I am studying differential geometry, and am unable to align what I have learnt in classical mechanics with how differential geometers describe the situation. Consider a particle travelling in a circle. We draw its position, its velocity, and its acceleration all on the same plane, and are given by the equations:

\begin{align*} &p, v, a: [0, 2 \pi) \to \mathbb R^2 \\ &p(\theta) \equiv (\cos \theta, \sin \theta) \\ &v(\theta) \equiv (-\sin \theta, \cos \theta) \\ &a(\theta) \equiv (-\cos \theta, - \sin \theta) \end{align*}

and is drawn with $p$ as a curve, $v, a $ as vector fields on the curve:

Let's now turn to differential geometry:

  • The manifold $M$ is the surface $\mathbb R^2$.
  • The path of the particle $p: [0, 2 \pi] \to M$ is a curve that lives on the surface of the manifold.
  • The velocity vectors are a vector field $v: [0, 2 \pi] \to TM$.
  • The normal vectors are a vector file $a: [0, 2 \pi] \to T^2M$, since they are derivatives of the vector field $v$.

However, at least in the physics picture, there is a "natural" identification between $M$, $TM$, and $T^2M$, since I can "draw" the velocity and acceleration next to each other, and even take the dot product between velocity and acceleration! So where am I missing the link between classical mechanics and differential geometry? In particular, I'd like to understand:

  1. Why can we draw the tangent and normal vectors next to each other on the circle? This must give some identification of $TM$ and $T^2 M$
  2. Is this special for curves? Does the same thing happen for vector fields of surfaces?
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    $\begingroup$ I'd say the main piece you're missing is Riemannian geometry, which allows you to take a "covariant derivative" of the velocity field to obtain an acceleration field with values in $TM$. $\endgroup$ Nov 18, 2021 at 6:51
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    $\begingroup$ To expand a bit on Anthony's (+1) comment, the answer to what does the integral of position with respect to time mean discusses some of the details about the double tangent bundle and its identification with the tangent bundle in the presence of a connection. $\endgroup$ Nov 18, 2021 at 13:27

1 Answer 1


Clarifying the Setting:

  • The path of a particle is described by a smooth curve $\gamma:I\to M$, where $I\subset \Bbb{R}$ is an open interval, and $M$ is a smooth manifold (in Physics, one often uses $Q$, and calls this the configuration manifold).
  • The velocity of a particle is described by the velocity field $\gamma':I\to TM$. This is a smooth vector field along $\gamma$ (meaning $\pi\circ \gamma'=\gamma$, where $\pi:TM\to M$ is the projection, or more explicitly, for every $t\in I$, we have a vector $\gamma'(t)\in T_{\gamma(t)}M$).
  • The acceleration we think of is NOT the second tangent map $\gamma'':I\to T^2M$. The second tangent bundle has a lot of extra information and structure, which isn't what we're thinking of when we draw those pictures.

Rather, the acceleration we have in mind comes from Riemannian geometry. We equip $M$ with a Riemannian metric, and equip $TM$ with the Levi-Civita connection. With this connection, we can take covariant derivatives to get $\frac{D\gamma'}{dt}\equiv \nabla_{\gamma'}\gamma':I\to TM$, and this is once again a smooth vector field along $\gamma$.

So, position is $p=\gamma:I\to M$ while velocity is $v=\gamma':I\to TM$ and acceleration (in the manifold $M$) is $a=\frac{D\gamma'}{dt}:I\to TM$. We can look at the higher derivative (I believe it's called "jerk") $j=\frac{D^2\gamma'}{dt^2}=\frac{Da}{dt}:I\to TM$ and so on. This is the nice thing about covariant derivatives: our target space always stays the same; and geometrically it also has a nice interpretation in terms of the parallel-transport defined by the connection.

Linking the Abstractness to the Pictures:

Note that one can never truly identify the tangent bundle with the base manifold; the dimensions just don't match up. So, what's going on with the pictures? Suppose $M$ is an embedded submanifold of $\Bbb{R}^n$. Then, for any point $x\in M$, we have $T_xM\subset T_x\Bbb{R}^n$, and we have a standard isomorphism $\Phi:T_x\Bbb{R}^n\to \Bbb{R}^n$ (this isomorphism is obtained by virtue of the identity chart). The precise definition of this isomorphism depends of course on the precise definition of tangent spaces you're working with. But in any case, the way to think of this is as "a copy of $\Bbb{R}^n$ attached to the point $x$". So, if we want to heuristically draw a vector in $T_xM$, we draw an arrow emanating from the point $x\in M$, and we draw it "tangent to $M$".

