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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?

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    @SamLisi: I see. Could you give me a short deduction of the local form? This is what I am mostly confused.2012-09-03

1 Answers 1

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$\newcommand{\C}{\mathbb{C}}$ 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 $E$.

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    This now is completely clear to me; thank you for your help.2012-09-12