Recent content by Taoy

Pah! :) In that context the wedge is just as evil as that evil \times :). No, it was just me using the wrong sign :). I was in a hurry to get to a class, so I wasn't as careful as I should have been. I'll fix the original post (so as to avoid unnecessary bad posts :).

(This post has been corrected.) Well expanded it looks like, 0 = \underrightarrow{dx^i} \underrightarrow{dx^j} \left( \partial_i (e_j)^\alpha \gamma_\alpha + \frac{1}{2} \omega_i{}^{\mu \nu} (e_j)^\alpha \gamma_{\mu \nu} \times \gamma_\alpha \right) So, what is the clifford...

Wait; how many Clifford bases are we using here? Are you using \sigma_B \equiv \gamma_B, or are these bases separate and distinct?

Hmm, so from the perspective of the metric the different between left and right invariant fields gets hidden; they are separate and distinct at the level of the vierbein though. What's the geometric meaning of this?

Yay :) That was bothering me too. p.s. have had great conversations with Hestenes and Lasenby (more the former); Doran had to cancel so no low-down on BHs.

It should be. I'm primarily here because tomorrow David Hestenes is hosting a parallel session on Geometric Algebra and Gravity; Doran and Lasenby are here too apparently. I'm hoping to find some people who are into the conformal projective framework... I'll definitely let you know how it goes :).

Hey Garrett, finally found some wireless connectivity here at Freie University, Berlin; and for my sins I stayed up last night until I'd done my homework... yes; I'm coked up on coffee to make up for it :). Ok, here's what I think. I think that the answer is yes, because {\cal...

Hey Garrett, no not yet; I'm off to Berlin today for a week (to the Marcel Grossman conference), so I won't get to post anything for a week or so. BUT I'm expecting to have it all worked out by next weekend :). Feel free to post the next bit though if you want; I'll catch up. BTW, my last...

Can you explain how to talk about the metric in your notation? The metric tends to be formulated as, g = g_{ij} dx^i \otimes dx^j, however your \underrightarrow{dx^i} basis elements are antisymmetric, not symmetric. How do you define this? Of course, operating on a scalar, the lie...

Awww, I like to get a good freak on from time to time o:) Thanks for the clarification. Ok, so it must be this (with a previous typo in this post fixed), \begin{align*} (L_{\vec{\xi_A}} \vec{\xi'_B}) \; \underrightarrow{\partial} g & = \vec{\xi_A} \underrightarrow{\partial}...

Ok, I'm sure I've not got the arrows in the right places :), but it looks like they commute, and the Lie derivate of one set of invariant fields with respect to the other is zero. Here's my reasoning: \begin{align*} (L_{\vec{\xi_A}} \vec{\xi'_B}) \; \underrightarrow{\partial} g & =...

I know that :wink:. I meant to say that vectors act on 1-forms and produce a 0-form (by contraction), and visa-versa. However here we have a vector acting on a 0-form also producing a 0-form; that seems strange to me; after all a 0-form acting (multiplying) a vector doesn't produce a 0-form...

Oooh, so vectors act on scalars the same as vectors act on one-forms? So how would you conserve the arrows in: ({\cal L}_{\vec{X}} \vec{Y}) f = (\vec{X} \vec{Y} f -\vec{X} \vec{Y} f)

Sure :). Working on it now. In the mean time another notation question that comes up when one expands the Lie bracket of two vector fields, ({\cal L}_{\vec{X}} \vec{Y}) f = (\vec{X} \vec{Y} f -\vec{X} \vec{Y} f) Taking the first term, \vec{X} \vec{Y} f = \vec{X} (Y^i \vec{\partial i} f)...

Hey Garrett, I need to clarify your notation a little more :), What does \underrightarrow{d} mean? There's an ambiguity here; this is not the one-form d_i \underrightarrow{dx^i} with components d_i, it's the exterior derivative operator, \underrightarrow{d} = \underrightarrow{dx^i}...