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Stream: learning: questions

Topic: T(T(M))

Bruno Gavranović (Feb 24 2022 at 23:43):

What's the intuition behind the tangent bundle of a tangent bundle $T(T(M))$?

If $T(M)$ has two coordinates (a point and a vector), then how many coordinates does $T(T(M))$ have? A lecture from Geoffrey Cruttwell on the tangent categories workshop seems to suggest there's four coordinates (at 13:14 timestamp), but I'm not sure how to visualise that. If $T(X)$ consists of a point $x$ in $X$ and a vector at that point, then shouldn't $T(T(M)$ consist of a point in $T(M)$ (i.e. already a pair $(m, v)$ of a point and a vector) and a tangent vector $w$ at $(m, v) : T(M)$ that point?

Though, even though the three-coordinate option seems more sensible, I'm still not sure how to visualise it, or get any intuition about what the second tangent vector represents

Zhen Lin Low (Feb 24 2022 at 23:44):

T(T(M)) has four times the dimensions of M, yes.

Bruno Gavranović (Feb 24 2022 at 23:49):

What do the last two components represent?
If I have an element $(m, v) : T(M)$, then $v$ can be thought of as a change in $m$.
If I have an element $(m, v, w, z) : T(T(M))$, then $v$ can be thought of as a change in $m$, but how do I think of $w$ and $z$?

Mike Shulman (Feb 24 2022 at 23:51):

$w$ is another change in $m$, while $z$ is a change in $v$.

Bruno Gavranović (Feb 24 2022 at 23:53):

Oh, so a tangent bundle of a pair type tells us how both of these elements change, not just how the "last one" changes? For some reason I was not expecting this... but it does seem more natural that I think about it now

Zhen Lin Low (Feb 24 2022 at 23:56):

The tangent bundle construction takes as input a manifold, not a bundle, so it doesn't care about that. Maybe you are thinking of some kind of fibrewise tangent bundle construction.

John Baez (Feb 25 2022 at 00:04):

Physicists think a lot about T(T(M)). A point in M can represent the position $q$ of a particle. A point $(q,v) \in$ T(M) then represents its position and velocity. In the Lagrangian approach to dynamics, the way position and velocity change over time is described by a vector field on T(T(M)). So, for any point $(q,v)$ in T(M) we get a point in T(T(M)) that describes the rate at which position and velocity change over time. Usually a consistency condition holds, which says that the rate at which position changes is equal to the velocity!

Bruno Gavranović (Feb 25 2022 at 00:05):

Right, I think I didn't internalise properly that $T(M$ is a manifold itself. The outside $T$ has no idea that we're giving it a manifold which already had $T$ applied to it.

Points in $T(M)$ are black boxes which have two components: the base point, and the change vector itself. Applying $T$ to it gives us a pair of black boxes, one original pair $(m ,v)$, and a new one telling us how both $m$ and $v$ itself can be changed.

John Baez (Feb 25 2022 at 00:09):

To think about these things mathematically, it's worth noting that like any bundle, there's a projection

$\pi_M : T(M) \to M$

Taking the differential of $\pi_M$, we get

$d \pi_M : T(T(M)) \to T(M)$.

(We could call this $d$ here "$T$" to indicate that it's just the functor $T$ applied to a morphism, but geometers don't do this.)

But also, because $T(T(M))$ is a bundle over $T(M)$, we have a map

$\pi_{T(M)} : T(T(M)) \to T(M)$

So, you should think about comparing these two ways to get a guy in $T(M)$ from a guy in $T(T(M))$. They're not equal!

This is connected to my remark

Usually a consistency condition holds, which says that the rate at which position changes is equal to the velocity!

and how you state this condition.

