Given the commutation relations: $$ [\alpha_m,\alpha_n]=m\delta_{m+n,0} $$ and $$ L_m=\frac{1}{2}\sum_\rho\alpha_{m+\rho}\alpha_{-\rho} $$ I am trying to calculate the commutator between $L_m$ and $L_n$ (the Witt algebra and the central extension) $$ [L_m,L_n] = (m-n)L_{m+n}+\frac{1}{12}m(m^2-1)\delta_{m+n,0} $$

Now when I substitute the relations I get the following $$ [L_m,L_n] = \frac{1}{4}\sum_\rho\sum_\lambda[\alpha_{m+\rho}\alpha_{-\rho},\alpha_{n+\lambda}\alpha_{-\lambda}]=\\ =\frac{1}{4}\sum_\rho\sum_\lambda\left(\alpha_{m+\rho}[\alpha_{-\rho},\alpha_{n+\lambda}]\alpha_{-\lambda}+[\alpha_{m+\rho},\alpha_{n+\lambda}]\alpha_{-\rho}\alpha_{-\lambda}+\alpha_{n+\lambda}\alpha_{m+\rho}[\alpha_{-\rho},\alpha_{-\lambda}]+\alpha_{n+\lambda}[\alpha_{m+\rho},\alpha_{-\lambda}]\alpha_{-\rho}\right)\\ =\frac{1}{4}\sum_\rho\sum_\lambda\left(\alpha_{m+\rho}(-\rho)\delta_{-\rho+n+\lambda}\alpha_{-\lambda}+(m+\rho)\delta_{m+\rho+n+\lambda}\alpha_{-\rho}\alpha_{-\lambda}+(-\rho)\delta_{-\rho-\lambda}\alpha_{n+\lambda}\alpha_{m+\rho}+(m+\rho)\delta_{m+\rho-\lambda}\alpha_{n+\lambda}\alpha_{-\rho}\right) $$ Next we fix the first sum(on $\lambda$) using the $\delta$'s $$ [L_m,L_n] =\frac{1}{4}\sum_\rho\left(-\rho\alpha_{m+\rho}\alpha_{n-\rho}+(m+\rho)\alpha_{-\rho}\alpha_{m+n+\rho}-\rho\alpha_{n-\rho}\alpha_{m+\rho}+(m+\rho)\alpha_{n+m+\rho}\alpha_{-\rho}\right) $$ from here I don't know how to proceed to turn this in a form as in the given algebra.


1 Answer 1


It might help to notice that in $$ L_m=\frac{1}{2}\sum_{\rho= -\infty}^\infty\alpha_{m+\rho}\alpha_{-\rho}\,, $$ all terms are of the form $\alpha_x\alpha_y$ satisfying $x+y = m$. Furthermore, since the sum goes over all $\mathbb{Z}$, we can rewrite it as $$ L_m=\frac{1}{2}\sum_{\rho= -\infty}^\infty\alpha_{m+\rho + a}\alpha_{-\rho -a}\,, $$ for any integer shift $a$, as that wouldn't change the summation range. This keeps the property $x+y=m$ shown above.

In your last expression all terms satisfy that property with $x+y=m+n$, so it should be possible to bring them to the form $L_{m+n}$ by applying some shifts on $\rho$. Of course that doesn't give you the $\delta_{m+n,0}$ term, which is a consquence of the fact that $L_0$ is not defined as the other $L_m$'s because of the normal ordering issue $$ L_0 = \frac12 \alpha_0^2 + \sum_{\rho=1}^\infty \alpha_{-\rho}\alpha_\rho\,, $$ note that the sum is only on the positive integers. Thus when $m+n=0$ you need to commute half of the sum (the one on negative integers) in order to put it in the order written above. This forces you to use the famous identity $$ \sum_{n=1}^\infty n\; ``=" \zeta(-1)-\frac1{12}\,. $$ Putting all pieces together should give the correct result.

I too don't like the identity above (the sum of all the integers is a negative fraction), hence the quotes on the equal sign. But I guess it comes down to a definition. Perhaps it should be best to express the $L_0$ as the other $L_m$'s and put the constant $-1/12$ in the definition of the mode expansion as a normal ordering constant, which has to be fixed by other, more rigorous, methods, outside the scope of this answer.

  • $\begingroup$ For the first term, I don't get the $(m-n)$ prefactor, I rather get just $m$ since the sums are equal, while the others cancel. $\endgroup$ Commented Apr 25, 2020 at 14:10
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    $\begingroup$ Nevermind, I figured it out. Once I derive the last part I'll give it an "answered" tag :) $\endgroup$ Commented Apr 25, 2020 at 14:17

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