There is a conjectural link between the Ising model and the $E_8$ exceptional Lie algebra, and this link becomes important when studying the massive scaling limit of the model with a magnetic field

There is a conjectural link between the Ising model and the $E_8$ exceptional Lie algebra, and this link becomes important when studying the massive scaling limit of the model with a magnetic field

To be specific, if we look at the model at inverse temperature $\beta_c$ on a discretization of a planar domain with mesh size $\delta>0$, one should consider a magnetic field $h=m\delta^{15/8}$ for some fixed $m\in\mathbb R$ and let $\delta\to0$

To be specific, if we look at the model at inverse temperature $\beta_c$ on a discretization of a planar domain with mesh size $\delta>0$, one should consider a magnetic field $h=m\delta^{15/8}$ for some fixed $m\in\mathbb R$ and let $\delta\to0$

The $15/8$ number is simply $2-1/8$, where $2$ is the dimension of the space and $1/8$ the scaling dimension of the spin field... this simply appears if one tried to make sure that the $h \sum_i\sigma_i$ stays bounded and nontrivial as $\delta\to0$ (we have $\delta^{-2}$ terms, which on average are $\delta^{1/8}$)

The $15/8$ number is simply $2-1/8$, where $2$ is the dimension of the space and $1/8$ the scaling dimension of the spin field... this simply appears if one tried to make sure that the $h \sum_i\sigma_i$ stays bounded and nontrivial as $\delta\to0$ (we have $\delta^{-2}$ terms, which on average are $\delta^{1/8}$)

The idea is that this will break most of the nice symmetries of the model (which would converge to a conformal field theory if $h=0$), but that something will stay and that this something is related to a structure related to the $E_8$ Lie algebra

The idea is that this will break most of the nice symmetries of the model (which would converge to a conformal field theory if $h=0$), but that something will stay and that this something is related to a structure related to the $E_8$ Lie algebra

What does $E_8$ mean here?

What does $E_8$ mean here?

A conservative view would be that there are some algebraic structure related to the $E_8$ algebra that is present in the conformal field theory that persists in the massive limit, and that this shows up in the mass spectrum of the spin-spin correlations

A conservative view would be that there are some algebraic structure related to the $E_8$ algebra that is present in the conformal field theory that persists in the massive limit, and that this shows up in the mass spectrum of the spin-spin correlations

What is the simplest appearance of $E_8$ for the Ising CFT?

What is the simplest appearance of $E_8$ for the Ising CFT?

The simplest thing that can be said about the link between the Ising CFT and the $E_8$ algebra is that the former can be written as a coset theory $(E_8^{(1)}\otimes E_8^{(1)})/E_8^{(2)}$, where the numbers in parentheses denote the level of the Affine Kac-Moody CFT with $E_8$ symmetry

The simplest thing that can be said about the link between the Ising CFT and the $E_8$ algebra is that the former can be written as a coset theory $(E_8^{(1)}\otimes E_8^{(1)})/E_8^{(2)}$, where the numbers in parentheses denote the level of the Affine Kac-Moody CFT with $E_8$ symmetry

But there is much more $E_8$ than the correlation mass spectrum

But there is much more $E_8$ than the correlation mass spectrum

What does it mean about the Ising model?

What does it mean about the Ising model?

A priori, it is just a question of representation theory, that in the Kac-Moody algebra $E_8^{(1)}\otimes E_8^{(1)}$, there is a diagonal sub-algebra $E_8^{(2)}$ and that if we mod out by this action, we find the three representations of the Virasoro algebra with $c=1/2$ corresponding to $\mathcal M_3$, the minimal model corresponding to the Ising model CFT

A priori, it is just a question of representation theory, that in the Kac-Moody algebra $E_8^{(1)}\otimes E_8^{(1)}$, there is a diagonal sub-algebra $E_8^{(2)}$ and that if we mod out by this action, we find the three representations of the Virasoro algebra with $c=1/2$ corresponding to $\mathcal M_3$, the minimal model corresponding to the Ising model CFT

Can we say that $\mathcal M_3\otimes E_8^{(2)}\simeq E_8^{(1)}\otimes E_8^{(1)}$?

Can we say that $\mathcal M_3\otimes E_8^{(2)}\simeq E_8^{(1)}\otimes E_8^{(1)}$?

At the level of representations of the Virasoro algebra, this is probably true, but the interesting question is whether this statement has any field-theoretic content at all

At the level of representations of the Virasoro algebra, this is probably true, but the interesting question is whether this statement has any field-theoretic content at all

There is a particular a paper by Forgács, Horváth, Palla, Vecsernyés titled 'Higher Level Kac-Moody Representations and Rank Reduction in String Models', where we see that there is a rich space of states in the $E_8^{(1)}\otimes E_8^{(1)}$ theory, and that the transformations of the Ising model appear in there, in particular the spin-field symmetry corresponds to $(\mathcal O,\mathcal O')\to(\mathcal O', \mathcal O)$; and so the spin sector gets matched with the antisymmetric pairs of fields and the identity and energy sectors get matched with the symmetric ones

