# Intuition behind Nekreasov's instanton partition function. What do the partitions represent exactly?

+ 4 like - 0 dislike
403 views

I am struggling to understand many things behind Nekrasov's solution. Firstly I want to understand the following

1. In this theory, $a$ represents VEVs the Higgs scalar. So, is the gauge field of the hyper broken? The theory is broken from $SU(N)$ to what?

and then

1. How are the Young diagrams related to the point like $k$-instantons? And why do we sum over these Young diagrams? Is there an intuitive way to understand it?

A readable reference is the Master thesis A Pedagogical Introduction to the AGT Conjecture by Robert J. Rodger, chapter 3. (PDF)

This post imported from StackExchange Physics at 2015-10-15 11:24 (UTC), posted by SE-user Marion

edited Oct 15, 2015
Tip: Consider adding references and context in order to get useful and focused answers.

This post imported from StackExchange Physics at 2015-10-15 11:24 (UTC), posted by SE-user Qmechanic
I did, but I don't think it will be of any help. Only someone working on this field can actually help (I think). Let's see.

This post imported from StackExchange Physics at 2015-10-15 11:24 (UTC), posted by SE-user Marion

+ 5 like - 0 dislike

When the complex scalar in the vector multiplet has a non-trivial VEV $a$, then indeed the gauge group is spontaneously broken and we have the Higgs mechanism. As any field of the vector multiplet, the complex scalar is adjoint valued and so, classically, the gauge group is broken to the stabilizer of $a$ in the adjoint representation. For generic $a$, it is simply the maximal torus of the gauge group but for special $a$ the unbroken subgroup can be larger. For $G=U(N)$, by gauge transformation one can assume $a$ diagonal, $a=diag(a_1,...,a_N)$. If all the $a_i$ are distinct then $U(N)$ is broken to its maximal subgroup $U(1)^{N}$.

The Young diagrams come from the localization at the fixed points of the instantons moduli space. The Nekrasov partition function is a priori defined as a path integral. Using supersymmetry and after computation of the perturbative part, this path intergral can be reduced to a sum over the instanton number of finite dimensional integrals over the moduli spaces of instantons. These integrals are then computed by localization with respect to the action of the group $T \times U(1)^2$ where $T=U(1)^N$ is the maximal torus of the gauge group $U(N)$ and where $U(1)^2$ is the maximal torus of the group $SO(4)$ of rotations of the Euclidean four dimensional space. $(a_1,...,a_N)$ will appear as the equivariant parameters with respect to $T$ and  $(\epsilon_1,\epsilon_2)$ as equivariant parameter with respect to $U(1)^2$. The key point is that that the fixed points of $T \times U(1)^2$ acting on the moduli space of $U(N)$ instantons of instanton charge $k$ are naturally parametrized by the data of $N$ Young diagrams whose sum of sizes is equal to $k$. To understand that, one has to realize a subtle point: the instanton moduli space is in general singular, due to point-like instantons, and to have well-defined integrals, one has to consider a smooth moduli space. A way to resolve the singularity is to turn on a noncommutative deformation of the spacetime and the Young diagrams parametrize  fixed points on the moduli space of instantons on this non-commutative spacetime. As we consider fixed points for the action of $U(1)^2$, each fixed point corresponds to a non-commutative instanton localized at the origin of spacetime. As we consider fixed points for the action of $T$, each fixed point corresponds to a non-commutative $U(N)$ instanton which is in fact a direct sum of $N$ non-commutative $U(1)$ instantons. The notion of $U(1)$ instanton may sound stange: in the commutative world, such things do not exist but they exist in the non-commutative world.

