# Notation in Ch. 10 of Nakahara

1. Apr 23, 2013

### 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.

2. Apr 23, 2013

### Ben Niehoff

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

3. Apr 23, 2013

### Bballer152

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

4. Apr 23, 2013

### 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?

5. Apr 23, 2013

### 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.

6. Apr 23, 2013

### 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.

7. Apr 24, 2013

### 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.

8. Apr 25, 2013

### 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!

9. Feb 3, 2015

### 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
$$\omega_i \equiv g_i^{-1}\pi^*A_ig_i + g_i^{-1}\mathrm{d}_Pg_i$$
really means
$$\omega_i \equiv \text{Ad}_{g_i^{-1}}\pi^*A_i+ L_{g_i^{-1}\star}g_{i\star}$$
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.

10. Jan 30, 2016

### 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 :)