Notation in Ch. 10 of Nakahara

[Bballer152]

Hi All,

I'm extremely confused by what's going on in section 10.1.3, pg 377 of the 2nd edition of Nakahara, in regards to his notation for lie algebra-valued one forms.

We let \{U_i\} be an open covering of a smooth manifold M and let \sigma_i be a local section of M into the principle G-bundle P defined on each element in the covering. We also let A_i be a Lie-algebra-valued one form on each U_i.

We then want to construct a Lie-algebra-valued one-form on P using this data, and we do so by defining \omega_i \equiv g_i^{-1}\pi^*A_ig_i + g_i^{-1}\mathrm{d}_Pg_i where d_P is the exterior derivative on P and g_i is the canonical local trivialization defined by \phi_i^{-1}(u)=(p,g_i) for u=\sigma_i(p)g_i.

What I don't understand is the second term in the definition of \omega_i. I don't understand what the exterior derivative is acting on there. He writes it as if it's acting directly on the group element g_i\in G and later writes terms like \mathrm{d}_Pg_i(\sigma_{i*}X), where X\in T_pM and \sigma_{i*} is the push-forward so that \sigma_{i*}X \in T_{\sigma_i(p)}P as if \mathrm{d}_Pg_i is itself a Lie algebra-valued one-form, which I simply don't understand at all. It just doesn't seem well defined and I see no way of making sense of that expression. Thanks so much in advance for any clarification as to what that term is doing, i.e., how it takes in a vector in the tangent space of some point in P and spits out an element of the lie algebra of G.

 

[Ben Niehoff]

$dg$ is not a Lie-algebra valued 1-form, but $g^{-1} dg$ is.

 

[Bballer152]

How so? I don't see how that expression specifies an action sending tangent vectors to Lie-algebra elements...

 

[Ben Niehoff]

Near any (constant) Lie group element $g_0$, a local neighborhood of the Lie group can be expanded via the exponential map

g = g_0 e^{tX}, \qquad X \in \mathfrak{g}
Why don't you start from there?

 

[Bballer152]

Well I understand how this gives a Lie-algebra valued one-form on G itself, but I don't see how it does on the full bundle P. For example, on the next page, Nakahara says that \mathrm{d}_Pg_i(\sigma_{i*}X)=0 \mathrm{\ since\ } g\equiv e \mathrm{\ along\ } \sigma_{i*}X, and again I have no clue how to make since of the expression \mathrm{d}_Pg_i(\sigma_{i*}X). I've scoured his book and found nothing like it, it really seems to come out of nowhere.

 

[Bballer152]

In particular, here is precisely what I don't understand. The argument of \mathrm{d}_Pg_i(\sigma_{i*}X) is a tangent vector on P, which means that the thing that is acting on it should have some kind of "1-form" quality to it, which is good because the thing acting on it is the exterior derivative of something, which smells a lot like a one-form. But how can the exterior derivative on P act on an element g_i in G?? There must be something hiding behind that notation that I'm missing, because this reasoning leads me to a complete and utter roadblock.

 

[Ben Niehoff]

Clearly $g_i$ is a map from somewhere into $G$. That is, $g_i : \mathcal{X} \to G$. I think your question will be answered if you work out what $\mathcal{X}$ is. It should turn out to be a space where the exterior derivative makes sense.

 

[Bballer152]

Okay, I think I've got it, now let's see if I can put it into words (and thanks, by the way, for the patience/hints).

We use the notation of the first post above. Namely, \phi_i:U_i\times G\rightarrow \pi^{-1}(U_i) is the local trivialization such that \phi_i(p,g)=\sigma_i(p)g with the usual right action of g on the right hand side there. Then g:\pi^{-1}(U_i)\rightarrow G (or equivalently but less rigorously g:U_i \times G\rightarrow G) where \sigma_i(p)g \mapsto g (the abuse of notation is purposeful to highlight the "canonical-ness" of this), and this is well defined because everything in \pi^{-1}(U_i) can be expressed as \sigma_i(p)g for some p and g.

Now we know that the exterior derivative works on the domain of g, but we have to modify it a bit because g is not \mathbb{R}-valued and therefore not a true function for the exterior derivative to act on. Thus, we say that \mathrm{d}_Pg:T_{\sigma_i(p)g}P\rightarrow T_gG defined by X\mapsto vert(X) where vert(X) is the projection into the vertical subspace defined by the section \sigma_i as the complement of the pushforward \sigma_{i*}Y of all Y\in T_pM. Then g^{-1}\mathrm{d}g is a Lie-algebra valued one-form because the left action of g^{-1} takes the element in T_gG to something in T_eG\simeq \mathfrak{g}.

That's really the only way I can make sense of this whole thing, so hopefully it's not too far off!

 

[danschub]

I realize this post is almost two years old, but this is a really confusing notation, and the 'hints' in this thread weren't too useful. Since I eventually figured out how to understand this I thought I would post to help those in the same situation who might stumble on this thread through a search.

Even though Nakahara says $d_Pg$ is an "exterior derivative" it is better to understand it as the differential of the map $g : P\rightarrow G$. In other words it is the pushforward dg \equiv g_\star : T_uP \rightarrow T_g G. Perhaps it indeed can be understood as an exterior derivative of a Lie algebra-valued form. But other sources are more clear it is a pushforward (mathematicians use the notation $d$ in both cases), and this is easier for me to understand. The original poster came to a similar conclusion himself.

So the definition
<br /> \omega_i \equiv g_i^{-1}\pi^*A_ig_i + g_i^{-1}\mathrm{d}_Pg_i
really means
<br /> \omega_i \equiv \text{Ad}_{g_i^{-1}}\pi^*A_i+ L_{g_i^{-1}\star}g_{i\star}
where $L_{g_i^{-1}\star}$ is the pushforward of the left-translation which takes $T_{g_i} G \rightarrow T_e G = \mathfrak{g}$. If you use this definition, Nakahara's proofs that $\omega$ has the required properties are easy to follow.

 

[Bballer152]

Well, even though my reply is close to a year late, I thought you should know that I've read this response and I love it, THANK YOU. It all makes sense now :)