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On page 74 of David Tong's Statistical Field Theory lecture notes, it is said that $(\partial_1\phi)^2 + (\partial_2\phi)^2 $ respects both $D_8$ (that includes discrete four-dimensional rotation symmetry such as $x_1 \rightarrow x_2$ etc., as well as reflection symmetry $x_1 \rightarrow -x_1$ etc.) and $SO(2)$. And that the lowest-dimensional term that preserves $D_8$ but not $SO(2)$ is $$\phi \partial_1^4 \phi + \phi\partial_2^4\phi.$$

I was just wondering why this is the case? How can we tell if a term respects $SO(2)$? And what about $$\phi\partial_1^2\phi + \phi\partial_2^2\phi~? $$

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Funny you ask, I was just reading this part in Tong's notes yesterday. As Cryo already mentioned $\vec{\nabla}$ is a vector. The point is for something to have the full rotational symmetry it needs to be a dot product. So $$(\partial_1 \phi)^2 + (\partial_1 \phi)^2 = (\vec{\nabla} \phi) \cdot (\vec{\nabla} \phi). $$

The other term you mentioned $\phi\partial_1^2\phi + \phi\partial_1^2\phi $ is actually not at all different from the above expression, up to a boundary term. But first notice,
$$\phi\partial_1^2\phi + \phi\partial_1^2\phi = \phi \vec{\nabla} \cdot \vec{\nabla} \phi $$ so again a scalar. The two are actually related since by integrating by parts, $$\int d^dx\ (\vec{\nabla} \phi) \cdot (\vec{\nabla} \phi) = -\int d^dx\ \phi \vec{\nabla} \cdot \vec{\nabla} \phi \ + BT, $$ where BT indicate some boundary term.

The term with the derivative to the forth power cannot be written as a dot product of any vectors, you can try $\phi ( \vec{\nabla} \cdot \vec{\nabla} \ \vec{\nabla} \cdot \vec{\nabla} ) \phi$, but it doesn't work. Yet that term is invariant under a subgroup of rotations, and the reflections. Not sure If I this ansers your question.

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  • $\begingroup$ You answered it perfectly. Thank you! $\endgroup$ – Lepnak Jul 17 at 20:45
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Regarding the $(\partial_1\phi)^2+(\partial_2\phi)^2$. Define vector $V^\mu = g^{\mu\nu}\partial_\nu\phi$ and co-vector $V_\mu=\partial_\mu \phi$, where $g^{\mu\nu}=diag(1,1)^{\mu\nu}$ is the inverse of the trivial 2d metric for Cartesian coordinates in Eucledian space.

Then $(\partial_1\phi)^2+(\partial_2\phi)^2=V_\mu V^{\mu}$ which is clearly a scalar and will not change under any isometries of the Eucledian space, including, in particular any rotations. More precisely, if $R^{\mu}_{\:\nu}\left(\theta\right)$ is the rotation matrix, i.e. rotation of vectors is given by $V^\mu\to R^\mu_{\:\nu}V^\nu$, then clearly for co-vectors $V_\mu \to \left(R^{-1}\right)^{\nu}_{\:\mu}V_\nu$, so that $V^\mu V_\mu \to V^\nu R^\mu_{\:\nu} \left(R^{-1}\right)^{\kappa}_{\:\mu} V_\kappa = V^\nu V_\nu$, i.e. no change.

Or, in simple terms: If $\mathbf{V}=\boldsymbol{\nabla}\phi=\left(\array{\partial_1\phi \\ \partial _2 \phi}\right)$ is a vector then $(\partial_1\phi)^2+(\partial_2\phi)^2=\mathbf{V}.\mathbf{V}=V^2$ which is invariant under rotations. (Try it by letting $\mathbf{V}\to\left(\array{\cos\theta & -sin\theta \\ \sin\theta & \cos \theta }\right)\mathbf{V}$).

Anything that does not change under arbitrary 2d rotations "respects SO(2)".

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