Why is the Polyakov action special for $p\!=\!1$ branes?

Question

Reading this article I ran into a doubt.

It derives the Polyakov action for a general $p$-brane $$S(X,g)=-\frac{T}{2}\int\text{d}^{p+1}\xi\,\sqrt{-g}\left(g^{ab}h_{ab}-(p-1)\right)\tag{7}$$ by integrating out the Lagrange multiplier from the action $$S(X,g,\Lambda)=-T\int\text{d}^{p+1}\xi\,\left(\sqrt{-g}+\Lambda^{ab}(h_{ab}-g_{ab})\right).\tag{5}$$ Indeed, the equations of motion obtained by varying $g_{ab}$ in the second action yield $$\Lambda^{ab}=\frac{1}{2}\sqrt{-g}g^{ab}.\tag{6}$$

Now, the equations of motion obtained from the first action demand that $$h_{ab}=\frac{1}{2}g_{ab}\left(g^{bc}h_{bc}+1-p\right).\tag{8}$$ Multiplying this equation by $g^{ab}$ one obtains for $p\neq 1$ that $$g^{ab}h_{ab}=p+1\tag{9}.$$ This in turn yields $$g_{ab}=h_{ab}\tag{4}$$ which was already present in the second action by varying the Lagrange multiplier. However, in $p=1$ (for a string) one can at most say that $g_{ab}\propto h_{ab}$. Thus, it seems that the second and first actions are not equivalent in this case. Why is that if we managed to prove the equivalence by varying $g_{ab}$?

Answer 1

This is related to that the Polyakov action$^1$ (7) is invariant under Weyl transformations $$g_{ab}\to \lambda g_{ab} \tag{15}$$ iff $p=1$. Therefore the EL eq. (8) for $g_{ab}$ must also be Weyl-covariant if $p=1$.

Now the punch-line: Since eq. (4) is not Weyl-covariant, it can not be derived from eq. (8) if $p=1$.

References:

1. J.A. Nieto, Remarks on Weyl invariant $p$-branes and $Dp$-branes, arXiv:hep-th/0110227; section 2.

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$^1$ Although the Polyakov action (7) follows directly from the action (5), it might not be as predictive, as we eliminated the Lagrange multipliers $\Lambda^{ab}$ without using their EL eqs. And in fact it is not if $p=1$ because eq. (4) can then no longer be derived.