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  Is partial derivative a vector or dual vector?

+ 7 like - 0 dislike

The textbook(Introduction to the Classical Theory of Particles and Fields, by Boris Kosyakov) defines a hypersurface by $$F(x)~=~c,$$ where $F\in C^\infty[\mathbb M_4,\mathbb R]$. Differentiating gives $$dF~=~(\partial_\mu F)dx^\mu~=~0.$$ The text then says $dx^\mu$ is a covector and $\partial_\mu F$ a vector. I learnt from another book that $dx^\mu$ are 4 dual vectors(in Minkowski space), $\mu$ indexes dual vector themselves, not components of a single dual vector. So I think $\partial_\mu F$ should also be 4 vectors, each being the directional derivative along a coordinate axis. But this book later states that $(\partial_\mu F)dx^\mu=0$ describes a hyperplane $\Sigma$ with normal $\partial_\mu F$ spanned by vectors $dx^\mu$, and calls $\Sigma$ a tangent plane (page 33-34). This time, it seems to treat $\partial_\mu F$ as a single vector and $dx^\mu$ as vectors. But I think $dx^\mu$ should span a cotangent space.

I need some help to clarify these things.

[edit by Ben Crowell] The following appears to be the text the question refers to, from Appendix A (which Amazon let me see through its peephole):

Elie Cartan proposed to use differential coordinates $dx^i$ as a convenient basis of 1-forms. The differentials $dx^i$ transform like covectors [...] Furthermore, when used in the directional derivative $dx^i \partial F/\partial x^i$, $dx^i$ may be viewed as a linear functional which takes real values on vectors $\partial F/\partial x^i$. The line elements $dx^i$ are called [...] 1-forms.

This post imported from StackExchange Physics at 2014-11-11 14:50 (UTC), posted by SE-user elflyao
asked Oct 31, 2014 in Mathematics by elflyao (35 points) [ no revision ]
Most voted comments show all comments
@MBN: "$\partial_\mu F$ is a function not a vector." You mean it's a function in the sense that it varies over spacetime? Sure, that's certainly true, but it's a vector-valued function. "Perhaps you mean the fourtuple of the $\partial_\mu F$'s?" What I would normally mean by this is that when the index $\mu$ is unbound, and no coordinate system has been specified, the index notation $\partial_\mu F$ indicates the entire gradient covector, as in abstract index notation. This is similar to "the function $f(x)$," where $x$ is unbound. Kosyakov's "vectors $\partial F/\partial x^i$," with [...]

This post imported from StackExchange Physics at 2014-11-11 14:50 (UTC), posted by SE-user Ben Crowell
plural "vectors," suggests that he thinks of something like $\partial F/\partial x^0$ as a vector. This doesn't make much sense to me, since it clearly can't have the transformation properties of a vector. It's a single real number, and, e.g., under a change of coordinates $x^0\rightarrow 7x^0$, $\partial F/\partial x^0 \rightarrow (1/7)\partial F/\partial x^0$, which is what a component of a covector would do. A component of a vector would get 7 times bigger.

This post imported from StackExchange Physics at 2014-11-11 14:50 (UTC), posted by SE-user Ben Crowell
@BenCrowell: OK, then it makes sense. To me it looked like one partial derivative.

This post imported from StackExchange Physics at 2014-11-11 14:50 (UTC), posted by SE-user MBN
Wasn't this question already answered? I see a lot of great answers below...

This post imported from StackExchange Physics at 2014-11-11 14:50 (UTC), posted by SE-user QuantumBrick
I compared Kosyakov's definitions with Wald, General Relativity, pp. 15 and 20ff. There are some inconsistencies. Wald defines $\mathscr{F}$ as the set of smooth scalar fields on a manifold $M$, and defines a tangent vector $v\in V_p$ as a map $v:\mathscr{F}\rightarrow \mathbb{R}$ that is linear and obeys the Leibniz rule at $p$, so that it can be interpreted as a directional derivative at $p$. This makes a partial derivative a vector as a matter of definition, and it also means that for $F\in\mathscr{F} $, $\partial F/\partial x^i$ is a real number, not a vector as Kosyakov describes it.

