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Given a generic smooth function $f:M\to N$, we know that its differential is a smooth function $\mathrm df:TM\to TN$ such that $$\mathrm df(p,[\gamma'(0)])\equiv (f(p),\underbrace{\big[\partial_t\big|_0 f(\gamma(t))\big]}_{\in \eta\circ T_{f(p)}N}),$$ for any path $\gamma:I\to M$ with $\gamma(0)=p$ and $\gamma\in[\gamma'(0)]$, and denoting with $[\gamma'(0)]$ the equivalence class of paths identified by their first derivative in some chart, and similarly denoting with $\partial_t\big|_0 [f(\gamma(t))]$ the corresponding equivalence class of paths $I\to N$. I'm also denoting with $\eta$ the map sending an element in $TN$ into its second component, to clarify that the second element in the expression above is an equivalence class of curves (the notation is from Tao's notes I believe).

On the other hand, a vector field is also a smooth function $X:M\to TM$, and it should therefore make sense to talk about its differential, which is then a map $\mathrm dX:TM\to TTM$, with $TTM$ tangent bundle of the tangent bundle of $M$. If I try to unravel the same definition used above in this case I get a bit lost in the notation, however. In particular, we should have $$\mathrm dX(p,[\gamma'(0)]) = \big(\underbrace{X(p)}_{\in T_pM}, \underbrace{\big[\partial_t\big|_0X(\gamma(t))\big]}_{\in \eta\circ (T_{X(p)}TM)} \big).$$ So far, so good. But then we also know that $X(p)=(p,[\partial_t\big|_0\Phi_X^t(p)])$, where $t\mapsto \Phi_X^t(p)$ is a curve representing the tangent curve corresponding to $X(p)$ (there might be a better notation for this, I'm not sure). Using this, I'd get $$\mathrm dX(p,[\gamma'(0)]) = \big( \underbrace{(p,\,\,[\partial_t|_0 \Phi_X^t(p)])}_{\in T_pM}, \,\, \big[\partial_t\big|_0\big\{\underbrace{(\gamma(t),\,\,[\partial_s|_0 \Phi_X^s(\gamma(t))])}_{X(\gamma(t))}\big\}\big] \big).$$ On the RHS we are now dealing with equivalence classes of paths in $TM$. My question is, is there a way to simplify this expression to have just a tuple of four numbers? Naively, I would be tempted to just rewrite this as $$\mathrm dX(p,[\gamma'(0)]) = \big( p,\,\, [\partial_t|_0 \Phi_X^t(p)],\,\, [\gamma'(0)],\,\, \Big[\partial_t|_0\big[\partial_s|_0 \Phi_X^s(\gamma(t))\big]\Big] \big),$$ but I'm not sure whether this is legit, as I'm pretending that I can simply add pointwise the components of the tangent bundle. I'm sort of taking a curve $\tilde\gamma:I\to TM$ and writing it as $\tilde\gamma(t)=(\gamma_1(t),\gamma_2(t))$ for some pair of curves $\gamma_1:I\to M$ and $\gamma_2:I\to\eta\circ TM$. Locally, we can do this via the trivialisation of the bundle, but is it legit to write this sort of expression more in general? Furthermore, is there a way to further rewrite the last bit of this expression, the one with the multiple derivatives, more explicitly?

Addendum:

Perhaps a more expressive, if less standard, notation for the above equations would be as follows. Given any curve $\gamma:I\to M$, define the map $D_t$ as sending smooth curves defined at $t\in\mathbb R$ to the corresponding equivalence class of curves defined by their slope at $t$, so that $D_t[\gamma]\subset\mathrm{Curves}(M)$ for any $\gamma\in\mathrm{Curves}(M)$. With this, we can write for any $f:M\to N$, $$\mathrm df(p,D_0[\gamma]) = (f(p), D_0[f\circ\gamma]).$$ For vector fields, we then have $$X(p)=(p,D_0[\Phi_X(p,\cdot)]), \qquad X(\gamma(t)) = (\gamma(t), D_0[\Phi_X(\gamma(t),\cdot)]), $$ and thus $$\mathrm dX(p,D_0[\gamma]) = ( X(p), D_0[X\circ\gamma] ) \\ = ( p, D_0[\Phi_X(p,\cdot)], \,\, D_{t=0}\big[(\gamma(t), D_{s=0}[\Phi_X(\gamma(t),s)])\big] \,\, ),$$ where I also had to write $D_{t=0}$ and $D_{s=0}$ to remark which curve/functional relation ought to be used as input to the map $D_0$.

glS
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  • I later found this answer to be quite related: https://math.stackexchange.com/a/3748861/173147 – glS Dec 14 '21 at 14:23

1 Answers1

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It's unclear exactly what you're asking, but here are a few points to consider:

The tangent space of a manifold and the differential of a smooth map can be defined in a number of ways. Equivalence classes of curves is one which can be useful in some situations, but it tends to be cumbersome, because the vector space structure of tangent spaces is highly obfuscated, and choosing representatives introduces superfluous information. If you're looking for an alternative "intrinsic" definition, derivations are the standard choice.

