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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).


    Background

    The Euclidean YM action

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

    yields

    [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)

    therefore:

    [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
    nonsense
     
    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

    fzero

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    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.
     
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