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I am using the reference Black hole entropy from an SU(2)-invariant formulation of Type I isolated horizons for this question. I am trying to understand the two equations (30) that give the variation of Ashtekar connection $A_+$ and its conjugate momenta $\Sigma^+$ in the self-dual Platini action under diffeomorphisms induced by $\alpha : \Delta \rightarrow \mathfrak{sl(2,\mathbb{C})}$. The equations give \begin{align} \delta_\alpha \Sigma^+ = [\alpha,\Sigma^+], \quad \delta_\alpha A_+ = -d_{A_+}\alpha. \end{align}

I know that the Lie derivative between two vector fields is calculated by the Lie commutator but I do not understand why is it that for conjugate momenta we get a commutator and for the connection one form we get $\delta_\alpha A_+ = -d_{A_+}\alpha$.

Thanks in advance for any help.

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    $\begingroup$ This has nothing to do specifically with Ashtrkar variables, it’s only a general rule of how the Lie derivative acts on tensors. $\endgroup$ Commented May 29 at 10:34
  • $\begingroup$ @Prof.Legolasov is there a reference that I can learn similar calculations from it with similar examples? $\endgroup$
    – mortimer
    Commented May 29 at 11:48
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    $\begingroup$ Just look up the definition of Lie derivative acting on tensors of various types. The connection is a rank-1 covariant tensor. That definition reduces to the expression you’re wondering about. $\endgroup$ Commented May 29 at 20:31
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    $\begingroup$ That’s correct, but that’s a global transformation. Now to derive the infinitesimal transformation, plug $g = 1 + \alpha + \mathcal{O}(\alpha^2)$ to get the formula with the covariant derivative. $\endgroup$ Commented Jun 3 at 23:09
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    $\begingroup$ I added an expanded answer to help with your confusion. PTAL. $\endgroup$ Commented Jun 3 at 23:44

1 Answer 1

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Let's derive these transformation laws from the first principles.

Start with the gauge transformation. Let's have a field $\psi(x)$ that transforms in some representation $\rho$ of the gauge group. Then we have,

$$ \psi(x) \mapsto \psi'(x) = \rho(g(x)) \psi(x), $$

parameterized by the group-valued map $g(x)$.

There's a mapping between (at least the neighborhood of the identity element in) the group and the algebra, it's called the exponential map. Probably the best way to see it is to consider the universal enveloping algebra (UEA) which contains both the Lie algebra and the Lie group.

Working in the UEA, we can write an equation

$$ g(x) = 1 + \varepsilon \alpha(x) + \mathcal{O}(\varepsilon^2), $$

which relates the finite gauge transformation parameterized by $g(x)$ and the infinitesimal transformation parameterised by $\alpha(x)$ and the infinitesimal parameter $\varepsilon$.

The infinitesimal transformation law for $\psi(x)$ is then,

$$ \psi(x) \mapsto \left(1 + \varepsilon \rho(\alpha(x)) \right) \psi(x), $$ $$ \frac{\delta \psi(x)}{\delta \varepsilon} = \rho(\alpha(x)) \psi(x). $$

Now note that the gradient of $\psi$ doesn't transform linearly:

$$ d \psi \mapsto \rho(g) d \psi + d \rho(g) \cdot \psi. $$

To compensate, we introduce the covariant derivative $D \psi = d \psi + \rho(A) \psi$. By construction,

$$ D \psi \mapsto \rho(g) D \psi. $$

As an exercise I will leave deriving the transformation law for $A$ from here. You already posted it in your comments, so I am sure you're familiar with it:

$$ A \mapsto g A g^{-1} + g dg^{-1}. $$

Note that we can write an equation without $\rho$ here, because the above reasoning is valid for any representation $\rho$. This equation is valid in the UEA, so it's not a problem that we have Lie group and algebra elements in the same equation.

Now to derive the infinitesimal form:

$$ g(x) = 1 + \varepsilon \alpha(x) + \mathcal{O}(\varepsilon^2), $$ $$ A \mapsto \left( 1 + \varepsilon \alpha \right) A \left( 1 - \varepsilon \alpha \right) + \left( 1 + \varepsilon \alpha \right) d \left( - \varepsilon \alpha \right) + \mathcal{O}(\varepsilon^2) = $$ $$ A - \varepsilon [A, \alpha] - \varepsilon d \alpha + \mathcal{O}(\varepsilon^2). $$ $$ \frac{\delta A}{\delta \varepsilon} = - d \alpha - [A, \alpha] = - d \alpha - \rho_{\text{adjoint}}(A) \alpha = - D_A \alpha. $$

Here $D_A \alpha$ is the covariant derivative taken in the adjoint representation.

That's your formula (30) from your post.

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  • $\begingroup$ Thank you so much. I got my answer. $\endgroup$
    – mortimer
    Commented Jun 4 at 7:04

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