ and the counit is ${ϵ}_X(s\otimes \phi )=\phi (s)$. To prove that they are isomorphisms on the admissible free/cofree categories, first transpose a homogeneous $A$ endomorphism of $D$. Since each Cartan component of $A$ is finite-dimensional, $D^{\mathrm{\#}}=A$, and the transpose is a homogeneous right-$A$ endomorphism of $A$. It is multiplication by its value at $1$, so ${\mathrm{End}}_A^{\mathrm{w}\mathrm{t}}(D)=A^{\mathrm{o}\mathrm{p}}$, with no completed power series. Choose homogeneous bases of the $V_{\mu }$. Project a homogeneous map $\phi :D\to D\otimes V[{\gamma }_y]$ of right-Cartan weight $\nu$ onto any one basis coordinate of weight $\mu$. This is a homogeneous endomorphism of $D$, hence right multiplication by a unique element of $A_{\nu -\mu }$. The first finite sum above says that only finitely many of all these possible coefficients exist. Therefore $\phi (d)={\sum }_j(d{a}_j)\otimes v_j$ with a finite sum and ${\sum }_j{a}_j\otimes v_j\in (A\otimes V{)}_{\nu }$. Conversely every such finite sum defines the required map. This proves $GF(M)=A\otimes V$, with the unit exactly this isomorphism. On a cofree object with the same admissible $V$, the counit is exactly $D{\otimes }_A(A\otimes V)\to D\otimes V$, $d\otimes a\otimes v\mapsto da\otimes v$, so is an isomorphism as well. The adjunction makes both maps $U$ linear. Exact sequences of free objects split as graded $A$ modules. Exact sequences of these finite-weight cofree objects also split as graded $A$ modules: restricted duality is exact weight by weight and turns them into free right-$A$ objects, whose homogeneous generators can be lifted. Hence $F,G$ are inverse exact equivalences on precisely these indicated exact categories. The actual Verma presentations satisfy both finite sums. The first is the ordinary affine PBW partition finiteness. For the second take $V$ to be the PBW words in the negative root generators outside ${\mathfrak{n}}_y$. A coordinate of $D$ has a positive inversion-root weight; $y$ sends that creator to a negative root. A creator in $V$ has weight $-\beta$ with $\beta$ positive and not inverted; $y$ sends it to $-y\beta$, again negative. These transformed creators give exactly the negative root multiset, with the original imaginary multiplicities. At every full weight the number of exponent choices is therefore finite by the same affine PBW partition count. The finite normal-order shift does not change finiteness: the top weight after relabeling is $y(\mu +{\gamma }_y)=y\cdot \mu$, hence is exactly $\nu$ when $\mu =y^{-1}\cdot \nu$. Finite sums and finite Verma flags remain admissible by the graded $A$ splitting and the same argument. The proof makes no inverse assertion for arbitrary infinite-rank free or cofree modules satisfying only one of the two conditions.

Canonical-map determinant. Apply it to a Verma and twist the root labels by y. Write $M^y(\nu )$ for the result obtained from $M(y^{-1}\cdot \nu )$. Its ordered vector basis has one creator of negative original-root weight for every positive root, with the actual imaginary multiplicities. Consequently its character and highest line are those of M(nu). The current/Cartan matrices in this basis are polynomial in nu: every reordering uses finitely many Lie brackets and the Cartan acts on the induced highest line by the corresponding linear weight plus the fixed normal-order shift. Let c_y(nu):M(nu)→M^y(nu)

be the highest-line map, normalized to1 on that line.

At a fixed depth beta the determinant of $c_y$ has highest Cartan-degree term, up to nonzero constant, product_(gamma in I(y)) (nu,gamma)^(sum_(m>=1) P(beta-m gamma)).

To verify this, use the PBW basis before and after the indicated finite dualizations. A creator whose root coordinate remains free passes through as the identical free coordinate, so contributes degree0 in nu. An inverted root creator is evaluated against its dual coordinate. Its top-degree contribution comes from pairing with the opposite root vector and retaining the Cartan commutator; for k copies its leading coefficient is a nonzero factorial times (nu,gamma)^k. A different exponent vector or a non-Cartan bracket loses at least one Cartan degree. Thus the top-degree matrix is diagonal on exponent vectors. Counting each occurrence of gamma gives the displayed exponent. Root spaces in an inversion set are real and one-dimensional. In particular there is no imaginary-direction top factor.

