# Questions about tangent spaces & the tangent bundle

This is a slightly physics oriented question, so apologies for that.

Basically, having started studying differential geometry it has started to become a little clearer to me why one can consider the Lagrangian as a function of position and velocity, but I don't feel I'm quite there yet.

My confusion arises with the tangent bundle. I understand that the state of a physical system at a given instant in time is completely specified by its position and velocity at that given instant, and as such, for the Lagrangian to characterise all possible dynamics of a physical system, it must therefore be a function on the tangent bundle (correct?!)

As I understand it, the tangent bundle $\mathcal{TM}$ is defined as the disjoint union of tangent spaces $T_{p}\mathcal{M}$ parametrised by the manifold $\mathcal{M}$ (of dimension $n$), $$\mathcal{TM}=\lbrace (p,\mathbf{X}_{p})\;\;\vert\quad p\in \mathcal{M}, \;\;\mathbf{X}_{p} \in T_{p}\mathcal{M}\rbrace$$
Would it then be correct to say that there is an $n$-dimensional tangent space $T_{p}\mathcal{M}$ for each fixed point $p\in\mathcal{M}$, and as such, for a fixed point, there are an infinite number of tangent vectors that one could choose from? On the face of it I see how, as the manifold formed by the tangent bundle is $2n$-dimensional, one can specify a point on the this manifold uniquely by specifying a point $p\in\mathcal{M}$ and a vector in the tangent space to that point, $\mathbf{X}_{p}\in T_{p}\mathcal{M}$, however, I'm finding it difficult to see how they can be treated as independent variables, as surely one is required to specify a point $p\in\mathcal{M}$ before one can determine the tangent space to that point (and thus, the possible tangent vectors that one can choose from)?

I have been reading this set of notes http://www.physics.usu.edu/torre/6010_Fall_2010/Lectures/01.pdf and they have been enlightening me a bit on the subject, but I'm a little unsure that I'm understanding it correctly?!

I understand that points $p$ on the manifold $\mathcal{M}$ exist independently of any chosen curve, as do the tangent vectors $\mathbf{X}_{p}$ in the tangent space $T_{p}\mathcal{M}$ to those points, so would it be correct to say that each point in the tangent bundle represents a point on $\mathcal{M}$ through which a curve could pass through at a given velocity? [In the middle of page 6 the lecturer pauses to discuss a "common point of confusion" about the notation $\dot{q}$ (for a vector in the tangent space), stating that it is not the derivative of anything, but just a vector that exists at the point $q$. Is this just the case because for each given point we can associate a set of vectors (which form a tangent space at that point); they are not derivatives of anything, but they do, in essence, contain information on what direction (and at what speed) curves can move through that point?!]

He then moves on to say that a given curve in the tangent bundle looks like $$q=q(t),\qquad\dot{q}=\dot{q}(t)$$ I assume by this that the curve is parametrised by some parameter $t$ and that a given value of $t$ corresponds to a given point $(q,\dot{q})$ in the tangent bundle? Also, he says that such a curve may not actually correspond to any motion of the system and we need $\dot{q}(t)$ to actually represent the tangent to $q(t)$, that is, we need to choose $\dot{q}(t) = \frac{dq(t)}{dt}$. By this does he mean that although a given value of $t$ will correspond to a point $q=q(t)$ and a tangent vector $\dot{q}=\dot{q}(t)$, the tangent vector associated with that value of $t$ may not be that of the curve passing through $q$, hence we require that, in actual fact, the given value of $t$ corresponds to a vector tangent to the point $q$ such that its value $\dot{q}(t)$ corresponds to the derivative of the curve evaluated at that point, i.e. $\dot{q}(t)=\frac{dq(t)}{dt}$?

Having done all this, do we then say that the Lagrangian is a one-parameter family of functions (parameterised by $t$) on the tangent bundle, i.e. $$\mathcal{L}=\mathcal{L}(q,\dot{q}, t)=\mathcal{L}(\lbrace q_{i}\rbrace, \lbrace\dot{q}_{i}\rbrace , t)$$ One may then choose a curve in configuration space $q(t)=\lbrace q_{i}(t) \rbrace$ such that when the Lagrangian is evaluated on that curve, for a given value of $t$ it returns a number $\mathcal{L}(q(t),\dot{q}(t),t)$ corresponding to the value of the Lagrangian evaluated at the point $q=q(t)$ on the curve, whose tangent vector is $\dot{q}(t)=\frac{dq(t)}{dt}$?

