Affine Lie Algebras and Loop Central Extensions — Examples

Multiplication gives $ef=\mathrm{diag}(1,0)$, $fe=\mathrm{diag}(0,1)$, so $[e,f]=h$ and $B(e,f)=1$; also $h^2=I$, so $B(h,h)=2$ and $[h,h]=0$. Trace cyclicity makes $B$ invariant; its Gram matrix on $e,f,h$ has determinant $-2$, so is nondegenerate. Thus F1 yields the first two displayed brackets. In particular $[e_1,f_{-1}]=h_0+c$, $[h_1,h_{-1}]=2c$, while $[e_0,f_n]=h_n$ and $[h_0,h_n]=0$.

F2 gives $[d,e_m]=m{e}_m$, including $[d,e_0]=0$ and $[d,e_{-1}]=-e_{-1}$. If $m+n\mathrm{\ne }0$, both central terms in step 1.1 vanish; if $m+n=0$, their coefficients are exactly $m$ and $2m$. Hence all integer signs, zero modes and the opposite-mode boundary obey the displayed formulas. No choices are needed.

Finite Cartan elements commute. Thus F1 gives the displayed bracket, entirely in $\mathbb{C}c$, proving closure. By F2 each nonzero mode has dimension $\mathrm{\ell }$ and the underlying vectors are precisely the stated Cartan modes. For $m\mathrm{\ne }0$, pairing it with mode $-m$ gives the nondegenerate pairing $mB$ because $B$ is nondegenerate on the finite Cartan.

Let $z=ac+{\sum }_{m\mathrm{\ne }0}{h}_m\otimes t^m$ have finite support and at least one $h_j\mathrm{\ne }0$. Choose $k\in \mathfrak{h}$ with $B(h_j,k)\mathrm{\ne }0$. Its bracket with $k\otimes t^{-j}$ is exactly $jB(h_j,k)c\mathrm{\ne }0$: all other modes have nonopposite degrees and contribute zero. Hence no such $z$ is central. Conversely $ac$ is central by F1. This proves center $\mathbb{C}c$ and nondegeneracy of the alternating form on $\mathcal{H}\mathrm{/}\mathbb{C}c$. If zero modes were included, they would commute with every Cartan mode and enlarge the center by $\mathfrak{h}$. The finite-support zero element and rank-one case require no exception.

The finite roots are $\alpha ,-\alpha$, so the highest root is $\theta =\alpha$. F1 gives the displayed data. Since $\delta (c)=\delta (h)=\alpha (c)=0$, we compute ${\alpha }_0(h_0)=0-\alpha (c-h)=2$, ${\alpha }_1(h_0)=\alpha (c-h)=-2$, ${\alpha }_0(h_1)=-\alpha (h)=-2$, and ${\alpha }_1(h_1)=2$. These are all four entries in the stated row-coroot, column-root convention.

By F2, $e_0=f\otimes t$ and $f_0=e\otimes t^{-1}$, with $[e_0,f_0]=c-h=h_0$; the finite generators lie in degree zero. The matrix has rank one since its second row is the negative of its first, and its null vector is $(1,1)$. Accordingly ${\alpha }_0+{\alpha }_1=\delta$ and $h_0+h_1=c$, both nonzero in the full realization. Thus the two off-diagonal double entries describe an affine, not finite rank-one, matrix. All data are explicit and choice-free.

Direct multiplication gives $[e,f]=h$, $[h,e]=2e$, $[h,f]=-2f$. Thus these matrices represent ${\mathfrak{s}\mathfrak{l}}_2$, and $[a^{m}x,a^{n}y]=a^{m+n}[x,y]$. This is the image of the affine mode bracket because the added central term acts zero. F1 consequently gives the claimed representation for all positive, zero and negative modes. The central operator is zero, so its level is zero.

The matrix $e$ is nonzero, so F2 excludes an extension. Explicitly, mode zero would give $[D,e]=0$, while mode one would give $[D,ae]=ae$. Bilinearity makes the latter left side zero, but $ae\mathrm{\ne }0$, a contradiction. This holds also at $a=1$, and negative powers are defined precisely because $a\mathrm{\ne }0$. All vectors and operators are explicit; no AC is used.

Take $e_0=f_{\theta }\otimes t$ and $f_0=e_{\theta }\otimes t^{-1}$. F1 gives $[e_0,f_0]=[f_{\theta },e_{\theta }]\otimes 1=-{\theta }^{\vee }$. If we adjoin $c$ as an independent central vector but leave this bracket unchanged, it still has zero $c$ coordinate. F2 instead requires $[e_0,f_0]=c-{\theta }^{\vee }$, whose $c$ coordinate is one. These vectors differ by the nonzero $c$.

Thus the required mixed relation fails and F3's realization cannot persist. In the unextended loop algebra there is not even a vector for this independent central coordinate. Sending $c$ to zero does produce a quotient representation of the derived affine algebra, but it cannot be the claimed faithful full realization. For finite ${\mathfrak{s}\mathfrak{l}}_2$ the same discrepancy is $-h$ versus $c-h$, already with degrees $1,-1$ and form value one. This explicit witness refutes the assertion without any choice assumption.

F1 gives ${\omega }_a=a\omega$. Choose $x,y$ with $B(x,y)=1$. For modes $x_1,y_{-1}$ the old scalar cocycle is $1$, while the new one is $a$. In particular $a=2$ is an explicit failure of invariance of the numerical cocycle. Nondegeneracy permits the finite pair choice and scalar normalization.

Let $c_a$ denote the central generator for $B_a$. A map fixing all loop modes and sending $c_a$ to $a^{-1}c$ preserves brackets, since the image of $[x_m,y_n{]}_a=[x,y{]}_{m+n}+am{\delta }_{m,-n}B(x,y)c_a$ is $[x,y{]}_{m+n}+m{\delta }_{m,-n}B(x,y)c$. It is invertible, with inverse $c\mapsto a{c}_a$, so is a Lie isomorphism. This central image is forced: apply any such map to the bracket in step 1.1 and subtract its fixed loop component to get $a\phi (c_a)=c$.

Pulling a level-$k$ module for the old extension back through this isomorphism makes $c_a$ act as $a^{-1}k$. Equivalently the transported weight has ${\mathrm{\Lambda }}_a(c_a)=a^{-1}\mathrm{\Lambda }(c)$ by F3. At $a=2$ this is $k\mathrm{/}2$, differing from $k$ whenever $k\mathrm{\ne }0$. The case $k=0$ stays zero and $a=1$ gives the identity normalization; $a=0$ is excluded because both nondegeneracy and the inverse would fail. Thus rescaling requires exactly the stated adjustment of the central generator and numerical levels. No AC is used.