# Coordinate basis vectors on tangent bundle (extrinsic view)

Short Version: when we say that $$(\pmb{q},\pmb{u}):TQ_{(q)}\to\mathbb{R}^{2n}$$ are local coordinates for the tangent bundle of $$Q$$, which can be viewed as an embedded submanifold of a higher dimensional euclidean space, what is the extrinsic view of the coordinate basis vectors $$\pmb{\partial}_{u^i}$$ associated with the velocity-level coordinate functions, $$u^i$$?

Long Version: Let $$Q\subset \mathbb{E}^{3N}$$ be an $$n$$-dimensional configuration manifold for some classical system of $$N$$ particles in 3-space subject to some $$K$$ holonomic constraints such that we can describe the configuration (position of all the particles) by some $$n=3N-K$$ generalized coordinates, $$\pmb{q}=(q^1,\dots, q^n)$$, giving a (local) diffeomorphism $$\pmb{q}:Q_{(q)}\to \mathbb{R}^{n}$$. The following is how I understand the extrinsic view of the associated coordinate basis on $$Q$$, defined in terms of a basis for $$\mathbb{E}^{3N}$$.

Let $$\vec{e}_{\alpha} \in\mathbb{E}^{3N}$$ be any global non-accelerating basis (not necessarily orthonormal) with dual basis $$\vec{\epsilon}^a \in(\mathbb{E}^{3N})^*$$. We can write a ''position vector'' for a point on $$Q$$, viewed as embedded in $$\mathbb{E}^{3N}$$:

$$\vec{r}(\pmb{q}) = r^a (\pmb{q}) \vec{e}_a \in Q \subset \mathbb{E}^{3N} \qquad, \qquad \pmb{q}=\pmb{q}(\vec{r}) \in \mathbb{R}^n \qquad,\qquad a=1,\dots ,3N$$

(I'm abusing notation; many would name the chart $$\varphi:Q_{(q)}\to\mathbb{R}^n$$ and write $$\varphi(\vec{r})=\pmb{q}$$ and $$\vec{r}=\varphi^{-1}(\pmb{q})= r^a (\pmb{q}) \vec{e}_a$$) where $$r^a:\mathbb{E}^{3N}\to\mathbb{R},\;\vec{r}\mapsto\vec{\epsilon}^a\cdot\vec{r}$$, are global linear coordinates for $$\mathbb{E}^{3N}$$. The $$\pmb{q}$$ coordinate basis is then simply

$$\pmb{\partial}_{i} = \frac{\partial r^a}{\partial q^i} \vec{e}_a \in \mathfrak{X}(Q) \qquad,\qquad \mathbf{d}q^i = \frac{\partial q^i}{\partial r^a} \vec{\epsilon}^a \in \Omega^1(Q) \qquad,\qquad i=1,\dots , n \qquad (1)$$

Is the above correct so far? I have defined a vector field $$\pmb{\partial}_{i}\in\mathfrak{X}(Q)$$ as a linear combination of vectors on $$\mathbb{E}^{3N}$$ (with functions as coefficients), and similarly for the basis 1-forms. Is this appropriate for the extrinsic view of things? I feel like $$\pmb{\partial}_{i}$$ is not actually equal to a linear combination of $$\vec{e}_a$$ as I have above, but is instead isomorphic in some trivial way?

Now the induced coordinate basis on $$TQ$$: The above $$\pmb{\partial}_i|_r = \frac{\partial r^a}{\partial q^i}\big|_r \vec{e}_a \in T_rQ$$ is a basis for each tangent space, with dual basis $$\mathbf{d}q^i|_r \in T^*_rQ$$. A basis gives a linear coordinate chart for a vector space such that for any $$\mathbf{v}_r \in T_rQ$$ we can write $$\mathbf{v}_r = (\mathbf{d}q^i \cdot \mathbf{v}_r)\pmb{\partial}_i=:v^i\pmb{\partial}_i$$ (i should technically write $$\mathbf{d}q^i|_r$$ and $$\pmb{\partial}_i|_r$$). We can then define a local coordinate chart on tangent bundle by the ''tangent lift'' $$T\pmb{q}=:(\pmb{q},\pmb{u}):TQ_{(q)}\to\mathbb{R}^{2n}$$ as

$$(\pmb{q},\pmb{u}):TQ_{(q)}\to\mathbb{R}^{2n} \quad,\quad (r,\mathbf{v}_r) \mapsto (\pmb{q}(r), \pmb{u}(r,\mathbf{v}_r) ) = ( q^1(r),\dots,q^n(r), v^1,\dots, v^n) \quad,\quad u^i(r,\mathbf{v}_r) = \mathbf{d}q^i|_r\cdot\mathbf{v}_r =:v^i$$

