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Notation in Ch. 10 of Nakahara

  1. Apr 23, 2013 #1
    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 [tex] \{U_i\} [/tex] be an open covering of a smooth manifold [tex] M [/tex] and let [tex] \sigma_i [/tex] be a local section of M into the principle G-bundle P defined on each element in the covering. We also let [tex]A_i[/tex] be a Lie-algebra-valued one form on each [tex]U_i.[/tex]

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

    What I don't understand is the second term in the definition of [tex]\omega_i.[/tex] 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 [tex]g_i\in G[/tex] and later writes terms like [tex]\mathrm{d}_Pg_i(\sigma_{i*}X),[/tex] where [tex]X\in T_pM[/tex] and [tex]\sigma_{i*}[/tex] is the push-forward so that [tex] \sigma_{i*}X \in T_{\sigma_i(p)}P[/tex] as if [tex]\mathrm{d}_Pg_i[/tex] 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.
     
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  3. Apr 23, 2013 #2

    Ben Niehoff

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    ##dg## is not a Lie-algebra valued 1-form, but ##g^{-1} dg## is.
     
  4. Apr 23, 2013 #3
    How so? I don't see how that expression specifies an action sending tangent vectors to Lie-algebra elements...
     
  5. Apr 23, 2013 #4

    Ben Niehoff

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

    [tex]g = g_0 e^{tX}, \qquad X \in \mathfrak{g}[/tex]
    Why don't you start from there?
     
  6. Apr 23, 2013 #5
    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 [tex] \mathrm{d}_Pg_i(\sigma_{i*}X)=0 \mathrm{\ since\ } g\equiv e \mathrm{\ along\ } \sigma_{i*}X,[/tex] and again I have no clue how to make since of the expression [tex] \mathrm{d}_Pg_i(\sigma_{i*}X).[/tex] I've scoured his book and found nothing like it, it really seems to come out of nowhere.
     
  7. Apr 23, 2013 #6
    In particular, here is precisely what I don't understand. The argument of [tex] \mathrm{d}_Pg_i(\sigma_{i*}X)[/tex] 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.
     
  8. Apr 24, 2013 #7

    Ben Niehoff

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    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.
     
  9. Apr 25, 2013 #8
    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, [itex]\phi_i:U_i\times G\rightarrow \pi^{-1}(U_i)[/itex] is the local trivialization such that [itex]\phi_i(p,g)=\sigma_i(p)g[/itex] with the usual right action of [itex]g[/itex] on the right hand side there. Then [itex]g:\pi^{-1}(U_i)\rightarrow G[/itex] (or equivalently but less rigorously [itex]g:U_i \times G\rightarrow G[/itex]) where [itex]\sigma_i(p)g \mapsto g[/itex] (the abuse of notation is purposeful to highlight the "canonical-ness" of this), and this is well defined because everything in [itex]\pi^{-1}(U_i)[/itex] can be expressed as [itex]\sigma_i(p)g[/itex] for some [itex]p[/itex] and [itex]g[/itex].

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

    That's really the only way I can make sense of this whole thing, so hopefully it's not too far off!
     
  10. Feb 3, 2015 #9
    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 [tex]dg \equiv g_\star : T_uP \rightarrow T_g G.[/tex] 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
    [tex]
    \omega_i \equiv g_i^{-1}\pi^*A_ig_i + g_i^{-1}\mathrm{d}_Pg_i[/tex]
    really means
    [tex]
    \omega_i \equiv \text{Ad}_{g_i^{-1}}\pi^*A_i+ L_{g_i^{-1}\star}g_{i\star}[/tex]
    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.
     
  11. Jan 30, 2016 #10
    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 :)
     
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