Bruno Gavranović (Feb 25 2022 at 01:08):

Ah, right, so we've got two maps of type $T(T(M)) \to T(M)$. Let me see if I can get this right. The first one $T\pi_M$ takes a quadruple $(m, v, m', v')$ to $(m, m'$). This is because $\pi_M$ just projects out the base point of $M$, so $T\pi_M$ should project out the base point and the change over it. The second one just projects out the base point of $T(T(M))$, i.e. $(m, v)$

Bruno Gavranović (Feb 25 2022 at 01:09):

Though, now that I wrote it, I'm suspicious whether $(m, m')$ can be of type $T(M)$

Bruno Gavranović (Feb 25 2022 at 01:10):

I suppose it can, but the question is, what is the type of $m'$? I suppose it should be the "space of changes over m", i.e. the same type as $v'$? In which case it's all fine and the result typechecks

John Baez (Feb 25 2022 at 01:18):

It's good to use the standard notation $T_m M$ for the tangent space of $M$ at the point $m$. So:

$m \in M$
$v \in T_m M$

and

$(m',v') \in T_{(m,v)} (TM)$

so what you're wondering about is whether you can interpret $m'$ as an element of $T_m M$.

John Baez (Feb 25 2022 at 01:24):

I think what's really going on is that you need to use the two maps I mentioned to deal with $m'$ and $v'$, and even show that an element of $T_{(m,v)} (TM)$ deserves to be written as a pair. Your intuitions are right, it just takes some work to formally justify them!

Drew Allen McNeely (Feb 25 2022 at 02:43):

I've always tried to keep a low dimensional example in my head to make things easier to visualize. I like this picture from Wikipedia:
https://en.wikipedia.org/wiki/Tangent_bundle#/media/File:Tangent_bundle.svg

Say $M$ is the circle whose coordinate is called $\alpha$, then the tangent spaces are the red lines. Let's just call a tangent vector $\dot{\alpha}$ since it could represent a path's velocity along $\alpha$. We rearrange the tangent spaces into a cylinder to get the tangent bundle $T(M)$, so the velocity is now represented by height along the cylinder.

A tangent space in $T(M)$ is going to be a plane touching the cylinder, so the bundle $T(T(M))$ will indeed be 4 dimensional. Unfortunately it's already hard to visualize the bundle since we've run out of spatial dimensions. But you can imagine the planes rearranged in the same way that we rearranged the tangent lines. So a vector in a tangent plane will have a component along $\alpha$ and a component along $\dot{\alpha}$. The component along $\dot{\alpha}$ could obviously represent acceleration $\ddot{\alpha}$, but I'm also unsure of what the $\alpha$-component could represent physically in terms of paths.

Paolo Perrone (Feb 25 2022 at 12:21):

Every curve on $M$ of coordinates $\{m_i(t)\}$ also gives a curve on $TM$ by looking at the tangent vectors, that is, coordinates $\{m_i(t), \dot{m}_j(t)\}$. Iterating the procedure we get a curve on $TTM$ but you see that, this way, the second and third components in $TTM$ are going to be equal. (This is due to the fact that we started from $M$.)
Indeed in geometry (differential, symplectic, contact geometry,...) is interesting to see which vector fields on $TTM$ come from motions on $M$ and which do not -- and one can ask for similar questions on other bundles as well.

Spencer Breiner (Feb 25 2022 at 13:51):

When these things are axiomatized as a [[tangent bundle category]] there are two operations assumed as non-trivial structure. The latter seems relevant here.

Vertical lift $T\Rightarrow T^2$:
$\ell(m,v) = (m,0,0,v)$

Canonical flip $T^2\Rightarrow T^2$:
$c(m,v,m',v')=(m,m',v,v')$

John Baez (Feb 25 2022 at 17:20):

Yes! I guess Bruno has a choice whether to first learn differential geometry in one of the usual ways and then see why the tangent bundle (which can be defined in various equivalent ways) has these structures, or first learn about tangent bundle categories and then turn to the category of manifolds as an example.

