Affine Lie Algebras and Loop Central Extensions

For any Laurent polynomial $a={\sum }_j{a}_j{t}^j$, the coefficient of $t^{-1}$ in $a^{\mathrm{\prime }}={\sum }_{j}j{a}_j{t}^{j-1}$ is $0a_0=0$. Thus $\mathrm{Res}(a^{\mathrm{\prime }}dt)=0$, including constant and zero $a$.

For pure tensors $u=x\otimes f$, $v=y\otimes q$, symmetry of $B$ and the product rule give $\omega (u,v)+\omega (v,u)=B(x,y)\mathrm{Res}((fq{)}^{\mathrm{\prime }}dt)=0$.

Expand arbitrary $u,v$ into their finite tensor sums and apply step 2.1 term by term. This proves skewness for all inputs, including empty sums. Setting $v=u$ gives $2\omega (u,u)=0$; since the field is $\mathbb{C}$, division by $2$ gives alternatingness.

For $x,y,z\in \mathfrak{g}$, invariance and symmetry give $B([x,y],z)=B(x,[y,z])=B([y,z],x)$, and the cyclic repetition gives $B([z,x],y)$ as the same scalar $b$.

For $u=x\otimes f$, $v=y\otimes q$, $w=z\otimes r$, the required cyclic sum is $b\mathrm{Res}(((fq{)}^{\mathrm{\prime }}r+(qr{)}^{\mathrm{\prime }}f+(rf{)}^{\mathrm{\prime }}q)dt)$. Expanding the three derivatives gives twice each of $f^{\mathrm{\prime }}qr,f{q}^{\mathrm{\prime }}r,fq{r}^{\mathrm{\prime }}$, so the sum is $2b\mathrm{Res}((fqr{)}^{\mathrm{\prime }}dt)$.

If $fqr={\sum }_j{a}_j{t}^j$, its derivative has $t^{-1}$ coefficient $0a_0=0$. Step 2.1 therefore vanishes. Each general input is a finite sum of pure tensors, and the cyclic expression is trilinear; distributing reduces it to these vanishing summands. Empty sums, zero arguments and repeated arguments are included.

Jacobi makes $[\mathfrak{g},\mathfrak{g}]$ an ideal. It is nonzero because $\mathfrak{g}$ is nonabelian, so simplicity gives $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$. Write any $x\in \mathfrak{g}$ as a finite sum ${\sum }_j[y_j,z_j]$. For any $m\in \mathbb{Z}$, ${\sum }_j[(y_j{)}_m,(z_j{)}_0]=x_m$: the central coefficient is $m{\delta }_{m,0}=0$. Thus every loop mode, including degree zero, lies in the derived algebra, using only brackets within $\overline{L\mathfrak{g}}$.

Choose $h,h^{\mathrm{\prime }}\in \mathfrak{h}$ with $B(h,h^{\mathrm{\prime }})=1$, possible by F2 and rescaling the nondegenerate form. The Cartan is nonzero since its roots span the dual and a nonzero simple algebra cannot have an empty root decomposition. Then $[h_1,h_{-1}^{\mathrm{\prime }}]=c$, because $[h,h^{\mathrm{\prime }}]=0$. Hence $c$ also lies in the derived algebra. This uses just two finite-dimensional vector choices, not AC.

Every defining bracket has zero $d$ coordinate. Bilinearity gives $[\widehat{\mathfrak{g}},\widehat{\mathfrak{g}}]\subseteq L\mathfrak{g}\oplus \mathbb{C}c$, while steps 1.1–1.2 give the reverse inclusion. Since $d$ is a separate nonzero basis vector, this subspace is proper. The same two steps also show the central extension itself is perfect.

The adjoint module is simple, since its invariant subspaces are ideals. F2 therefore gives a dominant integral highest weight $\theta$ and support in $\theta -Q_{+}$. Its weights by F1 are $0$ and the roots. The top weight cannot be zero: the support contains both a root $\beta$ and $-\beta$, whereas neither pair can both lie in $-Q_{+}$ by F3. Thus $\theta$ is a positive root dominating all roots. Any other root with that property dominates $\theta$ and is dominated by it, so equals it by independence of the simple roots. Write $\theta ={\sum }_i{m}_i{\alpha }_i$; since each ${\alpha }_i\le \theta$, every integer $m_i\ge 1$.

