Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble to understand how one may write down the connection in terms of local trivialization of the vector bundle.

Assume $\pi: E\rightarrow X$ is a vector bundle with rank $n$. A connection on $E$ is a map $$A:TE\rightarrow \pi^{*}E$$ such that $A$ is a splitting of the sequence $0\rightarrow \pi^{*}E\rightarrow TE\rightarrow \pi^{*}TX\rightarrow 0$ and commutes with multiplication by scalars. Here the map from $\pi^{*}E$ to $TE$ is defined by $$(e,v)\rightarrow (e,\frac{d}{dt}|_{t=0}(e+tv))$$

The author (Cliff Taubes) asserts that if we let $x_{i}$ be the $n$ corresponding trivialization functions $E_{U}\rightarrow U\times \mathbb{C}^{n}$ such that we have $$x(e)=(\pi(e),(x^{1}(e)...x^{n}(e))$$ Then the connection $A$ takes value in $\mathbb{C}^{n}$ (viewed as a one form on $E$ with values in $\pi^{*}E$). So far I can follow since $\pi^{*}E$ has extra $n$ dimensions. But I feel at loss with the following assertion because I do not know how to construct an inverse to form a splitting map:

Let $A^{a}$ be the coordinates, then we can write $$A^{a}=dx^{a}+A^{ab}x^{b}$$ Here $A^{ab}$ is a 1-form pulled back from $U$. We can think of $A$ as an End($\mathbb{C}^{n}$)valued 1-form on $U$.

My questions are:

1): How do I get this formula? How should I interpret it in terms of the exact sequence?

2): How can I show any connection $A$ can written in this form, and any $A^{ab}$ defined on $U$ is suffice to define $A$?

3): There is a marked difference between this definition and the definition on wikipedia, for example in wikipedia connection is defined to be a map from sections to sections. What is the reason for the discrepancy in the language?

share|improve this question
    
The notation $\pi^*E$ confuses me. Is $\pi: E\to X$? What does it mean to pullback $E$ to $E$? Is $\pi^*E$ just $E$? –  Matt Aug 24 '12 at 20:08
1  
You can pull back a vector bundle by forming the ordered pair whose coordinates project to the same element in $X$. In this case it is of the form $(e,v)$, where $v\in V=E_{\pi(e)}$. –  Bombyx mori Aug 24 '12 at 20:14
    
this en.wikipedia.org/wiki/… should help you out with seeing the conncetion (pun intended) between the two definitions. –  Eric O. Korman Aug 24 '12 at 20:35
    
Thanks, though I am still confused. –  Bombyx mori Aug 24 '12 at 20:55
    
I'll try to write more later, but the key observation is that the map from $\pi^*E \to TE$ has image equal to the vertical tangent space (i.e. $\ker d\pi \subset TE$). For $A$ to split the sequence means exactly what you think: it is a projection to the vertical tangent space along what will now be defined to be the horizontal subspace. Let $\sigma$ be a section of $E$. The wiki definition has $\nabla \sigma \colon TX \to E$. Yours allows you to take $d\sigma \colon TX \to TE$ and then compose with $A$. IMO, wikipedia's more convenient for vector bundles, Taubes's generalizes better. –  Sam Lisi Sep 2 '12 at 17:13

1 Answer 1

up vote 1 down vote accepted
+50

$\newcommand{\C}{\mathbb{C}}$ Consider $U = X$, and $E = U \times \mathbb{C}^n$, a trivial bundle. Then, we identify the following three bundles over $E$ as follows \[ \pi^* E = U \times \C^n \times \C^n \to E, \quad (u, x, v) \mapsto (u, x). \] \[ TE = TU \times \C^n \times \C^n \to E, \quad (u, u', x, v) \mapsto (u, x). \] \[ \pi^* TU = TU \times \C^n \to E, \quad (u, u', x) \mapsto (u, x). \] The two maps in your exact sequence of vector bundles over $E$ become $(u, x, v) \mapsto (u, 0, x, v)$ and $(u, u', x, v) \mapsto (u, u', x)$.

Note that with these identifications, my vectors $v$ are really to be thought of as $v = \sum x^a \partial_{x^a}$. (To be honest, I am not super comfortable with Einstein notation, so I apologize if I get the indices up or down incorrectly.)

We want $A$ to be a $\operatorname{Hom}(TE, \pi^*E)$, i.e.~a 1-form on $E$ with values in $\pi^*E$. In our coordinates, this means that we want $A(u, u', x, v) = F(u,x) (u', v)$, where $F(u,x)$ is a $(u,x)$ dependent family of matrices giving linear maps from $\C^n \times \C^n \to \C^n$. We can then write this as $F(u,x)(u', v) = F_1(u,x) u' + F_2(u,x)v$ where $F_i$ are square matrices.

The condition that $A$ splits the exact sequence of vector bundles means that $A(u, 0, x, v) = (u, x, v)$. In terms of this $F$, then, this means $F(u,x)(0, v) = v$, i.e. $F_2(u,x) = \operatorname{Id}$.

The condition that $A$ be a linear connection (commuting with scalars as you mention above) means that $F_1$ should also be linear in $x$. We may then write it as $F_1(u,x) = \sum F_1^b x^b$, where the $F_1^b$ can be thought of as 1-forms on $U$ with values in $\C^n$.

Now we observe that what I wrote is precisely what the formula you gave means. The $dx^a$ terms correspond to the thing I called $F_2$ and the $A^{ab}x^b$ correspond to this $F_1$.

OK, this shows that any connection form $A$ of the form you require can be written in coordinates this way. The converse, that any data of these forms $A^{ab}$ (and trivializations etc) give a connection satisfying your requirements is essentially the same idea.

Eric's link from his comment Ehresmann connections describes the relationship between these two definitions. If you chase through what your definition states, it describes a linear Ehresmann connection on a vector bundle, whereas the article you link in wikipedia describes the connection in terms of covariant derivatives. These two points of view are equivalent (as I also mention in my comment above). I personally find the covariant derivative definition in terms of eating sections to be easier to understand and easier to work with in calculations, but the Ehresmann connection generalizes more clearly. It also helps to make clear what a connection really is: it is a splitting of the tangent space to the bundle into a vertical subbundle (canonically given by the fibration structure) and a horizontal subbundle (involving the choice of the connection). Covariant derivatives in a vector bundle are using this idea together with the natural identification of the vertical subbundle of $TE$ with $E$.

share|improve this answer
    
Thank you a lot. I need sometime to read it carefully to do justice for your help. –  Bombyx mori Sep 4 '12 at 15:58
    
I feel there are some confusions at here, the $v$ you had written is really a vector in the fibre of $E_{x}$, and I do not get why you write $v=\sum x^{\alpha}\partial_{\alpha}$. It is an element in $E$! The bundle $\pi^{*}(E)$ should be written as $(u,x,v):u,v\in E_{x},x\in X$. I am reading the rest. –  Bombyx mori Sep 12 '12 at 0:55
    
This now is completely clear to me; thank you for your help. –  Bombyx mori Sep 12 '12 at 1:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.