Sorry for the long-windedness of this post, I'm just really keen to get these concepts firmly cemented (correctly) in my mind!

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## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
Perhaps it would help to think about a simple example. Let the manifold be a sphere. Then the "tangent plane" at any given point really is a plane tangent to the sphere at that point. And the "tangent bundle" for the sphere is the collection of all such tangent planes together with a function that, to every point of the sphere assigns its tangent plane.

A "curve" on sphere is a continuous function that assigns, to every real number (in some give interval) a point on the sphere. The "derivative", $\frac{dq}{dt}$, assigns to each point on the sphere, a vector in the tangent plane at that point. Some texts go the other way- defining a "tangent vector" at a given point to be an equivalence class of curves through that point with the equivalence relation being, of course, that the curves have the same derivative there.

So the answer to your last question
one may then choose a curve in configuration space q(t)={qi (t)} q(t)=\lbrace q_{i}(t) \rbrace such that when the Lagrangian is evaluated on that curve, for a given value of t t it returns a number L(q(t),q˙ (t),t) \mathcal{L}(q(t),\dot{q}(t),t) corresponding to the value of the Lagrangian evaluated at the point q=q(t) q=q(t) on the curve, whose tangent vector is q˙ (t)=dq(t)dt \dot{q}(t)=\frac{dq(t)}{dt} ?
is "yes".

So would it be correct to say that one can treat the variables $p\in\mathcal{M}$ and $\mathbf{v}\in T_{p}\mathcal{M}$ as independent variables on the tangent bundle, as for each given $p\in\mathcal{M}$ we are free to choose any tangent vector $\mathbf{v}\in T_{p}\mathcal{M}$ from the tangent space to that point (although I'm not sure how this would work the other way round, i.e. does one always have to specify the point $p\in\mathcal{M}$ first)? As such, a given point on the tangent bundle is completely described by the ordered pair $(p, \mathbf{v})$?

He then moves on to say that a given curve in the tangent bundle looks like
q=q(t),q˙=q˙(t)​
q=q(t),\qquad\dot{q}=\dot{q}(t) I assume by this that the curve is parametrised by some parameter tt and that a given value of tt corresponds to a given point (q,q˙)(q,\dot{q}) in the tangent bundle? Also, he says that such a curve may not actually correspond to any motion of the system and we need q˙(t)\dot{q}(t) to actually represent the tangent to q(t)q(t), that is, we need to choose q˙(t)=dq(t)dt\dot{q}(t) = \frac{dq(t)}{dt}. By this does he mean that although a given value of tt will correspond to a point q=q(t)q=q(t) and a tangent vector q˙=q˙(t)\dot{q}=\dot{q}(t), the tangent vector associated with that value of tt may not be that of the curve passing through qq, hence we require that, in actual fact, the given value of tt corresponds to a vector tangent to the point qq such that its value q˙(t)\dot{q}(t) corresponds to the derivative of the curve evaluated at that point, i.e. q˙(t)=dq(t)dt\dot{q}(t)=\frac{dq(t)}{dt}?

When one introduces a coordinate curve (as described on page 6 in the notes that I attached) is my understanding of this correct (as I talk about in the quoted text above)?

When one starts to talk about variations (upon introducing a curve $q(t)$) is it correct to say that a variation in the curve, $\delta q(t)$, will induce a variation in the tangent vector, $\delta\dot{q}(t)$, to the curve evaluated at the point $q=q(t)$ [which is equivalent to taking the derivative of the variation of the curve at the point $q=q(t)$ as $\delta\dot{q}(t) = \dot{q}'(t) - \dot{q}(t) =\frac{dq'(t)}{dt}- \frac{dq(t)}{dt} = \frac{d}{dt}(q'(t)-q(t)) = \frac{d}{dt}(\delta q(t))$ ]? I have some confusion here, as usually one can not compare vectors at two different points on a manifold? Is it just that by varying the position that the curve takes for a given value of $t$ will naturally vary the tangent space we are considering and hence the vary the tangent vectors, and so the tangent vector $\delta\dot{q}(t)$ is the vector tangent to the new (varied) curve $\delta q(t)$ and the requirement that this corresponds to the actually velocity of this new curve leads to the result that $\delta\dot{q}(t) = \frac{d}{dt}(\delta q(t))$ ?