This post imported from StackExchange Physics at 2014-11-11 14:50 (UTC), posted by SE-user Ben Crowell
Most recent comments show all comments
After writing an answer and then succeeding in getting a look at what Kosyakov wrote, I'm just as confused as elflyao. I would be interested in hearing from others who might have broader experience or be able to explain whether Kosyakov has an unusual point of view.

This post imported from StackExchange Physics at 2014-11-02 19:45 (UTC), posted by SE-user Ben Crowell
cross-posted here: physicsforums.com/threads/…

This post imported from StackExchange Physics at 2014-11-11 14:50 (UTC), posted by SE-user Ben Crowell

6 Answers

+ 7 like - 0 dislike

Below follows a handful of excerpts from the book Introduction to the Classical Theory of Particles and Fields (2007) by B. Kosyakov.

Controversial/misleading/wrong statements are marked in $\color{Red}{\rm red}$. We agree with OP that the statements marked in $\color{Red}{\rm red}$ are opposite standard terminology/conventions. Some (not all) correct statements are marked in $\color{Green}{\rm green}$.

1.2 Affine and Metric Structures

[...] Let ${\bf e}_1$, $\ldots$, ${\bf e}_n$ and ${\bf e}^{\prime}_1$, $\ldots$, ${\bf e}^{\prime}_n$ be two arbitrary bases. Each $\color{Green}{\rm vector}$ of the latter basis can be expanded in terms of $\color{Green}{\rm vectors}$ of the former basis: $$ {\bf e}^{\prime}_i ~=~ {\bf e}_j~L^j{}_i .\tag{1.37} $$

[...] Thus, linear functionals form the dual vector space $V^{\prime}$. If $V$ is $n$-dimensional, so is $V^{\prime}$. Indeed, let ${\bf e}_1$, $\ldots$, ${\bf e}_n$ be a basis in $V$. Then any $\omega\in V^{\prime}$ is specified by $n$ real numbers $\omega_1=\omega({\bf e}_1)$, $\ldots$, $\omega_n=\omega({\bf e}_n)$, and the value of $\omega$ on ${\bf a} = a^i {\bf e}_i$ is given by $$\omega({\bf a}) ~=~ \omega_i a^i .\tag{1.52} $$ We see that $V^{\prime}$ is isomorphic to $V$. That is why we sometimes refer to linear functionals as $\color{Green}{covectors}$. A closer look at (1.52) shows that a $\color{Green}{\rm vector}$ ${\bf a}$ can be regarded as a linear functional on $V^{\prime}$. One can show (Problem 1.2.3) that changing the basis (1.37) implies the transformation of $\omega_i$ according to the same law: $$ \omega^{\prime}_i ~=~\omega_j ~L^j{}_i .\tag{1.53} $$ We will usually suppress the argument of $\omega({\bf a})$, and identify $\omega$ with its components $\omega_i$. [...]

1.3 Vectors, Tensors, and $n$-Forms

[...] A simple generalization of vectors and covectors are tensors. Algebraically, a tensor $T$ of rank $\color{Green}{(m,n)}$ is a multilinear mapping $$\color{Green}{T: \underbrace{V^{\prime} \times\ldots\times V^{\prime}}_{m\text{ times}} \times \underbrace{V \times\ldots\times V}_{n\text{ times}} \to \mathbb{R}}. \tag{1.112} $$ We have already encountered examples of tensors in the previous section: a scalar is a rank $(0,0)$ tensor, a $\color{Green}{\rm vector}$ is a rank $\color{Green}{(1,0)}$ tensor, a $\color{Green}{\rm covector}$ is a rank $\color{Green}{(0,1)}$ tensor, the metric $g_{ij}$ is a rank $(0,2)$, while $g^{ij}$ is a rank $(2,0)$ tensor, and the Kronecker delta $\delta^i{}_j$ is a rank $(1,1)$ tensor. Just as $\color{Green}{\rm four~vectors}$ can be regarded as objects which transform according to the law $$ a^{\prime \mu} ~=~ \color{Green}{\Lambda^{\mu}{}_{\nu}} ~a^{\nu} ,\tag{1.113} $$ where $\Lambda^{\mu}{}_{\nu}$ is the Lorentz transformation matrix relating the two frames of reference, so tensors of rank $(m,n)$ can be described in terms of Lorentz group representations by the requirement that their transformation law be $$T^{\prime\mu_1\cdots \mu_m}{}_{\nu_1\cdots \nu_n} ~=~\color{Green}{\Lambda^{\mu_1}{}_{\alpha_1}\ldots\Lambda^{\mu_m}{}_{\alpha_m}}~ T^{\alpha_1\cdots \alpha_m}{}_{\beta_1\cdots \beta_n}~ \color{Red}{\Lambda^{\beta_1}{}_{\nu_1}\ldots\Lambda^{\beta_n}{}_{\nu_n}}. \tag{1.114} $$

