I have a two questions regarding this exercise:
- Are there any mistakes in my solution attempt?
- If one speaks of a vector field on the double tangent bundle, but does not specify which bundle structure is meant, does one refer to the standard structure or also the secondary structure or even all other possible bundle structures?
0. Preliminaries: Notation and definitions from the book...
Let $\mathcal M$ and $\mathcal N$ be differentiable manifolds.
Tangent Maps: If $f :\mathcal M \to \mathcal N$ is a smooth map then its differential / tangent map is denoted by $f_* :\mathcal {TM} \to \mathcal {TN}$.
The tangent space $\mathcal T_m \mathcal M$ at some point $m \in \mathcal M$ is defined to be the space of all derivations on the germ $\mathcal F_{\mathcal M}^{\infty}(m)$ of smooth functions which are defined on a neighbourhood of $m$.
By this construction all the tangent spaces attached to different points are disjoint - i.e. $\mathcal T_m \mathcal M \cap \mathcal T_n \mathcal M = \emptyset\; \; \forall m, n \in \mathcal M, \, m \neq n$, and the tangent bundle is then defined as the disjoint union $\mathcal{TM} = \bigcup_{m \in \mathcal M} \mathcal T_m \mathcal M$.
The canonical projection associated to $\mathcal {TM}$ is defined as the map $\pi: \mathcal{TM} \to \mathcal M, v \in \mathcal T_m \mathcal M \mapsto m$.
A vector field $X$ on $\mathcal M$ is a mapping $m \mapsto X(m) \in \mathcal T_m \mathcal M$.
From these definitions the following conclusions can be drawn:
- For $m \in \mathcal M$, $\mathcal T_m \mathcal M = \pi^{-1}(\{m\}).$
- $X$ vector field on $\mathcal M$ $\Leftrightarrow \; \pi \circ X = \mathrm{id}_{\mathcal M}$.
1. The problem:
Given is a smooth $d$-dimensional manifold $\mathcal M$ and on it a smooth vector-field $X$ with flow $\mu_t$.
On pp. 280-282 in Bishop's book the canonical lift of $X$ is introduced as a vector-field $W$ on $\mathcal T^*\mathcal M$ which is related to $X$ in a special sense and whose flow on $\mathcal T^*\mathcal M$ is given by $\mu_{-t}^*$. Then the following problem is given (in these precise words):
Problem 6.3.3: We can define the canonical lift of $X$ to $\mathcal{TM}$ as the vector field whose flow is the dual flow $\{\mu_{t *}\}$. Show that the canonical lift to $\mathcal{TM}$ is $X_* : \mathcal{TM} \to \mathcal{TTM}$.
2. My solution attempt:
Since $\mu_t$ is the flow of $X$ I know that $$\left.\frac{\mathrm d}{\mathrm d t}\right|_{t=0} \mu_t = X.$$ And since $[v, \frac{\mathrm d}{\mathrm d t}]=0 \; \forall v \in \mathcal {TM}$ it follows that $$\left.\frac{\mathrm d}{\mathrm d t}\right|_{t=0} \mu_{t*} = \left(\left.\frac{\mathrm d}{\mathrm d t}\right|_{t=0} \mu_t\right)_* = X_*.$$
So if $X_*$ were a vector-field on $\mathcal {TM}$, $\mu_{t*}$ would be its flow. And while trying to show the former I got more and more confused about...
- ...what it actually means for an object to be a vector-field on $\mathcal {TM}$.
- ...what the fibres of $\mathcal {TTM}$ actually look like.
- ...whether based on the definitions given in bishops book, problem 6.3.3 is just posed incorrectly (which I believe to be the case).
I might be mistaken, but as far as I know there is only one sensible bundle structure / projection map for ($\mathcal{TM}, \mathcal M)$, namely $\pi$ as defined above. But for $(\mathcal {TTM}, \mathcal {TM})$ there seem to be infinitely many. Two of these I would call natural bundle structures on $\mathcal {TM}$ because they seem to appear in a natural way, and these are the ones I have to deal with here.
The standard tangent bundle structure on $\mathcal {TM}$ (as I will call it here) is obtained by just repeating the construction of $\mathcal {TM}$ above with the substitution $\mathcal M \to \mathcal {TM}$. Hence for any $v \in \mathcal {TM}$ the fiber $\mathcal T_v \mathcal {TM}$ is now the space of derivations of the germ $\mathcal F^{\infty}_{\mathcal {TM}}(v)$, then one has another disjoint union $\mathcal {TTM} = \bigcup_{v \in \mathcal {TM}} \mathcal T_v \mathcal {TM}$, and finally the canonical projection is then given as $$ p: \left\{ \begin{align} \mathcal {TTM} &\to \mathcal{TM}\\ w \in \mathcal T_v \mathcal {TM} &\mapsto v \end{align} \right., $$ which behaves in close analogy to $\pi$ from above. And of course a mapping $Y: \mathcal {TM} \to \mathcal {TTM}$ is a vector-field on $\mathcal {TM}$ iff $p \circ Y = \mathrm{id}_{\mathcal {TM}}$. I'm going to call such a vector-field a $p$-vector field.