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Fredrik
Staff Emeritus
Science Advisor
Gold Member
Some of the confusion is probably caused by the fact that you (and the text you're reading?) denote both the curve q and its value at t by q(t). I also wouldn't use terminology like "parametrized by a variable t". It would be odd to say that, when we use different notations for the curve and its value at a point in its domain.

In what I'm about to say, the definition ##TM=\bigcup_{p\in M}T_pM## is convenient in one place, and the definition ##TM=\{(p,v)|p\in M,\ v\in T_pM\}## is convenient in another. Because of the bijective correspondence between these two sets, I will ignore the distinction and talk about them as if they're the same. (Yes, this is a mild abuse of the definitions).

The Lagrangian is a function ##\mathcal L:TM\to\mathbb R##. Ignoring some technical issues, the tangent vector field of a curve ##C:[a,b]\to M##, is defined as the function ##\dot C:[a,b]\to TM## such that
$$\dot C(t)(f)=(f\circ C)'(t),$$ for all ##t\in[a,,b]## and all smooth ##f:M\to\mathbb R##. So if ##q## is a curve in ##M##, then ##\dot q## is a curve in ##TM##, and if their common domain is ##[a,b]## (or rather, an open interval that contains ##[a,b]##), then for each ##t\in[a,b]##, we have ##q(t)\in M## and ##\dot q(t)\in T_{C(t)}M##. So ##(q(t),\dot q(t))## is a point in ##TM##, the domain of ##\mathcal L##.

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lavinia
Science Advisor
Gold Member
In regular vector calculus in Euclidean space, one thinks of a velocity vector as located at a point in space. If one imagines all possible velocity vectors at the point, these form a vector space.

The exact same thing is true on a manifold. No difference.

The union of all of these vector spaces is called the tangent bundle.

A lagrangian depends not only on positions but also on velocities. This means that it is defined in the tangent bundle.

Your problem with independence seems to have has nothing to do with the phase space itself but with the idea of a dynamical system. In a system velocity vectors are seemingly not independent since they occur at points along the path of the system. But this just means that a dynamical system is following a path. In the phase space itself, the positions and potential velocities are independent of each other just as the x and y coordinates are independent of each other in the plane.

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So if qq is a curve in MM, then q˙\dot q is a curve in TMTM, and if their common domain is [a,b][a,b] (or rather, an open interval that contains [a,b][a,b]), then for each t∈[a,b]t\in[a,b], we have q(t)∈Mq(t)\in M and q˙(t)∈TC(t)M\dot q(t)\in T_{C(t)}M. So (q(t),q˙(t))(q(t),\dot q(t)) is a point in TMTM, the domain of L\mathcal L.

So is the point that to each curve $q\in M$ there are associated a family of curves $\dot{q}\in T_{c(t)}M$ in the tangent space to this curve. If we wish to choose a particular curve, that has a particular velocity, then we require that the curve $\dot{q}\in T_{c(t)}M$ is equal to the velocity of that curve, i.e. $\dot{q}(t)= \frac{dq(t)}{dt}$?

A lagrangian depends not only on positions but also on velocities. This means that it is defined in the tangent bundle.

Is this because of the following:
We wish to characterise the dynamics of a given system (or set of systems) by a function (the Lagrangian). To do so, such a function will necessarily be dependent on the state of the system at each point in time. Empirically it is known that the state of a given system at a given instant in time is uniquely determined by specifying its position and velocity at that instant, hence the Lagrangian must be a function of positions and velocities?!

Your problem with independence seems to have has nothing to do with the phase space itself but with the idea of a dynamical system. In a system velocity vectors are seemingly not independent since they occur at points along the path of the system. But this just means that a dynamical system is following a path. In the phase space itself, the positions and potential velocities are independent of each other just as the x and y coordinates are independent of each other in the plane.

So is it correct to say that the Lagrangian is a function on this phase space, so we are free to choose the positions and velocities independently. It is not until we choose a particular curve that they lose their mutual independence, at which point the velocity becomes the time derivative of the curve at each point?!