[...]The differential operator $$ \partial_{\mu}~=~\frac{\partial}{\partial x^{\mu}} \tag{1.140} $$ transforms like a $\color{Green}{\rm covariant~vector}$. To see this, we use the chain rule for differentiation: $$ \frac{\partial}{\partial x^{\mu}}~=~\frac{\partial x^{\prime \nu}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\prime \nu}}, \tag{1.141} $$ and note that, for linear coordinate transformations $x^{\prime\mu} = \color{Green}{\Lambda^{\mu}{}_{\nu}}~x^{\nu} + a^{\mu}$ $$\frac{\partial x^{\prime \mu}}{\partial x^{\nu}} ~=~\color{Green}{\Lambda^{\mu}{}_{\nu}}. \tag{1.142} $$ We will always use the shorthand notation $\partial_{\mu}$, and treat this differential operator as an ordinary $\color{Green}{\rm vector}$. [...]

1.4 Lines and Surfaces

[...] We define a hypersurface $M_{n−1}$ by $$F(x) ~=~ C , \tag{1.176} $$ where $F$ is an arbitrary smooth function $\mathbb{M}_4 \to \mathbb{R}$. Differentiating (1.176) gives $$ (\partial_{\mu}F) dx^{\mu} ~=~ 0 . \tag{1.177} $$ One may view $dx^{\mu}$ as a $\color{Green}{\rm covector}$, and $\partial_{\mu}F$ as a $\color{Red}{\rm vector}$. Indeed, $dx^{\mu}$ transforms like a $\color{Red}{\rm covector}$ under linear coordinate transformations $x^{\prime\mu} = \color{Green}{\Lambda^{\mu}{}_{\nu}}~x^{\nu} + a^{\mu}$, $$dx^{\prime\mu} ~=~ \frac{\partial x^{\prime\mu}}{\partial x^{\nu}}dx^{\nu} ~=~\color{Green}{\Lambda^{\mu}{}_{\nu}}~dx^{\nu}, \tag{1.178} $$ and $\partial_{\mu}F$ transforms like a $\color{Red}{\rm vector}$: $$\frac{\partial F}{\partial x^{\prime\mu}} ~=~\frac{\partial F}{\partial x^{\nu}} \frac{\partial x^{\nu}}{\partial x^{\prime\mu}} ~=~\frac{\partial F}{\partial x^{\nu}}\color{Red}{\Lambda^{\nu}{}_{\mu}}. \tag{1.179} $$ In Minkowski space, vectors and covectors can be converted to each other according to (1.121). For this reason, we will often regard $dx^{\mu}$ as vectors. [...]

A. Differential Forms

[...] Elie Cartan proposed to use differential coordinates $dx^i$ as a convenient basis of $\color{Green}{\rm one~forms}$. The differentials $dx^i$ transform like $\color{Red}{\rm covectors}$ under a local coordinate change, $$dx^{\prime j}~=~ \frac{\partial x^{\prime j}}{\partial x^i}dx^i. \tag{A.1} $$ [If the coordinate change is specialized to Euclidean transformations $x^{\prime j} =\color{Red}{L^j{}_i} ~x^i + c^j$, then $\partial x^{\prime j} /\partial x^i$ reduces to $\color{Red}{L^j{}_i}$, an orthogonal matrix with constant entries, and (A.1) $\color{Red}{\rm becomes}$ (1.53), the transformation law for $\color{Green}{\rm covectors}$.] [...]


