Equations of Motion and Noether's Theorem in the Functional Formalism

First, let us recall the derivation of the equations of motion and Noether's theorem in classical field theory. We have some action functional $S[\phi]$ defined by some local Lagrangian:
$$S[\phi] = \int L(\phi, \partial \phi) dx.$$
The classical equations of motion are just the Euler-Lagrange equations
$$\frac{\delta S}{\delta \phi(x)} = 0  \iff \partial_\mu \left( \frac{\partial L}{\partial(\partial_\mu\phi)} \right) = \frac{\partial L}{\partial \phi}$$

Now suppose that $S$ is invariant under some transformation $\phi(x) \mapsto \phi(x) + \epsilon(x) \eta(x)$, so that $S[\phi] = S[\phi+\epsilon \eta]$. Here we treat $\eta$ as a fixed function but $\epsilon$ may be an arbitrary infinitesimal function. The Lagrangian is not necessarily invariant, but rather can transform with a total derivative:
$$L(\phi+\epsilon \eta) = L(\phi) + \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \partial_\mu \epsilon + \epsilon \partial_\mu f^\mu$$
For some unknown vector field $f^\mu$ (which we could compute given any particular Lagrangian). So let's compute
$$\begin{align}\delta_\epsilon S &= \int \delta_\epsilon L \\\ &= \int \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \partial_\mu \epsilon + \epsilon \partial_\mu f^\mu \\\ &= \int \partial_\mu \left(f^\mu - \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \right) \epsilon \end{align}$$
Let us define the Noether current $J^\mu$ by
$$J^\mu = \frac{\partial L}{\partial (\partial_\mu \phi)} \eta - f^\mu.$$
Then the previous computation showed that
$$\frac{\delta S}{\delta \epsilon} = -\partial_\mu J^\mu.$$
If $\phi$ is a solution to the Euler-Lagrange equations, then the variation $dS$ vanishes, hence we obtain:

Theorem (Noether's theorem) The Noether current is divergence free, i.e.
$$\partial_\mu J^\mu = 0.$$

Functional Version

First, we derive the functional analogue of the classical equations of motion. Consider an expectation value
$$\langle \mathcal{O(\phi)} \rangle = \int \mathcal{O}(\phi) e^{\frac{i}{\hbar} S} \mathcal{D}\phi$$
We'll assume that $\phi$ takes values in a vector space (or bundle). Then we can perform a change of variables $\psi = \phi + \epsilon$, and since $\mathcal{D}\phi = \mathcal{D}\psi$ we find that
$$\int \mathcal{O}(\phi+\epsilon) \exp\left(\frac{i}{\hbar} S[\phi] \right) \mathcal{D}\phi$$
is independent of $\epsilon$. Expanding to first order in $\epsilon$, we have
$$0 = \int \left(\frac{\delta\mathcal{O}}{\delta \phi}  + \frac{i \mathcal{O}}{\hbar} \frac{\delta S}{\delta \phi} \right)  \exp \left( \frac{i}{\hbar} S \right) \mathcal{D}\phi$$
So we find the quantum analogue of the equations of motion:
$$\left\langle \frac{\delta \mathcal{O}}{\delta \phi} \right\rangle + \frac{i}{\hbar} \left\langle \mathcal{O} \frac{\delta S}{\delta \phi} \right\rangle = 0$$

Next, we move on to the quantum version of Noether's theorem. Suppose there is a transformation $Q$ of the fields leaving the action invariant. Assuming the path integral measure is invariant, we obtain
$$\left\langle QF \right \rangle + \frac{i}{\hbar} \left\langle F QS \right\rangle = 0$$
To compare with the classical result, consider $Q$ to be the (singular) operator
$$Q = \frac{\delta}{\delta \epsilon(x)}$$
Then by the previous calculations,
$$Q S = -\delta_\mu J^\mu,$$
so we obtain
$$\left\langle \frac{\delta \mathcal{O}}{\delta \epsilon(x)} \right\rangle = \frac{i}{\hbar} \left\langle \mathcal{O} \partial_\mu J^\mu \right\rangle.$$
This is the Ward-Takahashi identity, the quantum analogue of Noether's theorem.

