# Relationship Between Connections on a Vector Bundle and a Riemannian Base

I'm starting to get acquainted with how to define affine connections on a vector bundle. Suppose $\pi: E \to M$ is a rank $k$ vector bundle over a Riemannian manifold $M$ with metric $g$, where we will denote $g(u,v)|_p = \langle u,v\rangle_p$ for all $p \in M$ and $u,v \in T_pM$.

I know we can define an affine connection on $E$ as a mapping $\nabla: \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ (where $\Gamma(X)$ denotes the vector space of smooth global sections of a bundle $X$) such that the following conditions hold:

1. Denoting $\nabla(X,s) = \nabla_Xs$ we have lower linearity: $\nabla_{fX+Y}s = f\nabla_Xs + \nabla_Ys$ where $f \in C^\infty(E)$.
2. Derivational property: given $f \in C^\infty(E)$ we have $\nabla_X(fs) = X(f)s + f\nabla_Xs$.

As a reference I'm looking at some informal papers; primarily this and some of this. I've also checked out the book The Geometry of Jet Bundles. So here's my question:

Given that the base of the bundle $M$ is a Riemannian manifold with a unique compatible connection $\nabla^M$, is there a canonical/obvious way to define a metric (and subsequently a connection $\nabla^E$) on the bundle space $E$ such that $\nabla^E$ is an extension of (or in some way compatible with) the connection $\nabla^M$?

Full explanations or reference suggestions are all welcome. Thanks in advance!

• If $E$ is a tensor bundle, then the metric on $TM$ induces one on $E$, otherwise, there is no natural metric on $E$. Furthermore, even after the choice of a metric, there is no natural connection on $E$. – Michael Albanese Mar 6 '16 at 0:47
• @MichaelAlbanese Can you elaborate further? What is inherently special about a tensor bundle that yields a "natural" connection on $E$, as opposed to arbitrary vector bundles? – Mnifldz Mar 8 '16 at 2:18
• A connection $\nabla^E$ on a vector bundle $E$ induces connections on all the tensor bundles. First of all, the induced connection $\nabla^{E^*}$ on $E^*$ satisfies $d(\sigma(e)) = (\nabla^{E^*}\sigma)(e) + \sigma(\nabla^Ee)$ where $\sigma \in \Gamma(E^*)$ and $e \in \Gamma(E)$. In a similar way, one can write down an induced connection on $\bigotimes^pE\otimes\bigotimes^qE^*$. See Lemma $4.6$ of Lee's Riemannian Manifolds: An Introduction to Curvature for more details. – Michael Albanese Mar 10 '16 at 3:16
• I don't know what an acceptable answer to this question should look like, given that the answer is no. Is there something in my comments you want me to expand on? I don't know how to convince you that there isn't a natural metric on $E$. – Michael Albanese Mar 14 '16 at 15:58

An affine connection on $M$ manifold is a connection on $TM$, so no affine connection exists on an arbitrary vector bundle. I'd just call it a connection on the vector bundle $E$.

Now, when we have a fixed metric $g$ on a vector bundle $E\to M$, there is a Riemannian connection $\nabla^E$, which actually depends on $g$.

The space of connections on a vector bundle is actually an affine space, that is, if $\nabla_1, \nabla_2$ are connections on $E$ then $\lambda \nabla_1+(1-\lambda)\nabla_2$ are connections on $E$. This is easy to check: $C^{\infty}(M)$-linearity on one variable, and Leibniz rule on the other, i.e. $\nabla_{fX}(e)=f\nabla_X(e)$, and $\nabla_X(fe)=f\nabla_X(e)+X(f)e$ (and good behaviour w.r.t. the sum of fields $X_1+X_2$ and sections of $E$, $e_1+e_2$, of course).

Likewise, the vector space associated to the affine space of connections on $E$ is precisely $\Gamma(\Omega^1_M \otimes E)$, i.e. the vector space of $E$-valued differential forms of degree one on $M$. This follows from the fact that, if $\nabla_i$ are connections, then $\nabla_2-\nabla_1$ is $C^{\infty}$-linear on both variables, i.e. $\nabla_2-\nabla_1$ is an $E$-valued tensor, as specified above.

So, you need only fix one metric on $E$ to ensure existence of a connection on $E$. Alternatively, you can do it by piecing together connections you make up in different charts, then use partitions of unity (that's how Chern, Chen, Lam do it, if I'm not mistaken). Then all others are obtained by summing $E$-valued differential forms. I prefer piecing together metrics you build on different charts into one metric (which is shown in most textbooks), also through partitions of unity, and then to use uniqueness of the metric connection (the proof for $E$ arbitrary is similar to that shown in Do Carmo, Riemannian Geometry in the case where $E=TM$).

One final remark. There are many, many sections to the bundle $\Omega^1_M\otimes E$. It suffices to choose a trivialising open set $U$, then another contractible, well-chosen open subset $V\subset U$ and then one can build sections that take a prescribed value on $V$ and are $0$ outside $U$ (standard partition-of-unity techniques).

To sum it all up: no, unless there is some extra feature in the geometry of $E$, there is no preferred metric on $E$, but you're actually dealing with an affine space of connections with no preferred point (i.e. no origin of your preference unless some other elements enter the picture). The associated vector space is that of the global sections of $\Omega^1_M\otimes E$, and so once you establish the existence of one connection you may produce all the others by the usual procedure $E\to A$ once you fix a point $p\in A$, i.e. $\overrightarrow{v}\mapsto p+\overrightarrow{v}$, as is the case with every affine space $A$ and associated vector space $E$.

Hope this helps.

• Thank you for your answer, and I'm sorry the bounty was not awarded. I was traveling abroad and couldn't access Stackexchange at that time. – Mnifldz Mar 19 '16 at 3:01
• Never mind. If there is some part that is too terse, just let me know. The book I mention is Chern, Chen, Lam, Lectures on Differential Geometry - the first two chapters contain the relevant information. The language I'm using is just like the one you use - I learnt this from Do Carmo's Riemannian Geometry. – Theon Alexander Mar 21 '16 at 10:17