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 $$ [[L_m,L_n],f(z)]=(m-n)[L_{m+n},f(z)].$$


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.

  • 2
    $\begingroup$ ...and your question is whether there's a sign error or not in the sources? $\endgroup$
    – ACuriousMind
    Oct 15 '15 at 19:48
  • $\begingroup$ yes, exactly. Or what's wrong with my reasoning (if anything). $\endgroup$
    – Gytis
    Oct 15 '15 at 20:30

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:

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

  • $\begingroup$ 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
    Oct 29 '15 at 9:20
  • 1
    $\begingroup$ 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
    Oct 29 '15 at 12:01
  • 1
    $\begingroup$ 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
    Oct 29 '15 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.