The way the concepts of tangent spaces, vector fields were introduced in the book and by the manner the problem was posed I was expecting that $X_*$ should be a $p$-vector-field. But this seems not to be the case:
Everything what follows has to be understood wrt. the local coordinate chart $\mu : U \subset \mathcal M \to \mathbb R^d, m \mapsto (x^i(m))$ with associated coordinate basis $\{\partial_{x^i}\}$.
Local coordinates on $\mathcal TU = \pi^{-1}(U) \subset \mathcal {TM}$ are then given by the coordinate chart $\nu: \mathcal TU \to \mathbb R^{2d}, v \mapsto ((y^i(v)), (z^i(v)))$, where $\{y^i = x^i \circ \pi\}, \{z^i = \mathrm d x^i\}$. The associated coordinate basis is then $\{\partial_{y^i}\} \cup \{\partial_{z^i}\}$.
And hence on $\mathcal {TT}U = p^{-1}(\pi^{-1}(U))$ a naturally given chart is $\kappa: \mathcal {TT}U \to \mathbb R^{4d}, w \mapsto ((y^i \circ p(w)), (z^i \circ p(w)), (a^i(w)),(b^i(w))),$ where $\{a^i = \mathrm d y^i \}, \{b^i = \mathrm d z^i\}$.
If $w \in \mathcal {TT} U$ with coordinates $\kappa(w) = ((m^i), (v^i),(a^i),(b^i))=:\mathbf{w}$, and $w \in \mathcal T_v \mathcal {TM}$, $v \in \mathcal T_m \mathcal M$, then $p(w) = v, \pi(v) = m, w = \kappa^{-1}(\mathbf{w}) = a^i \partial_{y^i}|_{v} + b^i \partial_{z^i}|_{v}, v = v^i \partial_{x^i}|_m, m = \mu^{-1}((m^i))$.
Doing some calculating I get $$ X_*: \left\{ \begin{align} \mathcal{TM} &\to \mathcal{TTM}\\ v &\mapsto v^i \partial_{y^i}|_{X \circ \pi(v)} + \partial_{x^j} X^i(\pi(v)) v^j \partial_{z^i}|_{X \circ \pi(v)} \end{align} \right., $$ whereas $p$ can be expressed as $$ p: \left\{ \begin{align} \mathcal {TTM} &\to \mathcal{TM}\\ a^i \partial_{y^i}|_{v} + b^i \partial_{z^i}|_{v} \in \mathcal T_v \mathcal{TM} & \mapsto v^i \partial_{x^i}|_{\pi(v)} \end{align} \right.. $$ Therefore $p \circ X_* = X \circ \pi \neq \mathrm{id}_{\mathcal{TM}}$, which means that $X_*$ is not a $p$-vector-field.
But using the fact that $X$ is a vector field on $\mathcal {M}$, i.e. $\pi \circ X = \mathrm{id}_{\mathcal M}$, and by applying the chain rule I get $$\pi_* \circ X_* = (\pi \circ X)_* = (\mathrm{id}_{\mathcal M})_* = \mathrm{id}_{\mathcal {TM}},$$ which means that $X_*$ is a $\pi_*$-vector field, i.e. a vector field wrt. the so called secondary bundle structure, given by $\pi_*: \mathcal {TTM} \to \mathcal{TM}$.
Remark on Notation: What I have denoted by $p$ and $\pi$ is called $\pi_{\mathcal{TTM}}$ and $\pi_{\mathcal{TM}}$ on Wikipedia.
Also expressed wrt. the local basis from above this projection has the form $$ \pi_*: \left\{ \begin{align} \mathcal {TTM} &\to \mathcal{TM}\\ a^i \partial_{y^i}|_{v} + b^i \partial_{z^i}|_{v} \in \mathcal T_v \mathcal{TM} & \mapsto a^i \partial_{x^i}|_{\pi(v)} \end{align} \right., $$
Hence for $v \in \mathcal {TM}$ one has $p^{-1}(\{v\}) = \mathcal T_v \mathcal{TM}$, but $\pi_*^{-1}(\{v\})$ is a different fiber.