Bruno Gavranović (Feb 26 2022 at 14:41):

John Baez said:

I think what's really going on is that you need to use the two maps I mentioned to deal with $m'$ and $v'$, and even show that an element of $T_{(m,v)} (TM)$ deserves to be written as a pair. Your intuitions are right, it just takes some work to formally justify them!

The two maps are of type $T(T(M)) \to T(M)$. It seems that the question of "what's the type of the third coordinate of $T(T(M))$" precedes defining these maps - in order to define how a map acts on something, I need to know what that something is.

But yeah, I'm not sure how to more formally say that $m'$ is an element of $T_mM$. If $v$ is an element of $T_mM$, and $m'$ too represents a "change of $m$", then $m'$ ought to be an element of $T_mM$ as well.

John Baez (Feb 26 2022 at 16:20):

Bruno Gavranovic said:

The two maps are of type $T(T(M)) \to T(M)$. It seems that the question of "what's the type of the third coordinate of $T(T(M))$" precedes defining these maps - in order to define how a map acts on something, I need to know what that something is.

I defined those maps without using any description of points of $T(T(M))$ as 4-tuples. We just need "a point in $T(X)$ is a pair $(x,v)$ consisting of $x \in X$ and $v \in T_x X$". We need to use this definition twice to describe points of $T(T(M))$.

We get that a point of $T(T(M))$ is a triple $(m,v,w)$ where $m \in M, v \in T_m M$ and $w \in T_{(m,v)} TM$.

So then the question becomes, how can we describe $w$ more concretely?

John Baez (Feb 26 2022 at 16:22):

For this it helps to consider the map

$\pi_M : T(M) \to M$

which sends $(m,v)$ to $m$, and its differential

$d \pi_M : T(T(M)) \to T(M)$.

defined using the fact that $T$ is a functor - not defined using any stuff about triples or quadruples.

John Baez (Feb 26 2022 at 16:31):

This latter map gives a way to extract an element of $T(M)$, and thus an element of $T_m M$, out of our triple $(m,v,w)$. This element is not $v$! It's something else.

Bruno Gavranović (Feb 28 2022 at 18:04):

I see. I suppose the main idea is to understand how $T$ acts on the projection map $\pi_M : TM \to M$. Am I supposed to look at how the action of $T$ is defined on morphisms? If I do so, the action of $T$ on a map $f : M \to N$ is defined as $(m, v) \mapsto (f(m), D(f)(m, v))$ (from here), where $D(f)(m, v)$ tells us how much changing $m$ and $v$ affects the output. So in this case, changing $v$ doesn't do anything, since it's not used. Changing $m$ does.

So I suppose that alone tells me that the $w$ from above is a pair: we need to account for changing $m$ and $v$. I.e. $T(\pi_M) : TTM \to TM := ((m, v), (m', v')) \mapsto (m, m')$.

Bruno Gavranović (Feb 28 2022 at 18:05):

But I'm not sure how satisifed I am with this formality. I suspect I might've used things I "already know" about how derivatives and changes work, and not the data of a tangent category itself.

John Baez (Feb 28 2022 at 19:26):

Wait a minute - you're trying to use just the tangent category axioms? I'm not - I don't even know what they are. I just know differential geometry, so I know how the functor $T$ works in the category of manifolds.

Bruno Gavranović (Feb 28 2022 at 20:01):

Ah, you said it takes work to formally justify these ideas. I thought your formal framework meant tangent categories

John Baez (Feb 28 2022 at 20:23):

No, I meant differential geometry.

John Baez (Feb 28 2022 at 20:25):

As far as I know, you need to use a bit of differential geometry to prove that smooth manifolds are an example of tangent categories. Well, maybe someone has built a big machine by now to prove that lots of categories are tangent categories. But I'm old-school: I know and love differential geometry, and I don't know the axioms for tangent categories.

John Baez (Feb 28 2022 at 20:26):

I''ve looked at them, and my impression is "yup, that's right" - but I haven't thought hard about them.