Every finite root can be moved to a dominant root of the same length: if $\beta (h_i)<0$, replace $\beta$ by $s_i\beta =\beta -\beta (h_i){\alpha }_i$. This increases its integer height and stays in the finite root set, so iteration terminates with all pairings nonnegative. For that dominant root $\gamma$, step 1.1 gives $\theta -\gamma ={\sum }_i{n}_i{\alpha }_i$ with $n_i\ge 0$. Dominance of $\theta ,\gamma$ then gives $(\theta ,\theta )-(\gamma ,\gamma )={\sum }_i{n}_i({\alpha }_i,\theta +\gamma )\ge 0.$ Hence $\theta$ has maximal root length and the normalization in F4 is $(\theta ,\theta )=2$.

Choose $[e_{\theta },f_{\theta }]={\theta }^{\vee }$ using F1. Invariance gives $B(e_{\theta },f_{\theta })=2\mathrm{/}(\theta ,\theta )=1$ as in F4. Thus the mode bracket gives $[f_{\theta }\otimes t,e_{\theta }\otimes t^{-1}]=-{\theta }^{\vee }+c=h_0$. The diagonal entry ${\alpha }_0(h_0)$ is $2$; for $i>0$, ${\alpha }_0(h_i)=-\theta (h_i)$ and ${\alpha }_i(h_0)=-{\alpha }_i({\theta }^{\vee })$. They are nonpositive integers by dominance of $\theta$ and crystallographic integrality, and vanish simultaneously since both are positive multiples of $-(\theta ,{\alpha }_i)$. The finite entries already satisfy the GCM axioms.

Put ${\beta }_0=-\theta$ and ${\beta }_i={\alpha }_i$ for $i>0$. The matrix of step 3.1 has entries $2({\beta }_i,{\beta }_j)\mathrm{/}({\beta }_i,{\beta }_i)$. Multiplication of row $i$ by $({\beta }_i,{\beta }_i)\mathrm{/}2$ gives the symmetric Gram matrix of these $\mathrm{\ell }+1$ vectors. It is positive semidefinite of rank $\mathrm{\ell }$, since the finite simple roots are a basis. Its kernel is spanned by $(1,m_1,\dots ,m_{\mathrm{\ell }})$, all entries positive. Each proper principal Gram matrix is positive definite: any dependence would extend to a kernel vector with a zero coordinate, impossible. This is the affine, rather than finite, symmetrizable matrix associated with the highest-root extension; it defines the untwisted affine convention here.

The $h_0,h_1,\dots ,h_{\mathrm{\ell }}$ are independent since only $h_0$ has a nonzero $c$ coordinate; the ${\alpha }_0,{\alpha }_1,\dots ,{\alpha }_{\mathrm{\ell }}$ are independent since only ${\alpha }_0$ is nonzero on $d$. Also $\dim \widehat{\mathfrak{h}}=\mathrm{\ell }+2=2(\mathrm{\ell }+1)-\mathrm{rank}\widehat{A}$. Together with step 3.1 these verify every minimal-realization condition. All root selections and height iterations were finite. Rank one is included: the same formulas give $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$.

F1 gives $[e_0,f_0]=h_0$ in the target. F2 gives the Cartan action with weights ${\alpha }_0=\delta -\theta$ and $-{\alpha }_0$, as well as commutativity of the Cartan. The finite simple brackets hold by F5. For $i>0$, $[e_0,f_i]$ lies at finite weight $-\theta -{\alpha }_i$ and $[e_i,f_0]$ at $\theta +{\alpha }_i$, both absent by highest-root maximality. Their mode degrees are nonzero, so no central term occurs. Thus every mixed relation $[e_i,f_j]={\delta }_{ij}{h}_i$ holds.

The finite positive Serre relations follow from finite root strings. For $i>0$, $f_{\theta }$ is a lowest vector for the $i$th finite triple, of weight $-\theta (h_i)={\widehat{a}}_{i0}$, because $-\theta -{\alpha }_i$ is absent. Its raising string is killed after $1-{\widehat{a}}_{i0}$ applications of $\mathrm{ad}{e}_i$. Conversely $e_i$ is a highest vector for the $\theta$ triple, of weight ${\alpha }_i({\theta }^{\vee })=-{\widehat{a}}_{0i}$, since $\theta +{\alpha }_i$ is absent. Its lowering string is killed after $1-{\widehat{a}}_{0i}$ applications of $\mathrm{ad}{f}_{\theta }$. If $\theta ={\alpha }_i$, this is the adjoint rank-one string $e_{\theta },h_{\theta },f_{\theta },0$ of length three. In the loop brackets the relevant positive mode degrees never produce a central term, so these are exactly the two Serre relations involving index zero. Interchanging raising and lowering and replacing every mode degree by its negative proves the negative Serre family by the same strings.

The matrix is symmetrizable by F1. Steps 1.1–1.2 and F3 therefore give a unique homomorphism $\phi :\mathfrak{g}(\widehat{A})\to \widehat{\mathfrak{g}}$ with the specified images. It fixes the embedded Cartan, so its kernel meets that Cartan trivially. F4 forces the kernel to be zero.

