Pushforward, tangent map, vertical endomorphism and all that

Question

I've already posted a similar question a couple of times, here and here, without receiving a fully clarifying answer, so I post it again, trying to be more specific.

Suppose you have a smooth surjective submersion between smooth manifolds: $f: N \rightarrow M \equiv f(N)$, with $\dim N >\dim M$. The tangent map $Tf: TN \rightarrow TM$ is defined pointwise in terms of $f_\ast $, the pushforward by $f$. In local coordinates, if $(x, v)$ is a point in $TN$, corresponding to the tangent vector $ v = \left. v^a \frac{\partial}{\partial x^a }\right|_x \in T_x N$, then

$$(Tf)(x,v) = (y, w) := (f(x),f_{\ast,x}v), $$ where $w := f_{\ast,x}v \in T_{f(x)}M$ is such that $(f_{\ast,x}v)g = v (g \circ f)$ for any smooth $g: M \rightarrow \mathbb{R}$. Of course, since $f$ is surjective but not injective, if $f(x)=f(x') =y$, then the images of the two points $(x,v)$ and $(x', v')$ correspond to two generally different tangent vectors to $M$ at the same point $y$. Thus, $Tf$ cannot map the tangent field in $X \in \mathcal X(N)$ to a tangent field in $Y \in \mathcal X(M)$, unless $f_{\ast,x}X_x = Y_{y}$ for any $x$ in the inverse image of $y$, i.e. $\forall x \in f^{-1}(y)$ $(Tf)(x, X(x)) = (y, Y)$. In this case, $Y = f_\ast X$ and the two fields are said to be $f$-related.

Now, let us specialize these considerations to the case of the tangent bundle $\pi_M: TM \rightarrow M$, that is let us pose $N \equiv TM$, $ f \equiv \pi_M$, with $\pi_M$ the bundle projection, which is indeed a surjective submersion. Obviously, assuming $\dim M = n$, we have $\dim TM = 2n$ and, if $x$ identifies a point in $M$ in local coordinates, a point $\xi \in TM$ can be locally parametrized as $\xi = (x, u)$, where $u \in T_x M$. The tangent map $T\pi_M: TTM \rightarrow TM$ maps a generic point $\Xi = (\xi, \eta) \equiv ((x,u),(v,w)) \in TTM$ to the point $\xi' = (x', u') = (\pi_M(\xi), \pi_{M*,\xi} \eta) = (x, \pi_{M*,(x,u)}(v,w)) \in TM$, where the tangent vector $\eta \in T_\xi TM$, $\eta = \eta^A \left.\frac{\partial}{\partial \xi^A}\right|_\xi $, can also be conveniently expressed, in terms of coordinates adapted to the local trivialization of $TTM$ as $U \subseteq M \times \mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}^n$, in the form:

$$ \eta = v^a\left.\frac{\partial}{\partial x^a}\right\vert_{(x,u)} + w^a \left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}. $$ The second term of $\xi'$, $u' = \pi_{M*,\xi} \eta$, can be determined by requiring that, for any function $g \in \mathcal F(M)$, $(\pi_{M*,\xi}\eta) g = \eta(\pi_M^\ast g) \equiv \eta(g \circ \pi_M)$. We note that $g \circ \pi_M$ is a generic function in $\mathcal F(TM)$ that is "constant along the fiber", i.e. depends only on $x$ and not on $u$: $(g \circ \pi_M)(x,u) = g(x)$. So, $u' = \pi_{M*,\xi}\eta \equiv v = v^a \left.\frac{\partial}{\partial x^a}\right\vert_x$ and the action of the tangent map can be written as $(T\pi_M)(x,u,v,w) = (x,v)$. This is to be contrasted with the action of the bundle projection map $\pi_{TM}$ that enters into the definition of the (second) tangent bundle $\pi_{TM}: TTM \rightarrow TM$. Indeed, $\pi_{TM}(x,u,v,w) = \pi_{TM}(\xi,\eta) = \xi = (x,u)$. As a consequence, $\pi_M \circ \pi_{TM} = \pi_M \circ T\pi_M: TTM \rightarrow M$.

