1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

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

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


    The Euclidean YM action

    [itex] \mathbb{S} = - \int_{\mathbb{R^4}} Tr (F \wedge \ast F) [/itex]


    [itex] D \ast F = 0 [/itex]

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

    [itex] \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}} [/itex]

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

    [itex] T_{a}, T_{b} [/itex] = [itex] - \epsilon_{abc} T_{c}, T_{a} = \frac{1}{2} i \sigma_{a} [/itex]

    where [itex] Tr(T_{a}, T_{b}) = - \frac {1}{2} \delta_{ab} [/itex]

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

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

    when [itex] D+1=4 [/itex] is a gauge theory in Minkowski space [itex] M [/itex], and [itex] A [/itex] is the gauge potential,

    [itex] ( \ast F)_{{\mu}v} = \frac{1}{2} \epsilon_{{\mu}v{\alpha}{\beta}}F^{{\alpha}{\beta}} [/itex]

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

    [itex] -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 [/itex]

    [itex] d^{4}x = \frac{1}{24} \epsilon_{{\mu}v{\alpha}{\beta}}dx^{\mu} \wedge dx^{v} \wedge dx^{\alpha} \wedge dx^{\beta} [/itex]

    with identities

    [itex] \epsilon_{{\mu}v{\alpha}{\beta}}dx^{4} = - dx^{4} \wedge dx^{v} \wedge dx^{\alpha} \wedge dx^{\beta} [/itex]

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

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

    [itex] F_{{\mu}v} (x)[/itex] ~ [itex] O (\frac{1}{r^3}) [/itex]

    [itex] A_{\mu} [/itex] ~ [itex] \partial_{\mu} gg^{-1} + O \frac{1}{r^2} [/itex] as [itex] r \rightarrow \infty [/itex]

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

    For example:

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

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

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

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

    deg (f) is an integer given by

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

    (proof withheld)


    [itex] f: X \rightarrow SU(2) = S^{3} [/itex] where X is closed.

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

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

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

    Scaling Argument:

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

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

    [itex] = E_{F} + E_{D_{\phi}} + E_{U} [/itex]

    if [itex] A(x) [/itex] and [itex] \phi (x) [/itex] are critical points:

    [itex] \phi_{c} (x) = \phi (cx) [/itex]

    [itex] A_{c} (x) = cA(cx) [/itex]

    [itex] F_{c} = C^{2} F(cx) [/itex]

    [itex] D_{c} \phi_{c} = c D\phi (cx) [/itex]

    which leads to

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

    [itex] (D-4) E_{F} + (D-2) E_{D_{\phi}} + DE_{U} = 0 [/itex]

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

    3. The attempt at a solution

    A YM action S within a given topological sector

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

    bounded from below by [itex] 8\pi^2c_{2} [/itex]

    [itex] F \wedge F = \ast F \wedge \ast F [/itex]

    [itex] \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} [/itex]

    when [itex] F = - \ast F [/itex] 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. jcsd
  3. Mar 12, 2012 #2
    Last edited: Mar 12, 2012
  4. Mar 12, 2012 #3
    more nonsense.
    Last edited: Mar 12, 2012
  5. Mar 12, 2012 #4
    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
  6. Mar 12, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 [itex] A_{\mu} (x) [/itex] is a solution. Consider [itex] A_{\mu} (cx) [/itex] and make a change of coordinates, taking into account that [itex]A_\mu[/itex] scales like a tensor of the appropriate degree.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook