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

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 $$(h,\bar h)=(1,1)$$?

• 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 $(0,0)$). Commented Apr 12, 2022 at 13:09
• PS - In $D>2$ CFT, there is no $h$ and ${\bar h}$. There is only $\Delta$. Marginal operators must have $\Delta = D$. Commented Apr 12, 2022 at 13:11