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The polyakov action in fixed gauge is given as $$S[X]=-\frac{1}{4\pi\alpha}\int d^2\sigma\partial_\alpha X\cdot\partial^\alpha X$$

is this invariant under diff transformations?

Intuitively I think it should not because it is in fixed gauge.

if I naively check the transformation I see that $$S[X]\neq S[X']$$ but instead if I also transform the flat Minkowski metric that I find $$S[X,\eta]= S[X',\eta']$$ so which one is correct, in the usual Polyakov action we transform both the metric and fields. But in gauge fixed form should I transform both or only fields?

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You've discovered that the original Polyakov action

$$ S[X, g] $$

is invariant under diffeomorphisms.

After fixing the gauge, you are working with a flat space action

$$ S[X] $$

(note the absence of $g$ in square brackets, it is now an external parameter not a field variable).

Diffeomorphisms act on the gauge-fixed action by changing the field variables $X$ but not the external parameter $\eta$, which is why it is not invariant under them.

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  • $\begingroup$ I agree completely but don't understand the reason. Why $\eta$ is not transformed in the gauge fixed case? Why is it an external parameter not a field variable in gauge fixed case? $\endgroup$
    – physshyp
    Commented Jan 29, 2022 at 4:34
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    $\begingroup$ @physshyp by definition pretty much — by choosing a coordinate frame where the metric has a very specific value in components you fix the gauge $\endgroup$
    – Prof. Legolasov
    Commented Jan 29, 2022 at 4:36
  • $\begingroup$ If I am given an arbitrary action how do I tell whether the metric is an external parameter or a field variable? $\endgroup$
    – physshyp
    Commented Jan 29, 2022 at 4:38
  • $\begingroup$ @physshyp there isn’t a deep fundamental reason here — this is just how we do calculations in gauge invariant theories. The Faddeev-Popov procedure is the formal justification $\endgroup$
    – Prof. Legolasov
    Commented Jan 29, 2022 at 4:38
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    $\begingroup$ correct me if I am wrong but I think whether some thing is background or variable field is not solely defined by action. it is determined by also if you integrate over it in path integral. so that's how we decide. in other words, a system is not solely defined by hamiltonian it is both determined by hamiltonian and hilbert space. and same hamiltonian can have different hilbert spaces. and the hilbert space also sets whether theory is gauge theory or not. so whether sth is in brackets or not is determined by the measure in path integral. loosely speaking by hilbert space. $\endgroup$
    – physshyp
    Commented Jan 29, 2022 at 4:46

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