New answers tagged gauge-theory
1
Let's compare the following train of thought for $U(1)$ vs $SU(3)$:
U(1)
$U(1)$ is an Abelian group.
This means, the commutator of two group elements is zero.
The field strength tensor that appears in the Lagrangian, $F^{\mu\nu}$, contains such a commutator: $$F_{\mu\nu}^{QED} = \partial_\mu A_\nu - \partial_\nu A_\mu + \text{i} g [A_\mu, A_\nu] = \...
2
The QCD Lagrangian is,
$\mathcal{L}=\sum_{\alpha} \overline{\psi}_{q, a}\left(i \gamma^{\mu} \partial_{\mu} \delta_{a b}-g_{s} \gamma^{\mu} t_{a b}^{C} \mathcal{A}_{\mu}^{C}-m_{q} \delta_{a b}\right) \psi_{q, b}-\frac{1}{4} F_{\mu \nu}^{A} F^{A \mu \nu}$.
The field strength tensor is given by,
$F_{\mu \nu}^{A}=\partial_{\mu} \mathcal{A}_{\nu}^{A}-\partial_{...
3
Maurer-Cartan (MC) equations are e.g. used in:
The coadjoint orbit method, see e.g. Refs. 1 & 2.
Various gauge field theories. The Batalin-Vilkovisky (BV) quantum master equation can be viewed as a generalized MC equation.
References:
J.E. Marsden and T.S. Ratiu, Intro to Mechanics and Symmetry, 2nd Eds, 1998.
B. Kolev, Lie Groups and mechanics: an ...
2
The another way from $0 = dF^a + f^{abc} A^b \wedge F^c\tag{1}$ to
$\epsilon^{\mu\nu\lambda\sigma}D_{\nu} F^a_{\lambda\sigma} = 0.\tag{2}$
From (1) we can write
\begin{eqnarray}
0&=&dF +A^b \wedge F^c [T^b,T^c]\\
&=&dF+A\wedge F\equiv DF=0 \tag 3
\end{eqnarray}
or
$DF^a=0\tag 4$
Then expand (4) in coordinate basis we have
$0= D_\nu F^a_{\...
1
As already mentioned in the comment, the exact version was proven in the link. However, if you insisted to show it infinitesimally in your notation, here is how.
First, consider the variation (warning: sloppy of notation use in the following but it is clear in the context I guess)
\begin{equation} \label{delL}
\delta L = -\frac{1}{2} \sum_a \left( \...
3
Notice that field strength transforms under adjoint representation. Moreover, recall that the connection piece of covariant derivative acts on an adjoint represented field in a commutator manner:
$D_\mu \phi = \partial_\mu \phi + [A_\mu, \phi] = \partial_\mu \phi^a T^a + f^{bca}A_\mu^b \phi^c T^a$.
With this understanding, the equivalence of the tensor ...
0
First, let me say that a good place to look at would be [1]. I haven't checked there so I'm not sure, but it contains technology for computing such things even for exceptional groups. It uses a so-called "birdtrack" notation that treats tensor contractions as graphs.
Here however is a way to brute force it. I didn't try it myself so I don't know if it's ...
2
To be concrete let spacetime be 2D with Minkowski signature.
On one hand, the infinite-dimensional algebra
$${\rm locconf}(1,1)~=~{\rm Vect}(\mathbb{S}^1)\oplus {\rm Vect}(\mathbb{S}^1)\tag{1a}$$ of locally defined conformal transformations in 2D is the algebra of conformal Killing vector fields (CKVF), which is 2 copies of the real Witt algebra, cf. e.g. ...
1
This is what I think makes the $U(1)$ electric 1-form symmetry global. The action of shift of the gauge field by a flat connection on a Wilson line operator is
$$
A \rightarrow A + \lambda\implies W_n(C) \rightarrow W_n(C) \exp \left( i n \oint_C \lambda \right).
$$
Let $\alpha(C)=\oint_C \lambda$, which is the group parameter for the above transformation. ...
1
I got it by grinding out the algebra
\begin{equation}
\Gamma^\nu_{\hphantom 0\mu\lambda}=e^\nu_{\hphantom 0 a}\partial_\mu e_\lambda^{\hphantom 0 a}+e^\nu_{\hphantom 0 a}e_\lambda^{\hphantom 0 b}\omega_{\mu\hphantom 0 b}^{\hphantom 0 a}
\end{equation}
\begin{equation}
R^\rho_{\hphantom 0\mu\sigma\nu}=\partial_\sigma\Gamma^\rho_{\hphantom 0\nu\mu}-\partial_\...
1
Hint: The Dirac conjecture states that the generators $K_i$ of the gauge group are given by the first-class constraints, i.e. Gauss law $(D\cdot E)_i$. OP's last equation then follows by recalling the CCR for the conjugate pair $A^i$ and $E_j$. (Here deWitt's condensed notation is assumed everywhere.)
References:
M. Henneaux & C. Teitelboim, ...
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