That is, $$q^i\in\mathcal{F}(Q)$$ is the same as before (although now treated as $$q^i\circ \tau\in\mathcal{F}(TQ)$$ where $$\tau:TQ\to Q$$ is the projection) and $$u^i\in\mathcal{F}(TQ)$$ just gives the components of any tangent vector in the $$\pmb{q}$$ coordinate basis from Eq(1).

My Main Question: Now, for some vector field $$\mathbf{X}\in\mathfrak{X}(TQ)$$, we can write it in the local basis for the above coordinates $$(\pmb{q},\pmb{u})$$ as

$$\mathbf{X} = V^i \pmb{\partial}_{q^i} \,+\, W^i \pmb{\partial}_{u^i} \; \in\mathfrak{X}(TQ)$$ where $$V^i,W^i$$ are just names I've given the components. As I understand it, if I re-name the $$q^i$$ coordinate basis vectors in Eq (1) as $$\mathbf{e}_{q^i}=\frac{\partial r^a}{\partial q^i} \vec{e}_a\in\mathfrak{X}(Q)$$, then the above $$\pmb{\partial}_{q^i}\in\mathfrak{X}(TQ)$$ can essentially be viewed as $$\pmb{\partial}_{q^i} = \mathbf{e}_{q^i} \oplus \mathbf{0}$$ (am I correct about this?). My main question is, what is the extrinsic interpretation of the the above $$\pmb{\partial}_{u^i}$$ basis vectors? Is there a way to define/view them in terms of the $$\vec{e}_a\in\mathbb{E}^{3N}$$ as in Eq(1)? Or in terms of the $$\mathbf{e}_{q^i}\in\mathfrak{X}(Q)$$?

Anther question: Same as above but for the basis 1-forms $$\mathbf{d}q^i\in\Omega^1(TQ)$$ and $$\mathbf{d}u^i\in\Omega^1(TQ)$$. If I re-name the basis 1-forms from Eq(1) as $$\pmb{\epsilon}^{q^i}\in\Omega^1(Q)$$, then I think that I view the $$\mathbf{d}q^i\in\Omega^1(TQ)$$ (which is really $$\mathbf{d}(q^i\circ \tau)$$) as $$\pmb{\epsilon}^{q^i}\oplus \pmb{0}$$? But what about the $$\mathbf{d}u^i$$?

We can all see the effort you put into writing this question.

However, I think trying to work with a global non-accelerating basis'' is itself a dead end. Consider in general theory of relativity, your spacetime has curvature, then it should be clear that such a thing cannot be assumed to exist.

This forces us to work solely in terms of tangent spaces. As long as we are not near a singularity, the derivatives of any arbitrarily-laid-but-still-sensible coördinates would define the coördinate basis $$\partial_q$$ at any point. This is necessarily a linear space (which is later seen to be what physics requires a vector space to be, as opposed to other possible mathematical linear spaces that do not behave correctly as a vector to physics). Addition, subtraction and scaling of vectors in this tangent space is well-defined, whereas the original coördinates do not allow for such operations when there is curvature.

The union of all the tangent spaces (at each coördinate point) is the tangent bundle. I am not sure why you are using $$TQ$$ for this, when in fact your base thing should be a manifold $$M$$, and $$TM$$ the tangent bundle of this manifold. For the stuff in this tangent bundle, you have to specify the coördinates $$q^i$$ they are to be found at, and contravariant $$v^i$$ for their values, where $$v^i \partial_i$$ is the actual invariant vectorial quantity. Do note that you are supposed to be able to make arbitrary higher ranked tensors on this.