After this very long and tedious introduction, let me explain what my concern is. The latter is related to the possibility to map vector fields in $TTM$ to vector fields in $TM$ via $T\pi_M$. The point is the following: since locally we can write $x = \pi_M(\xi)$, that is $x^a = \xi^a$, $a=1, \dots, n$, the Jacobian of this transformation is $\pi_{M*,\xi} = \frac{\partial x^a}{\partial \xi^A}(\xi) \equiv (\delta^a_b, 0^a_b)$, $a,b = 1, \dots, n$, $A=1, \dots, 2n$. This is a constant $n \times 2n$ matrix of rank $n$, consisting of the unit $n \times n$ matrix followed by $n \times n$ zeroes. By applying this Jacobian to a generic vector field $X \in \mathcal(TTM)$, which can be written locally as:

$$ X = v^a(x,u)\left.\frac{\partial}{\partial x^a}\right\vert_{(x,u)} + w^a (x,u)\left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}, $$

I get $v^a(x,u)\left.\frac{\partial}{\partial x^a}\right\vert_{x}$, which is $not$ a well-defined vector field on $M$, unless $v^a(x, u) \equiv v^a(x)$. In other words, it seems to me that $X$ is $\pi_M$-related to $Y = v^a(x)\left.\frac{\partial}{\partial x^a}\right\vert_{x}$ iff $ X = v^a(x)\left.\frac{\partial}{\partial x^a}\right\vert_{(x,u)} + w^a (x,u)\left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}.$

In summary, it looks like we can project onto $M$ via $T\pi_M$ only a well-specified subset of all possible vector fields on $TM$. Nonetheless, in the study of Lagrangian dynamical systems, one introduces the so-called vertical endomorphism $S$. This is a rank-2 mixed vector field on $TM$, $S \in \mathcal T_1^1(TM)$, defined pointwise as $S = \mathrm{vl} \circ T\pi_M$, where $\mathrm{vl}$ is the vertical lift, which maps vectors in $M$ to vertical vectors in $TM$. In local coordinates: $$ \mathrm{vl}\left(v^a\left.\frac{\partial}{\partial x^a}\right\vert_{x}\right) = v^a\left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}. $$ Without entering into much detail, the vertical endomorphism tensor field can be written locally as $S = dx^a \otimes \frac{\partial}{\partial u^a}$ and maps tensor fields on $TM$ into vertical tensor fields on $TM$, that is, as I understand it, the generic field $$ X = v^a(x,u)\left.\frac{\partial}{\partial x^a}\right\vert_{(x,u)} + w^a (x,u)\left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}, $$ is mapped by $S$ into the corresponding vertical field $$ X^{\mathrm{v}} = v^a(x,u)\left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}. $$

This result seems to imply that one first, by means of $T\pi_M$, projects $X$ down to $v^a(x,u)\left.\frac{\partial}{\partial x^a}\right\vert_{x}$, whatever it is, and then lifts this intermediate result up to $X^{\mathrm{v}}$. That this is the case is demonstrated by the following well-known result in Lagrangian analytical mechanics:

$$ S(\Gamma) = \Delta. $$ Here $\Delta \in \mathcal{X}(TM)$ is the Liouville or dilation field, expressed locally as $\Delta = u^a \left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}$ and $\Gamma \in \mathcal{X}(TM)$ is any so-called SODE (second-order differential equation) field, written in local coordinates as $\Gamma = u^a\left.\frac{\partial}{\partial x^a}\right\vert_{(x,u)} + \Gamma^a (x,u)\left.\frac{\partial}{\partial u^a}\right\vert_{(x,u)}$. For SODE fields the components multiplying $\left.\frac{\partial}{\partial x^a}\right\vert_{(x,u)}$ are not only dependent on $u$, but do as a matter of fact coincide with $u$. So, the action of $S$, and hence of $T\pi_M$, on unrestricted, generic vector fields $X \in \mathcal{X}(TM)$ is perfectly legitimate.

I don't understand how this is possible. Any insight is welcome.

UPDATE

Since I received no answers so far, I'll try to elaborate a little bit, so as to stimulate a reply.