To make comparison of the two different vector-bundle-structures easier I simplify the notation further by using boldface symbols for the $d$-dimensional sub-tuples of the coordinates, such that $\mathbf{w} = ((m^i), (v^i),(a^i),(b^i)) = (\mathbf m, \mathbf v,\mathbf a, \mathbf b).$
The coordinate expressions for the vector-fields $X, X_*$ then are $$ \begin{align} &\nu \circ X \circ \mu^{-1}(\mathbf m) &&= (\mathbf m, \mathbf X(\mathbf m)),\\ &\kappa \circ X_* \circ \nu^{-1}(\mathbf m, \mathbf v) &&= (\mathbf m, \mathbf X(\mathbf m), \mathbf v, \mathrm d \mathbf X(\mathbf m) \mathbf v) \end{align} $$ where $\mathbf X(\mathbf m) = (\mathrm d x^i(X \circ \mu^{-1}(\mathbf m))), \; \mathrm d \mathbf X(\mathbf m) = \mathrm d(\mathrm d x^i(X \circ \mu^{-1}))(\mathbf m),$ and those for the projections $\pi, p, \pi_*$ are $$ \begin{align} &\mu \circ \pi \circ \nu^{-1}(\mathbf m, \mathbf v) &&= \mathbf m,\\ &\nu \circ p \circ \kappa^{-1}(\mathbf m, \mathbf v, \mathbf a, \mathbf b) &&= (\mathbf m, \mathbf v),\\ &\nu \circ \pi_* \circ \kappa^{-1}(\mathbf m, \mathbf v, \mathbf a, \mathbf b) &&= (\mathbf m, \mathbf a). \end{align} $$
Excursus: Examples of different vector-field types on $\mathcal{TM}$
$p$-vector-field:
With respect to the previously chosen local coordinate bases for $\mathcal{TM}$ and $\mathcal{TTM}$ the vertical lift of $X$ can be expressed as $$ \mathrm{ver} X : \left\{ \begin{align} \mathcal{TM} &\to \mathcal{TTM}\\ v \in \mathcal{T}_m\mathcal{M} &\mapsto X^i(m) \partial_{z^i}|_{v} \end{align} \right., $$ which clearly satisfies $p \circ {\mathrm{ver} X} = \mathrm{id}_{\mathcal{TM}}$ and $\pi_* ({\mathrm{ver} X}) = (m \mapsto 0 \in \mathcal T_m \mathcal M)$. Hence $\mathrm{ver} X$ is a $p$- but not a $\pi_*$-vector-field.
Its coordinate expression is $\kappa \circ {\mathrm{ver} X} \circ \nu^{-1}(\mathbf m, \mathbf v) = (\mathbf m, \mathbf v, \mathbf 0, \mathbf X(\mathbf m)).$
$p$-, $\pi_*$-vector-field
A vector field $Y$ on $\mathcal {TM}$ which expresses a 2nd-order differential equation on $\mathcal M$ has the form: $$ Y : \left\{ \begin{align} \mathcal{TM} &\to \mathcal{TTM}\\ v &\mapsto v^i \partial_{y^i}|_{v} + Y^i(v) \partial_{z^i}|_{v}, \end{align} \right., $$ and therefore satisfies $p \circ Y = \mathrm{id}_{\mathcal{TM}} = \pi_* \circ Y$.
Its coordinate expression is $\kappa \circ Y \circ \nu^{-1}(\mathbf m, \mathbf v) = (\mathbf m, \mathbf v, \mathbf v, \mathbf Y(\mathbf m, \mathbf v)).$
Now I believe that what was called 'vector field' on $\mathcal {TM}$ in problem 6.3.3 is to be understood as $p$-vector-field. So in order to obtain such a field I would have to employ the canonical flip $j : \mathcal {TTM} \to \mathcal {TTM}$, which provides an involute isomorphism between $\pi_*$- and $p$-vector-fields on $\mathcal {TM}$ and then $j \circ X_*$ would be such a field. But by doing this I would get a flow different from $\mu_{t*}$. So either I made a mistake while deriving $X_*$, or within the context of this problem the term 'vector-field' was employed in a loose way which allows for all vector-field types to be solutions, or the problem is ill-posed.
Any thoughts on this? Did I make a mistake somewhere? Is my reasoning sound?
I would be happy for any comments on this. Thanks in advance.
Addendum:
Since I was not completely sure whether my derivation of $X_*$ as the vector-field with flow $\mu_{t*}$ was sound, I tried here a different approach:
Let $\mathcal M$, $\pi$, $X$ and $\mu_t$ be defined as above. For $m \in \mathcal M$ let $\mathcal{F}_{\mathcal M}^{\infty}(m)$ denote the germ of smooth functions which are defined on a neighbourhood around $m$. If $f \in \mathcal{F}_{\mathcal M}^{\infty}(m)$ then $\mathrm d f \in \mathcal{F}_{\mathcal TM}^{\infty}(v)$ for any $v \in \mathcal T_m \mathcal{M}$.
Let $v \in \mathcal{T}_m\mathcal{M}$. By using the definitions of the action of a vector-field on a function, the tangent map, the flow, and the fact that $[v, \partial_t]=0$ and $\partial_t f = 0$ I get
$$ \begin{align} (\dot{\mu}_{0*} v) \mathrm d f& = \partial_t|_{t=0}(\mathrm d f \circ \mu_{t*} v)\\ & = \partial_t|_{t=0}((\mu_{t*} v) f)\\ & = \partial_t|_{t=0} v (f \circ \mu_t)\\ & = v \partial_t|_{t=0} (f \circ \mu_t)\\ & = v \dot{\mu}_0 f\\ & = v X f\\ & = v (\mathrm d f \circ X)\\ & = (X_* v) d f, \end{align} $$ hence $$ (\dot{\mu}_{0*} v) \mathrm d f = (X_* v) d f \quad \forall v \in \mathcal T_m \mathcal M, \; \forall f \in \mathcal F_{\mathcal M}^{\infty}(\pi(v)),$$ which just means that $$\dot{\mu}_{0*} = X_*.$$