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In a QFT at finite temperature we consider the Euclidean time to be periodic, i.e. we consider a theory on the manifold $\mathbb{R}^{d - 1} \times S^1$, where the spatial coordinates are in $\mathbb{R}^{d - 1}$ and the time coordinate in $S^1$. The temperature of the system is then the inverse of the circumstance $T = \frac{1}{\beta} \ , \tau\in[0,\beta)$. Thus the zero temperature limit is $\beta\rightarrow\infty$ where $S^1\rightarrow\mathbb{R}$. Now what I'm wondering is whether the limit $S^1\rightarrow\mathbb{R}$ is single-valued or multi-valued?

Single-valued = one point in $S^1$ goes to one point in $\mathbb{R}$. In this case the entire $\mathbb{R}$ becomes the $S^1$ in the finite temperature limit.

Multi-valued = one point in $S^1$ maps to several points in $\mathbb{R}$. In this case the $\mathbb{R}$ wraps around the $S^1$ several times in the finite temperature limit.

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  • $\begingroup$ Is there a mistake at $T = \frac{1}{\beta} \ , \tau\in[0,\beta)$? Should it be $T = \frac{1}{\tau} \ , \tau\in[0,\beta)$? $\endgroup$
    – Ozz
    Commented Aug 29, 2018 at 9:54
  • $\begingroup$ There is not a mistake. $T$ is the temperature, $\beta$ is the circumstance of the $S^1$ and $\tau\in S^1$ is the Euclidean time. This gives $\tau\in[0,\beta)$ where $\tau$ is periodic $\tau + \beta \equiv \tau$, and a temperature of $T = \frac{1}{\beta}$. This formalism of finite temperature in QFT is btw called imaginary time formalism. $\endgroup$
    – A.Dunder
    Commented Aug 29, 2018 at 11:46

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Saying that the theory is defined on $S^1$ is equivalent to saying it is defined on $\mathbb{R}$ but with all functions being periodic with period $2 \pi \beta$. This way you see that the $\beta \rightarrow \infty$ limit just corresponds to sending the periodicity to infinity, or in other words not having periodicity at all.

That being said, I think your alternative is not very well defined. You have the identification $S^1 = \mathbb{R}/(2 \pi \beta \mathbb{Z})$ so there are always, by definition, an infinite number of points in $\mathbb{R}$ corresponding to a single point in $S^1$. But these points are very far apart in the large $\beta$ limit, and as far as local physics is concerned, you can cover any bounded region of $\mathbb{R}$ by a single patch of $S^1$ in such a way that the correspondence be bijective -- which I believe is what you want to express with your "single-valued" possibility.

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  • $\begingroup$ Thanks! That's the kind of answer I was looking for. $\endgroup$
    – A.Dunder
    Commented Apr 9, 2020 at 16:22

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