1
$\begingroup$

As explained in Axiom 2.3 on page 7 of https://arxiv.org/abs/1609.09523, the independence of the stress tensor $$T(y)=\sum_{n}\frac{L_n}{(y-z)^{n+2}}$$ on the choice of expansion point $z$ leads to the following relation:

$$ \frac{\partial L_n}{\partial z} = -(n+1)L_{n-1}$$

Now consider a primary field for which $L_0$ corresponds to multiplication by a number (the scaling dimension), then substituting $n=0$ into above I obtain the following contradictory statement (recalling $L_{-1}=\partial/\partial z$),

$$0 = - \frac{\partial}{\partial z}$$

$\endgroup$

2 Answers 2

1
$\begingroup$

$L_0$ is an operator, and can be replaced with a number only when acting on some field -- but the number depends on the field.

In your case, if $L_0 V = \Delta V$, then $L_0 \partial V = (\Delta+1)\partial V$. This agrees with the relation that you find problematic, i.e. $[\partial, L_0] = -\partial$.

$\endgroup$
1
$\begingroup$

It is important to distinguish between modes (or other objects, composite operators, etc.) defined in different conformal frames. On a general surface one cannot get very far at all unless one is very careful about the frame dependence of the underlying objects. We start with some basic groundwork.

Any conformal tensor (i.e. primary) of weight $h$ transforms as a scalar under rigid shifts: $$ \boxed{\mathscr{O}^{(z)}(z) = \mathscr{O}^{(w)}(z-v)\quad{\rm under}\quad\phantom{\big(}z\rightarrow w(z)=z-v\phantom{\big)}} \label{dagger}\tag{$\dagger$} $$ because for a general conformal transformation, $\mathscr{O}^{(z)}(z)\,\mathrm{d}z^h= \mathscr{O}^{(w)}(w)\,\mathrm{d}w^h$, and $\mathrm{d}w = \mathrm{d}z$ when $w(z)=z-v$. Both $\mathscr{O}^{(z)}(z)$ and $\mathscr{O}^{(w)}(w)$ in (\ref{dagger}) are evaluated at a point $p\in M$ in the underlying manifold, $M$, so that the arguments $z$ and $w$ are really shorthand for chart representations, $z(p)$ and $w(p)$, of the point $p$ respectively. Since the components, $\mathscr{O}^{(z)}(z)$ and $\mathscr{O}^{(w)}(w)$, are frame dependent, we have also placed superscripts, $^{(z)}$ and $^{(w)}$, that serve as a reminder about which frame is used to define them. Such superscripts will have this interpretation throughout.

The quantity $\mathscr{O}^{(z)}(z)$ in (\ref{dagger}), for fixed $z$, is clearly independent of the rigid shift, $v$, and therefore: \begin{equation} \partial_v \mathscr{O}^{(w)}(z-v) = 0. \label{*}\tag{*} \end{equation}

It is usual to define corresponding modes, $\mathscr{O}^{(z)}_n$ and $\mathscr{O}^{(w)}_n$, by a simple Laurent expansion about $z=0$ and $w=0$ respectively:
$$ \mathscr{O}^{(z)}(z) = \sum_{n=-\infty}^\infty \frac{\mathscr{O}^{(z)}_n}{z^{n+h}}\qquad{\rm and}\qquad \mathscr{O}^{(w)}(w) = \sum_{n=-\infty}^\infty \frac{\mathscr{O}^{(w)}_n}{w^{n+h}} $$ and these can in turn be inverted using the Cauchy integral formula, $$ \mathscr{O}_n^{(z)} = \frac{1}{2\pi i}\oint \mathrm{d}z\,z^{n+h-1}\mathscr{O}^{(z)}(z),\qquad{\rm and}\qquad \mathscr{O}_n^{(w)} = \frac{1}{2\pi i}\oint \mathrm{d}w\,w^{n+h-1}\mathscr{O}^{(w)}(w) \label{x}\tag{x} $$ We are interested in the $v$ dependence of these modes. The $^{(z)}$ frame modes, $\mathscr{O}_n^{(z)}$, are completely independent of $v$, whereas the $^{(w)}$ frame modes, $\mathscr{O}_n^{(w)}$, do depend on $v$. Let us make use of this to expose the $v$ dependence of $\mathscr{O}_n^{(w)}$.

From (\ref{x}), making use of (\ref{dagger}) and (\ref{*}), it follows that, $$ \begin{aligned} \mathscr{O}_n^{(w)} &=\frac{1}{2\pi i}\oint \mathrm{d}w\,w^{n+h-1}\mathscr{O}^{(w)}(w)\\ &= \frac{1}{2\pi i}\oint \mathrm{d}z\,(z-v)^{n+h-1}\mathscr{O}^{(w)}(z-v)\\ &= \frac{1}{2\pi i}\oint \mathrm{d}z\,\sum_{a=0}^{\infty}\binom{n+h-1}{a}(-v)^{a}z^{n-a+h-1}\mathscr{O}^{(z)}(z)\\ &= \sum_{a=0}^{\infty}\binom{n+h-1}{a}(-v)^{a}\frac{1}{2\pi i}\oint \mathrm{d}z\,z^{(n-a)+h-1}\mathscr{O}^{(z)}(z)\\ &= \sum_{a=0}^{\infty}\binom{n+h-1}{a}(-v)^{a}\mathscr{O}^{(z)}_{n-a}\\ \end{aligned} $$ This is how the modes of any conformal primary are related in the two frames. The energy-momentum tensor is not a conformal primary when the central charge does not vanish, but it does transform as a tensor under SL(2,$\mathbf{C}$), and rigid shifts are a subset thereof. Therefore, the above displayed equations apply to the energy-momentum tensor independently of the central charge.

