# Peskin's duality in XY model (Mandelstam-'t Hooft duality in abelian lattice models)

I am studying the old paper by Peskin (1978): Mandelstam-'t Hooft duality in abelian lattice models (https://doi.org/10.1016/0003-4916(78)90252-X). However, I am confused about some details of calculation. For example, in (3.4), it gives a gauge constraint $$0 = \sum_\mu (m_{n+\hat{\mu},\mu}-m_{n,\mu}), \tag{3.4}$$ where $n$ runs over lattice sites and $\mu$ over elementary vectors on a $d$-dimensional cubic lattice. This seems a little weird to me because the common gauge constraints are divergence $\sum_\mu (m_{n-\hat{\mu},\mu}-m_{n,\mu})$ or curl $\sum_{\mu,\nu} \epsilon_{\sigma \mu \nu} (m_{n+\hat{\nu},\mu}-m_{n,\mu})$ to be zero. I don't know how to interpret the constraint (3.4).

In the latter part of this paper, the author defines new variable in (3.12) $$M_{n,\sigma} = \sum_{\mu \nu}\epsilon_{\sigma \mu \nu} (m_{n+\hat{\nu}+\hat{\sigma},\mu}-m_{n+\hat{\sigma},\mu})$$ and says that it satisfies the constraint $$(M_{n-\hat{\sigma},\sigma}-M_{n,\sigma})=0.$$ I don't see why the above expression is zero. I am not sure if they are typos or I miss something important. I will appreciate very much if some experts can help me figure it out.

• The second part is resolved. It should be corrected into $\sum_\sigma (M_{n-\hat{\sigma},\sigma}-M_{n,\sigma})=0$. Now, I am only confused about (3.4). – Yu-An Chen Mar 27 '18 at 6:01

## 1 Answer

I think it's easier to imagine those equations by the field theory analogy:

Originally, $\theta_n$ ranges from $0$ to $2 \pi$ and $m_{n \mu} \in \mathbb{Z}$. However, due to the "gauge symmetry," one can do the gauge transformation until $\partial_{\mu} m_{\mu}=0$ (Lorentz gauge), and some degrees of freedom are moved from $m_{\mu}$ to $\theta$ at the same time so that $\theta \in \mathbb{R}$ (it's the reverse spirit of how the degree of freedom is passed from matter field to gauge field in Higg's mechanism). After that, in the path integral there is actually an implicit gauge fixing contraint $\delta(\partial_{\mu} m_{\mu})$.

By definition, $M=\nabla \times m \implies \nabla \cdot M=0$. When one change $\sum_m \rightarrow \sum_M$, it's like changing summing over gauge potential into summing over magnetic fields. All magentic field configurations are automatically gauge invariant, so one can forget about gauge fixing. However, the price to pay is that magnetic field (defined like this) should be automatically divergentless, thus subject to the constraint $\nabla \cdot M=0$.