# Connection on a manifold (or a principal bundle on the manifold) from notion of “parallel transport”

Let $$M$$ be a manifold.

A connection on a manifold $$M$$ (connection on vector bundle $$TM\rightarrow M$$) gives, among other things, an isomorphism $$T_aM\rightarrow T_bM$$ for each path $$\gamma$$ in $$M$$ with $$\gamma(0)=a$$ and $$\gamma(1)=b$$.

For some of physics students I know, a connection is precisely some isomorphism $$T_aM\rightarrow T_bM$$ for each path $$\gamma$$ in $$M$$ with $$\gamma(0)=a$$ and $$\gamma(1)=b$$. I am not able to convince/motivate them that this is not the whole data.

One justification they almost got convinced is that, with this data, I can not "pullback" connections even along surjective submersions.

Can some one suggest some way to motivate/convince about the full data that comes with connection on a manifold?

One can look for same in case of principal $$G$$-bundles.

Let $$\omega:P\rightarrow \Lambda^1_{\mathfrak{g}}T^*P$$ be a connection on the Principal $$G$$-bundle. Given a path $$\gamma$$ in $$M$$, we have an isomorphism $$\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}$$.

Suppose we are given a collection of isomorphisms $$\mathcal{C}=\{\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}\}$$ in $$P$$, indexed by paths $$\gamma$$ in $$M$$. Can we then think of some connection whose "parallel transport" is given by $$\mathcal{C}$$? Some obvious restrictions are, these $$\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}$$ are actually $$G$$-equivariant diffeomorphisms. What other restrictions are reasonable (from view of some one seeing for first time) to impose on this collection $$\mathcal{C}$$ to hope for a connection on $$P(M,G)$$.

• Can a connection not be uniquely recovered from this set of isomorphisms? – user7530 Aug 15 at 18:58
• @user7530 May be I am missing something.. Can you clarify how does one uniquely recover a connection? – user537667 Aug 15 at 19:00
• Suppose $\pi:N\rightarrow M$ be a smooth map which is a surjective submersion... Let $\{T_{\gamma(a)}\rightarrow T_{\gamma(b)}|\gamma:[0,1]\rightarrow M\}$... For a curve $\gamma$ in $N$, how does one associate an isomorphism $T_{\gamma(0)}N\rightarrow T_{\gamma(1)}N$? One can take the image $F(\alpha)$ in $M$, then take the isomorphism $T_{F(\alpha)(0)}M\rightarrow T_{F(\alpha)(1)}M$.. But this would not give any obvious choice of isomorphism $T_{\alpha(0)}N\rightarrow T_{\alpha(1)}N$ as $T_{\alpha(0)}N\rightarrow T_{F(\alpha(0))}M$ is only surjective and not an isomorphism.. – user537667 Aug 15 at 20:21
• Also you recover a connection from its parallel transport via $$D_v X = \left.\frac{d}{dt}\right|_{t=0} P_{-\gamma} (X \circ \gamma (t) )$$ where $X$ is a vector field and $\gamma$ a curve with $\gamma'(0) = v$. I guess it boils down to checking how perverse you can be in choosing parallel transport such that this still gives you a connection. – Carlos Esparza Aug 16 at 22:16
• @Piquito what should I do with it? – user537667 Aug 18 at 4:23

It is also clear to me that not every such function is equal to the parallel transport of a connection - at the very least, one would need to start imposing compositional conditions (the isomorphism from $$a$$ to $$b$$ by following $$\gamma$$, composed with the isomorphism from $$b$$ to $$c$$ following $$\gamma'$$, should coincide with the isomorphism from $$a$$ to $$c$$ following the concatenation of $$\gamma$$ and $$\gamma'$$).
Moreover, I believe that once things are explained in this way (existence vs uniqueness), it does not matter if the audience is mathematicians or physicists - the idea is clear and intuitive either way. It is essentially the same as saying: a function from a group $$G$$ to a group $$H$$ is enough to determine a group homomorphism, but not every such function is equal to a group homomorphism.
• Thanks for your answer... My question was how to motivate to fill the structure that is not present in an arbitrary collection of isomorphisms $\{\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}\}$ indexed by paths $\gamma$ in $M$ to give a connection on the principal/vector bundle over $M$... – user537667 Aug 20 at 6:07
• I'm not sure what motivation is needed. Obviously if I parallel transport along a curve, it cannot be any different than if I stop the transport partway then allow it to continue. But that is precisely the same thing as the concatenation rule I wrote. And it is obvious that if you just have arbitrary isomorphisms, they won't satisfy this condition - for the same reason if I write down three arbitrary matrices, the product of two of them will not equal the third in general. The group homom ex. - think of $G$ via path concatenation, $H$ via composition of isomorphisms. – pre-kidney Aug 20 at 6:16