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Let's take as an example Di Francesco et al. but every source I am aware of is doing the same.

First of all, the Virasoro algebra is usually defined as

$$[L_m,L_n] = (m - n)L_{m+n} + \frac{c}{12} m (m^2 -1) \delta_{m+n,0}.\tag{6.24}$$

A field is primary if $$[L_n, \Phi_i (z, \bar{z})]= z^{n+1} \partial_z \Phi_i(z,\bar{z}) + h_i (n+1) z^n \Phi_i(z,\bar{z})\tag{6.28}$$ and similarly for the antiholomorphic part.

Then the authors continue:

"The generators $L_m$ $(m > 0)$ also increase the conformal dimension, by virtue of the Virasoro algebra (6.24):

$$[L_0, L_{-m}] = m L_{-m}. \quad\quad (6.35)"$$

However, given $(6.28)$, I get $$ [[L_m,L_n], f(z)]= - [ f(z),[L_m,L_n]] = [L_m,[L_n,f(z)]]+[L_n,[f(z),L_m]]\\ =[L_m,[L_n,f(z)]]-[L_n,[L_m,f(z)]] $$ $$[L_m,[L_n,f(z)]]=[L_m,z^{n+1}\partial_z f(z)+h(n+1) z^n f(z)]\\=z^{m+n+1}(n+1)\partial f + z^{m+n+2}\partial^2 f + h(n+1)n z^{m+n} f +h(n+1) z^{m+n+1}\partial f +h(m+1)z^{m+n+1}\partial f + h^2(m+1)(n+1) z^{m+n}f. $$

Thus, $$ [[L_m,L_n],f(z)]=(n-m)z^{m+n+1}\partial f+ (n-m)h(m+n+1)z^{m+n}f=(n-m)[L_{m+n},f(z)].$$ But if $(6.24)$ is correct, then $$ [[L_m,L_n],f(z)]=(m-n)[L_{m+n},f(z)].$$

EDIT

So my question is the following:

is my reasoning correct and Di Francesco and other sources should be corrected, or there is a flaw with such a consistency check which I have given above.

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    $\begingroup$ ...and your question is whether there's a sign error or not in the sources? $\endgroup$
    – ACuriousMind ♦
    Commented Oct 15, 2015 at 19:48
  • $\begingroup$ yes, exactly. Or what's wrong with my reasoning (if anything). $\endgroup$
    – Gytis
    Commented Oct 15, 2015 at 20:30

1 Answer 1

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You have an algebraic error in the step after the Jacobi Identity. The nested commutator $[L_m,[L_n,f(z)]]$ is given by: \begin{align} [L_m,[L_n,f(z)]] &= [L_m,z^{n+1}\partial_zf(z)+h(n+1)z^nf(z)]\\ &=z^{n+1}[L_m,\partial_zf(z)]+h(n+1)z^n[L_m,f(z)]\\ &=z^{n+1}[z^{m+1}\partial_z^2f+(m+1)z^m\partial_zf+h(m+1)z^m\partial_zf\\&+h(m+1)mz^{m-1}f] +h(n+1)z^n[z^{m+1}\partial_zf + h(m+1)z^mf]\\ &=z^{m+n+2}\partial_z^2f+(m+1)z^{m+n+1}\partial_zf+h(m+1)z^{m+n+1}\partial_zf \\ &+h(m^2+m)z^{m+n}f +h(n+1)z^{m+n+1}\partial_zf +h^2(n+1)(m+1)z^{m+n}f \end{align}

From this one gets:

$$ \require{cancel} [L_m,[L_n,f(z)]]-[L_n,[L_m,f(z)]] = (m-n)z^{m+n+1}\partial_zf+\cancel{h(m-n)z^{m+n+1}\partial_zf} + h(m-n)(m+n+1)z^{m+n}f-\cancel{h(m-n)z^{m+n+1}\partial_zf} \\=(m-n)[L_{m+n},f] $$

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  • $\begingroup$ Thank you, but are you certain that we can just leave out the powers of $z$ as you did in your second equality? I mean in the usual representation $L_m= -z^{m+1} \partial_z$. $\endgroup$
    – Gytis
    Commented Oct 29, 2015 at 9:20
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    $\begingroup$ I think you're confusing the quantum generators [$L_m$ of the Virasoro algebra] and the classical generators [$l_m=-z^{m+1}\partial_z$ of the Witt algebra]. When one quantises a classical field theory to get a quantum field theory, the fields are promoted to operators. The fields then may be mode expanded in terms of functions of space-time variables and "creation/annihilation" operators (ref. Di Francesco eqn $(6.7)$). Now, in this case, the virasoro generators $L_m$ will only "talk" to other operators (ex. $f_{m,n}$ of mode expansion of $f(z)$). $\endgroup$
    – Anshuman
    Commented Oct 29, 2015 at 12:01
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    $\begingroup$ Concisely, $[\hat{L}_m,z^k]=[\hat{L}_m,z^k\hat{I}]=0$ where the hat denotes operators explicitly and $\hat{I}$ is the identity operator. $\endgroup$
    – Anshuman
    Commented Oct 29, 2015 at 12:14

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