The Moyal Product

Today I want to understand the Moyal product, as we will need to understand it in order to construct quantizations of symplectic quotients. (More precisely, to incorporate stability conditions.)



Let $A$ be the algebra of polynomial functions on $T^\ast \mathbb{C}^n$. This algebra has a natural Poisson bracket, given by
$$\{p_i, x_j\} = \delta_{ij}.$$
We would like to define a new associative product $\ast$ on $A((\hbar))$ satisfying:

1. $f  \ast g = fg + O(\hbar)$
2. $f \ast g - g \ast f = \hbar \{f, g\} + O(\hbar^2)$
3. $1 \ast f = f \ast 1 = f$
4. $(f \ast g)^\ast = -g^\ast \ast f^\ast$

In the last line, the map $(\cdot)^\ast$ takes $x_i \mapsto x_i$ and $p_i \mapsto -p_i$. To figure out what this new product should be, let's take $f,g \in A$ and expand $f \ast g$ in power series:

$$f \ast g = \sum_{n=0}^\infty c_n(f,g) \hbar^n$$

Now, equations (1) and (2) will be satisfied by taking $c_0(f,g) = fg$ and $c_1(f,g) = \{f,g\}/2$. Let $\sigma$ be the Poisson bivector defining the Poisson bracket. This defines a differential operator $\Pi$ on $A \otimes A$ by

$$\Pi = \sigma^{ij} (\partial_i \otimes \partial_j)$$

Let $B = \sum_{n=0}^\infty B_n \hbar^n$ and write the product as

$$f \ast g = m \circ B(f \otimes g).$$
Now, condition (2) tells us that $B(0) = 1$ and that
$$\left. \frac{dB}{d\hbar} \right|_{\hbar=0} = \frac{\Pi}{2}$$
So
$$B = 1 + \frac{\hbar \Pi}{2} + O(\hbar^2)$$
It is natural to guess that $B$ should be built out of powers of $\Pi$, and a natural guess is
$$B = \exp(\frac{\hbar \Pi}{2}),$$
which certainly reproduces the first two terms of our expansion. Let's see that this choice actually works, i.e. defines an associative $\ast$-product. Let $m: A \otimes A \to A$ be the multiplication, and
$m_{12}, m_{23}: A \otimes A \otimes A \to A \otimes A$, $m_{123}: A \otimes A \otimes A \to A$ the induced multiplication maps. Then
$$\begin{align} f \ast (g \ast h) &= m \circ(B( f \otimes m \circ B(g \otimes h) ) ) \\\ &= m \circ B( m_{23} \circ (1 \otimes B)(f \otimes g \otimes h) ) \\\ &= m_{123} (B \otimes 1)(1 \otimes B)(f \otimes g \otimes h) \end{align}$$
On the other hand, we have

$$\begin{align} (f \ast g) \ast h) &= m \circ(B( m \circ B(f \otimes g) \otimes h) ) ) \\\ &= m \circ B( m_{12} \circ (B \otimes 1)(f \otimes g \otimes h) ) \\\ &= m_{123} (1 \otimes B)(B \otimes 1)(f \otimes g \otimes h) \end{align}$$

Hence, associativity is the condition
$$m_{123} \circ [1\otimes B, B \otimes 1] = 0.$$

On $A \otimes A \otimes A$, write $\partial_i^1$ for the partial derivative acting on the first factor, $\partial_i^2$ on the second, etc. Then
$$1 \otimes B = \sum_n \frac {\hbar^n}{2^n n!}  \Pi^{i_1 j_1} \cdots \Pi^{i_n j_n} \partial^2_{i_1} \partial^3_{j_1} \cdots \partial^2_{i_n} \partial^3_{j_n}$$
and similarly for $B \otimes 1$. So we have
$$\begin{align} m_{123} (B\otimes 1)(1 \otimes B) &= \sum_n \sum_{k=0}^n \frac {\hbar^n}{2^n k! (n-k)!}  \Pi^{k_1 l_1} \cdots \Pi^{k_k l_k} \partial_{k_1} \partial_{l_1} \cdots \partial_{k_k} \partial_{l_k} \\\  & \ \times  \Pi^{i_1 j_1} \cdots \Pi^{i_{n-k} j_{n-k}} \partial_{i_1} \partial_{j_1} \cdots \partial_{i_{n-k}} \partial_{j_{n-k}} \\\ &= m_{123}(1 \otimes B)(B \otimes 1) \end{align}$$
Hence we obtain an associative $\ast$-product. This is called Moyal product.