Every possible determinant zero lies on a Casimir hyperplane. If the map has a kernel at depth beta, that proper submodule has a singular vector at nu-eta, $0<\eta \le \beta$, and centrality of the generalized Casimir gives 2(nu+rhohat,eta)=(eta,eta). The polynomial factor argument from the adjacent Shapovalov proof applies verbatim. Comparison with the displayed leading term forces $\eta =m\gamma ,\gamma$ in I(y), m positive integral. A critical factor is impossible here: its leading direction is delta, absent from the top term. Thus only the real factors $⟨\nu +\widehat{\rho },{\gamma }^{\vee }⟩-m$ with gamma in I(y) can occur.

Each present factor has multiplicity P(beta-m gamma). At a generic point of its hyperplane the source Verma has exactly the simple radical $R=M(\nu -m\gamma )$, as proved in the adjacent note. The kernel of $c_y$ is nonzero but avoids the top line, hence is exactly R. Source and target characters agree, so the cokernel has the character of R. It too is the simple R: its highest line generates a nonzero quotient of the simple Verma R and exhausts every weight dimension.

Differentiate $c_y$ transversely to this hyperplane. The induced first derivative $R\to coker(c_y)$ is a module map: differentiate equivariance, and the extra terms vanish after passing to the cokernel because the original kernel maps to zero. If that derivative vanished, lift the highest kernel vector to u0+t u1 with $c_y(t)(u0+tu1)=0$ modulo t². Higher adjacent weights have no kernel, so equivariance makes the lift singular modulo t². The generalized Casimir now gives incompatible first-order scalars, whose difference is m(gamma,gamma), exactly as in the full Gram argument. Therefore the derivative is an isomorphism between the two simple copies of R. A matrix block reduction adapted to kernel/image gives determinant order dim R_(nu-beta)=P(beta-m gamma).

For every fixed gamma in I(y), the total leading degree in its linear direction is sum_m P(beta-m gamma). Present factors have precisely the stated nonnegative exponents. Equality of that finite sum forces every allowed m to occur. Hence det(c_y|depth beta) = nonzero constant times product_(gamma in I(y)) product_(m>=1) (<nu+rhohat,gamma-coroot>-m)^(P(beta-m gamma)).

This proves the finite-polarization determinant. It does not infer any complex exactness from a determinant.

The half-Jantzen kernel bound. Deform nu generically over Ct. Since $c_y(t)$ is a module map between weightwise finite free modules of the same character, the preimages of $t^j$ times the target define a module filtration. Its first specialized positive layer is $ker(c_y)$. DVR diagonal reduction identifies the sum of layer dimensions with the displayed determinant valuation. Therefore sum_(j>=1) ch K_y^j M(nu) =sum_(gamma in I(y), m=<nu+rhohat,gamma-coroot> in Z>0) ch M(s_gamma dot nu).

This is an actual valuation identity for the canonical map. Suppose $\mu =s_{\beta }$ dot nu is a Bruhat cover in the integral orbit, and beta is not in I(y). The simple L(mu) has multiplicity1 in M(nu) by the ordinary Jantzen formula and the cover condition. None of the terms on the right above can contain L(mu): strong linkage would put $s_{\gamma }$ dot nu between nu and mu in Bruhat order; for a cover this forces $\gamma =\beta$. Thus $ker(c_y)$ has no L(mu) subquotient, and the unique singular vector at mu has nonzero image under $c_y$. This is the precise nonvanishing needed later, including at intersections of many real hyperplanes.

The derived action on the admissible simple. For one original simple root i, the bimodule is $U_{fi}\mathrm{/}U$, so the flat Ore sequence gives only two derived terms: L_1 T_i(M)=Tits_i ker(M→M_fi), L_0 T_i(M)=Tits_i coker(M→M_fi).

If the simple highest module L(mu) has $\mu (h_i)$ nonnegative integral, it is integrable for that ORIGINAL sl2_i. Localization kills it. The finite Tits operator $exp(e_i)exp(-f_i)exp(e_i)$ on the module implements the root relabeling, so $L_1{T}_i(L(\mu ))=L(\mu )$ and the other terms vanish.

