Timeline for Define a (lattice) Yang-Mills theory on $\mathbb{T}^4$ v.s. $\mathbb{R}^4$
Current License: CC BY-SA 4.0
11 events
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| Jul 22, 2018 at 16:16 | comment | added | wonderich | For pure Yang-Mills, there are only glueballs (as some excited states), this shall relate to the mass gap energy scale. There are no other TQFTs (causing dimensions of Hilbert space $dim \mathcal{H}$ >1, such as no dynamical BF theory etc). This applies to $T^4$ or $R^4$ if I am not mistaken. | |
| Jul 22, 2018 at 6:11 | comment | added | John Baez | I agree except for your claim that there is no other observable for Yang-Mills theory on $\mathbb{T}^4$. You seem to be using the word "observable" in a strange way. | |
| Jul 21, 2018 at 19:33 | comment | added | wonderich | @John Baez, thank you. for a pure YM (without the topological $\theta$-term), there is only one ground state on $T^4$, and the only energy scale is the mass gap. There is no other observable on $T^4$. This is the similar situation also on $R^4$. There is only one ground state on $R^4$, and the only energy scale is the mass gap. You agree or not? | |
| Jul 21, 2018 at 16:49 | comment | added | John Baez | 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. | |
| Jul 21, 2018 at 0:06 | history | edited | wonderich | CC BY-SA 4.0 |
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| Jul 21, 2018 at 0:04 | comment | added | wonderich | 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?). | |
| Jul 20, 2018 at 21:53 | comment | added | Phil Tosteson | 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. | |
| Jul 20, 2018 at 18:21 | history | edited | Abdelmalek Abdesselam |
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| Jul 20, 2018 at 15:38 | comment | added | wonderich | 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? | |
| Jul 20, 2018 at 6:40 | comment | added | John Baez | "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". | |
| Jul 20, 2018 at 0:20 | history | asked | wonderich | CC BY-SA 4.0 |