First of all, IMHO it is certain that in general the map $T\pi_M$, that is the differential of the canonical projection $\pi_M$, does not not send vector fields in $TM$ to vector fields in $M$. Indeed, given the generic vector field $X \in \mathcal{X}(TM)$ in induced coordinates as introduced before, the two vectors $$ X_1 = v^a(\bar{x},\bar{u}_1)\left.\frac{\partial}{\partial x^a}\right\vert_{(\bar{x},\bar{u}_1)} + w^a (\bar{x},\bar{u}_1)\left.\frac{\partial}{\partial u^a}\right\vert_{(\bar{x},\bar{u}_1)} \equiv {v_1}^a\left.\frac{\partial}{\partial x^a}\right\vert_{(\bar{x},\bar{u}_1)} + {w_1}^a \left.\frac{\partial}{\partial u^a}\right\vert_{(\bar{x},\bar{u}_1)} $$ and $$ X_2 = v^a(\bar{x},\bar{u}_2)\left.\frac{\partial}{\partial x^a}\right\vert_{(\bar{x},\bar{u}_2)} + w^a (\bar{x},\bar{u}_2)\left.\frac{\partial}{\partial u^a}\right\vert_{(\bar{x},\bar{u}_2)} \equiv {v_2}^a\left.\frac{\partial}{\partial x^a}\right\vert_{(\bar{x},\bar{u}_2)} + {w_2}^a \left.\frac{\partial}{\partial u^a}\right\vert_{(\bar{x},\bar{u}_2)}, $$ belonging to two different points of the same fiber ${\pi_M}^{-1}(\bar{x})$, are mapped by $T\pi_M$ onto two different vectors at the same point $\bar{x}$ in $M$, namely $$ X_1 \mapsto {v_1}^a\left.\frac{\partial}{\partial x^a}\right\vert_{\bar{x}}, X_2 \mapsto {v_2}^a\left.\frac{\partial}{\partial x^a}\right\vert_{\bar{x}}. $$ Of course, any vector on the same fiber is mapped onto a different vector at the same point in the base space, unless the vector field is projectable, that is the components $v^a$ do not depend on $u$ (are constant on the fiber). Only in this case $T\pi_M$ maps (projectable, i.e. not general) vector fields onto vector fields.

Does the definition of the vertical endomorphism, $S:=\mathrm{vl}\circ T\pi_M$ make sense even for general vector fields $X \in \mathcal{X}(TM)$ if applied pointwise? My temporary answer is: yes, provided the vertical lift of the vectors mapped onto $M$ by means of $T\pi_M$ keeps track of the point on the fiber where the original vectors in $TM$ resided.

What I mean is the following. Considering my previous example, in principle the two vectors ${v_1}^a\left.\frac{\partial}{\partial x^a}\right\vert_{\bar{x}}$ and ${v_2}^a\left.\frac{\partial}{\partial x^a}\right\vert_{\bar{x}}$ can be mapped by $\mathrm{vl}$ to vertical vectors to any point $(\bar{x},u)$ on the fiber over $\bar{x}$. The application of the vertical endomorphism $S$ to $X$ only results in a vertical vector $field$ $X^V \in TM$ if the first vector is mapped by $\mathrm{vl}$ to the point $(\bar{x},\bar{u}_1)$ and the second to $(\bar{x},\bar{u}_2)$ - and so on for all other vectors of the form $v^a(x,u)\left.\frac{\partial}{\partial x^a}\right\vert_{x}$ (for given $x$ and $u$) obtained by applying $T\pi_M$ to $X$. Yet, how can the map $\mathrm{vl}$ know which point on the fiber the vector it's lifting comes from? If we work pointwise, components like ${v_1}^a$ and ${v_2}^a$ are no functions, just numbers.

The problem is that all these considerations are absent from the discussion of the vertical endomorphism tensor in any of the reference sources I could consult. This makes me suspect I'm wrong somewhere.

But where am I wrong?

