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The usual definition of an affine connection $\nabla_X Y$ requires $X,Y$ to be smooth vector fields on the ambiant differential manifold $M$, i.e. $C^\infty$. However at any point $p\in M$, $(\nabla_X Y)_p$ means the differential of $Y$ at $p$ in the direction $X_p$. So it seems enough that $Y$ is just differentiable, and we don't need any hypothesis on $X$, not even continuity.

Likewise, geodesics are required to be smooth curves, but their definition $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ suggests that they are only $C^2$ (their acceleration is zero).

And regarding the parallel transport of a tangent vector $u\in T_p M$ along a curve $\gamma$, it seems that $\gamma$ just has to be piecewise differentiable : compose the parallel transports on each segment where $\gamma$ is differentiable. The speed $\dot{\gamma}$ does not need to exist at the junction points.

Are there generalized definitions of connections for these non smooth cases ?

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    $\begingroup$ One can define these things, of course, but n.b. some of the objects we construction with connections requires some degree of regularity. For example, to define the torsion $T(X, Y) := \nabla_X Y - \nabla_Y X - [X, Y]$ of an affine connection $\nabla$ we need $X$ and $Y$ both to be $C^1$. The computation of curvature is similar. More generally most of the definitions in differential geometry work just fine if we lower the requisite regularity of objects to $C^k$ for some $k$, but in general any computation that uses derivatives (hence many proofs) requires some minimum value of $k$ to work. $\endgroup$ Jul 25, 2023 at 21:17
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    $\begingroup$ For most purposes it's reasonable to learn the theory assuming that all objects are $C^\infty$ and then go back to a proof and verify that it still works when you absolutely need the result for some weaker class $C^k$. $\endgroup$ Jul 25, 2023 at 21:19

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One can define affine connections like you write, of course, but in practice we often need objects derived from affine connections, including tensorial ones like curvature and torsion, to have at least a certain degree of regularity, and so we typically need the ingredients themselves to have some regularity (roughly speaking, $1$ additional derivative for each covariant derivative used to define them).

More generally most of the definitions in differential geometry work just fine if we lower the requisite regularity of objects to $C^k$ for some $k$, but in general any computation that uses derivatives (hence many proofs) requires some minimum value of $k$ to work. For most purposes it's reasonable to learn the theory assuming that all objects are $C^\infty$ and then go back to a proof and verify that it still works when you need the result for some weaker regularity class, e.g., $C^k$.

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    $\begingroup$ +1 for last paragraph, but I don’t really think the torsion example is good (or atleast that’s not exactly how I’d phrase the matter). Sure, that formula may require the $X,Y$ to have some regularity, but ultimately, torsion is a tensor field, so it can be evaluated ‘pointwise’, and we should instead be asking how smooth is the torsion tensor field. So I guess my point is that rather than looking at the regularity needed for a particular intermediate formula to make sense, it’s better to look at the regularity of the final resulting object. $\endgroup$
    – peek-a-boo
    Jul 26, 2023 at 3:06
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    $\begingroup$ For example, if $M$ is a $C^k$ manifold, then $TM$ is a $C^{k-1}$ manifold, and we can define $C^r$ connections on $TM$ with $0\leq r\leq k-2$. We can then define covariant derivative $\nabla_{\xi_p}Y\in T_pM$ for each $Y$ which is once differentiable at $p$. One can then define its torsion as a $C^r$ tensor field of type $(1,2)$, $T:TM\oplus TM\to TM$. The curvature (defined if $r\geq 1$) is then a $C^{r-1}$ tensor field of type $(1,3)$, $R:TM^{\oplus 3}\to TM$. These tensor fields can then be evaluated pointwise on arbitrarily rough vector fields to produce suitably rough vector fields. $\endgroup$
    – peek-a-boo
    Jul 26, 2023 at 3:07
  • $\begingroup$ @peek-a-boo You're right, I can see how that discussion could be misleading. I've adjusted my answer to prevent (I hope) a reader running into that issue. $\endgroup$ Jul 26, 2023 at 4:33

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