There is a particular a paper by Forgács, Horváth, Palla, Vecsernyés titled 'Higher Level Kac-Moody Representations and Rank Reduction in String Models', where we see that there is a rich space of states in the $E_8^{(1)}\otimes E_8^{(1)}$ theory, and that the transformations of the Ising model appear in there, in particular the spin-field symmetry corresponds to $(\mathcal O,\mathcal O')\to(\mathcal O', \mathcal O)$; and so the spin sector gets matched with the antisymmetric pairs of fields and the identity and energy sectors get matched with the symmetric ones

What suggests particularly strongly that the $M_3\otimes E_8^{(2)}\simeq E_8^{(1)}\otimes E_8^{(1)}$ may have some probabilistic nature is that we can really rewrite the stress-energy tensor of the right-hand side as a sum of two independent stress-energy tensors, that resemble the Ising and the $E_8^{(2)}$ ones

What suggests particularly strongly that the $M_3\otimes E_8^{(2)}\simeq E_8^{(1)}\otimes E_8^{(1)}$ may have some probabilistic nature is that we can really rewrite the stress-energy tensor of the right-hand side as a sum of two independent stress-energy tensors, that resemble the Ising and the $E_8^{(2)}$ ones

And something that should at least be noted is that we really 'double the $E_8$' symmetry when we tensor with $E_8^{(2)}$

And something that should at least be noted is that we really 'double the $E_8$' symmetry when we tensor with $E_8^{(2)}$

But at the same time, this is what I don't really understand; there is clearly a double $E_8$ symmetry in $E_8^{(1)}\otimes E_8^{(1)}$, but I don't really understand how the Ising model doubles it

But at the same time, this is what I don't really understand; there is clearly a double $E_8$ symmetry in $E_8^{(1)}\otimes E_8^{(1)}$, but I don't really understand how the Ising model doubles it

It tells us how to 'twist-inject' $E_8^{(2)}$ into $E_8^{(1)}\otimes E_8^{(1)}$, and since the injection has a doubled symmetry, it means that if we act on one of the copies, there should be a way to take this into account both on $E_8^{(2)}$ and on the Ising side... if we swap the two $E_8^{(1)}$ copies, then it is completely the job of the Ising spin field to take that

It tells us how to 'twist-inject' $E_8^{(2)}$ into $E_8^{(1)}\otimes E_8^{(1)}$, and since the injection has a doubled symmetry, it means that if we act on one of the copies, there should be a way to take this into account both on $E_8^{(2)}$ and on the Ising side... if we swap the two $E_8^{(1)}$ copies, then it is completely the job of the Ising spin field to take that

Ok, to be a bit clearer about all of this, there are Affine Kac-Moody actions on the fields; and in the case of $E_8^{(1)}\otimes E_8^{(1)}$, there are currents associated with the two copies, and these currents commute, and the claim is (probably) that if we arrange things by degrees, then we can really identify the two sides

Ok, to be a bit clearer about all of this, there are Affine Kac-Moody actions on the fields; and in the case of $E_8^{(1)}\otimes E_8^{(1)}$, there are currents associated with the two copies, and these currents commute, and the claim is (probably) that if we arrange things by degrees, then we can really identify the two sides

Something That is Important To Note

Something That is Important To Note

In the Affine Kac-Moody business, there are (not many) primary fields which are called because they are primary with respect to the the Kac-Moody mode action; but these are not the same as the _Virasoro_ primaries, which is the 'geometric' notion of conformal covariance

In the Affine Kac-Moody business, there are (not many) primary fields which are called because they are primary with respect to the the Kac-Moody mode action; but these are not the same as the Virasoro primaries, which is the 'geometric' notion of conformal covariance

So, it is not surprising that we don't find too many AKM-primaries in $E_8^{(1)}$ (there is only the identity) and three AKM-primaries in $E_8^{(2)} $ (the identity, one of scaling dimension $15/16$, and one of scaling dimension $3/2$)

So, it is not surprising that we don't find too many AKM-primaries in $E_8^{(1)}$ (there is only the identity) and three AKM-primaries in $E_8^{(2)}$ (the identity, one of scaling dimension $15/16$, and one of scaling dimension $3/2$)

Anyway, all of this suggests there are really 'more interesting fields' in $E_8^{(2)}$ than there are in $E_8^{(1)}$ (where there are only the currents)... and somehow the addition of the Ising model to $E_8^{(2)}$ (at least if we have the chiral fields, or something like this... which remains to be clarified) 'frees' these interesting fields and allows them to be represented as two 'independent copies'

Anyway, all of this suggests there are really 'more interesting fields' in $E_8^{(2)}$ than there are in $E_8^{(1)}$ (where there are only the currents)... and somehow the addition of the Ising model to $E_8^{(2)}$ (at least if we have the chiral fields, or something like this... which remains to be clarified) 'frees' these interesting fields and allows them to be represented as two 'independent copies'

Jan 28, 2025

Jan 28, 2025

Does it mean we have an E8 Ising symmetry?

Does it mean we have an E8 Ising symmetry?