The picture which you posted is thus doing two things simultaneously. We have one copy of $\Bbb{R}^2$, which we think of as $M$. Here, we're thinking of $\Bbb{R}^2$ as a manifold; this is where the blue curve lives. Next, we come to the red arrows. Notice how the red arrows are emanating from points on the curve; this is keeping track of the base-point of the vector. The arrow keeps track of the vector aspect; refer to the previous paragraph about $T_x\Bbb{R}^2\cong \Bbb{R}^2$. This time we think of $\Bbb{R}^2$ as a vector space. So, really it's like there are several copies of $\Bbb{R}^2$ lying around. The first and primary one is $M=\Bbb{R}^2$, the base manifold. Next is for each $x\in M=\Bbb{R}^2$, we have the tangent space $T_x\Bbb{R}^2\cong \Bbb{R}^2$, which we think of as the vector space "attached to the point $x$", and so that's where all our arrows emanate from. Of course, since everything was just $\Bbb{R}^2$, we are able to condense all of these subtleties into a single picture.

Likewise, the blue acceleration vectors (which should, for consistency sake, emanate from points along the curve, just as the red velocity vectors, but I guess due to lack of space, the picture is drawn the way it is) also belong to the same vector space as the red velocity vectors.

"Geometric Description" of Covariant derivatives (Maybe you want to skip on a first read)

Covariant derivatives take maps (with values in a vector bundle) and differentiate somehow to still give us something in that same vector bundle (unlike the usual tangent mapping, which requires us to look at a bigger target manifold). This is part of why we like it so much. Also, covariant derivatives allow us to "differentiate along maps". This is very important for us, because going back to your picture, we wish to be able to differentiate vector fields (or more generally tensor fields, or maps into vector bundles) defined along the particle's trajectory.

The usual descriptions of connections and covariant derivatives proceeds rather algebraically (using the Koszul definition for a connection). While this algebraic method is very slick and efficient, I never understood the basic "geometry" in "Riemannian geometry".

Instead, I find that thinking of connections as a specification of Horizontal vectors (i.e Ehresmann's definition) is much more intuitive (more abstract yes, but the idea is simple enough); see the picture at the bottom of this question of mine. So, the connection $C$ specifies horizontal vectors. Now, coming to the part about how to differentiate along maps. Suppose $(E,\pi, M)$ is a vector bundle with a linear connection $C$, and suppose $N$ is a smooth manifold, $f:N\to M$ a smooth map, and $\phi:N\to E$ a smooth lifting of $f$ (this just means $\pi\circ \phi=f$, i.e that for each $x\in N$, $\phi(x)$ is an element of the fiber $E_{f(x)}$, so this generalizes the notion of a vector field along a map).

How can we differentiate $\phi$? Well, we have for each $x\in N$, the usual tangent mapping $T\phi_x:T_xN\to T_{\phi(x)}E$. So, if I have a vector $h_x\in T_xN$, then $T\phi_x(h_x)\in T_{\phi(x)}E$. The fact that this lives in the tangent space to $E$ at a point means roughly that "it points outside of the fibers". So, that's where the connection comes in. We can now subtract from this tangent mapping the horizontal part: \begin{align} T\phi_x(h_x)-C_x(Tf_x(h_x), \phi(x)) \end{align} Once we subtract off this horizontal part, we're left with a vertical vector, i.e an element of $\ker(T\pi_{\phi(x)})= T_{\phi(x)}(E_{f(x)})$, i.e the tangent space to the fiber itself, rather than the entire manifold $E$. Since $E_x$ is itself a vector space, its tangent space can be canonically identified with itself. So, in this manner, we obtain a vector $\nabla_{h_x}\phi\in E_{f(x)}$. In short, we have defined a mapping $\nabla\phi:TN\to E$, $h_z\mapsto \nabla_{h_z}\phi$.

If you open up a book in Riemannian geometry and look at the coordinate formulas, you'll see that the covariant derivative of vector field is given by first taking partial derivatives of components of the vector field and then correcting by an additional term involving $\Gamma^i_{jk}$. These Christoffel symbols are part of the chart-representative of the connection map $C$, and they offer a "correction term" which classically people describe as saying that it gives rise to the correct transformation law. Well, now geometrically, you can see this as taking the usual tangent mapping $T\phi$, and then subtracting off the excessive horizontal part. Once we subtract, we're only left with the vertical part, i.e something tangent to the fibers, and hence can be identified with an element of the fiber.