Sheafifying the Construction

Now suppose that $U$ is a (Zariski) open subset of $X = T^\ast \mathbb{C}^n$. Then the star product induces a well-defined map
$$\ast: O_X(U)((\hbar)) \otimes_\mathbb{C} O_X(U)((\hbar)) \to O_X(U)((\hbar))$$
In this way we obtain a sheaf $\mathcal{D}$ of $O_X$ modules with a non-commutative $\ast$-product defined as above.

Define a $\mathbb{C}^\ast$ action on $T^\ast \mathbb{C}^n$ by acting on $x_i$ and $p_i$ with weight 1. Extend this to an action on $\mathcal{D}$ by acting on $hbar$ with weight -1.

Proposition: The algebra $C^\ast$-invariant global sections of $\mathcal{D}$ is naturally identified with the algebra of differential operators on $\mathbb{C}^n$.

Proof: The $\mathbb{C}^\ast$-invariant global sections are generated by $\hbar^{-1} x_i$ and $\hbar^{-1} p_i$. So define a map $\Gamma(\mathcal{D})^{\mathbb{C}^\ast} \to \mathbb{D}$ by
$$\hbar^{-1} x_i \mapsto x_i$$
$$\hbar^{-1} p_i \mapsto \partial_i$$
From the definition of the star product, it is clear that this is an algebra map, and that it is both injective and surjective.

An Exercise in Quantum Hamiltonian Reduction

Semiclassical Setup

Let the group $GL(2)$ act on $V = \mathrm{Mat}_{2\times n}$ and consider the induced symplectic action on $T^\ast V$. If we use variables $(x,p)$ with $x$ a $2 \times n$ matrix and $p$ an $n \times 2$ matrix, then the classical moment map $\mu$ is given by
$$\mu(x,p) = xp$$
This is equivariant with respect to the adjoint action, so we can form the $GL(2)$-invariant functions
$$Z_1 = \mathrm{Tr} \mu$$
$$Z_2 = \mathrm{Tr} (\mu)^2$$
If we think of $x$ as being made of column vectors
$$x = ( x_1 \cdots x_n )$$
and similarly think of $p$ as being made of row vectors, then there are actually many more $GL(2)$ invariants, given by
$$f_{ij} = \mathrm{Tr} x_i p_j = p_j x_i$$
In terms of the invariants, the $Z$ functions are
$$Z_1 = \sum_k f_{kk}$$
$$Z_2 = \sum_{jk} f_{jk} f_{kj}$$
Let us compute Poisson brackets:
$$\begin{align}  \{f_{ij}, f_{kl}\} &= \{p_j^\mu x_i^\mu, p_l^\nu x_k^\nu\} \\\ &= x_i^\mu p_l^\nu \delta_{jk} \delta^{\mu\nu} - p_j^\mu x_k^\nu \delta_{il} \delta^{\mu\nu} \\\ &= f_{il} \delta_{jk} - f_{kj} \delta_{il}. \end{align}$$
So we see that the invariants form a Poisson subalgebra (as they should!). Let's compute:
$$\begin{align} \{Z_1, f_{ij} &= \sum_k \{ f_{kk}, f_{ij} \} \\\ &= \sum_k \left( f_{kj} \delta_{ki} - f_{ik} \delta_{kj} \right) \\\ &= f_{ij} - f_{ij} = 0. \end{align}$$
Hence $Z_1$ is central with respect to the invariant functions $f_{ij}$. Similarly,
$$\begin{align} \{Z_2, f_{kl}\} &= \sum_{ij} \{f_{ij} f_{ji}, f_{kl}\} \\\ &= \sum_{ij} f_{ij} \left(f_{jl} \delta_{ik} - f_{ki} \delta_{jl} \right) + f_{ji} \left(f_{il} \delta_{jk} - f_{kj} \delta_{il} \right) \\\ &= \sum_j f_{kj} f_{jl} - \sum_i f_{il} f_{ki} + \sum_i f_{ki} f_{il} - \sum_j f_{jl} f_{kj} \\\ &= 0. \end{align}$$
So we see that the $Z_i$ are in the center of the invariant algebra. In fact, they generate it, so we'll denote by $Z$ the algebra generated by $Z_1, Z_2$. Let $A$ be the algebra generated by the $f_{ij}$. The inclusion $Z \hookrightarrow A$ can be thought of as a purely algebraic version of the moment map. In particular, given any character $\lambda: Z \to \mathbb{C}$, we can define the Hamiltonian reduction of $A$ to be
$$A_\lambda := A / A\langle \ker \lambda \rangle$$
The corresponding space is of course $\mathrm{Spec} A$.