That is still unclear, because I don't have a simple example of a situation like the one described above to understand that... it is pretty clear that acting on the Ising model alone with $E_8$ is going to be hard, but the fact that we could act with two copies of $E_8$ (somehow) and get as a result an action on $E_8^{(2)}$ and an action on Ising is kind of interesting

That is still unclear, because I don't have a simple example of a situation like the one described above to understand that... it is pretty clear that acting on the Ising model alone with $E_8$ is going to be hard, but the fact that we could act with two copies of $E_8$ (somehow) and get as a result an action on $E_8^{(2)}$ and an action on Ising is kind of interesting

At least in terms of dimension counting, this should mean something about Ising

At least in terms of dimension counting, this should mean something about Ising

And in terms of perturbation, do we get anything? If we perturb by the spin field, does it e.g. mean we try to break the 'left-right' symmetry (i.e. we try to break the $(\mathcal O,\mathcal O')\to(\mathcal O',\mathcal O)$ symmetry, and not the 'holomorphic-antiholomorphic' symmetry, which is also often referred to as left-right symmetry), and that we can see something from that? Some residual of that symmetry left in the dimension counting or in the integrability?

And in terms of perturbation, do we get anything? If we perturb by the spin field, does it e.g. mean we try to break the 'left-right' symmetry (i.e. we try to break the $(\mathcal O,\mathcal O')\to(\mathcal O',\mathcal O)$ symmetry, and not the 'holomorphic-antiholomorphic' symmetry, which is also often referred to as left-right symmetry), and that we can see something from that? Some residual of that symmetry left in the dimension counting or in the integrability?

An Interesting Point to Clarify

An Interesting Point to Clarify

February 3rd, 2025

February 3rd, 2025

If we look at the $E_8^{(1)}\otimes E_8^{(1)}$ formulation, it is well-understood that the spin-flip involution is related to the swap of the copies, and hence that as a result the 'spin-even' sector consists of the 'symmetric' sector of the $E_8^{(1)}\otimes E_8^{(1)}$ theory

If we look at the $E_8^{(1)}\otimes E_8^{(1)}$ formulation, it is well-understood that the spin-flip involution is related to the swap of the copies, and hence that as a result the 'spin-even' sector consists of the 'symmetric' sector of the $E_8^{(1)}\otimes E_8^{(1)}$ theory

Now, the question that would well be worth clarifying is whether in that symmetric sector there is an analogue of the Kramers-Wannier duality (and there should be one... at least in the 'Framed Vertex Operator Algebra' papers, there is definitely an analogue of Kramers-Wannier), and this would have $-1$ eigenvalue on the energy sector and $+1$ eigenvalue of the identity sector

Now, the question that would well be worth clarifying is whether in that symmetric sector there is an analogue of the Kramers-Wannier duality (and there should be one... at least in the 'Framed Vertex Operator Algebra' papers, there is definitely an analogue of Kramers-Wannier), and this would have $-1$ eigenvalue on the energy sector and $+1$ eigenvalue of the identity sector

The reason this would be interesting is that if we can identify the identity sector, this would make it a little clearer why the $\bar \partial \mathcal I=m\partial \mathcal S$ equation has an infinite number of independent solutions, where $\mathcal I$ is a quasi-primary ('nonderivative') descendent of the identity and $\mathcal S$ a descendent of the spin

The reason this would be interesting is that if we can identify the identity sector, this would make it a little clearer why the $\bar \partial \mathcal I=m\partial \mathcal S$ equation has an infinite number of independent solutions, where $\mathcal I$ is a quasi-primary ('nonderivative') descendent of the identity and $\mathcal S$ a descendent of the spin

The question could at least be formulated with the $E_8^{(1)}\otimes E_8^{(1)}$ theory, but in principle, the answer about the equation could be found there... what makes the 'massive' $\bar\partial$ operator on the identity sector definition is an OPE with the spin operator, ultimately

The question could at least be formulated with the $E_8^{(1)}\otimes E_8^{(1)}$ theory, but in principle, the answer about the equation could be found there... what makes the 'massive' $\bar\partial$ operator on the identity sector definition is an OPE with the spin operator, ultimately

So, besides understanding Kramers-Wannier, understanding what the $\partial=L_{-1}$ operator means is what is important here

So, besides understanding Kramers-Wannier, understanding what the $\partial=L_{-1}$ operator means is what is important here

If we differentiate a product of independent CFTs (and think the correlations are products of the said CFTs), then the question is clearly that we should think of $\partial (\mathcal O_1\otimes \mathcal O_2)=\partial \mathcal O_1 \otimes \mathcal O_2+\mathcal O_1 \otimes\partial \mathcal O_2$... it's clear that Leibniz rule applies to product of correlations

If we differentiate a product of independent CFTs (and think the correlations are products of the said CFTs), then the question is clearly that we should think of $\partial (\mathcal O_1\otimes \mathcal O_2)=\partial \mathcal O_1 \otimes \mathcal O_2+\mathcal O_1 \otimes\partial \mathcal O_2$... it's clear that Leibniz rule applies to product of correlations