If you want to "explicitly write down" the differential of a vector field $X$ on an $n$-manifold $M$, you can choose local coordinates on $M$ (and the induced coordinates on $TM$), and $dX$ will corresponds to the derivative of $X$ in the multivariable calculus sense, which is just a set of $n+n^2$ functions (the components of $X$ and all of their partial derivatives). When changing coordinates, however, these functions will not transform in a striaghtforward way, so if you with to formulate an intrinsic object which captures this information, there are a few options:

One option is to simply define a new object which obeys this transformation law. These are known as jets. The differential $dX$ is then equivalent to the jet $j^1(X)$, which is a section of the jet bundle $J^1(\pi_{TM})$. In fact, jects are closely related to the definition of the tangent space as equivalence classes of curves, and they are correspondingly difficult to work with.

Alternately, one can endow $TM$ with an extra piece of structure which allows us to "split" $dX$ into a 0th order part and a 1st order part in an intrinsic way. These are known as connections, and they can be formulated in a variety of ways.

Kajelad
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  • thank you. I'm aware of the definition of tangent vectors via derivations, but I was trying to purposely work out how things work with curves, to better compare the two formalism. The question actually originates from thinking about the commutator of vector fields, which is easy via derivations, but if I try to work it out via curves I end up having to deal with differentials of vector fields, hence the question. I suppose my main question here is whether it makes sense to write the last equation, "splitting" a curve in $TM$ in two curves, or whether there is a better/more correct way – glS Dec 14 '21 at 09:19
  • about working in local coordinates: the resulting expression would be $\mathrm dX(x,v)=(x,X^i e_i,v, v^i\partial_i (X^j e_j))$, with the basis vectors $e_i$ constants, so that the last component reads $v^i (\partial_i X^j)e_j$, correct? Also, apologies for the too many follow-ups, but how do connections allow the "splitting" you are referring to? I'm aware of connections on a manifold to define parallel transport etc, but I don't quite see how they enter the discussion here – glS Dec 14 '21 at 09:31
  • @glS I don't know of a convenient way of defining the Lie bracket in terms of curves, but I doubt it would involve the differential of vector fields. A prescription for "splitting" curves in the way you describe is essentially equivalent to a local trivialization, but there is no canonical choice of trivialization, and different choices will not agree on the "vertical" and "horizontal" part of the derivative of a curve. – Kajelad Dec 15 '21 at 00:02
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    As for connections, the basic idea is that $TTM$ has a canonical vertical subbundle $VTM=\ker(d\pi)$, but in order to decompose elements of $TTM$ into vertical and horizontal parts, we need a complementary horizontal subbundle $HTM\subset TTM$. There is no canonical choice of horizontal subbundle, so instead we can add one as an extra piece of structure on our bundle. This is called an Ehresmann connection, and it is in some sense the most general notion of "connection", though the correspondences can be rather involved. – Kajelad Dec 15 '21 at 00:09
  • what I meant about the differential of vector fields from the Lie bracket is that if we define vector fields as derivations $X:C^\infty(M)\to C^\infty(M)$, with $X(f)\equiv\mathrm df(X)$ for $f\in C^\infty(M,\mathbb R)$, then $[X,Y]\equiv X\circ Y-Y\circ X$, and $X(Y(f))=\mathrm d(Y(f))(X)$, which is where the differential comes out (unless I'm getting something wrong here). Although admittedly this is the differential of $Y(f):M\to M$, so it might not really be that connected to the question at hand – glS Dec 15 '21 at 00:12
  • That's all correct, but note that $Y(f)\in C^\infty M$, so $d(Y(f))$ is the differential of a function, not a vector field. The basic issue is that the $X\circ Y$ is a second order differential operator rather than a vector field, and such an object is rather difficult to define using the curve model of tangent spaces. – Kajelad Dec 15 '21 at 00:21