  1. The corrected eq. (1.114) reads $$T^{\prime\mu_1\cdots \mu_m}{}_{\nu_1\cdots \nu_n} ~=~\Lambda^{\mu_1}{}_{\alpha_1}\ldots\Lambda^{\mu_m}{}_{\alpha_m}~ T^{\alpha_1\cdots \alpha_m}{}_{\beta_1\cdots \beta_n}~ (\Lambda^{-1})^{\beta_1}{}_{\nu_1}\ldots(\Lambda^{-1})^{\beta_n}{}_{\nu_n}. \tag{1.114} $$

  2. The corrected eq. (1.179) reads $$\frac{\partial F}{\partial x^{\prime\mu}} ~=~\frac{\partial F}{\partial x^{\nu}} \frac{\partial x^{\nu}}{\partial x^{\prime\mu}} ~=~\frac{\partial F}{\partial x^{\nu}}(\Lambda^{-1})^{\nu}{}_{\mu}. \tag{1.179} $$

  3. To explain why (A.1) does not becomes (1.53), let ${\bf e}^1$, $\ldots$, ${\bf e}^n$, be a (dual) basis in $V^{\prime}$. In light of (1.53), in order for a covector $\omega=\omega_i{\bf e}^i\in V^{\prime}$ to be independent of the choice of basis, the dual basis must transform as $$ {\bf e}^{\prime i} ~=~ M^i{}_j ~{\bf e}^j, \tag{*}$$ where $$ M~=~L^{-1}. \tag{1.45}$$ Identifying the dual bases ${\bf e}^i\leftrightarrow dx^i$, the above eq.(*) becomes (A.1). Moreover, in the sentence below eq. (A.1), the $L$ matrix should be replaced with the $M$ matrix in two places.

  4. Finally, let us answer OP's title question: A partial derivative $\partial_{\mu}F$ (of a scalar function $F$) is a component of a cotangent vector $dF=(\partial_\mu F)dx^\mu$, while the un-applied partial derivative $\partial_{\mu}$ is a local basis element of a tangent vector. Both $\partial_{\mu}F$ and $\partial_{\mu}$ transform as covectors.

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Qmechanic
answered Nov 3, 2014 by Qmechanic (3,120 points) [ no revision ]
well point 4. is nicely put, but i dont like so much the red/green tags, as this is convention than anything else (of course any book errata excepted)

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Nikos M.
@NikosM.: Qmechanic's points 1 and 2 are not matters of convention; Kosyakov has made mistakes in those spots.

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Ben Crowell
+ 6 like - 0 dislike

I took a quick look at pages 59 and 60 of "Gravitation", section 2.6 "Gradients and Directional Derivatives", to see if there's anything there we can use to clarify this issue.

In this section, the gradient of $f$ is $\mathbf df$, the directional derivative along the vector $\mathbf v$ is $\partial_{\mathbf v}f$ and the following relationship holds:

$$\partial_{\mathbf v}f = \langle\mathbf df, \mathbf v \rangle$$

Then assuming a set of basis forms $\mathbf dx^{\mu}$ and dual basis vectors $\mathbf e_{\mu}$ we have

$$\partial_{\mu} f \equiv \partial_{\mathbf e_{\mu}}f = \langle\mathbf df, \mathbf e_{\mu} \rangle = \frac{\partial f}{\partial x^{\mu}}$$

So, according to MTW in this section, $\partial_{\mu} f$ are the components of $\mathbf df$ on this basis.

Thus, it must be that, per the 2nd equation in the question,

$$\mathbf df = (\partial_{\mu} f) \mathbf dx^{\mu} $$

which is just the expansion of the form $\mathbf df$ on the basis forms $\mathbf dx^{\mu}$

As to why Kosyakov would identify this as a contraction of a form and a vector I haven't a clue.

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Alfred Centauri
answered Nov 2, 2014 by Alfred Centauri (110 points) [ no revision ]
+ 4 like - 0 dislike

I believe this is just imprecise use of language by the author - there is nothing mysterious happening, it is just not well stated:

As stated in the question, for a hypersurface $\Sigma$ defined by

$$ F(x) = c \in \mathbb{R}$$

we find that

$$ \mathrm{d}F = 0$$

must hold on $\Sigma$. This is crucial - it means that the 1-form $\mathrm{d}F$ acting upon tangent vectors of $\Sigma$ must vanish identically:

$$ \forall v \in T_x\Sigma : \mathrm{d}F(v) = (\partial_\mu F)v^\mu = 0$$

But we can recognize $(\partial_\mu F)v^\mu$ as the scalar product of the vectors $v$ and $g(\mathrm{d}F,\dot{})$, the latter being the usual dual of $\mathrm{d}F$ with components $\partial^\mu F$. Since $T_x\Sigma \subset T_x\mathbb{M}^4$ naturally, this means that $\mathrm{d}F = 0$ indeed sweeps out a hypersurface in the tangent space that has, in sloppy diction, the gradient as its normal (although it is really its dual).