The Cartan Algebra and the Center

Define functions

$$h_1 = Z_1 = \sum_i f_{ii}$$
$$h_2 = Z_2 = \sum_{ij} f_{ij} f_{ji}$$
$$h_3 = \sum_{ijk} f_{ij} f_{jk} f_{ki}$$
$$h_k = \sum_{i_1, i_2, \ldots, i_k} f_{i_1 i_2} f_{i_2 i_3} \cdots f_{i_k i_1}$$

These are just the traces of various powers of the $n \times n$ matrix $px$. In particular, $h_k$ for $k>n$ may be expressed as a function of the $h_i$ for $i \leq n$. The algebra generated by the $H$ plays the role of a Cartan subalgebra. So we have inclusions
$$Z \subset H \subset A$$

Quantization

Now we wish to construct a quantization of $A$ and $A_\lambda$. The quantization of $A$ is obvious: we quantize $T^\ast V$ by taking the algebraic differential operators on $V$. Denote this algebra by $\mathbb{D}$. It is generated by $x_i$ and (\partial_i\) satisfying the relation
$$[\partial_i, x_j] = \delta_{ij}$$
Then we simply the subalgebra of $GL(2)$-invariant differential operators as our quantization of $A$. Call this subalgebra $U$. We can define Hamiltonian reduction analogously by taking central quotients. So we need to understand the center $Z(U)$, but this is just the subalgebra generated by quantizations of $Z_1$ and $Z_2$, i.e. the subalgebra of all elements whose associated graded lies in $Z(A)$.

More to come: stability conditions, $\mathbb{D}$-affineness, and maybe proofs of some of my claims.

The 1PI Effective Action

In this post I'd like to try to understand the 1PI effective action that is often of interest. Suppose we have a QFT in some bosonic field $\phi(x)$ taking values in a vector space (this is important). Then its vev $\phi_{cl}(x) := \langle \phi(x) \rangle$ is just an ordinary (but possibly distributional) field on spacetime. The question is, what is the field equation satisfied by $\phi_{cl}$? I.e., if we average over quantum effects by replacing all fields by their vevs, what is the action that governs this (now completely classical) theory? The 1PI effective action answers exactly this question.

Consider the generating functional
$$Z[J] = \int e^{-S[\phi]+\langle \phi, J \rangle} \mathcal{D}\phi$$
Then for a given source $J$, define the $J$-vev of $\phi(x)$ to be
$$\phi_J(x) = \frac{\partial \log Z[J]}{\partial J}.$$
Now let's take $\Gamma$ to be the Legendre transform of $\log Z$ with respect to $J$:
$$\Gamma[\phi_J] = \langle J, \phi_J \rangle - \log Z[J]$$
Then we compute:
$$\frac{\partial \Gamma}{\partial \phi_J} = J + \frac{\partial J}{\partial \phi_J} \phi_J -\frac{\partial \log Z[J]}{\partial J} \frac{\partial J}{\partial \phi_J} = J.$$

Now consider the situation without a background source, i.e. $J = 0$. Then $\phi_0 = \phi_{cl}$ and we find
$$\frac{\partial \Gamma}{\partial \phi_{cl}} = 0$$
Hence, $\phi_{cl}$ satisfies the Euler-Lagrange equations associated to the functional $\Gamma$. Note that from the Legendre transform, $\Gamma$ takes quantum effects (i.e. Feynman diagrams with loops) into account, even though the field and the equations are purely classical!