Also, going back to an earlier point, is it correct to say that if we choose a coordinate chart $\varphi : U_{i}\rightarrow \mathbb{R}^{n}$ (with $U_{i}\subset M$) in the neighbourhood of some point $p\in U_{i}$ on the manifold, then $TM\supset TU_{i}\cong\mathbb{R}^{2n}$. As such, each point $(p, \mathbf{v})\in TU_{i}$ can be represented by a string of 2n-coordinates, i.e. $(p, \mathbf{v})\mapsto (x^{1},\ldots ,x^{n},v^{1},\ldots , v^{n})\in\mathbb{R}^{2n}$. Now, as the coordinate vectors $(v^{1},\ldots , v^{n})\in\mathbb{R}^{n}$ they are not attached to any particular tangent space to any particular point, as such we are free to choose the coordinates $(x^{1},\ldots , x^{n})$ and $(v^{1},\ldots , v^{n})$ independently. If we choose the space coordinates $(x^{1},\ldots , x^{n})$ first, then this corresponds to choosing a point $p\in U_{i}$, we are still free to choose the vector coordinates $(v^{1},\ldots , v^{n})$ independently as there are an infinite number of vectors in the tangent space to that point that we can choose from. If we instead choose the vector coordinates $(v^{1},\ldots , v^{n})$ first, then by choosing a particular point with space coordinates $(x^{1},\ldots , x^{n})$ assigns this vector to a particular tangent space, such that any two vectors that have the same coordinates $(v^{1},\ldots , v^{n})$ in $\mathbb{R}^{n}$ are distinguished by which tangent space they belong to (i.e. they are not equal, despite having the same component values in $\mathbb{R}^{n}$).

lavinia
Science Advisor
Gold Member
Is this because of the following:
We wish to characterise the dynamics of a given system (or set of systems) by a function (the Lagrangian). To do so, such a function will necessarily be dependent on the state of the system at each point in time. Empirically it is known that the state of a given system at a given instant in time is uniquely determined by specifying its position and velocity at that instant, hence the Lagrangian must be a function of positions and velocities?!

yes

Thanks Lavinia :-)

Is this right as well?

So is it correct to say that the Lagrangian is a function on this phase space, so we are free to choose the positions and velocities independently. It is not until we choose a particular curve that they lose their mutual independence, at which point the velocity becomes the time derivative of the curve at each point?!

Also, going back to an earlier point, is it correct to say that if we choose a coordinate chart φ:Ui→Rn\varphi : U_{i}\rightarrow \mathbb{R}^{n} (with UiMU_{i}\subset M) in the neighbourhood of some point pUip\in U_{i} on the manifold, then TMTUi≅R2nTM\supset TU_{i}\cong\mathbb{R}^{2n} . As such, each point (p,v)∈TUi(p, \mathbf{v})\in TU_{i} can be represented by a string of 2n-coordinates, i.e. (p,v)↦(x1,…,xn,v1,…,vn)∈R2n(p, \mathbf{v})\mapsto (x^{1},\ldots ,x^{n},v^{1},\ldots , v^{n})\in\mathbb{R}^{2n}. Now, as the coordinate vectors (v1,…,vn)∈Rn(v^{1},\ldots , v^{n})\in\mathbb{R}^{n} they are not attached to any particular tangent space to any particular point, as such we are free to choose the coordinates (x1,…,xn)(x^{1},\ldots , x^{n}) and (v1,…,vn)(v^{1},\ldots , v^{n}) independently. If we choose the space coordinates (x1,…,xn)(x^{1},\ldots , x^{n}) first, then this corresponds to choosing a point pUip\in U_{i}, we are still free to choose the vector coordinates (v1,…,vn)(v^{1},\ldots , v^{n}) independently as there are an infinite number of vectors in the tangent space to that point that we can choose from. If we instead choose the vector coordinates (v1,…,vn)(v^{1},\ldots , v^{n}) first, then by choosing a particular point with space coordinates (x1,…,xn)(x^{1},\ldots , x^{n}) assigns this vector to a particular tangent space, such that any two vectors that have the same coordinates (v1,…,vn)(v^{1},\ldots , v^{n}) in Rn\mathbb{R}^{n} are distinguished by which tangent space they belong to (i.e. they are not equal, despite having the same component values in Rn\mathbb{R}^{n}).

lavinia
Science Advisor
Gold Member
So is the point that to each curve $q\in M$ there are associated a family of curves $\dot{q}\in T_{c(t)}M$ in the tangent space to this curve. If we wish to choose a particular curve, that has a particular velocity, then we require that the curve $\dot{q}\in T_{c(t)}M$ is equal to the velocity of that curve, i.e. $\dot{q}(t)= \frac{dq(t)}{dt}$?

