The Virasoro algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 20 April 2012

The Virasoro algebra

Let $A$ be an algebra over $ℂ.$ A derivation of $A$ is a linear map $d:A\to A$ such that $$d\left(a_1{a}_2\right)=d\left(a_1\right)a_2+a_{1}d\left(a_2\right),for\; all\; a_1,a_2\in A.$$ The vector space of derivations on $A$ is a Lie algebra with bracket $$[d_1,d_2]=d_1{d}_2-d_2{d}_1.$$

Let $𝔤$ be a Lie algebra. A central extension of $𝔤$ is a short exact sequence of Lie algebras $$0\to 𝔠\to {𝔤}_1\stackrel{{\phi }_1}{\to }𝔤\to 0such\; that𝔠\subseteq Z\left({𝔤}_1\right),$$ the center of $\widetilde{𝔤}.$ A morphism of central extensions is a Lie algebra homomorphism $\psi :{𝔤}_1\to {𝔤}_2$ such that ${\phi }_2\psi ={\phi }_1.$ A universal central extension is a central extension $\widetilde{𝔤}$ such that there is a unique morphism from $\widetilde{𝔤}$ to every other central extension of $𝔤.$ It classifies the projective representations of $𝔤$ (at least this is right for GROUPS, see Steinberg). Isomorphism classes of one-dimensional central extensions are in bijection with elements $c\in H^2(𝔤;𝔽)$ via the formula $${[x,y]}^{˜}=[x,y]+c(x,y)\mathbf{c},for\; x,y\in 𝔤,$$ where $\mathbf{c}$ is a basis element of $Z\left(\widetilde{𝔤}\right).$

The Witt algebra is the Lie algebra of derivations of $ℂ[t,t^{-1}].$ If $d:ℂ[t,t^{-1}]\to ℂ[t,t^{-1}]$ is a derivation then $$d\left(1\right)=0,d\left(t^k\right)=k{t}^{k-1}d\left(t\right),for\; all\; k\in \mathbb{Z},$$ and hence $d$ is determined by the value $d\left(t\right).$ Thus $$Whas\; basis\left\{d_j\right|j\in \mathbb{Z}\},where{d}_j=-t^{j+1}\frac{d}{dt},$$ and $$[d_n,d_m]=(n-m)d_{n+m}.$$ Note that $ℂ[t,t^{-1}]$ is the complexification of the ring of smooth functions on the circle $S^1.$

The Virasoro algebra is the universal central extension of $W.$ It has basis $$\{c,d_i|i\in \mathbb{Z}\}with[c,d_i]=0,[d_n,d_m]=(n-m)d_{n+m}+{\delta }_{n,-m}\frac{n^3-n}{12}c.$$ To try to prove this note that if $$[d_n,d_m]=(n-m)d_{n+m}+c(n,m)z,$$ then $$[d_n,d_m]=-[d_m,d_n]forcesc(n,m)=-c(m,n),$$ and the Jacobi identity forces $$c(n+m,l)+c(l+n,m)+c(m+l,n)=0.$$

The Virasoro algebra has triangular structure and skew linear $(\theta \left(\xi x\right)=\stackrel{\_}{\xi }\theta \left(x\right),for\xi \in ℂandx\in \mathrm{Vir})$ Cartan involution given by $$\begin{array}{rcl}{\mathrm{Vir}}_{<0}& =& \mathrm{span}\left\{d_i\right|i\in {\mathbb{Z}}_{<0}\},\\ {\mathrm{Vir}}_0& =& \mathrm{span}\{c,d_0\},\\ {\mathrm{Vir}}_{>0}& =& \mathrm{span}\left\{d_i\right|i\in {\mathbb{Z}}_{>0}\},\end{array}with\begin{array}{rrcl}\theta :& \mathrm{Vir}& \to & \mathrm{Vir}\\ & d_n& \mapsto & d_{-n}\\ & c& \mapsto & c\end{array}$$

