Coordinate bases: in what sense is $\partial_i=\frac{\partial}{\partial q^i}$ a basis vector?

Question

I have gotten comfortable working with coordinate bases but I still do not fully understand them. Let $\mathbb{Q}$ be a smooth $n$-manifold and $q=(q^1,\dots,q^n):Q\to \mathbb{R}^n$ some local coordinates on some region $Q\subseteq\mathbb{Q}$. For this question, I will denote the associated $q^i$ basis vector fields as $\mathbf{e}_i\in\mathfrak{X}(\mathbb{Q})$ with dual 1-form fields $\pmb{\epsilon}^i\in\Omega^1(\mathbb{Q})$. We may write (locally) any $\mathbf{u}\in\mathfrak{X}(\mathbb{Q})$ in this basis in the usual way: $\mathbf{u}= \pmb{\epsilon}^i(\mathbf{u})\mathbf{e}_i = u^i\mathbf{e}_i$.

Rather than $\mathbf{e}_i$, many sources would instead write $\partial_i$ or $\frac{\partial}{\partial q^i}$. I get that, as a differential operator acting on functions, we have $\mathbf{u}[f] = \mathbf{d}f(\mathbf{u}) = u^i\frac{\partial}{\partial q^i} f$. This is then often used to justify writing $\mathbf{u}$ itself as $u^i\frac{\partial}{\partial q^i}$ and this is what confuses me. In the expression $u^i\frac{\partial}{\partial q^i} f\in\mathcal{F}(\mathbb{Q})$, the role of $\frac{\partial}{\partial q^i}$ is not confusing to me; I view it as the "standard" partial derivative in essentially the same way that any undergraduate student would view it in intro calculus. But I do not understand what type of object $u^i\frac{\partial}{\partial q^i}$ itself is supposed to be.

my question: If $\mathbf{u}=u^i\mathbf{e}_i\in\mathfrak{X}(\mathbb{Q})$ is the actual tensor (expressed in a particular basis), is it accurate to write $\mathbf{u}=u^i\frac{\partial}{\partial q^i}$? If so, this implies $\frac{\partial}{\partial q^i} = \mathbf{e}_i \in \mathfrak{X}(\mathbb{Q})$ are the exact same object? Or, is the use of $\frac{\partial}{\partial q^i}$ as a basis vector merely a slight abuse of notation alluding to the fact that $\mathbf{e}_i[f] = \frac{\partial}{\partial q^i}f$?

In other words, if I insist on reserving the notation $\frac{\partial}{\partial q^i}$ for the "standard" partial derivative one encounters in basic calculus, is this same object also literally a vector field on the domain of the function $q^i$? Or, instead, does there exist some vector field $\mathbf{e}_i\in\mathfrak{X}(\mathbb{Q})$ on the domain of the coordinate $q^i$ whose action on a function is given by $\mathbf{e}_i[f] = \frac{\partial}{\partial q^i}f$, but where $\mathbf{e}_i$ and $\frac{\partial}{\partial q^i}$ are not literally the same object?

Answer 1

One way of defining the tangent space on a manifold is to define its vectors as derivations.

Further, note, that $\partial_i$ on a manifold is not the "standard" elementary partial derivative to begin with – it is rather a chart dependent notion of differentiation on the manifold. Given a chart $(U, \psi)$, a function $f \colon M \to \mathbb{R}$ and a point $p \in U$ we have $$ \partial_i f(p) = \big( \partial_i (f \circ \psi^{-1}) \big)\big(\psi(p)\big), $$ where only the $\partial_i$ on the right hand side is an elementary partial derivative.

If you take this point of view, identifying tangent space vectors with derivations, then it is natural to identify the $\partial_i$ on the left-hand side with a vector field (just defined on the chart – not necessarily on the whole manifold). Futhermore, the $\partial_i$ at $p$ will form a basis of the tangent space $T_pM$.

The connection to other representations of the tangent space is given by vector space isomorphisms. This is a common theme in mathematics: Construct something in different manners, show that the constructions are isomorphic, and the identify them by this equivalence as the same underlying object in different guises. Then you can freely use the one or the other picture as required or convenient for the problem at hand.

We consider a manifold $M$ smoothly embedded into $\mathbb{R}^n$, the tangent vectors at $p$ in the tangent hyperplane are denoted by $\tilde T_p M$, while the tangent space formulated in terms of derivations is $T_p M$.

Any function $f \in C^\infty(\mathbb{R}^n)$ induces a function $f \in C^\infty(M)$ (by restricting the domain). Given a vector $\mathbf{e} \in \tilde T_p M$, we can define a functional $D_p \colon C^\infty(M) \to \mathbb{R}$ $$ D_p f = \mathbf{e} \cdot \nabla f(p) $$ (where $\nabla f$ is the usual gradient of $f$ in $\mathbb{R}^n$)

We can easily check, this functional $D_p$ is a derivation at $p$, so it is an element of $T_pM$. Further it is easy to prove, that this mapping from $\tilde T_pM \to T_pM$ does preserve vector addition and scaling – so it is a vector space homomorphism. (And you can also carry over additional properties, like the scalar product, if you consider the Riemannian structure induced on the embedded manifold from the metric in the embedding space.)

Conversely, you can show that any derivation $D_p$ corresponds to a vector in $\tilde T_p M$. (This is a bit more involved – you need to compute the dimension of $T_p M$ and show that the mapping $\mathbf{e} \mapsto D_p$ above is injective.)