Considering an energy-momentum tensor then, where $h=2$, the above relates the corresponding Virasoro generators, $$ \boxed{ L_n^{(w)}=\sum_{a=0}^{\infty}\binom{n+1}{a}(-v)^{a}L^{(z)}_{n-a} }\label{***}\tag{***} $$ For example, $$ \begin{aligned} L_{-1}^{(w)} &= L_{-1}^{(z)} \\ L_{0}^{(w)} &= L_{0}^{(z)} - v L_{-1}^{(z)}\\ L_1^{(w)} %&= \binom{2}{0}L^{(z)}_{1} + \binom{2}{1}(-v)L^{(z)}_{0}+ \binom{2}{2}(-v)^{2}L^{(z)}_{-1}\\ &= L^{(z)}_{1} -2vL^{(z)}_{0} + v^{2}L^{(z)}_{-1}\\ L_{2}^{(w)} %&= L^{(z)}_{2} + \binom{2+1}{1}(-v)L^{(z)}_{2-1}+ \binom{2+1}{2}(-v)^2L^{(z)}_{2-2}+\binom{2+1}{3}(-v)^{3}L^{(z)}_{2-3}\\ &= L^{(z)}_2 - 3vL^{(z)}_1+ 3v^2L^{(z)}_{0} - v^3L^{(z)}_{-1}\\ &\,\,\,\vdots \end{aligned} \label{daggerdagger}\tag{$\dagger\dagger$} $$

Now that the $v$ dependence of the $^{(w)}$ frame Virasoro generators has been made manifest, it is immediate to compute the derivative of $L_0^{(w)}$ that the OP was puzzling about: $$ \begin{aligned} \partial_v L_0^{(w)} &= \partial_v\big(L_{0}^{(z)} - vL_{-1}^{(z)}\big)\\ &= -L_{-1}^{(z)}\\ &= -L_{-1}^{(w)}\\ \end{aligned} $$ That is, $$ \boxed{[\partial_v, L_0^{(w)}] = -L_{-1}^{(w)},\qquad{\rm and}\qquad [\partial_v, L_0^{(z)}] = 0} $$ Similarly, from (\ref{daggerdagger}), $\partial_v L_1^{(w)}=-2L_0^{(w)}$, and $\partial_v L_2^{(w)}=-3L_1^{(w)}$. More generally, from (\ref{***}) one derives that (as the OP quotes), $$ [\partial_v, L_m^{(w)}] = (-1-m)L_{-1+m}^{(w)} \label{daggerdaggerdagger}\tag{$\dagger\dagger\dagger$} $$

Note that this relation is only true in the $^{(w)}$ frame. Taking for granted that, $[L_n^{(z)}, L_m^{(z)}] = (n-m)L_{n+m}^{(z)}$, we learn that it must be the case that: $$ [L_n^{(w)}, L_m^{(w)}] = (n-m)L_{n+m}^{(w)} \label{fancystarstar}\tag{$\star\star$} $$ since the labels $z$ are $w$ are freely interchangeable in (\ref{x}). One can check this commutator explicitly using (\ref{***}), or (\ref{daggerdagger}); for example, $$ \begin{aligned} \phantom{a}[L_{-1}^{(w)}, L_0^{(w)}] &= -L_{-1}^{(w)}\\ [L_{-1}^{(w)}, L_{1}^{(w)}] &= -2 L_{0}^{(w)}\\ [L_{0}^{(w)}, L_{1}^{(w)}] &= -L_1^{(w)}\\ [L_{-1}^{(w)}, L_2^{(w)}] &= -3L_{1}^{(w)}\\ &\,\,\vdots \end{aligned} $$ Comparing (\ref{fancystarstar}) to (\ref{daggerdaggerdagger}) we also see why $L_{-1}^{(w)}$ has the representation $\partial_v$, $$ L_{-1}^{(w)}\sim \partial_v, $$ but there is clearly no correspondence between $\partial_v$ and $L_{-1}^{(z)}$. For $L_{-1}^{(z)}$, what exists instead is the representation: $$ L_{-1}^{(z)} \sim -\partial_z, $$ with the minus sign, and the derivative is with respect to the frame, $z$. The relative minus sign between the two Virasoro generator derivatives (in the two preceding displayed equations) arises due to the difference between 'active' and 'passive' viewpoints.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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