6
$\begingroup$

I have often heard statements like, Continuum limit of Lattice Yang-Mills doesnot exist in dimensions higher then four. Is there a neat way to see this? or could some one point me to some relevant literature where this has been shown. thanks

$\endgroup$
9
$\begingroup$

This may look like a deep question and in some sense, it is. But the explanation is completely trivial, doesn't really depend on the "lattice" at all, and boils down to dimensional analysis.

The action of Yang-Mills theory $$ S =\int d^D x \frac{1}{2 g^2} {\rm Tr} \, F_{\mu\nu} F^{\mu\nu} $$ Now, $F_{\mu\nu}\sim [D_\mu,D_\nu]$ has dimension of squared mass - because the (covariant) derivatives have the dimension of mass in these conventions. That's why $F^2$ has dimension $mass^4$. Because $d^D x$ has $mass^{-D}$, it's clear that the dimension of $g$ has to be $$ mass^{(4-D)/2}$$ For $D=4$, the coupling constant is classically dimensionless. For $D>4$, the coupling constant has the dimension of a negative power of mass (positive power of length); for $D<4$, it's the other way around.

The strength of interactions - e.g. the probability of scattering - goes like a positive power of $g$. However, to get e.g. the probability I mentioned, we must make it dimensionless. Dimensional analysis tells us that the right "dimensionless coupling" is $$g \cdot L^{(4-D)/2}$$ where $L$ is a characteristic distance scale of the experiment. For $D>4$, the interactions are therefore getting stronger - and infinitely strong in the $L\to 0$ limit. In particular, if $L=a$ is the lattice spacing which is send to zero, you will need to adjust the coupling constant at the lattice scale to be infinite if you want to get a finite $g \cdot L^{(4-D)/2}$ for a finite $L$. However, there's really no way to define a gauge theory with an infinite coupling, even at the level of a lattice.

In fact, for $D>4$, the gauge theory action above only works as an effective action for long enough distances $L$ where the "dimensionless coupling" is small enough. This is the only starting point where "a theory" exists and if you try to extrapolate it to much shorter distances, you will find out that the theory becomes ill-defined once the dimensionless coupling constant exceeds one or so - at the distance scale corresponding to a power of $g$.

So no theory that could be identified with the gauge theory exists at supershort distances if $D>4$. Alternatively, you may naively define a lattice gauge theory for $D>4$ with a finite coupling constant. However, the scaling laws guarantee that at distances that are infinitely longer - as in the continuum limit - the interactions will vanish. At any rate, you can't get finite yet non-vanishing interactions at both distances $a$ that are sent to zero as well as finite distances $L$.

The case of $D<4$ is different because the interactions become stronger at longer distances. So one may start with a lattice with "infinitesimal" interactions and they go stronger at longer distances.

$\endgroup$
  • $\begingroup$ This is very interesting, mainly because I learnt of the $D>4$ when my stat mech prof taught us the Ginzburg criterion, and did a procedure of when fluctuations can be neglected in mean field theories. Yet I am not familiar with the context of this Q/A (you start with action of the Lattice Yang Mills)... can you just drop a line on what connects? THanks. $\endgroup$ – yayu Apr 24 '11 at 16:32
  • $\begingroup$ Do you guys know the reason to why to choose $$\frac{1}{2g^2}$$ normalization in contrast to $\frac{1}{4g^2}$ ? In Clay math, they use $$\frac{1}{4g^2}$$ for Million prize: claymath.org/sites/default/files/yangmills.pdf I am curious what is the reason behind. $\endgroup$ – wonderich Jun 21 '18 at 23:09
  • $\begingroup$ Dear @wonderich - yup, for individual components you have 1/4 in front of f-mu-nu-squared in the action. But in my form, I have the "trace". And if you compute the trace of "some SU(2) matrix squared", the trace of the squared generators of SU(2) gives you 1/2, because these generators have two matrix elements of +1/2 or -1/2. Square it, you get a diagonal matrix with 1/4, 1/4, the trace of it is 1/2. Therefore, 1/2 already comes from the trace, and you only need to add 1/2g^2 to get the usual 1/4g^2 prefactor. Makes sense? $\endgroup$ – Luboš Motl Jun 23 '18 at 10:12
  • $\begingroup$ that is true what you said I knew. But I meant the different convention $1/(4g^2)$ in the Clay math, see also mathoverflow.net/q/303450/27004 -- not sure is that just a typo or is that convenient for something in Clay math? mathoverflow.net/q/303450/27004 $\endgroup$ – wonderich Jun 23 '18 at 17:32
  • $\begingroup$ If you read that page a little bit carefully, you will see that the form that has 1/2 in front of it is "F wedge F*". Indeed, with the wedge notation, there's another coefficient needed that depends on the conventions how to normalize the differential forms, the Hodge dual, and the integrals, and with the most typical convention, one needs 1/2 because the 4-form that is integrated naturally has "one component" and nothing is summed twice. The wedge products are usually normalized so that every inequivalent product appears exactly once in each component. And when it's so, you need 1/2. $\endgroup$ – Luboš Motl Jun 26 '18 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.