Let $U$ be the universal enveloping algebra of $\mathrm{Vir}.$ The action of $𝔥={\mathrm{Vir}}_0$ on $U_{<0}$ given $U_{<0}$ a ${\mathbb{Z}}_{<0}$ grading such that $$U_{-n}has\; basis\left\{d_{-\lambda }\right|\lambda is\; a\; partition\; ofn\}where{d}_{\mathrm{PICTURE}}=d_{-\lambda }=d_{-{\lambda }_1}\cdots d_{-{\lambda }_l},$$ if $\lambda =({\lambda }_1,...,{\lambda }_l).$ This is the Poincaré-Birkhoff-Witt basis of $U_{<0}.$

The action of admissible $\widehat{𝔤}$ modules

Because the Witt algebra is the space of derivations of $ℂ[t,t^{-1}]$ the Witt algebra acts on the loop algebra $𝔤\otimes ℂ[t,t^{-1}],$ and the Virasoro algebra also acts on $𝔤\otimes ℂ[t,t^{-1}]$ by $$[{\tilde{d}}_k,t^n\otimes x]=t^{k+1}\frac{d}{dt}(t^n\otimes x)=n{t}^{n+k}\otimes x$$ and $c$ acting by $0??$ We can "extend" this action to an action on admissible $\widehat{𝔤}$ modules.

Let $h$ be the Coxeter number of $𝔤$ and let $$T_k=\frac{1}{2}\sum _{j\in \mathbb{Z}}\sum _i:u_i(-j)u^i(j+k):$$ where the normal ordering is $$:u_i(-j)u^i(j+k):=\{\begin{array}{ll}{u}_i(-j)u^i(j+k),& if\; -j\le j+k,\\ u^i(j+k)u_i(-j),& if\; -j>j+k.\end{array}\phantom{\}}$$

If $V$ is a restricted $\widehat{𝔤}-$module of level $l$ and $l\ne -h$ then $$d_k\mapsto \frac{1}{l+h}{T}_{k}andz\mapsto \frac{l}{l+h}\dim \left(𝔤\right)$$ define an action of $\mathrm{Vir}$ on $V.$

Let $𝔤={\mathrm{𝔰𝔩}}_2$ and use the imbedding $$\begin{array}{rrcl}\iota :& {\mathrm{𝔰𝔩}}_2& \to & {\mathrm{𝔰𝔩}}_2\oplus {\mathrm{𝔰𝔩}}_2\\ & x& \mapsto & (x,x)\end{array}$$ to define an action of $\mathrm{Vir}$ on $L\left(\xi \right)\otimes L(m\xi +\frac{n}{2}\alpha )$ by $$d_k\mapsto \frac{1}{l+h}(T_k\otimes 1+1\otimes T_k)-\frac{1}{l+h}\iota \left(T_k\right).$$ This action of $\mathrm{Vir}$ commutes with the action of ${\widehat{\mathrm{𝔰𝔩}}\prime }_2.$ By a character computation $$\begin{array}{rcll}L\left(\xi \right)\otimes L(m\xi +\frac{n}{2}\alpha )& \cong & \underset{k\in I}{⨁}{L}_{{\widehat{\mathrm{𝔰𝔩}}\prime }_2}(\xi +\lambda -k\alpha )\otimes U_{m,n,k},& as\; ({\widehat{\mathrm{𝔰𝔩}}\prime }_2,\; <mi>Vir</mi>) \; bimodules,\; and\\ L\left(\xi \right)\otimes L(m\xi +\frac{n}{2}\alpha )& \cong & \underset{\genfrac{}{}{0ex}{}{k\in I}{j\ge k^2}}{⨁}{L}_{{\widehat{\mathrm{𝔰𝔩}}}_2}(\xi +\lambda -k\alpha -j\delta )\otimes U_{m,n,k}^j,& as\; {\widehat{\mathrm{𝔰𝔩}}}_2 \; modules,\end{array}$$ where $$I=\{k\in \mathbb{Z}|\frac{n}{2}-\frac{m+1}{2}\le k\le \frac{n}{2}\},$$ and $$\mathrm{char}\left(U_{m,n,k}\right)=\sum _{j\in {\mathbb{Z}}_{\ge 0}}\dim \left(U_{m,n,k}^j\right)q^j=(f_{m,n,k}-f_{m,n,n+1-k})\prod _{j\in {\mathbb{Z}}_{\ge 0}}\frac{1}{1-q^j},$$ with $$f_{m,n,k}=\sum _{j\in \mathbb{Z}}{q}^{(m+2)(m+3)j^2+(n+1+2k(m+2))j+k^2}.$$ Then $z$ acts on $U_{m,n,k}$ by the constant $$\frac{n(n+2)}{4(m+2)}-\frac{(n-2k)(n-2k+2)}{4(m+3)}+j$$ and the minimum value of $j$ for which $U_{m,n,k}^j\ne 0$ is $$j=k^{2}when\; d_0 \; acts\; by{h}_{r,s}=\frac{{((m+3)r-(m+2)s)}^2-1}{4(m+2)(m+3)}$$ where $$\begin{array}{lll}r=n+2,& s=n+1-2k,& if\; k\ge 0,\\ r=m-n+1,& s=m-n+2+2k,& if\; k<0.\end{array}$$

