# How can I understand instantons as sheaves?

In specific, instantons are considered or interpeted as torsion free coherent sheaves. Why is that the case? Is there a nice way to understand this relation and of course also understand how the two moduli spaces (instantons and torsion free coherent sheaves) are identified?

More background: the moduli space of instantons $M_{r,k}$ where $r$ is the rank of the gauge group or of the vector bundle and $k$ is the instanton number also defined as the second Chern class of the vector bundle. Now, these instantons are said to be framed which corresponds to requiring the field strength $F$ to vanish at infinity or, in other words, the instantons to be in pure gauge at infinity. According to Donaldson this space of "framed" instantons is identified with the moduli space of framed rank $r$-vector bundles on $\mathbb{P}^2 = \mathbb{C}^2 \cup l_{\infty}$ where $l_{\infty} = \mathbb{P}^1 = \mathbb{C}\cup \{ \infty \}$. Framing means that there exists a local trivilization at the line at the infinity ( $l_{\infty}$) such that the vector bundle is trivial there.

Now the space $M_{r,k}$ is also identified with the moduli space of torsion free sheaves $E$ on $\mathbb{P}^2$ such that rank($E$)$=r$ and $c_2(E)=k$ satisfying two conditions

• $E$ is torsion free on $\mathbb{P}^2$ and locally free (projective) in a neighborhood of $l_{\infty}$ (that is that at a neighborhood of infinity the sheaf looks like a vector bundle as above) and
• There exists a framing just like above (for the vector bundle).

What I ask for is intuition on the above objects. In what sense I can see the instantons as sheaves. If the instanon moduli space had no singularities and was smooth it is quite straight forward to understand the vector bundle construction. Sheafs are needed in order to take into consideration these singularities, right? And how is the sheaf theoretic construction related to $\text{Hilb}^n(X)$. I know that the problem is when two point like instantons approach each other a singularity appears and thus the vector bundle construction is not well defined but I do not quite understand how the sheaf theoretic construction resolves the problem.

• This could use a lot more context - what is your precise defintion of instanton in this context, and on what are you considering the sheaves? For the words "torsion-free" and "coherent" to make sense, you need a (preferably locally) ringed space, which one do you consider, and what is its structure sheaf? I suspect this is nothing more than viewing an instanton (or rather the instanton number) as defining a principal bundle, and then identifying the bundle with its sheaf of sections. Conversely, a sheaf with the right properties (which may be "torsion-free coherent") gives a bundle. – ACuriousMind Nov 12 '15 at 13:23
• I am not sure how to add context. I try to understand why the moduli space of framed instantons can be identified with the moduli space of coherent torsion free sheaves thus identifying instantons with the sheaves. The moduli space is indeed a locally ringed space and all the above properties you mention are valid. I would like a step by step explanation on how to start with the physics language of instantons as solutions of the ASD equations and arrive to the coherent sheaves. – Marion Nov 12 '15 at 20:54
• Well, that's a start! For one, include your definition of "framed instanton" (the instanton literature is horribly confusing precisely because people tend to talk about related, but different things). Then, say sheaves on what. You say the space of instanton bundles is to be the space of sheaves, which indicates for me that both the bundles and the sheaves are over the same space (spacetime, perhaps?), and it is this space - not the moduli spaces - which needs to be locally ringed. I.e. say what $X$ in the bundle maps $\pi:P\to X$ ia and open subsets of what space the sheaves live on. – ACuriousMind Nov 12 '15 at 21:01
• The space that needs to be locally ringed is the moduli space, right? I mean, the spacetime is anyway since it is a smooth manifold and you get the ring of continuous functions. The sheaves are "on" the moduli space I assume, just when two point like instantons happen to be at the same place allowing the rank of the bundle to not be constant, thus not being a vector bundle. I will try to edit my question in a bit, but since you seem to know already much more than me I would appreciate it if you could shed some light! – Marion Nov 13 '15 at 9:16
• I have tried to give a partial answer, but I am far from an expert in this area. For specific questions about the algebro-geometric details, you might also try math.SE/MathOverflow. – ACuriousMind Nov 13 '15 at 15:03

Okay, I cannot give you a full understanding of what is going on, but I can make the objects we are dealing with more precise:

There are two spaces here:

1. The moduli space $M_\text{sh}(r,k)$ of framed torsion-free coherent sheaves of rank $r$ and second Chern class $k$ on the projective scheme $\mathbb{P}^2$ viewed as a complex analytic space with its structure sheaf of analytic functions.