If $\mu (h_i)$ is nonintegral, $T_i$ is exact and sends $L(\mu )$ to $L(s_i\cdot \mu )$. The needed equivalence here is the simple-root finite-full-Cartan-slice equivalence, not an inference that every locally finite module satisfies the general two-sided support condition in step 12.1. Precisely, use the subquotient category generated by finite sums of the source Vermas and its reflected counterpart, with nonintegral $h_i$ eigenvalues. In a full Cartan coset modulo $\mathbb{Z}{\alpha }_i$, the negative PBW words other than $f_i$ have fixed other simple-root coefficients; these bound their affine loop degree and leave finitely many words. Thus each source Verma slice is finite free over $\mathbb{C}[f_i]$. The explicit nonintegral highest-string decomposition in [F1, Proof 30.1] proves the same for all its subquotients: the coefficients $r!{\prod }_{j=0}^{r-1}(a-j)$ never vanish, and distinct highest strings in the same weight coset have distinct Casimir eigenvalues unless the highest weight is integral. After inverse Tits relabeling, the corresponding reflected slices are finite cofree. The Ore models $U_{f_i}\mathrm{/}U$ and the evaluation unit/counit are exactly those explicitly constructed in [F1, Proof 31.1–32.1]; their rank-one computation uses finitely many strings in each such full-Cartan slice, so it requires no infinite product. These exact inverse functors preserve the indicated subquotient categories. Therefore they send a simple object to a simple object. The quotient presentation $M(\mu )\twoheadrightarrow L(\mu )$ puts the present simple in that category; its image is nonzero by the unit, and $f_i^{-1}v$ is its highest vector of weight $s_i\cdot \mu$. This proves the actual simple image. No equivalence at a zero or negative integral shifted pairing is asserted by this nonintegral step.

Now take a reduced original alcove gallery for y and fold it at each wall of the integral-root arrangement. Every folded alcove stays in the positive integral chamber. Crossing a nonintegral wall changes the simple highest weight by its dot reflection. Hitting an integral wall from inside means that, after the preceding nonintegral transports, the wall is an ORIGINAL simple root with positive integral shifted pairing. Thus the corresponding derived step is exactly the integrable case just proved, contributes one homological degree, and leaves the simple object unchanged up to its actual Tits isomorphism. This is shifted transported integrability; it never asserts integrability for the unshifted beta-root sl2.

More explicitly, let $v_j$ be the successive suffix products of the chosen reduced word, so $v_{j+1}=s_i{v}_j$. Choose $a_j$ in the integral Weyl group with $a_j{v}_j^{-1}(A)$ inside the fundamental integral chamber, and put $u_j=v_j{a}_j^{-1}$. The current simple is $L(u_j\cdot \lambda )$. At a nonintegral crossing $a_{j+1}=a_j$ and $u_{j+1}=s_i{u}_j$. At an integral crossing the wall reflection is $r=v_j^{-1}{s}_i{v}_j$; folding gives $a_{j+1}=a_{j}r$, so $u_{j+1}=s_i{v}_{j}r{a}_j^{-1}=u_j$. Its shifted pairing is $⟨\lambda +\widehat{\rho },a_j{v}_j^{-1}({\alpha }_i{)}^{\vee }⟩$, positive integral because that oriented wall bounds the positive integral chamber. These identities verify the object and degree update at every actual step.

If $y$ belongs to the integral Weyl group, the folded final alcove is the original one. The product of the nonintegral transports therefore returns the highest weight to $\lambda$. At every nonintegral crossing the current simple is a quotient of its current Verma, so it lies in the finite-slice subquotient category of step 26.1; that step gives an actual simple image and exactness. At an integral crossing the positive shifted pairing of step 28.1 makes the original simple-root highest weight nonnegative integral, so step 25.1 gives the sole derived object in degree one by its actual Tits isomorphism. Thus no inverse Hom equivalence for a general infinite-rank ${\mathfrak{n}}_y$ module, and no nonintegral inverse at an integral wall, is used in this gallery. The number of integral walls is the number separating the initial and final integral alcoves, namely ${\mathrm{\ell }}_{\mathrm{i}\mathrm{n}\mathrm{t}}(y)$. Apply the derived length-additive bimodule factorization of step 10.1 at each actual step. Only one homology object remains each time; hence $L_j{T}_{y}L(\lambda )=L(\lambda )$ for $j={\mathrm{\ell }}_{\mathrm{i}\mathrm{n}\mathrm{t}}(y)$ and is zero otherwise.

There is no spectral-sequence ambiguity: at each successive step only one simple homology object in one degree remains.

Applying $T_y$ to the actual ordinary BGG resolution is legitimate because every Verma term is acyclic. Relabel its terms by $v=yw$ and shift by ell_int(y). This yields an exact twisted BGG complex with terms $M^y(v\cdot \lambda )$, indexed by twisted integral length $el{l}_{int}(y^{-1}v)-el{l}_{int}(y)$, and only L(lambda) in degree0.