The Shapovalov determinant

The highest power of $h$ in $\det \left(M{(h,c)}^{(h+n,c)}\right)$ is $$\sum _{\lambda ⊢n}l\left(\lambda \right),with\; coefficient\prod _{\lambda ⊢n}{z}_{2\lambda },$$ where, for a partition $\lambda$ of $n,$ $\frac{n!}{z_{\lambda }}$ is the cardinality of the conjugacy class of the symmetric group $S_n$ labeled by $\lambda .$

		Proof.
	Let us first analyze the entries $⟨d_{-\mu }{v}^{+},d_{\lambda }{v}^{+}⟩$ in the matrix. Then $$⟨d_{-\mu }{v}^{+},d_{-\lambda }{v}^{+}⟩=⟨v^{+},d_{mu}{d}_{-\lambda }{v}^{+}⟩=p_{0,0}(h,c),$$ where $p_{0,0}(d_0,z)$ is the polynomial in $d_0$ and $z$ in the PBW basis expansion $$d_{\mu }{d}_{-\lambda }=\sum _{\nu ,\tau }{d}_{-\nu }{p}_{\nu ,\tau }(d_0,z)d_{\tau }.$$ This expansion is obtained by using the relations $$\begin{array}{rcll}{d}_k{d}_j& =& d_j{d}_k+(k-j)d_{j+k},& if\; j+k\ne 0,\\ d_k{d}_{-k}& =& d_{-k}{d}_k+(2k{d}_0+\frac{k^2(k-1)}{12}z),& for\; k>0,\\ d_1{d}_{-1}& =& d_{-1}{d}_1+2d_0,& \\ d_0{d}_{-k}& =& d_{-k}{d}_0+k{d}_{-k}=d_{-k}(d_0+k),& \\ z{d}_{-k}& =& d_{-k}z,& \end{array}$$ to put the $d_i$ in increasing order. The first relation "combines" $j$ and $k$ into $j+k.$ If $d_{-\nu }{p}_{\nu ,\tau }(d_0,z)d_{\tau }$ is a term in the PBW expansion then the parts of $-\nu$ and $\tau$ are combinations of parts of $\mu$ and $-\lambda$ and the degree in $d_0$ of the polynomial $p_{\nu ,\tau }(d_0,z)$ is the maximal number of 0 parts that can be obtained by combinations of the remaining parts of $\mu$ and $-\lambda$ (those that do not contribute to $\nu$ and $-\tau$). Thus the degree (in $d_0$) of $p_{0,0}(d_0,z)$ is the maximal number of 0 parts that can be obtained by combinations of the parts of $\mu$ and $-\lambda$ and is at most $l\left(\mu \right)$ and at most $l\left(\lambda \right).$ Since both $\lambda$ and $\mu$ are partitions of $n,$ a term of degree $l\left(\lambda \right)$ is produced only when $\lambda =\mu$ and each part of $\lambda$ is combined with a single part of $-\lambda .$ Thus the maximal degree term in row $\lambda$ of $A{(h,c)}^{(h+n,c)}$ appears in column $\lambda ,$ i.e. on the diagonal. The identity $$d_r{d}_{-r}^s=d_{-r}^s{d}_r+d_{-r}^{s-1}(2rs{d}_0+2r^2\left(\genfrac{}{}{0ex}{}{s}{2}\right)+s\left(\frac{r^3-r}{12}\right)z),$$ is verified by induction on $s,$ the induction step being Suppose that $$d_r^k{d}_{-r}^s=d_{-r}^s{d}_r^k+d_{-r}^{s-1}{p}_1^{k,s}{d}_r^{k-1}+d_{-\mathrm{r}}^2{p}_2^{k,s}{d}_r^{k-2}+\cdots +d_{-r}^{s-k}{p}_k^{k,s},$$ where $p_i^{k,s}$ are polynomials in $d_0$ and $z.$ Then $$\begin{array}{rcl}{d}_r^{k+1}{d}_{-r}^s& =& d_r\sum _{j=0}^k{d}_{-r}^{s-j}{p}_j^{k,s}(d_0,z)d_r^{k-j}\\ & =& \sum _{j=0}^k{d}_{-r}^{s-j}{d}_r{p}_j^{k,s}(d_0,z)d_r^{k-j}+d_{-r}^{s-j-1}(2r(s-j)d_0+2r^2\left(\genfrac{}{}{0ex}{}{s-j}{2}\right)+(s-j)\left(\frac{r^3-r}{12}\right)z)p_j^{k,s}(d_0,z)d_r^{k-j}\\ & =& \sum _{j=0}^k{d}_{-r}^{s-j}{p}_j^{k,s}(d_0-r,z)d_r^{k-j+1}+d_{-r}^{s-j-1}(2r(s-j)d_0+2r^2\left(\genfrac{}{}{0ex}{}{s-j}{2}\right)+(s-j)\left(\frac{r^3-r}{12}\right)z)p_j^{k,s}(d_0,z)d_r^{k-j},\end{array}$$ from which it follows that $$p_l^{k+1,s}(d_0,z)=p_l^{k,s}(d_0-r,z)+(2r(s-l+1)d_0+2r^2\left(\genfrac{}{}{0ex}{}{s-l+1}{2}\right)+(s-j)\left(\frac{r^3-r}{12}\right)z)p_{l+1}^{k,s}(d_0,z).$$ (I'm not quite sure if this calculation is exactly right, I need to do some checks for $s=2$ and $s=3$ to make sure.) In particular, $$p_{k+1}^{k+1,s}=\prod _{j=1}^{k+1}(2r(s-j)d_0+2r^2\left(\genfrac{}{}{0ex}{}{s-j}{2}\right)+(s-j)\left(\frac{r^3-r}{12}\right)z)={\left(2r\right)}^{k+1}(k+1)!d_0^{k+1}+ \; lower\; degree\; terms\; in\; d_0.$$ $\square$