By studying these equations, we might find instanton solutions (or solitons in Lorentz signature).

Now for the name. Some combinatorics and algebra (which I will skip!) show that $\Gamma[\phi_{cl}]$ is itself a generating functional for certain correlation functions, then 1PI correlation functions:
$$\frac{\partial^n \Gamma}{\partial \phi(x_1) \cdots \partial \phi(x_n)} = \langle \phi(x_1) \cdots \phi(x_n) \rangle_{1PI}.$$
The 1PI subscript means that the RHS is computed in perturbation theory by summing over only the connection 1PI (1 particle irreducible) Feynman diagrams.

Warning: As usual, there are regularization issues, both in the UV and IR. UV divergences can be solved by a cutoff (if we only care about effective field theory), but IR divergences are much more technical. For this reason (and others), it is sometimes preferable to try to understand the low energy dynamics by studying the Wilsonian effective action. As the Wilsonian effective action does not take IR modes into account, it can avoid many of the difficulties of the 1PI effective action.

Spontaneous Symmetry Breaking in QFT

In this post I want to try to understand symmetry breaking and the origin of the moduli space of vacua. Most of this can be found in the lectures by Witten in vol. 2 of Quantum Fields and Strings.

Non-Example: Quantum Mechanics

The main point of confusion for me is that my quantum intuition comes from ordinary quantum mechanics. However, this turns out to be incredibly misleading because for most reasonable quantum mechanical systems, spontaneous symmetry breaking cannot occur. In fact, we'll see that even in field theory, the question of whether spontaneous symmetry breaking can occur is intimately related to the geometry of spacetime. Since quantum mechanics is QFT in $0+1$ dimensions (i.e., the spatial part of spacetime is just a point), spontaneous symmetry breaking is forbidden.

Consider a particle in one spatial dimension, with Hamiltonian
$$H = -\frac{\hbar^2}{2} + (a^2-x^2)^2.$$
The classical ground states are given by the stationary solutions $x(t) = \pm a$. Hence we might expect that the quantum Hamiltonian has a degenerate ground state, i.e. the eigenspace of the lowest eigenvalue has dimension greater than one. However, this is not the case!

Sketch of proof: Define a function $E(\phi)$ on the unit sphere in $L^2(\mathbb{R})$. If $\phi$ is a global minimum, then it necessarily satisfies $H\psi = E_0 \phi$ where $E_0$ is the lowest eigenvalue of $H$. On the other hand, $E(\phi) = E(|\phi|)$ so $|\phi|$ is also a global minimum, hence satisfies $H|\phi| = E_0|\phi|$. This equation is elliptic, so by elliptic regularity $|\phi|$ must be at least $C^1$, and hence $\phi(x)$ has constant phase, so we might as well take $\phi(x)$ to be real and positive. Any other ground state $\psi$ would have these properties, and hence $(\phi,\psi) \neq 0$. Hence the eigenspace is 1-dimensional and the ground state is not degenerate.

Now by the Stone-von Neumann theorem, there is a unique irreducible representation of the canonical commutation relations on a separable Hilbert space, and by the argument above there is a unique (up to scale) vector $|\Omega\rangle$ which is the ground state of the Hamiltonian, called the vacuum vector. So for QM, we find a unique representation on $\mathcal{H}$ together with a unique vacuum vector $|\Omega\rangle \in \mathcal{H}$. The point I want to stress here is that the Poisson algebra of observables together with the Hamiltonian determine the data $\mathcal{H}, |\Omega\rangle)$ in an essentially unique way, so there is no ambiguity in quantization and no further choices need to be made.

