In bosonic string theory the Polyakov action can be put in into conformal gauge. It is then possible to show that the resulting gauge fixed action is conformally invariant. Actually it's shown that it's invariant under the combined actions of conformal transformations and Weyl transformations, but it's called 'conformal invariance'.

Since this invariance applies to the gauge fixed action, does this mean that the pre-gauge-fixed Polyakov action is conformally invariant also? It's just easiest to demonstrate in conformal gauge since the metric is flat and the determinant of the metric becomes unity?

In general I would expect that all symmetries of a gauge fixed action are also symmetries of the pre-gauge fixed action. On the other hand not all symmetries of the pre-gauge-fixed action are symmetries of the gauge fixed action. This is because the process of gauge fixing means fixing values that would otherwise freely vary according to those symmetries that are gauge fixed.

Is the above reasoning correct? Can we say the pre-gauge-fixed Polyakov action is conformally invariant?

  • $\begingroup$ I may have found the answer to my own question: It is possible to show that the pre-gauge-fixed Polyakov action is invariant under Weyl transformations and diffeomorphisms (reparametrisations). Since conformal transformations are equivalent to combined reparametrisations and Weyl transformations it must follow that the pre-gauge fixed Polyakov action is conformally invariant. Some sort of feedback would be greatly appreciated however. $\endgroup$ May 22, 2015 at 16:34
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    $\begingroup$ On second thoughts, I think the answer is that the conformal transformations are just a special case of the reparametrisations. Since the pre-gauge-fixed action is reparametrisation invariant, it's also conformally invariant. $\endgroup$ May 23, 2015 at 2:02

1 Answer 1


Going to the conformal gauge is nothing but using coordinates in which the metric is diagonal (in euclidean space this is called isothermal coordinates ).

Therefore in order to show that the Polyakov action is Weyl-invariant without using the conformal gauge, it is sufficient to show that the action does not dependent on the coordinate choice at all.

Chances are in whichever derivation for the Weyl invariance of the Polyakov action you saw, you started with some version of the one shown at the top of this Wikipedia article:

From Wikipedia

Notice all the indices here are contracted, and it is well-known and easy to show that $\sqrt{-h} d^2 \sigma$ is the invariant volume element, i.e. it doesn't change under a coordinate change. Therefore the above action is manifestly independent of coordinate (gauge) choice.

Therefore showing the above $S$ to be Weyl invariant is a result that is independent of coordinates choice (or 'conformal gauge fixing).

  • $\begingroup$ Thanks for the reply @bechira. I think you might be using the term 'Weyl invariance' where I would use 'conformal invariance'. Just to be clear I use 'conformal' to mean reparametrisations of the coordinates that also result in a scaling of the metric. I take Weyl transformations to just be direct scalings of the metric. So, with these definitions, I see that invariance under general reparametrisations must mean conformal invariance also (as you describe). Is this correct? $\endgroup$ May 24, 2015 at 22:18
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    $\begingroup$ @SirajRKhan 'invariance under general reparametrisations must mean conformal invariance' no this is not correct. Note that the integrand on the RHS is an expression involving tensors and vectors defined on tangent space over some point- coordinate invariance (as I described) means you can choose any basis for this tangent space - any well-defined vector or tensor has this property, which is why all indices being contracted is sufficient to show coordinate invariance. $\endgroup$
    – zzz
    May 24, 2015 at 22:25
  • $\begingroup$ But doesn't coordinate invariance mean the same thing as reparametrisation invariance (i.e: $\sigma \rightarrow \tilde{\sigma}(\sigma)$)? And then, as you said in your original answer, the conformal invariance follows from this invariance under coordinate choice/reparametrisation? $\endgroup$ May 24, 2015 at 22:29
  • $\begingroup$ Sorry that's not what I said in my answer- I said "showing the above S to be Weyl invariant is a result that is independent of coordinates choice ". I can see now the wording is a little confusing - I meant to say showing Weyl invariance for $S$ in one coordinate choice is sufficient to show it in all coordinates, because $S$ is coordinate independent. $\endgroup$
    – zzz
    May 24, 2015 at 22:35
  • $\begingroup$ Ah okay I think I get your meaning. Sorry to extend this discussion but I'm not sure my definition of 'conformal gauge' matches yours. In my definition a coordinate transformation (reparametrisation) is used to obtain the isothermal coordinates you describe, however in addition, a Weyl rescaling of the metric is used to take that metric to the flat space metric $\eta_{\alpha \beta}$. I think the Weyl rescaling means that the resulting action isn't simply the original action with a different choice of coordinates. $\endgroup$ May 24, 2015 at 22:48

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