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A related post might be What are marginal fields in CFT? where Qmechanic♦ pointed to Ginsparg secion 8.6.

However, I heard about two argument.

Claim 1:In a $D$ dimension CFT, the marginal operator must satisfy $h+\bar h=D$.

Though, the Gingsparg's paper provided the definition of the relevant and the irreverent operator, it did not rule out the possibility such as $(h,\bar h)=(2,0)$

Claim 2: In 2D, the marginal operator must satisfy $(h,\bar h)=(1,1)$, not other combination?

How to prove the claim 1 and the claim 2, that in 2D CFT the marginal operator must have $(h,\bar h)=(1,1)$?

Qmechanic
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ShoutOutAndCalculate
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    A marginal operator $V$ must be such that $\int d^2 z V$ is conformally invariant. It immediately follows that $V$ must have weight $(1,1)$ (because $d^2z\sim dz \wedge d{\bar z}$ has weights $(-1,-1)$ and conformally invariant quantities have weights $(0,0)$). – Prahar Apr 12 '22 at 13:09
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    PS - In $D>2$ CFT, there is no $h$ and ${\bar h}$. There is only $\Delta$. Marginal operators must have $\Delta = D$. – Prahar Apr 12 '22 at 13:11

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