There is a bijection $$\begin{array}{ccc}\left\{(\lambda ,i)\right|\lambda ⊢n,1\le i\le l\left(\lambda \right)\}& \leftrightarrow & \left\{(\mu ,\left(r^s\right))\right|r^s=\varnothing ,\left|\mu \right|+rs=n\}\\ (\lambda ,i)& \to & (\lambda -\left({\lambda }_i^{s_i}\right),\left({\lambda }_i^{s_i}\right))\\ (\mu \cup \left(r^s\right),j)& \leftarrow & (\mu ,\left(r^s\right))\end{array}PICTURE$$ where $\lambda -\left({\lambda }_i^{s_i}\right)$ is the partition obtained by removing all rows of length ${\lambda }_i$ which are in rows with number $\ge i,$ $s_i$ is the number of $j\ge i$ such that ${\lambda }_j={\lambda }_i$ and $j-1$ is the row number of the largest part $\le r$ in the partition $\mu .$

This bijection proves the identity $$\sum _{\lambda ⊢n}l\left(\lambda \right)=\sum _{\genfrac{}{}{0ex}{}{\left(r^s\right)\ne \varnothing }{n-rs\ge 0}}p(n-rs),$$ where $p\left(k\right)$ is the number of partitions with $k$ boxes.

If $k<n$ and $d_0-h$ divides the determinant $\det \left(M_{+k}\right)$ then ${(d_0-h)}^{p(n-k)}$ divides the determinant $\det \left(M_{+n}\right).$

$C_{rs}(h,c)$ divides the determinant $\det \left(M_{+rs}\right).$

		Proof.
	First proof: Second proof: Define a Vir action on the space of semi-infinite forms $$\mathscr{H}(\alpha ,\beta )=\mathrm{span}\{\cdots \wedge f_{i_k}\wedge \cdots \wedge f_{i_1}|with{i}_1<i_2<\cdots and{i}_k=-kforklarge\},$$ by setting $$d_n\left(f_j\right)=(j+\beta -(1-n)\alpha )f_{j-n}.$$ Then, for appropriate choice of $\alpha$ and $\beta ,$ the Vir module $\mathscr{H}(\alpha ,\beta )$ becomes a highest weight module of highest weight $(h,c).$ One can construct a number of highest weight vectors in $\mathscr{H}(\alpha ,\beta ),$ see ???. $\square$

Blocks