Is this because of the following:
We wish to characterise the dynamics of a given system (or set of systems) by a function (the Lagrangian). To do so, such a function will necessarily be dependent on the state of the system at each point in time. Empirically it is known that the state of a given system at a given instant in time is uniquely determined by specifying its position and velocity at that instant, hence the Lagrangian must be a function of positions and velocities?!

So is it correct to say that the Lagrangian is a function on this phase space, so we are free to choose the positions and velocities independently. It is not until we choose a particular curve that they lose their mutual independence, at which point the velocity becomes the time derivative of the curve at each point?!

Also, going back to an earlier point, is it correct to say that if we choose a coordinate chart $\varphi : U_{i}\rightarrow \mathbb{R}^{n}$ (with $U_{i}\subset M$) in the neighbourhood of some point $p\in U_{i}$ on the manifold, then $TM\supset TU_{i}\cong\mathbb{R}^{2n}$. As such, each point $(p, \mathbf{v})\in TU_{i}$ can be represented by a string of 2n-coordinates, i.e. $(p, \mathbf{v})\mapsto (x^{1},\ldots ,x^{n},v^{1},\ldots , v^{n})\in\mathbb{R}^{2n}$. Now, as the coordinate vectors $(v^{1},\ldots , v^{n})\in\mathbb{R}^{n}$ they are not attached to any particular tangent space to any particular point, as such we are free to choose the coordinates $(x^{1},\ldots , x^{n})$ and $(v^{1},\ldots , v^{n})$ independently. If we choose the space coordinates $(x^{1},\ldots , x^{n})$ first, then this corresponds to choosing a point $p\in U_{i}$, we are still free to choose the vector coordinates $(v^{1},\ldots , v^{n})$ independently as there are an infinite number of vectors in the tangent space to that point that we can choose from. If we instead choose the vector coordinates $(v^{1},\ldots , v^{n})$ first, then by choosing a particular point with space coordinates $(x^{1},\ldots , x^{n})$ assigns this vector to a particular tangent space, such that any two vectors that have the same coordinates $(v^{1},\ldots , v^{n})$ in $\mathbb{R}^{n}$ are distinguished by which tangent space they belong to (i.e. they are not equal, despite having the same component values in $\mathbb{R}^{n}$).

I think you have it.

Try working through some simple examples.

Fredrik
Staff Emeritus
Science Advisor
Gold Member
So is the point that to each curve $q\in M$ there are associated a family of curves $\dot{q}\in T_{c(t)}M$ in the tangent space to this curve. If we wish to choose a particular curve, that has a particular velocity, then we require that the curve $\dot{q}\in T_{c(t)}M$ is equal to the velocity of that curve, i.e. $\dot{q}(t)= \frac{dq(t)}{dt}$?
##\dot q(t)## is by definition the velocity at time t (and therefore the velocity at the point ##q(t)##) of a point particle moving as described by ##q##. The Lagrangian must be defined on all of ##TM## even though we will only be interested in inputs of the form ##(q(t),\dot q(t))##, because the fact that ##q## can be almost any curve ensures that ##\dot q(t)## can be any tangent vector.

I think you have it.

Try working through some simple examples.

Thanks for your help. I shall try to do so.

I think you have it.

Try working through some simple examples.

Thanks for your help, I shall try to do so.

##\dot q(t)## is by definition the velocity at time t (and therefore the velocity at the point ##q(t)##) of a point particle moving as described by ##q##. The Lagrangian must be defined on all of ##TM## even though we will only be interested in inputs of the form ##(q(t),\dot q(t))##, because the fact that ##q## can be almost any curve ensures that ##\dot q(t)## can be any tangent vector.

Appreciate all your help Fredrik, I'm a bit confused by this though, as in the notes that I linked in an earlier post the lecturer talks about $q(t)$ and $\dot{q(t)}$ as being a curve on the manifold and its associated curve in the tangent space, respectively, but he then goes on to say that for it to be a physically realisable path for the system we require that $\dot{q} (t) =\frac{dq(t)} {dt}$ (i.e. It is the time derivative of the curve)?