- #1
JorisL
- 492
- 189
Hello all,
I'm having a minor annoyance in proving an identity.
The identity is the following
[tex]
\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}}{(p-1)!}\nabla_\mu A^\mu_{\,\, \mu_1 \cdots \mu_{p-1}}\text{d}^\mu_1\wedge \cdots \wedge \text{d}^\mu_{p-1}
[/tex]
I'm stuck at the first step of proving this
[tex]\text{d}\star A_p = ?[/tex]
Because I'm in arbitrary dimension it's difficult to write out the full expression.
Not in the least because this can be written as
[tex]\text{d}(\star A_p) = \text{d}\left(\frac{1}{p! (D-p)!}A_{\mu_1\cdots \mu_p}\epsilon_{\nu_1\cdots \nu_{D-p}}^{\quad\quad\mu_1\cdots\mu_p} \text{d}^\nu_1\wedge \cdots \wedge \text{d}^\nu_{D-p}\right)[/tex]
If I immediately apply the definition of the exterior derivative to this I get a lot of ugliness.
I would need to introduce the metric ##p## times, once for each of the upper indices.
Then I need to use ##\nabla_\alpha \epsilon_{\mu_1\cdots \mu_D} = 0## to rewrite the partial in terms of the connection. The same happens for the metric factors.
Is there any way I can write this down in a clean way? If need be I can grind my way through but if there is an alternative I'm not going to do that (at first sight) useless exercise in index gymnastics.
Joris
PS.
The next step would be to write the identity as
[tex]
\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}\partial_\mu\left(\sqrt{|g|}A^{\mu\nu_1\cdots\nu_{p-1}}\right)}{(p-1)!\sqrt{|g|}}g_{\nu_1\mu_1}\cdots g_{\nu_{p-1}\mu_{p-1}}\text{d}^\mu_1\wedge \cdots \wedge \text{d}^\mu_{p-1}
[/tex]
I'm having a minor annoyance in proving an identity.
The identity is the following
[tex]
\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}}{(p-1)!}\nabla_\mu A^\mu_{\,\, \mu_1 \cdots \mu_{p-1}}\text{d}^\mu_1\wedge \cdots \wedge \text{d}^\mu_{p-1}
[/tex]
I'm stuck at the first step of proving this
[tex]\text{d}\star A_p = ?[/tex]
Because I'm in arbitrary dimension it's difficult to write out the full expression.
Not in the least because this can be written as
[tex]\text{d}(\star A_p) = \text{d}\left(\frac{1}{p! (D-p)!}A_{\mu_1\cdots \mu_p}\epsilon_{\nu_1\cdots \nu_{D-p}}^{\quad\quad\mu_1\cdots\mu_p} \text{d}^\nu_1\wedge \cdots \wedge \text{d}^\nu_{D-p}\right)[/tex]
If I immediately apply the definition of the exterior derivative to this I get a lot of ugliness.
I would need to introduce the metric ##p## times, once for each of the upper indices.
Then I need to use ##\nabla_\alpha \epsilon_{\mu_1\cdots \mu_D} = 0## to rewrite the partial in terms of the connection. The same happens for the metric factors.
Is there any way I can write this down in a clean way? If need be I can grind my way through but if there is an alternative I'm not going to do that (at first sight) useless exercise in index gymnastics.
Joris
PS.
The next step would be to write the identity as
[tex]
\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}\partial_\mu\left(\sqrt{|g|}A^{\mu\nu_1\cdots\nu_{p-1}}\right)}{(p-1)!\sqrt{|g|}}g_{\nu_1\mu_1}\cdots g_{\nu_{p-1}\mu_{p-1}}\text{d}^\mu_1\wedge \cdots \wedge \text{d}^\mu_{p-1}
[/tex]