Given $(h,c)$ the equation $$\mu +\frac{1}{\mu }=\frac{13-c}{6}determines\{\mu ,\frac{1}{\mu }\},$$ and for each choice of $\mu$ in this set, $$y^2=4\mu (\frac{1-c}{24}-h)determines\{y,-y\}$$ giving 4 lines $$s=\mu r+y,s=\mu r-y,s=\frac{1}{\mu }r-\frac{1}{\mu }y,s=\frac{1}{\mu }r+\frac{1}{\mu }y.$$ Conversely, given $(\mu ,y)$ then $$\frac{13-c}{6}=\mu +\frac{1}{\mu }determinesc,$$ and $$h=\frac{-y^2}{4\mu }+\frac{1-c}{24}determinesh.$$ Define $$\begin{array}{rcl}{C}_{rs}(h,c)& =& \frac{1}{4^2}(s-\mu r+y)(s-\mu r-y)(s-\frac{1}{\mu }r-\frac{1}{\mu }y)(s-\frac{1}{\mu }r+\frac{1}{\mu }y)\\ & =& \frac{1}{4^2}({(s-\mu r)}^2-y^2)({(s-\frac{1}{\mu }r)}^2-\frac{y^2}{{\mu }^2})\\ & =& ((s-\mu r)\frac{(\frac{1}{\mu }s-r)}{4}-\frac{y^2}{4\mu })((s-\frac{1}{\mu }r)\frac{(\mu s-r)}{4}-\frac{y^2}{4\mu })\\ & =& (\frac{\mu r^2-2rs+\frac{1}{\mu }{s}^2}{4}-\frac{y^2}{4\mu })(\frac{\mu r^2-2rs+\mu s^2}{4}-\frac{y^2}{4\mu })\\ & =& (\frac{1}{4}(\mu r^2+\frac{1}{\mu }{s}^2)-\frac{rs}{2}+h-\frac{1-c}{24})(\frac{1}{4}(\frac{1}{\mu }{r}^2+\mu s^2)-\frac{rs}{2}+h-\frac{1-c}{24}).\end{array}$$ If $$x=\frac{1}{2}\sqrt{\frac{25-c}{1-c}}$$ then $$\begin{array}{rcl}(x-\frac{1}{2})(x+\frac{1}{2})& =& x^2-\frac{1}{4}\\ & =& \frac{1}{4}\left(\frac{25-c}{1-c}\right)-\frac{1}{4}\\ & =& \frac{1}{4}\left(\frac{25-c-1+c}{1-c}\right)\\ & =& \frac{1}{4}\left(\frac{24}{1-c}\right)\\ & =& \frac{6}{1-c}\end{array}$$ so that $$\frac{1-c}{6}=\frac{1}{(x-\frac{1}{2})(x+\frac{1}{2})}and\frac{13-c}{6}=2+\frac{1-c}{6}=2+\frac{1}{(x-\frac{1}{2})(x+\frac{1}{2})}.$$ Then the solutions to $\mu +\frac{1}{\mu }=\frac{13-c}{6}$ are $$\mu =\frac{x+\frac{1}{2}}{x-\frac{1}{2}}and\frac{1}{\mu }=\frac{x-\frac{1}{2}}{x+\frac{1}{2}},$$ since $$\frac{x+\frac{1}{2}}{x-\frac{1}{2}}+\frac{x-\frac{1}{2}}{x+\frac{1}{2}}=\frac{x^2-x+\frac{1}{4}+x^2+x+\frac{1}{4}}{(x-\frac{1}{2})(x+\frac{1}{2})}=\frac{2x^2+\frac{1}{2}}{x^2-\frac{1}{4}}=\frac{2x^2-\frac{1}{2}+1}{x^2-\frac{1}{4}}=2+\frac{1}{(x-\frac{1}{2})(x+\frac{1}{2})}.