So in summary, there is a vector space isomorphism identifying the tangent vectors as derivations to tangent vectors in a hyperplane of the embedding space – and this allows you to think of $\mathbf{e}_\phi$ as $\partial_\phi$. The most intuitive way you can formulate this, is that vectors exactly correspond to directional derivatives.

Answer 2

Because $\vec u=u^i\vec e_i$ is a real vector (field) that is physically there and invariant, the contravariant components $u^i$ of this vector is actually dependent upon the choice of the covariant basis vectors $\vec e_i$, and it is often the case that we would deliberately choose not to deal with coördinate bases. In particular, the classical physics expressions are most often written in vierbeins rather than in terms of coördinate bases, and that means a lot of the wonderful differential geometry formulæ cannot be directly used.

It is always clearer to speak of stuff when we have direct examples, so here goes: Especially when spacetime is curved, we are allowed to simply lay down any coördinates we want. The thing we lose is that position is no longer a vector. The usual Cartesian niceness of $\vec r=(x,y,z)^T$ no longer works, not even in the flat curvilinear case, and all earlier schemes whereby we use polar coördinates to parametrise vectors in the Cartesian form, will stop working in curved spacetime. I think you might be familiar with the argument of how this fails on the spherical surface of the Earth.

In the neighbourhood of any point, we may define velocity vectors. Picking spherical polar coördinates, we have $$\tag1\vec v=\dot r\vec\partial_r+\dot\vartheta\vec\partial_\vartheta+\dot\varphi\,\vec\partial_\varphi$$ and its interpretation should be obvious. However, this is not what is usual in classical physics. Instead, classical physics likes to deal with vierbeins form, which look like $$\tag2\vec v =\dot r\vec\partial_r+r\,\dot\vartheta\left(\frac1r\vec\partial_\vartheta\right)+r\sin\vartheta\,\dot\varphi\left(\frac1{r\sin\vartheta}\vec\partial_\varphi\right) =\dot r\hat{\vec\partial_r}+r\,\dot\vartheta\,\hat{\vec\partial_\vartheta}+r\sin\vartheta\,\dot\varphi\,\hat{\vec\partial_\varphi}$$ and with this, it is immediately clear the difference: coördinate bases tend to not be unit length, while vierbeins are defined to be unit length. This rescaling means that the nice formulæ in tensors and differential geometry will disagree with the formulæ found on Wikipedia and in textbooks, because they are all following classical physics conventions.

The state of affairs is rather annoying, but here, it is the physics that is annoying, rather than any person teaching that is omitting details. This is because, per tensor analysis, all these contravariant and covariant business is actually physically meaningful. Nature simply wants us to think in terms of contravariant and covariant business. The most important example is that, in Noether's theorem, the position coördinates HAS to be expressed in contravariant form, whereas the conserved quantities are covariant. In particular, consider angular momentum, and for simplicity, I will consider just the equatorial plane, i.e. $\vartheta=\dfrac\pi2$. In the Lagrangian, you have to write that the RKE $=\dfrac12mr^2\dot\varphi^2$, whereas in the Hamiltonian, you have to write that it is RKE$=\dfrac{L^2}{2mr^2}$. In particular, they differ in that it is \begin{align}\tag{3 in Lagrangian}\text{KE}&=\frac12m \begin{bmatrix}\dot r&\dot\varphi \end {bmatrix} \begin{bmatrix}1&\\&r^2 \end {bmatrix} \begin{bmatrix}\dot r\\\dot\varphi \end {bmatrix}=\frac12mv^ag_{ab}v^b \\&=\frac1{2m} \begin{bmatrix}p_r&L \end {bmatrix} \begin{bmatrix}1&\\&1/r^2 \end {bmatrix} \begin{bmatrix}p_r\\L \end {bmatrix}=\frac1{2m}p_ag^{ab}p_b \tag{4 in Hamiltonian} \end{align} and you can see that the metric are inverses of each other.

Note that this is happening even in flat spacetime. We were just writing down the operators in curvilinear coördinates. The thing that is conserved is not the full, invariant momentum 1-form, but rather only the covariant components thereof. That is, contravariant momentum or vierbein momentum is NOT conserved. This is why we say that momentum is naturally a covariant quantity, and why velocity is naturally a contravariant quantity (interpreted as being inherited from position being contravariant).

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In view of the above, the fact that coördinate bases are so special and easy to use, warrants that, if a formula depends upon it, we should use partial derivatives to denote it. If we have a formula that works in general bases, then $\vec e_i$ would be more appropriate.

Now, if $u^i=\delta^{ik}$ for some fixed $k$, and you let this roam free over all points on the coördinate patch, then $\vec u=\vec\partial_k$ is a vector field associated with the $k^\text{th}$ coördinate. Again, interpretation is easy. So, yes, the partial derivative operator can be trivially considered as a vector field. It might not be very useful, but you can do it that way.

Do note that although contravariance and covariance (and note again that the conventional lingo is opposite between maths and physics on this point) is physically meaningful, vierbeins are the way to go. This is only because we want to deal with spin-half, and if we only have the metric, it would not be clear how to orient the spins. Instead, having vierbeins define alignment of directions, it becomes easier to define the spinors in curved spacetime, and so it is important to learn that too. Because of all these complications, it is basically impossible to write a GR text that covers everything completely in one go.