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user ACuriousMind
answered Nov 2, 2014 by ACuriousMind (910 points) [ no revision ]
I think the crucial point is in the sentence "But we can recognize...," and it's here that I don't follow you. Since the gradient $dF$ is a 1-form, its dual $\partial^\mu F$ is a vector. If so, then we can't take the scalar product of the vector $\partial^\mu F$ with a vector $v^\mu$. I don't see any reason for taking duals anywhere at all. Even if we didn't have a metric, and therefore couldn't take duals, we could simply have $(\partial_\mu F)v^\mu$, the scalar product of a covector with a vector.

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Ben Crowell
@BenCrowell: Yes, indeed. That will still define a hypersurface in the tangent space, but we will not have a "normal vector" to describe it. I agree that there is no need to take duals - but I think Kosyakov implicitly does exactly that when he talks of $\partial_\mu F$ being a normal vector. Your nomenclature seems a bit unorthodox to me though - a scalar product is between two vectors or two covectors (and usually induced by the metric) - applying a covector to a vector is not a scalar product.

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user ACuriousMind
OK, by "scalar product" I simply meant a product that transforms as a scalar. So taking "scalar product" in your answer to mean $g(\cdot,\cdot)$, I don't understand why one would describe $(\partial_\mu F)v^\mu$ as a scalar product of $v^\mu$ with $\partial^\mu$. That might indicate that we take the gradient, raise its index, lower its index, and then contract. I don't see the point of raising an index and then immediately lowering it again. Or we could raise the gradient's index, lower $v^\mu$'s index, and contract. Again, why raise or lower at all?

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Ben Crowell
@BenCrowell: Because it yields the geometric interpretation of $\partial^\mu F$ as the normal vector to the hyperplane of tangent vectors of $\Sigma$. I don't think there's anything deeper than that here.

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user ACuriousMind
+ 3 like - 0 dislike

On a manifold without extra structure, $dF$ makes sense only as a covector  (= 1-form = a section of the cotangent space). For example, like any 1-form, one can integrate $dF$ along a path (but unlike for a general 1-form, the integral only depends on the endpoints). To interpret $dF$ as a vector, one needs a metric that can be used to identify vectors and covectors.

You can see that $dF$ must be a covector from your formula, since as a linear combination of the covectors $dx^\mu$,  $dF$ is itself a covector - with components $dF_\mu$.  The term $dF_\mu(x)$ itself is just a number, and $dF_\mu$ a noncovariant function. 

Physicists using the Einstein summation convention refer to a tensor by its indices, then $dF_\mu$ is what the mathematicians call $dF$ - a covector. (But the physicists notation gets confusing when it is applied to $dx^\mu$, which looks like a vectoraccording to its indeices, but in fact is for each fixed $\mu$ a covector.)

answered Nov 9, 2014 by Arnold Neumaier (15,787 points) [ revision history ]
+ 3 like - 1 dislike

Hmm I guess the that the simplest way to think about it is the following. If you have a differentiable manifold \(X\)(which for the shake of argument let us suppose it enjoys all nice properties we like) then it is possible to define the tangent and co-tangent space at this point such  that the partial derivatives \(\partial_i \Big|_{p \in X} \)form an orthonormal basis in the tangent space \(T_pX\) and the 1-forms \(dx^{i}\)are elements of the co-tangent space \(T_p^*X\). Both of these quantities form a real vector space and they are isomorphic to each other. By the way, your first formula gives zero because F is constant. F is a 4-dimensional plane In general we have 

\(dF = \sum_i \partial_i F \, dx^i\)

You can understand the partial derivative as a vector if you really think of what it is doing upon acting on a scalar. It gives it a direction! Draw a graph of a smooth scalar function and take the tangent at a point. Intuitively you have drawn a vector (dir. derivative as you mention) from the point of contact to some direction.  Finally if your space is Minkowski then you have \(i,j=0,1,2,3\). Do not get confused on the notation. Each element \(dx^{i}\)denotes a cotangent vector. For example in 4d Minkowski  you have \(dx^{0}, dx^{1}, dx^{2}, dx^{3}\). You can find some nice explanations in Nakahara's book Geometry Topology and Physics and also in almost any General Relativity textbook. Also check out this.