$$ Then $$\begin{array}{rcl}\frac{1-c}{24}-\frac{1}{4}(\mu r^2+\frac{1}{\mu }{s}^2)+\frac{rs}{2}& =& \frac{1-c}{24}-\frac{1}{4}(\frac{x+\frac{1}{2}}{x-\frac{1}{2}}{r}^2+\frac{x-\frac{1}{2}}{x+\frac{1}{2}}{s}^2)+\frac{rs}{2}\\ & =& \frac{1-c}{24}-\frac{1}{4}\left(\frac{(x^2+x+\frac{1}{4})r^2+(x^2-x+\frac{1}{4})s^2}{(x-\frac{1}{2})(x+\frac{1}{2})}\right)+\frac{rs}{2}\\ & =& \frac{1-c}{24}-\frac{1}{4}(\frac{2x^2+\frac{1}{2}}{(x-\frac{12}{})(x+\frac{1}{2})}(r^2+s^2)+\frac{x}{(x-\frac{1}{2})(x+\frac{1}{2})}(r^2-s^2))+\frac{rs}{2}\\ & =& \frac{1-c}{24}-\frac{1}{4}(\left(\frac{13-c}{6}\right)(r^2+s^2)+\frac{1}{2}\sqrt{\frac{25-c}{1-c}}\left(\frac{1-c}{6}\right)(r^2-s^2))+\frac{rs}{2}\\ & =& \frac{1-c}{24}-\frac{1}{4}(\left(\frac{13-c}{6}\right)(r^2+s^2)+\frac{1}{12}\sqrt{(25-c)(1-c)}(r^2-s^2))+\frac{rs}{2}\end{array}$$ and $$\begin{array}{rcl}\frac{1-c}{24}-\frac{1}{4}(\mu r^2+\frac{1}{\mu }{s}^2)+\frac{rs}{2}& =& \frac{1-c}{24}-\frac{1}{4}\frac{{(s-\mu r)}^2}{\mu }\\ & =& \frac{1-c}{24}-\frac{1}{4}{(s-\frac{x+\frac{1}{2}}{x-\frac{1}{2}}r)}^2\frac{x-\frac{1}{2}}{x+\frac{1}{2}}\\ & =& \frac{1}{4}(\frac{1}{(x-\frac{1}{2})(x+\frac{1}{2})}-\frac{{((x-\frac{1}{2})s-(x+\frac{1}{2})r)}^2}{(x-\frac{1}{2})(x+\frac{1}{2})})\\ & =& \frac{{((x+\frac{1}{2})r-(x-\frac{1}{2})s)}^2-1}{-4(x-\frac{1}{2})(x+\frac{1}{2})}.\end{array}$$ Now put $m+\frac{5}{2}=x$ so that $x+\frac{1}{2}=m+3$ and $x-\frac{1}{2}=m+2.$

$$\det \left(A_{-n}\right)=\prod _{1\le r\le s\le n}{({\left(2r\right)}^{s}s!)}^{p(n-rs)-p(n-r(s+1))}\prod _{\genfrac{}{}{0ex}{}{r,s\in {\mathbb{Z}}_{\ge 0}}{rs\le n}}{(h-h_{rs})}^{p(n-rs)},$$ where $$h_{rs}=\frac{1}{48}((13-c)(r^2+s^2)+\sqrt{(c-1)(c-25)}(r^2-s^2)-24rs-2+2c).$$ Then $$C_{r,s}(h,c)=\{\begin{array}{ll}(h-h_{rs})(h-h_{sr}),& ifr\ne s,\\ h-h_{rr},& ifr=s.\end{array}\phantom{\}}$$

Notes and References

These notes are from lecture notes of Arun Ram from 2005.

References

[Cu1] C. Curtis, "Representations of Hecke algebras." Astérisque, 9 (1988), 13-60.

[Li1] P. Littelmann, Paths and root operators in representation theory, Ann. Math. 142 (1995), 499-525.

[Li2] P. Littelmann, Bases for representations, LS-paths and Verma flags, Volume in honor of the 70th birthday of C.S. Seshadri, ?????.

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