2. The moduli space $M_\text{in}(r,k)$ of framed instanton bundles of a gauge group of rank $r$ and second Chern class $k$ on the 4-sphere $S^4 = \mathbb{R}^4\cup\{\infty\}$, where $\infty$ is added to make the notion of framing precise.

Framed in the first case means that the sheaf is locally free at the line at infinity, framed in the second case means the gauge field configuration is pure gauge at the point at infinity. These two spaces are not the same. In particular, $M_\text{sh}$ is non-singular while $M_\text{in}$ is singular, and this is precisely the motivation to find the following equivalence apparently proved by Donaldson:

The moduli space of framed instantons $M_\text{in}$ is in bijection to the subset $M^\text{reg}_{0,\text{sh}}\subset M_\text{sh}$ of locally free sheaves.

Heuristically, it is not difficult to see that a vector bundle defines a sheaf. Given a bundle $P\to X$, the corresponding sheaf is defined by $U\mapsto \Gamma(U,P)$, i.e. the sheaf which just associates to every open set its local sections. The proof of the above assertion, however, is far more complicated: $S^4$ is just the $\mathbb{R}^4\cong\mathbb{C}^2$ with a point at infinity, but $\mathbb{P}^2$ is the same with a *line at infinity. You have to show that every locally free sheaf on the latter really defines a bundle on the former, and that the bundles on the former really give a proper sheaf on the latter that also plays nice with the analytic structure sheaf. Lastly, a principal bundle isn't a vector bundle, and the proof of the assertion relies on a proof that the instanton bundles on $S^4$ correspond to framed holomorphic vector bundles on $\mathbb{P}^2$.

Now, given the theorem by Donaldson, we see why the passage from instantons to sheaves resolves the issue of singularities - the space of all torsion-free sheaves is non-singular, so slightly generalizing the notion of instanton/locally free sheaf/holomorphic vector bundle to that of torsion-free sheaf gets rid of the issue.

The relation to Hilbert schemes arises because $$M_\text{sh}(1,k)\cong \mathrm{Hilb}^k(\mathbb{C}^2)$$

Donaldson's proof is in

Donaldson, "Instantons and geometric invariant theory", Comm. Math. Phys. 93 (1984)

and relies on prior work by Atiyah and Ward.

Atiyah and Ward, "Instantons and algebraic geometry", Comm. Math. Phys. 55 (1977)

The last correspondence can for example be found in

Nakajima, "Lectures on Hilbert Schemes of Points on Surfaces", AMS

• This is a very enlightening answer! However, how does one make this compatible with the fact that rank 1 principal or vector bundles have vanishing second Chern class? I mean, if you were content to remain outside of the algebraic setting, you would say rank 1 instantons have $k=0$, but somehow embedding the problem into the world of sheaves, you get rank 1 instantons with $k \neq 0$. Is this just an artifact of wanting to compactify a moduli space of something? – Benighted Jun 25 '17 at 21:32
• @StephenPietromonaco I'm afraid I do not understand the question - the rank 1 Lie group $\mathrm{SU}(2)$ certainly admits principal bundles with non-zero second Chern class. – ACuriousMind Jun 25 '17 at 21:52
• Ah I'm sorry. I was referring to $U(1)$ principal bundles whose associated vector bundles are line bundles. These I believe will not have a second Chern class $k$, and yet embedding into algebraic geometry, you get a moduli space like $\text{Hilb}^{k}(\mathbb{C}^{2})$ for $k \geq 0$. I apologize if I am confused about something even more basic. But assuming I'm not speaking nonsense....is this just for compactification reasons? I indeed understand how rank 1 sheaves can have a second Chern class. – Benighted Jun 25 '17 at 22:54
• @StephenPietromonaco Although I see now that neither the answer nor the question state this anywhere explicitly, the gauge group here is supposed to be one of the classical groups $\mathrm{SU},\mathrm{Sp},\mathrm{SO}$, see also the beginning of Donaldson's paper. In particular the specific ADHM construction for $\mathrm{SU}(N)$ is used, I do not claim anything about $\mathrm{U}(1)$ gauge theories in this answer. – ACuriousMind Jun 26 '17 at 0:18
• The moduli space of sheaves is not singular but once we use Gieseker-Mayurama compactification singularities appear. On the other hand the moduli space of instantons can be smooth, there is some condition I do not remember at the moment. My point is that for most sensible computations one needs to compactify these spaces. My impression was that when we include the point like instantons in the moduli space of instantons we actually do this GM compactification to obtain the singular but compact moduli space of sheaves. – Marion Jul 24 '17 at 8:00