Another Subtlety Regarding the Definition of Acceleration:

Suppose we have a Riemannian manifold $M$ and an embedded Riemannian submanifold $N\subset M$ (meaning if $\iota:N\to M$ is the inclusion and $g_M$ is the RIemannian metric of $M$, then $g_N:= \iota^*g_M$). Suppose we have a smooth curve $\gamma:I\to N$. Then, the velocity is $\gamma':I\to TN\subset TM$. When we talk about "what's the acceleration of the curve $\gamma$", this is really an ambiguous question. We should really specify the manifold with respect to which we wish to calculate the acceleration. i.e if $\nabla^M$ and $\nabla^N$ are the Levi-Civita connections on $TM$ and $TN$, then we have to decide whether we calculate $\frac{D^N\gamma'}{dt}=\nabla^N_{\gamma'}\gamma'$ by thinking of $\gamma':I\to TN$, or whether we calculate $\frac{D^M\gamma'}{dt}=\nabla^M_{\gamma'}\gamma'$ by thinking of the velocity as $\gamma':I\to TM$, mapping into the larger tangent bundle.

Fun fact: one can show that for any vector field $\xi$ along $\gamma$, $\frac{D^N\xi}{dt}(t_0)$ is the orthogonal projection of $\frac{D^M\xi}{dt}(t_0)$, i.e relative to the orthogonal direct sum decomposition $T_{\gamma(t_0)}M= T_{\gamma(t_0)}N \oplus (T_{\gamma(t_0)}N)^{\perp}\to T_{\gamma(t_0)}N$.

It should hopefully be reasonable that these two accelerations are different in general. One measures how much we're accelerating in $M$ (and when I say "accelerate in $M$", I mean such that the acceleration is tangent to $M$), and the other measures how much we accelerate in $N$. For example, if you have a particle in uniform circular motion in the plane $M=\Bbb{R}^2$, then it is definitely accelerating in $M$. However, this acceleration is orthogonal to the circle $N=S^1$, which means the amount of acceleration tangent to $N$ is $0$ (this is the 1-dimensional argument for why great circles are geodesics).

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    $\begingroup$ oh I didn't elaborate on this, but I mentioned in the beginning the second tangent bundl has a lot of information and has a lot of structure. What I mean by this is that one can consider $T^2M$ as a vector bundle over $TM$ in two different ways. The first is the obvious way: define $P=TM$, and then we have the usual tangent bundle with usual projection $\pi_P=\pi_{TM}:TP=T^2M\to P=TM$. The second is by taking the projection $\pi_M:TM\to M$ and taking its tangent map to get $T\pi_M:T^2M\to TM$. $\endgroup$
    – peek-a-boo
    Nov 19, 2021 at 6:04
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    $\begingroup$ There's a lot we can say about this, but my point is just that $T^2M$ has a lot of stuff going on, and elements of this space are not what we typically think of as acceleration (we like to think of acceleration as a vector tangent to our manifold, i.e an element of $TM$). $\endgroup$
    – peek-a-boo
    Nov 19, 2021 at 6:05
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    $\begingroup$ @Siddharth Bhat take a look at my linked question. The definition of $C$ (and a corresponding picture describing the wordy definition) is given there. For your second point, I'm not sure why it's counterintuitive. You're right to imagine the fibers as vertical, so tangent to fibers is also vertical. For example think of the cylinder as a real line bundle over the circle. The fibers are copies of real line attached vertically to the circle. If I have a curve in a single fiber then surely you can convince yourself the vector points either up/down(i.e vertically). $\endgroup$
    – peek-a-boo
    Nov 19, 2021 at 13:15
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    $\begingroup$ Said another way: the tangent space of the fiber is again canonically isomorphic to the fiber since we're dealing with a vector space. But we just said the fiber lies "vertically over the circle". So the vertical vectors are identified with elements of the fiber. $\endgroup$
    – peek-a-boo
    Nov 19, 2021 at 13:18
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    $\begingroup$ maybe later I'll draw an extra picture when I have the time to indicate how everything fits together $\endgroup$
    – peek-a-boo
    Nov 19, 2021 at 13:23

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