In Introduction to Conformal Field Theory by Blumenhagen and Plauschinn (springer link) the Virasoro algebra is introduced the central extension of the Witt algebra. They give the central extension $$\widetilde{\mathfrak{g}} = \mathfrak{g}\oplus \mathbb{C}$$ of a Lie algebra $\mathfrak{g}$ by $\mathbb{C}$ is characterised by the following commutations:

\begin{align}[\widetilde{x},\widetilde{y}]_\widetilde{\mathfrak{g}} &= [x,y]_\mathfrak{g} + c p(x,y),\tag{2.14a}\\ [\widetilde{x}, c]_\widetilde{\mathfrak{g}} &= 0,\tag{2.14b}\\ [c,c]_\widetilde{\mathfrak{g}} &= 0,\tag{2.14c}\end{align}

with $\widetilde{x},\widetilde{y}\in \widetilde{\mathfrak{g}}$, $x,y\in\mathfrak{g}$, $c \in \mathbb{C}$ and $p:\mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{C}$.

One of my questions is what does the last commutation relation mean? What do the subscripts after the brackets denote? Without it, it looks like this relation is held for any $c \rightarrow cc - cc = 0$.

  • 2
    $\begingroup$ Yes, OP is right: The last commutation relation should read $[c,c^{\prime}]_\widetilde{\mathfrak{g}} = 0$ with 2 different elements in order to not be a tautology. $\endgroup$
    – Qmechanic
    Commented May 29, 2019 at 16:46
  • $\begingroup$ @Qmechanic Is $c$ arbitrary? I.e. for any $c$ value there is a different "extension", with a different $p$? Another very basic question, what does $c\in\mathbb{C}$ mean exactly? Is it that $c$ is a complex number ($a+ib$)[in that case the commutator still makes no sense]? Or that it has complex entries, e.g. a matrix (wouldn't you denote this in a different way, e.g. as $c\in\mathbb{C}^{k,l}$)? Is $i\partial_{\mu}\in\mathbb{C}$? $\endgroup$
    – Vangi
    Commented May 30, 2019 at 16:16
  • 1
    $\begingroup$ Yes, $c$ is a complex number. However, $[,]$ is not a commutator, but a Lie bracket in the extended algebra. The authors stress this by using the subscript $\tilde{\mathfrak{g}}$. Lie bracket is an abstraction of the commutator. Because of that, if you do not define, say $[\tilde x, c]=0$, you have no means of computing this bracket. However, the $[c,c]$ bracket is zero by the axiom $[a,b]=-[b,a]$ of Lie bracket, so the last equation is redundant. $\endgroup$ Commented May 30, 2019 at 16:42
  • $\begingroup$ They also write $\tilde{\mathfrak{g}}=\mathfrak{g}\oplus \mathbb{C}$. This means that elements of $\tilde{\mathfrak{g}}$ are formal expressions $g+c$ with $g\in \mathfrak{g}$ and $c\in \mathbb{C}$. They then go on to define the Lie bracket on these elements. $\endgroup$ Commented May 30, 2019 at 16:44
  • $\begingroup$ @PeterKravchuk Thank you. Could you help me with the last bit as well? (2.15) says $(m-n)L_{m+n}$, why is it not $(m-n)l_{m+n}$? (2.14a) says that the Lie bracket is in the original algebra, $\mathfrak{g}$ (so it should be (2.12)?), isn't the one given in $\widetilde{\mathfrak{g} }$? $\endgroup$
    – Vangi
    Commented Jun 3, 2019 at 12:44

1 Answer 1


Yes, OP is right. It seems B & P use the symbol $c$ in up to 3 different meanings. For starters, it is (as it written) unclear why the rhs. of eq. (2.15a) is supposed to depend on $c$. It makes more sense to let $c$ be a fiducial central element/generator $\hat{c}$; write

$$\widetilde{\mathfrak{g}} = \mathfrak{g}\oplus \mathbb{C}\hat{c};$$ and replace $c\to\hat{c}$ in eqs. (2.14a-c). This is e.g. done in Kac & Raina. By $\mathbb{C}$-bilinarity of the Lie bracket it is enough to define it on its generators. Eq. (2.14c) is a tautology.

The abelian Lie subalgebra $\mathbb{C}\hat{c}\cong \mathfrak{u}(1)\cong \mathbb{C}$ is isomorphic to the complex numbers.

The Lie algebra element $\hat{c}$ is called a central charge. Be aware that by abuse of notation, the Lie algebra element $\hat{c}$ is often identified with its value in a representation.

  • $\begingroup$ Shouldn't the RHS of eqn. (2.15) contain $(m+n)l_{m+n}$ instead of $(m+n)L_{m+n}$? Eqn. (2.14) says that $[x,y]$ is a Lie bracket in $\mathfrak{g}$, isn't $(m+n)L_{m+n}$ an element of $\widetilde{\mathfrak{g}}$? $\endgroup$
    – Vangi
    Commented May 30, 2019 at 17:35
  • $\begingroup$ I will accept this answer and ask the other question separately. $\endgroup$
    – Vangi
    Commented Jun 6, 2019 at 12:29

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