Its image contains $\mathfrak{g}\otimes 1$ by F5. The set $J_{+}=\{x\in \mathfrak{g}:x\otimes t\in \mathrm{im}\phi \}$ is an ideal in $\mathfrak{g}$, since $[y\otimes 1,x\otimes t]=[y,x]\otimes t$. It contains the nonzero $f_{\theta }$, hence is all of $\mathfrak{g}$ by simplicity. Likewise $J_{-}=\{x:x\otimes t^{-1}\in \mathrm{im}\phi \}$ contains $e_{\theta }$ and is all of $\mathfrak{g}$.

Nonabelian simplicity gives $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, because the derived algebra is a nonzero ideal. If all positive modes of degree $k-1$ lie in the image, then $[x\otimes t,y\otimes t^{k-1}]=[x,y]\otimes t^k$ for $k\ge 2$; finite sums of these brackets span the degree-$k$ mode. Induction from step 3.1 gives all positive modes. Bracketing degree $-1$ with degree $-(k-1)$ gives all negative modes in the same way. The image already contains $c,d$ through the Cartan. Hence it is the full affine algebra. Combined with step 2.1 this proves the isomorphism. All sums, string calculations and selections at a fixed mode are finite, with no AC.

Extend the $\nu$ in the statement to vanish on $c,d$. The displayed operators fix $\delta$ and preserve $\lambda (c)$. In a composition the additional cross term in the $\delta$ coefficient is $B(\gamma ,\eta )\lambda (c)$, so symmetry gives $t_{\gamma }{t}_{\eta }=t_{\gamma +\eta }$. Thus $t_0=1$ and $t_{-\gamma }$ is the inverse. If $t_{\gamma }=1$, applying it to every level-zero functional gives $\lambda (\gamma )=0$ for all finite $\lambda$, hence $\gamma =0$.

In the Euclidean coroot system let $V$ be the real span of $W_0{\theta }^{\vee }$. It is nonzero and $W_0$-invariant. For a coroot $\beta$ not orthogonal to $V$, some $v\in V$ has $(v,\beta )\mathrm{\ne }0$; the reflection formula makes $v-s_{\beta }v$ a nonzero multiple of $\beta$, so $\beta \in V$. Thus every coroot belongs to either $V$ or $V^{\perp }$. If both classes occurred, finite roots would likewise split into two nonempty orthogonal classes. The corresponding root spaces and coroot spans would form commuting ideals: a mixed root sum lies in neither span, so is not a root; mixed Cartan actions vanish; opposite-root brackets lie in their corresponding coroot spans. This contradicts simplicity of $\mathfrak{g}$. Consequently $V$ is the entire coroot space.

Finite $W_0$ fixes $c,d,\delta$, preserves $B$ and satisfies $w\nu (\gamma )=\nu (w\gamma )$. Substitution proves $w{t}_{\gamma }{w}^{-1}=t_{w\gamma }$. Since $\nu ({\theta }^{\vee })=\theta$ and $B({\theta }^{\vee },{\theta }^{\vee })=2$, substitution also gives $t_{{\theta }^{\vee }}{s}_{\theta }(\lambda )=\lambda -(\lambda (c)-\lambda ({\theta }^{\vee }))(\delta -\theta )=s_0\lambda \mathrm{.}$ Here $s_{\theta }\in W_0$ by F5. Hence $t_{{\theta }^{\vee }}$ and all its finite Weyl conjugates belong to $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$.

Let $M$ be the integer span of $W_0{\theta }^{\vee }$. For any coroot $\beta$, step 1.2 supplies an orbit coroot $\alpha$ with $(\alpha ,\beta )\mathrm{\ne }0$. It has minimal coroot length because $\theta$ is long. If proportional, reducedness gives $\beta =\pm \alpha \in M$. Otherwise integrality and strict Cauchy–Schwarz give $0<\mid \frac{2(\alpha ,\beta )}{(\beta ,\beta )}\mid <\frac{2\mathrm{\parallel }\alpha \mathrm{\parallel }}{\mathrm{\parallel }\beta \mathrm{\parallel }}\le 2.$ The integer has absolute value one. Since $M$ is Weyl invariant, $\alpha -s_{\beta }\alpha =\pm \beta \in M$. Every coroot belongs to $M$, so $M=Q^{\vee }$. Steps 1.1 and 2.1 therefore put every $t_{\gamma }$, $\gamma \in Q^{\vee }$, in $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$.

