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On the site Vector Bundle Connection, it gives two definitions of a connection.

One is view a connection as a linear map from a section of $E\otimes TM$ to a section of $E$: $$ D:\Gamma(E\otimes TM)\rightarrow\Gamma(E) $$

I can understand this definition, thinking a connection as a directional directive: $$ v\otimes w\mapsto D_vw $$

However, I cannot understand the other definition, and which is seemingly more common: $$ D:\Gamma(E)\mapsto\Gamma(E\otimes T^*M)=\Gamma(E)\otimes\Omega^1 $$

Can anyone explain to me how such map works? Given a vector in $\Gamma(E)$, what is the image of it?

In addition, in the site above there is an example about the connection in a trivial bundle, saying that

Any connection on the trivial bundle $E=M\times\mathbb{R}^k$is of the form $\nabla s=ds+s\otimes\alpha$ where $\alpha$ is a one-form with value in Hom($E,E$).

However, I don't understand. I think $ds$ is an element in the dual bundle of $E$, but $s\otimes\alpha$ is not, although I cannot even point out to what space $s\otimes\alpha$ belongs, how can they be added?

hxhxhx88
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2 Answers2

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This definition tells you that a connection takes a vector field in $\Gamma(E)$ and produces an $E$-valued one-form. An $E$-valued one form (a section of $\Gamma(E)\otimes \Omega^1$) is a beast that takes in a vector field and produces a section of $E$ in a tensorial fashion (i.e., it's linear in smooth functions).

So in particular, if you have a covariant derivative $v\otimes s\to D_vs$, you should think of this definition as a "differential": $$ s \mapsto Ds: \big( v \mapsto D_vs\big).$$

As to the local coordinate expression, $\alpha$ is a one-form with values in $\operatorname{Hom}(E,E)$. That is, it is an endomorphism-valued one-form: It takes in a vector at a point $p$ and produces a linear map $E_p\to E_p$. Think of $s\otimes\alpha$ as a multiplication operator containing Christoffel symbols in the same way that a covariant derivative can be locally expressed as $Dw = dw + \sum w_k\Gamma^k_{ij}$.

Neal
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  • So, a connection is a map $\Gamma(E)\rightarrow\operatorname{Hom}(\Gamma(TM),\Gamma(E))\cong \Omega^1\otimes \Gamma(E)$, right? – hxhxhx88 Oct 17 '13 at 12:25
  • I'm still confused about the notation $s\otimes\alpha$, is it just a convenient formal notation? Because I'm analysing the space containing it. We have $s\in\Gamma(E)$, and $\alpha$ is a endomorphismed one-form, so $\alpha\in\Omega^1\otimes\operatorname{Hom}(E,E)\cong\Omega^1\otimes E^*\otimes E$, then $s\otimes\alpha$ seems not belongs to $\Omega^1\otimes\Gamma(E)$, right? – hxhxhx88 Oct 17 '13 at 13:41
  • Yeah, something's weird. I think $s\otimes\alpha$ should be the endomorphism field, not $\alpha$ itself. I'll check my notes. – Neal Oct 17 '13 at 13:44
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It's not really correct to think of connections as maps $\Gamma\left(E\otimes TM\right)\to\Gamma\left(E\right)$, because the interplay between the $E$ and $TM$ inputs is not $C^{\infty}\left(M\right)$-multilinear. More precisely, $s\otimes\left(fu\right)=\left(fs\right)\otimes u$, so if the connection factored via the tensor product we should have $D_{fu}s=D_{u}fs$; but $D_{fu}s=fD_{u}s$ is not equal to $D_{u}fs=\left(uf\right)s+f$$D_{u}s$ in general. Thus if you want to "keep the inputs together", you can't go any further than thinking of the connection as a map $\Gamma\left(E\right)\times\Gamma\left(TM\right)\to\Gamma\left(E\right)$.

Now, the other interpretation is indeed correct and common - given a section $s\in\Gamma\left(E\right)$, the derivative of $s$ is a $E$-valued one-form. For a section $s\in\Gamma\left(E\right)$, its image is the map $Ds:\Gamma\left(TM\right)\to\Gamma\left(E\right):u\mapsto D_{u}s$. This map is $C^{\infty}\left(M\right)$-linear, so we can in fact interpret it as an element of $\Gamma\left(E\otimes T^{*}M\right)$.

Anthony Carapetis
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    You help me a lot!! BTW, I am used to write $\Omega^1\otimes \Gamma(E)$ because I'm thinking $\operatorname{Hom}(A,B)\cong A^\otimes B$. Is it the same as $\Gamma(E\otimes T^M)$? – hxhxhx88 Oct 17 '13 at 12:47
  • @hxhxhx88: yes, they are the same. This is a consequence of the fact that all $C^\infty (M)$-linear maps between sections of bundles are induced by homomorphisms of the bundles; that is, $\mathrm{Hom}{C^\infty}(\Gamma(F),\Gamma(E)) = \Gamma(\mathrm{Hom\mathbb R}(F,E))=\Gamma(E \otimes F^*)$. – Anthony Carapetis Oct 17 '13 at 14:20
  • Got it, Thanks! – hxhxhx88 Oct 17 '13 at 14:22