# Emergence of rotational symmetry on 2D square lattice

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 $$\phi\partial_1^2\phi + \phi\partial_2^2\phi~?$$

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.

• You answered it perfectly. Thank you! – Lepnak Jul 17 at 20:45

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)".