QFT in Finite Volume

Now we'll argue that a similar symmetry breaking phenomenon should be expected whenever the spatial part of spacetime has finite volume. We'll have to use formal path integral arguments, so of course this won't be totally rigorous. Suppose we have two representations $\mathcal{H}_\pm$ with vacua $|\Omega\rangle_\pm$. Then we consider the direct sum $\mathcal{H} = \mathcal{H}_+ \oplus \mathcal{H}_-$. Then by construction, we should have
$$(\Omega_+, e^{-tH} \Omega_-) = 0$$
since $|\Omega\rangle_\pm$ are orthogonal eigenstates of $H$. On the other hand, we can compute this inner product using the Feynman-Kac formula. The semi-classical approximation to the path integral yields
$$(\Omega_+, e^{-tH} \Omega_-) = C \exp\left(- \frac{S(t) V}{\hbar} \right)$$
Where $S(t)$ is the classical least action and $V$ is the volume of space. If $V < \infty$, the right hand side is non-zero, contradicting our assumptions! Hence the vacuum is non-degenerate. So we find that QFT with finite spatial volume is much like QM, at least as far as symmetry breaking is concerned.

Note that this argument is essentially just a formal manipulation of the path integral, so you should expect a result of this form independent of the particular regularization scheme used to define the path integral.

QFT in Infinite Volume

Now we consider the case $V = \infty$, i.e. a non-compact space. Then the preceding argument fails spectacularly, as does the Stone-von Neumann theorem. So there is no guarantee of a unique irreducible representation of the algebra, and no guarantee of a unique vacuum vector.
So we see that the situation is significantly more complicated. We can expect a moduli space $\mathcal{M}$ of vacua of the theory, and the low energy effective theory is described by a $\sigma$-model with target $\mathcal{M}$. I'll try to discuss this in more detail in follow-up posts.

Seiberg-Witten Theory and the Riemann-Hilbert Problem

References:

• Introduction to Seiberg-Witten theory and its stringy origin
• Seiberg-Witten and follow-up

The Classical Moduli Space of Vacua

For definiteness, we'll consider just the case of $SU(2)$ considered by Seiberg and Witten. There is a scalar Higgs field $\phi$. The classical vacua of the theory are given by the absolute minima of the potential energy, which in this case is proportional to
$$\mathrm{Tr}[\phi, \phi^\dagger]^2$$
Hence at the minimum, $[\phi,\phi^\dagger]=0$ and $\phi$ is diagonalizable. Hence the classical moduli space of vacua $\mathcal{M}_{cl}$ is just $\mathbb{C}$, with complex coodinate $a$, corresponding to the Higgs field
$$\phi = \left( \begin{array}{rr}a & 0 \\ 0 & -a\end{array} \right)$$
Actually, due to gauge invariance, it is better to introduce another copy of the complex plane $\mathcal{B}$ with local coordinate $u = \frac{1}{2} Tr \phi^2 = a^2$. Then we can think of $\mathcal{M}_{cl}$ as a branched cover of $\mathcal{B}$, with $a$ a (local) choice of square root of $u$.

The goal is to understand the low energy effective theory. We introduce a cutoff $\Lambda$ to define the quantum theory, and integrate out all degrees of freedom except for the low momentum modes of $\phi$ (in particular, we integrate out the gauge field d.o.f.). The result is a $\sigma$-model with target $\mathcal{M}_{cl}$. The kinetic term of the $\sigma$-model is governed by the metric on $\mathcal{M}_{cl}$, hence the low energy effective action determines a metric on $\mathcal{M}_{cl}$.

We'll see that 1-loop calculations introduce monodromy, so that in the quantum theory, "functions" on $\mathcal{M}_{cl}$ are actually sections of non-trivial bundles over $\mathcal{M}_{cl}$, and furthermore that the metric receives corrections from instantons (or BPS states). So what we really would like to understand/construct is the quantum moduli space of vacua $\mathcal{M}$, which will be some non-trivial modification of $\mathcal{M}_{cl}$. The key to the Seiberg-Witten solution is that susy allows us to reduce the problem to finding a specified set of holomorphic functions (in the $u$ coordinate\) satsfying certain monodromies, and that once we know the monodromies the solution is given to us by the Riemann-Hilbert correspondence.