So it remains to realize the following: to give a $U(1)^2$-invariant non-commutative $U(1)$ instanton of charge $k$ at the origin of $\mathbb{R}^4$ is the same thing that to give a Young diagram with $k$ boxes. To do that, one has to understand what is a non-commutative $U(1)$ instanton of charge $k$ sitting at the origin. Let me just say that it is a $U(1)$ gauge field configuration with a singular behavior at the origin and that if the instanton is $U(1)^2$ invariant, this singular behavior is completely determined by the shape of a Young diagram.  (A mathematically precise formulation of a non-commutative $U(1)$ instanton is as the ideal sheaf of a non-reduced point in $\mathbb{C}^2$. A Young diagram is exactly a way to describe a possibly non-reduced $U(1)^2$-invariant point sitting at the origin of $\mathbb{C}^2=\mathbb{R}^4$).

answered Oct 15, 2015 by (5,120 points)
edited Oct 15, 2015 by 40227

@40227

(1) In what sense do you mean "special value of $a$"? Can you give an example? I was under the impression that $\phi$ (the complex scalar) always belongs to the Cartan subalgebra of $G$. What is different in the case of $SU(N)$?

(2) In what sense the instanton moduli space is singular? I.e. having two point-like instantons sitting at the same place? Would you be able to tell us how this is resolved by considering the Uhlenbeck compactification (which is mentioned in the article above but not clearly to my opinion). Of course this is related to the Hilber scheme of points. Is there a nice way to understand it?

(3) Also, if it is not too much, how does the $U(1)^2 \subset SO(4)$ you talk about related to the $\Omega$ background? Can you give us some nice intuition on how the $\Omega$ background it comes into the whole thing? I seem to understand the idea that we can consider a 5 or 6 dimensional theory and reduce it down to 4, but strictly from a 4 dimensional point of view, how do we understand it?

I apologize for all those questions, but I was about to ask a very very similar question. Basically I might actually ask a new question at PO very related to this one.

(1)The complex scalar $\phi$ is valued in the (complexification) of the Lie algebra of the gauge group $G$ and so its expectation value $a$ too. But up to gauge transformation, one can indeed assume that $a$ belong to the Cartan Lie algebra. In the case of $G=U(N)$, it is what I have written: $a$ is diagonal, $a=diag(a_1,...,a_n)$. For $G=U(N)$, non-special means all the $a_i$s distinct ans special means some $a_i$s equal. For a general semisimple Lie algebra, what I call "special" means in the union of the root hyperplanes.

(2) If by instanton one means a smooth configuration of the gauge field then the moduli space of instantons on $\mathbb{R}^4$ is smooth but it is non-compact. Of the reason for that is the possibility for the instanton size to shrink. The natural way to cure that is the Uhlenbeck compactification, which includes point-like instantons (It is what I call instanton moduli space is my answer). But this space is singular and indeed it happens when point like instantons come together. For example, consider the space of the Uhlenbeck compactification describing two point-like instantons, it is $Sym^2(\mathbb{R}^4)=(\mathbb{R}^4 \times \mathbb{R}^4)/(\mathbb{Z}/2)$ which is singular along the fixed point of the $\mathbb{Z}/2$-action, i.e. the diagonal in $\mathbb{R}^4 \times \mathbb{R}^4$. One obtains a smooth moduli space only by going to noncommutative instantons, i.e. by replacing symmetric products of copies of $\mathbb{C}^2$ by Hilbert schemes of points in $\mathbb{C}^2$.

3)Indeed working equivariantly with respect to $U(1)^2$ is the $\Omega$-background: $\epsilon_1$ and $\epsilon_2$ are the corresponding equivariant parameters. It is a natural way to localize the physics near the origin and so to cure the problems related to the non-compactness of $\mathbb{C}^2$.

@40227

Hi,

Why is the moduli space of two instants isomorphic to $Sym^2(\mathbb{R}^4)$? Can you please provide with some physical interpretation of the Uhlenbeck compactification? Am I correct to say that the moduli space of two point-like instantons is a copy of $\mathbb{R}^4$ for each except that we have to mod out the points that the two instantons might be sitting at the same position.

This of course brings me to the following question. The moduli space of a $k$-instanton configuration can be described by taking $k$ point-like instantons, right? Then this space is singular due to the fact that you can two instantons living at the same point. But, why is this considered a singularity? And also, what is the problem of the space being non-compact (being copies of $\mathbb{R}^4$? Why does the moduli space of the instantons need to be compact?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.