# Homework Help: SU(2) Pure YM on R^4

1. Mar 10, 2012

### Charles_Henry

1. The problem statement, all variables and given/known data

Derive the pure SU(2) YM theory on $\mathbb{R}^4$ from the action. Let $A_{\mu} (x)$ be a solution to these equations. Show:

$\tilde{A}_{\mu} (cx)$ is also a solution (with the same action).

Background

The Euclidean YM action

$\mathbb{S} = - \int_{\mathbb{R^4}} Tr (F \wedge \ast F)$

yields

$D \ast F = 0$

Let $\ast: \wedge^{p} \rightarrow \wedge^{D+1-p}$ be a linear map, such that

$\ast (dx^{\mu_{1}} \wedge ... \wedge dx^{\mu_{p}} = \frac{\sqrt{|det(n)!|}}{(D+1-p)!} \epsilon^{\mu_{1}...\mu_{p}}_{{\mu}_{p+1}...{\mu}_{D+1}} dx^{\mu}_{p+1} \wedge ... \wedge dx^{{\mu}_{p+1}}$

if $G = SU(2)$ we choose a basis $T_{a}, a = 1, 2, 3$ for an Anti-Hermitian 2 x 2 matrix

$T_{a}, T_{b}$ = $- \epsilon_{abc} T_{c}, T_{a} = \frac{1}{2} i \sigma_{a}$

where $Tr(T_{a}, T_{b}) = - \frac {1}{2} \delta_{ab}$

where $\sigma_{a}$ are pauli matrices, and a general group element $g = exp (\alpha^{a} T_{a} )$ with $\alpha^a$ real. Whence,

$(D_{\mu} \phi)^{a} = \partial_{\mu} \phi^{a} - \epsilon^{abc} A^{b}_{\mu} \phi^{c}$ and $F^{a}_{{\mu} v} = \partial_{\mu} A^{a}_{v} - \partial_{v}A^{a}_{\mu} - \epsilon^{abc}A^{b}_{\mu}A^{c}_{v}$

when $D+1=4$ is a gauge theory in Minkowski space $M$, and $A$ is the gauge potential,

$( \ast F)_{{\mu}v} = \frac{1}{2} \epsilon_{{\mu}v{\alpha}{\beta}}F^{{\alpha}{\beta}}$

A two form $F= \frac{1}{2} F_{{\mu}v}dx^{\mu} \wedge dx^{v}$ is self dual or ASD when $\ast F = F$ and $\ast F = - F$ respectively

$-Tr ( F \wedge \ast F) = - \frac{1}{2} Tr (F_{{\mu}v}F^{{\mu}v}) d^{4}x = \frac{1}{4}F^{a}_{{\mu}v}F^{{\mu}va} d^{a}x$

$d^{4}x = \frac{1}{24} \epsilon_{{\mu}v{\alpha}{\beta}}dx^{\mu} \wedge dx^{v} \wedge dx^{\alpha} \wedge dx^{\beta}$

with identities

$\epsilon_{{\mu}v{\alpha}{\beta}}dx^{4} = - dx^{4} \wedge dx^{v} \wedge dx^{\alpha} \wedge dx^{\beta}$

and $\epsilon_{{\alpha}{\beta}{\rho}{\sigma}} \epsilon^{{\mu}{v}{\rho}{\sigma}} = - 4 (\delta^{\mu}_{\alpha} \delta^{v}_{\beta})$

Instantons are non-singular solutions of classical equations of motion in Euclidean space whose Action is finite.

$F_{{\mu}v} (x)$ ~ $O (\frac{1}{r^3})$

$A_{\mu}$ ~ $\partial_{\mu} gg^{-1} + O \frac{1}{r^2}$ as $r \rightarrow \infty$

note: I understand that the gauge transformations g(x) needs to be defined only asymptotically, so $g: \mathbb{S}^{3}_{\infty} \rightarrow SU(2)$ is extended to $\mathbb{R}^4$ if its degree vanishes:

For example:

If $M_{1}$ and $M_{2}$ are oriented, compact, D-dimensional manifolds without boundary, and $w$ is a volume-form on $M_{2}$. where $deg (f)$ of a smooth map $f: M_{1} \rightarrow M_{2}$ is given by