Hope it helps.

answered Nov 8, 2014 by conformal_gk (3,625 points) [ no revision ]

$dF_\mu$ cannot have the meaning of a vector unless one has a (pseudo-)Riemannian manifold! See my answer.

This should be included in "all the nice properties". To my best knowledge we, physicists, almost always discuss (pseudo-)Riemannian manifolds. To a physicists, I think, my answer is quite right. I guess a mathematician would not like it though. In the end of the day, it seems we are talking about simple GR nor I said anywhere that $dF_{\mu}$ is a vector. I said it for the partial derivative. As for the metric, well, M4 is clearly a metric space, so not sure what is the target of your comment although your answer is very nice!

But the question itself distinguishes between vectors and covectors, hence assumes no metric. Otherwise there wouldn't be any confusion to begin with.

In GR many times we call the partial derivatives as vectors and and the one-forms as co-vectors in the sense that they belong in the cotangent bundle of the specified point. Furthermore the whole question assumes Minkowski 4, at least this is what I got.  I completely agree with your answer and in specific that sometimes we mix notation but I think that, at least intuitively, there is no error in my answer ) 

+ 1 like - 0 dislike

I believe you are confused because you are mixing up related but slightly different quantities.

Yes, a partial derivative is a vector and yes, a vector is an object with an upper index.

The above statement may seem contradictory, but in fact it is not for the following reason. A vector is an abstract quantity that is an element of a "vector space". In this case, the vector space that is being discussed is the tangent space. On a vector space, one can choose a basis, any basis. Once a basis has been chosen, any other vector in the vector space can be described by simply prescribing a set of numbers. For instance in ${\mathbb R}^2$ (rather the corresponding affine space), one can choose a basis of vectors as ${\hat x}$ and ${\hat y}$. Once this has been done any other vector can be described simply by 2 numbers. For instance the numbers $(1,2)$ really implies that we are talking about the vector ${\hat x} + 2 {\hat y}$.

How does the discussion above apply here?

On the tangent space, a natural choice of basis are the set of partial derivatives $\partial_\mu = \{ \partial_0 , \partial_1 , \partial_2 , \partial_3 \}$ (assuming we are in $M_4$. Each partial derivative is in itself a vector.

Now, once this basis has been chosen, every other vector can be described by a set of 4 numbers $v^\mu = (v^0 , v^1 , v^2 , v^3)$ which corresponds to the vector $v^\mu \partial_\mu$. It is this sense, that the bold statement above is true. Often, since the basis of partial derivatives is obvious, one simply describes a vector as an object with an upper index $v^\mu$.

Next, let us discuss co-vectors (quantities with a lower index). These are lements of the dual vector space (which is the space of linear functions on the vector space) of the tangent space. Given the partial derivative basis on the tangent space, one then has a natural basis in the cotangent space denoted by $dx^\mu = \{ dx^0 , dx^1 , dx^2 , dx^3 \}$. Note that each differential itself is a covector. This natural basis is defined by the relation $dx^\mu (\partial_\nu ) = \delta^\mu_\nu$. As before, once this natural basis has been chosen, any element of the cotangent space can be described by 4 numbers, namely $v_\mu = \{ v_0, v_1 , v_2 , v_3 \}$ which corresponds to the covector $v_\mu dx^\mu$.

In summary, $\partial_\mu$ for each $\mu$ corresponds to a 4-dimensional vector whereas $v^\mu$ for each $\mu$ corresponds to 4 components of a single vector. Similarly, $dx^\mu$ for each $\mu$ corresponds to a 4-dimensional covector whereas $v_\mu$ for each $\mu$ corresponds to 4 components of a single covector.

PS 1 - Sometimes people like to use bases other than $\partial_\mu$ and $dx^\mu$ on the tangent and cotangent spaces respectively. These are known as non-coordinate bases.

PS 2 - Just to be clear, $\partial_\mu$ is a vector, but $\partial_\mu F$ is a function

This post imported from StackExchange Physics at 2014-11-11 14:51 (UTC), posted by SE-user Prahar
answered Nov 3, 2014 by prahar21 (545 points) [ no revision ]

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