SU(2) Pure YM on R^4: Derivation and Solutions

[Charles_Henry]

Homework Statement

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)

which leads to

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

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

 

[Charles_Henry]

nonsense

 

[Charles_Henry]

more nonsense.

 

[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.

 

[fzero]

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

Derive the pure SU(2) YM theory on \mathbb{R}^4 from the action.

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

Let A_{\mu} (x) be a solution to these equations. Show:

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

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.