Because even in flat spacetime, curvilinear coördinates means that the tangent spaces at neighbouring points are actually not the same as each other, notions of parallel transport and connections come into play. That is where Christoffel symbols already come into play even in flat spacetimes.

More interesting is to consider the n-forms in such a scheme. Then you need the cotangent bundle $$T^*M$$, and its basis would be $$\mathrm d q^i$$. The actual invariant 1-form would be $$p_i \mathrm d q^i$$ at the location $$q^i$$, where, sadly, the ambiguity of using the same coördinate symbols as the 1-form basis is the norm. Covariant momentum $$p_i$$ are the ones that are conserved by Noether's theorem.

So, no, I am saying that you should consider coördinates as completely separate from vectorial quantities, and instead that the vectors and forms as living on tangent and cotangent spaces.

Just be aware that if you want to work with tensors, differential forms, and so forth in such nice coördinate bases, yes, you can derive beautiful theorems and so on. Everything would have sensible physical interpretation. However, they disagree with most of the formulӕ written in textbooks. Textbooks generally work in the tetrad formalism, so that you have to be wary of using the wrong formulӕ every time you want to actually do a computation.

• My $Q$ is just a smooth configuration manifold. E.g., for a double pendulum it would be a 2-torus, $Q=T^2$, for a spherical pendulum it would be a 2-sphere $Q=S^2$, etc. So $Q$ is a possibly-curved space that is embedded in a linear space. If no constraints are imposed then $Q$ is the linear space itself. If we take $Q=S^2\subset \mathbb{E}^3$, by ''global non-accelerating basis'' I basically mean the thing many undergrad textbooks label $\hat{x},\hat{y},\hat{z}$ but I didn't impose the restriction they be orthonormal. (cont.) Apr 13 at 15:01
• My first equation is then $\vec{r}(\pmb{q}) = a\sin\theta\cos\phi\hat{x} + a\sin\theta\sin\phi\hat{y}+a\cos\theta\hat{z}$ where $\pmb{q}=(\theta,\phi)$ are the usual coordinate for $S^2$ ($a$ the radius of the 2-sphere). The coordinate basis I have in Eq(1) is then the usual $$\mathbf{e}_\phi = -a\sin\theta\sin\phi \hat{x} + a\sin\theta\cos\phi \hat{y}$$ $$\mathbf{e}_\theta = a\cos\theta\cos\phi \hat{x} + a\cos\theta\sin\phi \hat{y} -a\sin\theta \hat{z}$$ Apr 13 at 15:04
• So, the clarify, Im not trying to work with a global non-accelerating basis for the manifold $Q$ itself; i'm trying to understand how a local coordinate basis on $TQ$ is defined in terms of a global basis for the linear space i which $Q$ is embedded. Apr 13 at 15:12
• I am specifically saying that the global basis may not exist. I know you are trying to learn about it by considering a curved submanifold of a flat manifold. That definitely makes things easier, but what is the point of learning only about this specific case when the general case is just right at hand? Apr 14 at 6:39
• I agree that in GR this framework is not all that useful and I prefer the intrinsic picture anyway, but it should be noted that the statement "Consider in general theory of relativity, your spacetime has curvature, then it should be clear that such a thing cannot be assumed to exist" is technically false. By Whitney's embedding theorem any smooth manifold can be embedded in a larger euclidean space. So replacing $3N$ with a large enough integer, the rest of the question should remain valid. However, I do advise you to take the time to learn to work in the intrinsic picture as that is nicer! Apr 19 at 12:40

The mathematics of your question is sound

I feel like $$\pmb{\partial}_{i}$$ is not actually equal to a linear combination of $$\vec{e}_a$$ as I have above, but is instead isomorphic in some trivial way?

You are right, for $$V$$ a vector space ($$\mathbb{E}^{3N}$$ in this case) and $$v\in V$$, there is a canonical isomorphism $$V\cong T_vV$$ by associating to $$w\in V$$ the derivative at $$v$$ in the direction of $$w$$, $$\partial_{w}|_v\colon f\mapsto \lim_{\delta\to0}\frac{f(v+\delta w)}{v+\delta w}.$$ So in your case, you view $$T_qQ\subset T_q\mathbb{E}^{3N}\cong\mathbb{E}^{3N}$$.

what is the extrinsic interpretation of the the above $$\pmb{\partial}_{u^i}$$ basis vectors?