The Riemann-Hilbert Correspondence

Let $X$ be $\mathbb{P}^1$ with punctures at the points $z_1, \ldots, z_n$. Let $U$ be the universal cover of $X$ and let $G$ be the fundamental group of $X$ (pick some basepoint away from the punctures). A set of monodromy matrices is exactly what is needed to specify a representation $V$ of $G$. Since $U / G = X$, we can form the associated bundle $E = U \times_G V$ over $X$. The (trivial) $G$-connection on $U \to X$ induces a flat connection $\nabla$ on $E$. This gives a map from representations of $G$ to flat connections on $X$.

Conversely, given a flat connection on $X$, the monodromy about the punctures determines a representation of $G$. Hence monodromy is a map from flat connections on $X$ to representations of $G$. The Riemann-Hilbert correspondence is that these two maps are bijections, modulo the natural notions of equivalence (conjugacy and gauge transformations).

Gross Overview of the Seiberg-Witten Approach

We are now ready to sketch the "big picture" idea of Seiberg and Witten, which applies not only to their $N=2, d=4$ example but also to certain other compactifications of the $N=1, d=6$ theory (in particular, the one considered by Gaiotto-Moore-Neitzke).

As discussed above, the theory will have a classical moduli space of vacua $\mathcal{M}$, which turns out to be a complex manifold (or variety, and possibly with singularities). We'll let $u$ be an abstract local complex coordinate on $\mathcal{M}$. Supersymmetry then tells us that the main quantities we are interested in (to compute the low energy effective action) are holomorphic in $u$ (away from the singularities/punctures of $\mathcal{M}$!). The general outline is as follows:

1. Identify functions $f_i(u)$ which by susy are holomorphic in $u$.
2. Compute the 1-loop corrections to $f_i(u)$.
3. Compute monodromies of the corrected $f_i(u)$.
4. Find the desired $f_i(u)$ by solving the Riemann-Hilbert problem for these monodromies.

Now, to be more clear, it is a consequence of susy that the renormalized quantities $f_i(u)$ are given schematically by

$$f_{i, \mathrm{ren}}(u) = f_{i, \mathrm{cl}}(u) + f_{i,1}(\frac{u}{\Lambda}) + \sum_{k=0}^\infty c_{i,k} \left(\frac{\Lambda}{u} \right)^k$$

Here, $f_{i,\mathrm{cl}}(u)$ is the classical function, $f_{i,1}(u)$ is the one-loop correction, and the terms in the series are corrections coming from instantions (BPS states). Non-renormalization theorems due to susy guarantee that there are no higher loop corrections. One expects the instanton series to converge, and hence the monodromy is completely determined by the one-loop calculation. This is the key: by Riemann-Hilbert, the monodromy determines the $f_i(u)$ uniquely--solving the Riemann-Hilbert problem is equivalent to computing the infinitely-many instantion corrections!

Now in general, solving the Riemann-Hilbert problem is difficult, so this reduction is of a theoretical but not necessarily practical nature. The second main idea of Seiberg and Witten is that we can solve this Riemann-Hilbert problem explicitly by introducing a family of curves $\{C_u\}_{u\in\mathcal{B}}$, called Seiberg-Witten curves (or spectral curves).

Electric-Magnetic Duality

An absolutely key requirement of the Seiberg-Witten construction is electric-magnetic duality. Maxwell's equations in vacuum are
$$dF = 0, \ \ \ d\ast F = 0.$$
Here $F$ is a 2-form, and $\ast F$ is its Hodge dual, a $(d-2)$-form in $d$-dimensions. The first equation implies that $F = dA$ for some 1-form $A$, and we normally think of the second equation as the Euler-Lagrange equations for the action written in terms of $A$. However, we could equally well take the starting point to be the second equation, taking $\ast F = dB$, and take the first equation to be the Euler-Lagrange equations for $B$. The problem with either of these approaches is that they allow particles of either electric or magnetic charge, but not both.

