Does the q-states Potts become the XY model in large q state? I have met several times in papers, the order of the phase transition of the $q$-state Potts model depends on $q$. E.g., in two dimensions, for $q = 2$ (the Ising model), $3$, $4$ the order-disorder phase transition is second order while for $q >4 $ the phase transition becomes first order. In three dimensions, only when $q = 2$ the phase transition is second order, while for $q > 3$ the phase transition is first order.
But If $q$ becomes infinite, say in the continuum limit, does a Potts model becomes the XY model? However, it is well know that the XY model has a KT transition in two dimensions and a second order phase transition in three dimensions. 
 A: I think that you are really interested in the $q$-state clock model, which is similar to the Potts model, and is defined as follows. Fix an integer $q\geq2$. 
For each $i\in\mathbb{Z}^d$, let
$$
\theta_i \in \bigl\{\frac{2\pi}{q} k\,:\, k\in\{0,1,\ldots,q-1\}\bigr\},
$$
and define the spin at site $i$ by
$$
\mathbf{S}_i = (\cos\theta_i,\sin\theta_i) .
$$
The Hamiltonian is given by
$$
H = -\beta \sum_{i\sim j} \mathbf{S}_i \cdot \mathbf{S}_j,
$$
where the sum is over nearest neighbors and $\mathbf{x}\cdot \mathbf{y}$ denotes the scalar product in $\mathbb R^2$.
Note that this model coincides with the Potts model only when $q=2$ or $q=3$. For higher values of $q$, the symmetry group is smaller. It is however a real discretization of the XY model, so much closer to what you had in mind.
In a very precise sense, the $q$-state clock model does behave like the XY model as soon as $q$ is large enough; this is the phenomenon of enhancement of symmetry. In particular, for large enough values of $q$, this model enjoys in two dimensions:


*

*uniqueness at high temperature, with exponential decay of 2-point function

*a massless phase at intermediate temperature, with algebraic decay of 2-point function

*ordered phases at low temperatures, with no decay of 2-point function


See this celebrated paper by Fröhlich and Spencer for a rigorous proof. This should actually hold for any $q\geq 5$, but is only proven for $q$ large enough.
In the limit $q\to\infty$, the ordered low temperature phase disappears, and you're left with the XY model.
