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Q1) Something about the idea of a connection on the tangent bundle of a manifold confuses me. Naïvely, given an abstract smooth connected manifold $M$ of dimension $m$ we want to "connect" the tangent spaces so that we can define, say, a covariant derivative at a point $p$. If we're being naïve, after we realize we can't directly imitate the covariant derivative on $\mathbb{R}^m$ (whose bases in $T\mathbb{R}^m$ are constant), why don't we simply pull that connection back via the coordinate chart at $p$ and use the transition functions to define it at all charts in an atlas?

Instead, given a chart at $p$, we notice that the connection coefficients ("Christoffel symbols"? though we don't have a metric) determine the connection, and we make a choice of how to define the connection via a choice of these coefficients in the chart. Why bother? If the answer to Q1 is that "we can't, in general" then this all makes sense but leads me to a second question:

Q2) If we embed $M$ into $\mathbb{R}^k$, then we get a connection induced from the embedding (via pullback). Is it the case that a choice of connection coefficients at $p$ is equivalent to choosing some embedding of $M$ and using the induced connection?

J. Dong
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  • What do you mean by pulling back a connection via the embedding? If you mean pulling back the metric and then taking the Levi-Civita connection, the not every connection arises this way (for instance, the Levi-Civita connection is always torsion free). – Pedro Nov 16 '16 at 00:10
  • @Pedro I mean pulling back the affine connection (covariant derivative) in this sense – J. Dong Nov 16 '16 at 05:14
  • This doesn't give a connection on the tangent bundle of $M$, it gives a connection on the pullback of the tangent bundle of $\mathbb{R}^n$, which is very different (for one thing, its rank is different!) – Pedro Nov 16 '16 at 12:31
  • But you can imagine the tangent bundle of $M$ as a subbundle of that, and then you can project onto it, and that does give a connection. But, as remarked by Malik B., these connections will always be torsion free, so not every connection shows up that way. – Pedro Nov 16 '16 at 12:33
  • If your connection is the Levi-Civita connection of a metric, though, then I believe it does show up that way (it should be a consequence of the Nash embedding theorem). – Pedro Nov 16 '16 at 12:33

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Given M as above, you can indeed pullback the flat connection on $\mathbb{R}^m$ via the chart. If $(U,\phi)$ is the chart at $p$, this will produce a flat connection on $U$, i.e., one with connection coefficients all 0. Note that this will only globalize to a flat connection on $M$ if there exists an atlas whose transition functions are constants*. So in general you won't get a flat connection on $M$. Also, this isn't canonical since it depends on what chart you choose.

*You can see this by computing the transformation rule for the connection coefficients.

The naive approach (read: intuitive) is to try and mimic the derivative of vector fields on $\mathbb{R}^n$ using the derivatives of the coordinate functions locally. This shouldn't work as there is no canonical association between any two tangent spaces along a path contained in a chart, but you can "force" it to work formally. This means you've adopted the flat connection on the chart. Of course this won't change covariantly to different coordinates unless you add in the appropriate thing (the connection coefficients times the coordinate functions of the vector field). What you realize is that you should account for the fact that the basis of $T_pM$ for $p\subset U$ isn't constant so you should be using the product rule in your differentiation above. This leads to the notion of the connection coefficients. These essentially determine how you've chosen to "align" your tangent spaces or how you've decided to match up their bases to "connect" them and they clearly don't come for free in any way (this is a definition, after all). Another way of thinking of them is that they measure how the bases are changing locally due to the choice of coordinates and NOT the geometry of the manifold itself.

Of course, if $M$ is a submanifold of $\mathbb{R}^n$, then $T_pM \subset T_p\mathbb{R}^n$ and $\mathbb{R}^n$ flat gives you a free connection on $M$ as you say, but I'm pretty sure these are only torsion free connections which is a negative answer to Q2. In either case, note that it's still not canonical since it depends on how you embed $M$. So ultimately, without additional requirements, like torsion-freeness and compatibility with a metric, there is no natural choice of connection.

SBRJCT
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