2

I'm a physics student and I'm studying differential geometry, my main reference for the topics about manifolds and Lie groups is Nakahara textbook (chap.5). My problem is about Maurer-Cartan one-form, Nakahara defines it as a Lie algebra valued one form $\theta$: \begin{equation} \theta: X \rightarrow {(L_{g})}_{*}^{-1}X \end{equation} where $L_{g}$ is the left action and X a left invariant vector field on the Lie Group G. Nakahara shows that the Maurer-Cartan form satisfies two relations:

  1. $\theta= V_{\mu} \otimes \theta^{\mu}$ (it can we written as a tensor product of generators $V_{\mu}$ and left invariant one forms ${\theta}^{\mu}$)
  2. Satisfies the equation: \begin{equation} d\theta+ \frac{1}{2} [V_{\mu}, V_{\nu}] \otimes \theta^{\mu} \wedge \theta^{\nu}=0 \end{equation} where d is the exterior derivative.

My problem is that my professor always uses the fact that $\theta= g^{-1}dg$ ($*$) to perform different calculations, and I don't understand how this definition is connected to Nakahara. Besides, I find that the definition ($*$) is not a good one: in my understanding it doesn't make sense to consider an exterior derivative on a group element g, since it is defined only on one-forms. I think I'm missing something and I hope someone can help.

  • 1
    Have you searched this site for other questions in this very topic? You’ll find a number of answers to your question. – Ted Shifrin Dec 12 '22 at 15:19
  • Yes I read your answer to a similar question, but it doesn't solve my doubts, which are explained in this question I posted. I think that the symbol dg does not represent an exterior derivative, and I'm asking for a formal way to introduce it in the context I wrote above – Explosiveness Dec 12 '22 at 16:39
  • 1
    It does represent the exterior derivative. Why do you think it does not? That's the issue. I explained in one or two answers that $g$ represents the identity map, not an element of $G$. Write this out explicitly with matrices so that you can identify $L_g$ with actual matrix multiplication. Differentiating $\theta = g^{-1}dg$ gives (essentially) the same result (you have to worry about signs and constants when you work with vector-valued forms) when you write it in coordinates: $d\theta = -g^{-1}dg,g^{-1}\wedge dg = -\theta\wedge\theta$. BTW, to say $d$ is defined only on $1$-forms? – Ted Shifrin Dec 12 '22 at 18:46
  • Ok, but the identity map g is not a scalar function (0-form), in this case I would agree that dg is a one-form. It maps G in G and, if I introduce a chart, I am mapping matrices in matrices, right? In my homework I did a calculation with SU(2) matrices and I found the left invariant vector fields by using $\theta=g^{-1}dg$, so I can see how the "d" should work, but I feel like theoretically I am not fully convinced. (Sorry I edited the question, exterior derivative is defined on n-forms). I really appreciate your help and if you want to suggest a book for reference it would be great – Explosiveness Dec 12 '22 at 20:32
  • Here are two possible sources. Chern/Chen/Lam Lectures on Differential Geometry, Chapter 6. Phillip Griffiths, "On Cartan's Method of Lie Groups and Moving Frames As Applied to Uniqueness and Existence Questions in Differential Geometry," Duke Math J. 41 (1974), 775-814. – Ted Shifrin Dec 12 '22 at 20:45
  • I wrote this answer some days ago: https://math.stackexchange.com/a/4539224/601797. Maybe it is useful to you – A. J. Pan-Collantes Jan 09 '23 at 17:36

0 Answers0