Answer 1

I think the basic answer to your question is that yes, the vertical map $\operatorname{vl}$ does keep track of the point in $TM$ where the original vector lies. More precisely, $S:T(TM)\to T(TM)$ is defined for $X_u \in T_u(TM)$, by $$ S(X_u) := \operatorname{vl}_u(T_u\pi_M(X_u)) $$ where $T_u\pi_M:T_u(TM)\to T_{\pi(u)} M$ and $\operatorname{vl}_u:T_{\pi(u)}M \to T_u(TM)$ is defined by $$ \operatorname{vl}_u(v) := \frac{d}{ds}\bigg\vert_{s=0}(u+sv). $$

Then a vector field $X\in\mathcal{X}(TM)$ is said to be a second-order differential equation if $S(X_u) = \Gamma_u$ for all $u\in TM$. See for example

Nester, James M., Invariant derivation of the Euler-Langrange equation, J. Phys. A 21, No. 21, L1013-L1017 (1988). ZBL0658.53071., p.L1014,

where $S$ is denoted $J$, and is called the almost tangent structure on $TTM$. Or alternatively

de León, Manuel; Rodrigues, Paulo R., Methods of differential geometry in analytical mechanics, North-Holland Mathematics Studies, 158. Amsterdam etc.: North-Holland. x, 483 p. {$} 110.50; Dfl 210.00 (1989). ZBL0687.53001., Sections 2.2 and 4.3,

which uses similar notation and terminology to Nester.

Answer 2

Let me formulate everything for you coordinate-independently, so there is no question whether the maps $T\pi_M$, $\operatorname{vl}$, and $S$ are well-defined globally without restriction.

Global formulas

First $\pi_M : TM \to M$ is the structure map of the fiber bundle $TM$. Since $TM$ and $M$ are smooth manifolds, we can take the derivative of this map just like any other to obtain a map of vector bundles $$ \require{AMScd} \begin{CD} TTM @>{T\pi_M}>> TM\\ @V{\pi_{TM}}VV @VVV\\ TM @>{\pi_M}>> M \end{CD}. $$

Next, for any vector bundle $p : E \to M$ we can define a map of vector bundles $$ \operatorname{vl} : E \times_M E \to TE. $$ Let's be very careful about what this means: here we are thinking of $E \times_M E$ as a vector bundle over $E$ via the projection $\operatorname{pr}_1$ onto the first factor, and $TE$ as a vector bundle over $E$ via $\pi_E$. (The notation $E \times_M E$ means the subset of all $(e, e') \in E \times_M E$ such that $p(e) = p(e')$.) Now, what's the definition? Thinking of tangent vectors as equivalence classes of curves (to avoid choosing any coordinates), we define $$ \operatorname{vl} : (e, e') \mapsto [t \mapsto e + t e']. $$ That is, for each $(e, e') \in E \times_M E$ there is a curve $\gamma : \mathbb{R} \to E$ defined by $\gamma(t) = e + t e'$, which by taking the equivalence class $[\gamma]$ defines a tangent vector in $T_e E \subset T E$. Of course, in our situation the map $\operatorname{vl}$ is the special case where $E = T M$.

Now, what is the definition of the map $S$? It is not really $S := \operatorname{vl} \circ T \pi_M$ as you say; this forgets too much information. Instead, note that we can form the product of the maps $\pi_{TM} : TTM \to TM$ and $T \pi_M : TTM \to TM$ to obtain a map $(\pi_{TM}, T \pi_M) : TTM \to TM \times_M TM$. This product is now composable with $\operatorname{vl}$ to obtain $$ S : TTM \xrightarrow{(\pi_{TM}, T \pi_M)} TM \times_M TM \xrightarrow{\operatorname{vl}} TTM. $$

Now everything is well-defined. Let me know if you have any questions.

Local formulas

If we put coordinates $(x, X)$ on $TM$ and $(x, X, \dot{x}, \dot{X})$ on $TTM$ so that everything is locally trivial, then tracing these definitions through we get $$ \begin{align*} \pi_{TM}(x, X, \dot{x}, \dot{X}) &= (x, X)\\ T\pi_{M}(x, X, \dot{x}, \dot{X}) &= (x, \dot{x})\\ \operatorname{vl}((x, X), (x, Y)) &= (x, X, 0, Y)\\ S(x, X, \dot{x}, \dot{X}) &= (x, X, 0, \dot{x}), \end{align*} $$ which of course agrees with your formulas.