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Pure Yang-Mills theory (YM) can be easily defined on $\mathbb{T}^4$ on a periodic lattice, using the Wilson lattice gauge theory approach. In reality, we know some of these mathematical results on $\mathbb{T}^4$, e.g. See T. Bałaban works in Comm Math Physics and also the descriptions and citations given in the claymath.org note.

So what makes the difference to define a lattice Yang-Mills theory on $\mathbb{T}^4$ v.s. $\mathbb{R}^4$?

  • Define a lattice Yang-Mills theory on $\mathbb{T}^4$ is easily doable since a $\mathbb{T}^4$ can have a finite length $L$, given the lattice constant $a$, there are about $(L/a)^4$ lattice links. In some sense, define a lattice Yang-Mills theory on $\mathbb{T}^4$ then taking the continuum limit (view the lattice constant $a$ to be small), also define a continuum Yang-Mills theory on $\mathbb{T}^4$. Most people believe that all the properties will carry over from the discrete to the continuum limit.

  • Define a lattice Yang-Mills theory on $\mathbb{R}^4$ seems harder, since the length of $\mathbb{R}$ is infinite $L_\mathbb{R}=\infty$, so there are about $(L_\mathbb{R}/a)^4\to \infty$ infinite lattice links.

However, the pure Yang-Mills theory should have the same property and the same mass gap for a large size $\mathbb{T}^4$ as good as $\mathbb{R}^4$ (is it correct?). If so, what makes the big deal difference to define lattice Yang-Mills theory on $\mathbb{T}^4$ v.s. $\mathbb{R}^4$? (It shall be a trivially gapped TQFT with a mass gap.)

  • Add: The topology of $\mathbb{T}^4$ and $\mathbb{R}^4$ is obviously different from each other. I am asking that why not just defines YM on $\mathbb{T}^4$ (at least on the lattice), which seems to be much tractable but captures every property we wanted for YM on $\mathbb{R}^4$.
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    $\begingroup$ "Easily defined"? The $a \to 0$ limit of lattice Yang-Mills $\mathbb{T}^4$ is very hard to construct - see Magnen's paper "Construction of YM4 with an infrared cutoff". $\endgroup$ – John Baez Jul 20 '18 at 6:40
  • $\begingroup$ My understanding is that for lattice gauge theory, the ultraviolet high energy, short distance cutoff does not concern us - it is set by the $a$, as long as the number of lattice sites is finite. My point is that the number of lattice sites is finite on $T^4$ but not on $R^4$, so why do we bother $R^4$ if every property (?) is the same in terms of physical observables? $\endgroup$ – wonderich Jul 20 '18 at 15:38
  • $\begingroup$ I think the confusion is coming from your use of "on a lattice on T^4" instead of "on Z/n^4" which limits to what John was thinking of as n goes to infinity. $\endgroup$ – Phil Tosteson Jul 20 '18 at 21:53
  • $\begingroup$ The topology of $\mathbb{T}^4$ and $\mathbb{R}^4$ is obviously different from each other. I am asking that why not just defines YM on $\mathbb{T}^4$, which seems to be much tractable but captures every property we wanted for YM on $\mathbb{R}^4$ (yes?). $\endgroup$ – wonderich Jul 21 '18 at 0:04
  • $\begingroup$ Sure, go ahead and "just" define Yang-Mills theory on $\mathbb{T}^4$. That would be very nice. You'd become famous - it's a famously difficult problem. If the expected values of physical quantities keep changing a lot as you increase the size of the $\mathbb{T}^4$, it will not have a theory on $\mathbb{R}^4$ as a limit and people will be disappointed - but still, getting a $\mathbb{T}^4$ theory would be a huge advance. $\endgroup$ – John Baez Jul 21 '18 at 16:49

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