In a recent paper (page 47, below eq (4.173)) they make a passing claim that the Schwarzian derivative action can be derived using effective low energy field theory reasoning. I imagine they mean that if I want to construct a least derivative action which is invariant under global $SL(2,\mathbb{R})$ transformations of the coordinates, then I will end up with Schwarzian derivative. I was wondering if this has been worked out anywhere. Also, using the same approach, what are the higher derivative invariants that I can possibly construct as 'less relevant' terms.

  • $\begingroup$ The authors worked it out previously arxiv.org/pdf/1606.01857.pdf $\endgroup$ – Dwagg Aug 26 '20 at 22:15
  • $\begingroup$ They start with the Einstein-Hilbert action with positive cc in $d=2$ and the Schwarzian effective action comes from the Gibbons-Hawking boundary term upon imposing the leading order bulk EOM. Of course it had to be Schwarzian since it is the lowest-derivative action with SL(2,R) symmetry $\endgroup$ – Dwagg Aug 26 '20 at 22:17

The answer to this question is rather simple. Let us assume that our field is $f(x)$ and we want to find an effective low energy action invariant under the $SL(2,\mathbb R)$ transformations,$$f(x)\to\frac{a\, f(x)+b}{c\, f(x)+d}~,\tag{1}$$with $a,b,c,d\in\mathbb R$ and $ad-bc=1$. Let us look at these transformations one by one:

  1. Translations: $f(x)\to f(x)+b$ imply that effective low energy action can only be a function of derivatives of the $f(x)$ field, $f'(x),f''(x),f^{(3)}(x),\ldots$.
  2. Scaling: $f(x)\to a\, f(x)$ implies that the effective action is a function of ratios of the fields, $\dfrac{f^{(n)}(x)}{f^{(m)}(x)}$.

Since we are working with effective action we want an action with the least number of derivatives. Since $\dfrac{f''(x)}{f'(x)}$ is a total derivative our lowest derivative candidates are $\dfrac{f'''(x)}{f'(x)}$ and $\left(\dfrac{f''(x)}{f'(x)}\right)^2$. Let's assume our action is, $$\dfrac{f'''(x)}{f'(x)}-\mu \left(\dfrac{f''(x)}{f'(x)}\right)^2~.\tag{2}$$

  1. Finally imposing invariance of the action under inversion: $f(x)\to\dfrac1{f(x)}$ one can easily fix the value of $\mu=\frac32$

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