The products $t_{\gamma }w$ form a group by steps 1.1 and 2.1 and lattice preservation. They contain all finite simple reflections and $s_0=t_{{\theta }^{\vee }}{s}_{\theta }$, so form all of $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$. Let $\mathrm{\Lambda }$ vanish on $\mathfrak{h},d$ and have $\mathrm{\Lambda }(c)=1$. Finite $W_0$ fixes it, whereas $t_{\gamma }\mathrm{\Lambda }=\mathrm{\Lambda }+\nu (\gamma )-B(\gamma ,\gamma )\delta \mathrm{/}2$. Thus $t_{\gamma }\in W_0$ forces $\nu (\gamma )=0$, hence $\gamma =0$. Trivial intersection gives uniqueness by comparing two products. Zero translation, identity finite factor and rank one are included. Every lattice expression is a finite sum and no AC is used.

For $h\in \mathfrak{h}$, $[h,x_{\alpha }\otimes t^n]=\alpha (h)x_{\alpha }\otimes t^n$, while $d$ acts with eigenvalue $n$ and $c$ acts as zero in the adjoint module. The finite decomposition therefore gives precisely the stated nonzero weights and spaces. Distinct $n$ are distinguished by $d$, and distinct finite weights by $\mathfrak{h}$. The zero-weight space is $\widehat{\mathfrak{h}}$ and is not a root space. Dimensions are one for finite-root modes and $\mathrm{\ell }$ for nonzero Cartan modes.

On the root span define $(\alpha +n\delta ,\beta +m\delta )=(\alpha ,\beta )$ using the positive definite finite-root form. F3's Gram-matrix calculation identifies this with the symmetrized form in F4. Thus $\delta$ is in its radical on the root span, every finite-root mode has positive square, and every simple root, including ${\alpha }_0$, has positive square. The Weyl action preserves this form and fixes $\delta$. Consequently $n\delta$ for $n\mathrm{\ne }0$ cannot be the image of a simple root and is imaginary. This says radical on the root span, not on the full dual Cartan form.

Every root in step 1.1 has simple coordinates of one sign. For $n>0$, $\alpha +n\delta =n{\alpha }_0+(n\theta +\alpha )$ has nonnegative coordinates: F3 gives $\theta +\alpha \in Q_{+}$ and $\theta \in Q_{+}$. The negative-$n$ case follows by negation, and $n=0$ is the finite-root sign rule. Multiples $n\delta$ have the same sign as $n$. In particular, reflecting a positive root not proportional to the reflecting simple root keeps it positive: the reflection changes only that simple coordinate, leaving another positive coordinate unchanged, and its image is a root by F4.

Let $\beta ={\sum }_{i=0}^{\mathrm{\ell }}{b}_i{\alpha }_i$ be a positive root of positive square. The equality $(\beta ,\beta )={\sum }_i{b}_i(\beta ,{\alpha }_i)>0$ yields an $i$ with $(\beta ,{\alpha }_i)>0$. Its positive integral coroot pairing is $(\beta ,{\alpha }_i)\mathrm{/}{d}_i$, with $d_i=({\alpha }_i,{\alpha }_i)\mathrm{/}2>0$. If $\beta$ is proportional to ${\alpha }_i$, F5 gives $\beta ={\alpha }_i$. Otherwise step 2.2 shows that $s_i\beta$ stays positive and has smaller positive integer height. Repeat this descent; height cannot decrease indefinitely, so it reaches a simple root. Reversing the finite reflection word proves $\beta$ real. A negative positive-square root is real as well, since its negative is real and simple roots have real negatives. Thus all $\alpha +n\delta$ are real. Combined with steps 1.1 and 2.1 this proves the complete list and multiplicities. The proof includes rank one and all integer modes; no choice over an infinite index set occurs.

F1 gives the zero central action, hence level zero in F3's scalar-action sense. A proposed operator $D$ for $d$ must satisfy $[D,\rho (x)]=0$ by F2 at mode $m=0$. At mode $m=1$ it must satisfy $[D,a\rho (x)]=a\rho (x)$. The left side is $a[D,\rho (x)]=0$, so $a\mathrm{\ne }0$ forces $\rho (x)=0$ for every $x$. Thus a nonzero finite action cannot extend.

Conversely, if $\rho =0$, all loop and central actions are zero. For every $D$ and every $m$, both sides of $[D,a^m\rho (x)]=m{a}^m\rho (x)$ vanish; also $[D,0]=0$ for $c$ and $[D,D]=0$ for $d$. The original loop relations already hold by F1, so this verifies every full-algebra relation. When $V\mathrm{\ne }0$, the operators $0$ and ${\mathrm{id}}_V$ are distinct compatible actions, proving nonuniqueness. When $V=0$ its only endomorphism is zero. These computations prove both directions of the exact extension criterion and use no AC.