Yang Mills

Let $M$ be a 4-d manifold equipped with metric $g$, and denote the determinant of $g$ by $\lvert g \rvert$. Let $G \subset SO \left( l \right)$ be the compact structure group for the Yang-Mills action we will construct on M. Let $\mathfrak{g} \subset so \left( l \right)$ denote the Lie algebra of $G$, and $\left( A , B \right) \equiv \mathrm{tr} A B^{\dagger}$ the trace inner product on $\mathfrak{g}$. Let $*$ represent the Hodge dual induced by the metric $g$. For ordinary real-valued $p$-forms in $\Lambda^p M$, the Hodge dual is defined in terms of the inner product on forms, which in turn is linearly extended from the definition

(1)

\begin{align} \langle e^1 \wedge \cdots \wedge e^p , f^1 \wedge \cdots \wedge f^p \rangle = \mathrm{det} \left[ g^{\mu \nu} e^i_{\mu} f^j_{\nu} \right]. \end{align}

In terms of this inner product, the Hodge dual is defined by the expression

(2)

\begin{align} \omega \wedge * \mu = \langle \omega , \mu \rangle \cdot \mathrm{vol}, \end{align}

where $\mathrm{vol}$ is the volume form induced by the metric $g$. In coordinates, this reduces to

(3)

\begin{align} * \omega = \frac{\sqrt{ \lvert g \rvert } }{p! \left( n - p \right)!} \omega_{\alpha_1 \dots \alpha_p} \left. \epsilon^{\alpha_1 \dots \alpha_p} \right. _{\beta_{p+1} \dots \beta_n} dx^{\beta_{p+1}} \wedge \dots \wedge dx^{\beta_n}, \end{align}

where $\epsilon_{\alpha_1 \dots \alpha_n}$ is the totally antisymmetric symbol and $\left. \epsilon^{\alpha_1 \dots \alpha_p}\right. _{\beta_{p+1} \dots \beta_n} = \epsilon_{\gamma_1 \dots \gamma_p \beta_{p+1} \dots \beta_n} g^{\alpha_1 \gamma_1} \cdots g^{\alpha_p \gamma_p}$.

For a Lie-algebra-valued $p$-form $\omega \otimes T \in \Lambda^p M \otimes \mathfrak{g}$, the definition of the Hodge dual is extended so that

(4)

\begin{align} * \left( T \otimes \omega \right) \equiv T^{\dagger} \otimes * \omega \end{align}

The inner product of two Lie-algebra valued forms is

(5)

\begin{align} \langle T \otimes \omega , S \otimes \mu \rangle &= \mathrm{tr} \left[ \left( T \otimes \omega \right) \wedge * \left( S \otimes \mu \right) \right] = \mathrm{tr} \left[ \left( T \otimes \omega \right) \wedge \left( S^{\dagger} \otimes * \mu \right) \right] \\ &= \mathrm{tr} \left( T S^{\dagger} \right) \left( \omega \wedge * \mu \right) = \mathrm{tr} \left( T S^{\dagger} \right) \langle \omega , \mu \rangle \cdot \mathrm{vol} \end{align}

The Lorentzian Yang-Mills action is given by

(6)

\begin{align} S_{YM} \left( A \right) = - \frac{1}{2} \int_M \mathrm{tr} \left( F \wedge * F \right), \end{align}

while the Euclidean Yang-Mills action is

(7)

\begin{align} S^E_{YM} \left( A \right) = \frac{1}{2} \int_M \mathrm{tr} \left( F \wedge * F \right). \end{align}

Note that the signs would be reversed, as in Nakahara, Baez/Muniain, or Jackiw, if we did not incorporate the adjoint into the definition of the Hodge dual. In terms of the definitions give above, we can rewrite the Lorentzian action as