To put electric and magnetic charge on equal footing, we introduce fields $F$ and $F_D$ (a 2-form and a (d-2)-form\). Then the Lagrangian is (up to factors that I'm too lazy to care about)
$$\mathcal{L} = \mathrm{Tr} F \wedge F_D$$
However, to recover Maxwell's equations, we need to impose $\ast F_D = F$ as a constraint. So to get the right equations of motion, introduce an auxiliary field $\lambda$ (a Lagrange multiplier), and modify the Lagrangian:
$$\mathcal{L} = \mathrm{Tr} F \wedge F_D + \lambda(F - \ast F_D)$$
Variation with respect to $(F, F_D, \lambda)$ will reproduce Maxwell's equations exactly, but in this form the EM duality is manifest. Since EM duality exchanges electric and magnetic charges, we should consider how to modify the Lagrangian to couple the field to EM sources. Let $J_e, J_m$ be the electric and magnetic currents, respectively. Up to conventions, Maxwell's equations read
$$dF = J_m, \ \ \ d F_D = J_e.$$
Then we take the Lagrangian to be
$$\mathcal{L} = \mathrm{Tr} F \wedge F_D + \lambda(F - \ast F_D) + F \wedge J_e + J_m \wedge F_D$$
to reproduce the right equations of motion.

In this form, we can consider particles with electric or magnetic charge (or both--dyons). If our gauge group has rank $r$, then the lattice of electric charges is $\mathbb{Z}^r$, while the lattice of magnetic charges is $(\mathbb{Z}^\ast)^r$. Hence the lattice of electromagnetic charges is
$$\Gamma = \mathbb{Z}^r \oplus (\mathbb{Z}^\ast)^r$$
which comes with a natural symplectic pairing
$$\langle \cdot, \cdot \rangle: \Gamma \otimes \Gamma \to \mathbb{Z}.$$

(You might ask why we take the natural sympletic pairing as opposed to the natrual symmetric pairing. This is because there is actually a larger $SL(2,\mathbb{Z})$ symmetry of the theory which preserves the symplectic pairing but not the symmetric pairing.)

Now there is an obvious source of symplectic lattices. Simply let $C$ be a genus $r$ compact Riemann surface. Then $\Gamma = H_1(C, \mathbb{Z})$ is a symplectic lattice of rank $2r$, where the symplectic pairing is now given by the intersection pairing. In fact, we can say more--if we take $a$- and $b$-cycles as generators, these form a Darboux (symplectic) basis of $\Gamma$.

Back to the gauge theory problem. Recall that the 1-loop calculation and consideration of BPS states leads to a set of monodromy data on $\mathcal{B}$. Suppose now that we could find a complex surface $C \to \mathcal{B}$ whose fibers $C_u$ are (possibly singular) genus $r$ curves, and such that the monodromies of $\Gamma_u := H_1(C_u, \mathbb{Z})$ agree with the given monodromies. Then we can solve the Riemann-Hilbert problem by doing geometry on this family, i.e. by finding holomorphic sections of certain associated bundles.

The SU(2) Seiberg-Witten Solution

We will now specialize to the case considered in the original paper of Seiberg and Witten. I will only construct the family--the details of the solution will follow in a subsequent post.

In this case, the group has rank $r=1$, so we should be looking for a family of elliptic curves. In this case, the solution is almost obvious, given what I've said above. Seiberg and Witten argue that the moduli space $\mathcal{B}$ must be $\mathbb{C} \setminus \{\Lambda^2, -\Lambda^2\}$. The punctures at $\pm \Lambda^2$ come from BPS states whose mass goes to zero at those values of $u$. So the monodromy consists of three matrices, $M_\infty, M_\pm$, the monodromies computed around $\infty$ and $\pm \Lambda^2$. These generate a certain modular subgroup $G$ of $SL(2, \mathbb{Z})$, allowing us to realize $\mathcal{B}$ as the modular curve $H / G$ (where $H$ is the upper half-plane). Now, the space of elliptic curves is just $H / SL(2, \mathbb{Z})$. So given any $u \in \mathbb{B}$, we pick a lift $\tilde{u}$ in $H$ and let $C_u$ be the corresponding elliptic curve. This is exactly the family needed to solve the Riemann-Hilbert problem!

Next time: details of this construction, including exact formulas, and some words about instanton counting.