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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 $P(M,G)$.

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  • $\begingroup$ Can a connection not be uniquely recovered from this set of isomorphisms? $\endgroup$ – user7530 Aug 15 at 18:58
  • $\begingroup$ @user7530 May be I am missing something.. Can you clarify how does one uniquely recover a connection? $\endgroup$ – user537667 Aug 15 at 19:00
  • $\begingroup$ 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.. $\endgroup$ – user537667 Aug 15 at 20:21
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    $\begingroup$ 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. $\endgroup$ – Carlos Esparza Aug 16 at 22:16
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    $\begingroup$ @Piquito what should I do with it? $\endgroup$ – user537667 Aug 18 at 4:23
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I find the wording of your question sufficiently ambiguous that it could lead to confusion already among mathematicians, without even going so far as to interfacing with physicists!

It is clear to me that any connection is determined by its parallel transport: in other words, for every function which assigns to each (smooth) path a (smoothly varying) isomorphism from the initial to the terminal tangent spaces, there exists at most one connection having this function as its parallel transport. (Indeed, the connection - if it exists - can be determined by differentiating this smoothly varying isomorphism, in some sense.)

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 $\gamma'$).

When you say the word "precisely", I think you mean what I wrote in the last paragraph: that just having a function from paths to isomorphisms does not necessarily mean that it can be written as the parallel transport of some connection. But then later when you say "this is not the whole data", it is misleading - since (as explained in my second paragraph) there is sufficient data to recover the connection if it exists - and if it does not exist, there is sufficient data to prove that it does not exist.

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.

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  • $\begingroup$ 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 $M$... $\endgroup$ – user537667 Aug 20 at 6:07
  • $\begingroup$ 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. $\endgroup$ – pre-kidney Aug 20 at 6:16

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