(8)

\begin{align} - \frac{1}{2} \int_M \langle F , F \rangle \, \mathrm{vol} &= - \frac{1}{2} \int_M \langle \frac{1}{2} F_{\mu \nu} dx^{\mu} \wedge dx^{\nu} , \frac{1}{2} F_{\rho \sigma} dx^{\rho} \wedge dx^{\sigma} \rangle \, \mathrm{vol} \\ &= - \frac{1}{8} \int_M \mathrm{tr} \left( F_{\mu \nu} F^{\dagger}_{\rho \sigma} \right) \Bigl\langle dx^{\mu} \wedge dx^{\nu} , dx^{\rho} \wedge dx^{\sigma} \Bigr\rangle \mathrm{vol} \\ &= - \frac{1}{8} \int_M \left( F_{\mu \nu} , F_{\rho \sigma} \right) \det \left( \begin{array}{ll} g^{\mu \rho} & g^{\mu \sigma} \\ g^{\nu \rho} & g^{\nu \sigma} \end{array} \right) \mathrm{vol} \\ &= - \frac{1}{4} \int_M g^{\mu \rho} g^{\nu \sigma} \left(F_{\mu \nu} , F_{\rho \sigma} \right) \sqrt{\lvert g \rvert} dx_M \end{align}

or in terms of a normalized (with respect to the trace inner product) basis of the Lie algebra,

(9)

\begin{align} - \frac{1}{4} \int_M F^I_{\mu \nu} F^{\mu \nu}_I \sqrt{\lvert g \rvert} dx_M, \end{align}

where $I$ runs over the Lie algebra basis, and up and down $I$ indices are the same. Thus for Minkowski space, the Lagrangian density for Yang-Mills is $\mathcal{L} = - \frac{1}{4} F^{I}_{\rho \sigma} F_{I}^{\rho \sigma}$. Alternatively, we could have reached this coordinate expression for the Yang-Mills action by substituting the coordinate form of $*F$ (use 4 and 3) into 7.

We find the momenta canonically conjugate to the components $A_{\mu}$ of the connection by

(10)

\begin{align} \Pi^{\mu} = \frac{\delta \mathcal{L}}{\delta \dot{A}_{\mu}}, \end{align}

but since the Lagrangian does not depend on $\dot{A}_0$, we cannot assign a momentum canonically conjugate to $A_0$. We deal with this issue by defining our Hamiltonian formalism with the Weyl gauge, $A_0 = 0$, imposed, so that we now only have three configuration variables $A_1, A_2, A_3$. We find their canonical momenta to be

(11)

\begin{align} \Pi^{i} = E^i = \dot{A}_i, \end{align}

where the Latin indices run over space ($i = 1, 2, 3$) only. Note that $E$ is the negative of the variable that in Maxwell theory would be called the electric field. We can also define the analog of the magnetic field $B^i = \frac{1}{2} \epsilon^{ijk} F_{jk}$.

Note that the squared vector norms of $E$ and $B$ are given by

(12)

\begin{align} \lvert E \rvert ^2 = - F^{I}_{0i} F_{I}^{0i} \\ \lvert B \rvert ^2 = \frac{1}{2} F^{I}_{jk} F_{I}^{jk}, \end{align}

so in terms of these we can write

(13)

\begin{align} S_{YM} \left( A \right) &= - \frac{1}{4} \int_{M} F^{I}_{\rho \sigma} F_{I}^{\rho \sigma} dx^0 \cdots dx^3 \\ &= \frac{1}{2} \int_M \lvert E \rvert ^2 - \lvert B \rvert ^2 dx^0 \cdots dx^3 \end{align}

and a Legendre transform $\mathcal{H} = \dot{A}^I_i E_I^i - \mathcal{L}$ gives the Hamiltonian density

(14)

\begin{align} \mathcal{H} = \frac{1}{2} \left( \lvert E \rvert ^2 + \lvert B \rvert ^2 \right) \end{align}

A similar analysis of the Euclidean action would give

(15)

\begin{align} S^E_{YM} = \frac{1}{2} \int_M \lvert E \lvert ^2 + \lvert B \rvert ^2 dx^0 \dots dx^3 \end{align}

because of the differing sign in the definition of the action and the Euclidean signature of the metric.