$\int_{M_{1}} f\ast w = [ deg(f) ] \int_{M_{2}} w$

let $y \in M_{2}$ when $f^{-1}(y) = {x; f(x) = y}$ is finite, and the Jacobian $J(f)$ is not zero (if $x \in U$ with local coordinates $x^{i}$ and $y \in f(u)$ with local co-ordinates $y^{i}$, then we can assume:

$\mathbb{J} = det \frac{\partial y^{i}}{\partial x^{J}}$ if $y^{i} (x^{1}, ... x^{D})$

deg (f) is an integer given by

$deg (f) = \Sigma_{x \in f^{-1} (y)} sign [ \mathbb{J} (x) ]$

(proof withheld)

therefore:

$f: X \rightarrow SU(2) = S^{3}$ where X is closed.

$deg(f) = \frac{1}{24\pi^2} \int_{X} Tr[(f^{-1} df)^3]$

the boundary conditions are understood in terms of one-point compactifications $S^4 = \mathbb{R}^4 \cup {\infty}$ which has a conformally equivalent metric to that of a flat metric in $\mathbb{R}^4$

A solution of YM equations on $S^4$ project stereographically to a connection on $\mathbb{R}^4$ with a curvature which vanishes at infinity.

Scaling Argument:

A Field(s) $(A, \phi)$ given by a potential one-form and a scalar higgs-field:

$E = \int_{\mathbb{R}} d^{D}x [|F|^{2} + |D \phi |^{2} + U(\phi)$

$= E_{F} + E_{D_{\phi}} + E_{U}$

if $A(x)$ and $\phi (x)$ are critical points:

$\phi_{c} (x) = \phi (cx)$

$A_{c} (x) = cA(cx)$

$F_{c} = C^{2} F(cx)$

$D_{c} \phi_{c} = c D\phi (cx)$

$E_{(c)} = \frac{1}{C^{D-4}}E_{F} + \frac{1}{C^{D+2}}E_{D_{\phi}} + \frac{1}{C^{D}}E_{U}$

$(D-4) E_{F} + (D-2) E_{D_{\phi}} + DE_{U} = 0$

note: I believe I am looking for a solution where $E_{D_{\phi}} = E_{U} = 0$ in D=4

3. The attempt at a solution

A YM action S within a given topological sector

$c_{2} = \frac{1}{8 \pi^2} \int_{\mathbb{R}} Tr( F \wedge F) > 0$

bounded from below by $8\pi^2c_{2}$

$F \wedge F = \ast F \wedge \ast F$

$\mathbb{S} = - \frac{1}{2} \int_{\mathbb{R}^4} Tr[(F + \ast F) \wedge (F + \ast F)] + \int_{\mathbb{R}^4} Tr (F \wedge F) = - \frac{1}{2} \int_{\mathbb{R}^4} Tr [(F + \ast F) \wedge \ast (F + \ast F) + 8 \pi^2c_{2} \geq 8 \pi^2c_{2}$

when $F = - \ast F$ hold

some bib: Atiyah, M.F and Ward, R.S (1977) Instantons and Algebraic Geometry, Commun. Math. Phy. 55, 117-124

Sacks, L. and uhlenbeck, K (1981) The existence of minimal immersions of 2-spheres, Ann. Math 113, 1-24

Last edited: Mar 10, 2012
2. Mar 12, 2012

### Charles_Henry

nonsense

Last edited: Mar 12, 2012
3. Mar 12, 2012

### Charles_Henry

more nonsense.

Last edited: Mar 12, 2012
4. Mar 12, 2012

### Charles_Henry

ok, i see it now...derive the euler-lagrange equations from the action that leads to gauge potential. Assume we could derive a solution of pure YM IN R^4 from the Vector Potential by defining invariance along any coordinate of our choosing.

Last edited: Mar 12, 2012
5. Mar 12, 2012

### fzero

This is meant to be much simpler than you're making it out to be.

Literally use the Euler-Lagrange equations to obtain the equation of motion for the gauge potential.

This is meant to follow from the scale-invariance of pure YM. Assume $A_{\mu} (x)$ is a solution. Consider $A_{\mu} (cx)$ and make a change of coordinates, taking into account that $A_\mu$ scales like a tensor of the appropriate degree.