Since we canonically have $$T_q\mathbb{E}^{3N} = \mathbb{E}^{3N}$$, we find $$T\mathbb{E}^{3N} = \mathbb{E}^{3N}\oplus\mathbb{E}^{3N} = \mathbb{E}^{6N}$$ can view $$TQ$$ extrinsically as being the embedding $$TQ := \{ q \oplus u ~|~ q\in Q,~ u\in T_qQ \}\subset \mathbb{E}^{3N}\oplus \mathbb{E}^{3N} = T\mathbb{E}^{3N}.$$ Then, again, we have $$T_{q,u}(TQ)\subset T_{q,u}(\mathbb{E}^{3N}\oplus \mathbb{E}^{3N})\cong \mathbb{E}^{3N}\oplus \mathbb{E}^{3N}$$. The $$\partial_{q^i}$$ live in the left-hand summand and can be thought of as the directional derivatives in the directions of $$Q$$ while the $$\partial_{u^i}$$ live in the right-hand summand and can be thought of as the directional derivatives in the direction of the fiber $$T_qQ$$. Thinking of $$TQ\to Q$$ as 'projecting down' to $$Q$$, we call the vectors in the first summand 'horizontal' and in the second 'vertical', similar to the terminology one might encounter when studying (principal) fiber bundles.

The picture for one-form is (by duality, see also https://math.stackexchange.com/a/1305849/1026213) exactly the same.

• Ok this helps but I'm still a bit lost. If we view the tangent bundle coordinate basis vectors as $\partial_{q^i},\partial_{u^i}\in\mathbb{E}^{3N}\oplus\mathbb{E}^{3N}$, then are the $q^i$ vectors ‘horizontal’ and of the form $\partial_{q^i}= \vec{v}_{q^i}\oplus 0$ (where $\vec{v}_{q^i}=\frac{\partial r^a}{\partial q^i}\vec{e}_a$ are as in my Eq. 1)? ...[cont.] Apr 20 at 19:05
• And, my main question, are the $u^i$ vectors ‘vertical’ and of the form $\partial_{u^i}=0\oplus \vec{v}_{u^i}$ for some set of $n$ linearly independent $\vec{v}_{u^i}\in\mathbb{E}^{3N}$? If so, what is the relation between these $\vec{v}_{u^i}$ and the $\vec{v}_{q^i}$? Apr 20 at 19:07
• For example, if $\pmb{q}=(\phi,\theta)$ are the usual coordinates on a 2-sphere, $S^2\subset\mathbb{E}^3$, of radius $a$ then what I just called $\vec{v}_{q^i}$ are, $$\vec{v}_\phi = -a\sin\theta\sin\phi \hat{x} + a\sin\theta\cos\phi \hat{y}$$ $$\vec{v}_\theta = a\cos\theta\cos\phi \hat{x} + a\cos\theta\sin\phi \hat{y} -a\sin\theta \hat{z}$$. And I would say $\partial_\phi=\vec{v}_\phi\oplus 0$ and $\partial_\theta=\vec{v}_\theta\oplus 0$. How do we (can we) construct the $\partial_{u^\phi}$ and $\partial_{u^\theta}$ basis vectors in a similar manner? Apr 20 at 19:14
• Recall for a coordinate patch $(q^1,\dots,q^n)$ the directional derivatives $(\partial_{q^1}, \dots, \partial_{q^n})$ are a basis of the tangent space $T_qQ$ (for $q$ in the patch). In particular, $(q^1,\dots,q^n,\partial_{q^1}, \dots, \partial_{q^n})$ is a coordinate patch in the tangent space $TQ$. Renaming $\vec{u^i}:=\partial_{q^i}$ (and by abuse of notation letting the $q^i$ also refer to the coordinates in $TQ$), you again find $(\partial_{q^1}, \dots, \partial_{q^n},\partial_{u^1}, \dots, \partial_{u^n})$ to be a basis of the tangent space $T_v(TQ)$ (for $v\in T_qQ$, $q$ in your patch) Apr 21 at 15:58