# How to use basis vectors defined as partial derivative operators?

Edit to add clarification and improve notation:

This question was originally written for physicists, not mathematicians. I have differing expectations of the reader in those different contexts. My question pertains to the practice of identifying vectors and directional derivatives which occurs in mathematical physics.

Gravitation, by Charles W. Misner, Kip S. Thorne & John Archibald Wheeler, Chapters 8 and 9 (Equation 8.2); and Geometrical Methods of Mathematical Physics, by Bernard F. Schutz, Chapter 2.

As is often the case, I misunderstood it wrong. That is, the authors who introduced the concept expected me to balk at the justification for the identity. That is not a problem for me. My problem is one of understanding the syntax of application. I am told that a vector $\mathfrak{u}$ is identical to a directional derivative operator $\mathfrak{u}\equiv\partial_{\mathfrak{u}}.$

Modified Einstein summation notation is used, where coordinate systems are distinguished by the presence or absence of a bar over an index. $\alpha$ and $\overline{\alpha}$ are distinct index variables, one indicating components in the "unbarred" system, the other components in the "barred" system.

In traditional linear algebra, let $\{\mathfrak{e}_{\alpha}\}$ be a spanning basis of a vector space, and let $\{L_{\overline{\alpha}}^{\alpha}\}$ be the matrix associated with a nonsingular linear transformation of the basis. So

$$L^{\alpha}{}_{\overline{\alpha}}\mathfrak{e}_{\alpha}=\mathfrak{e}_{\alpha}L^{\alpha}{}_{\overline{\alpha}}=\mathfrak{e}_{\overline{\alpha}}$$

represents a change of basis.

Given some scalar function $f$ of position, write $\{M^{\alpha}{}_{\overline{\alpha}}\}=f\{L^{\alpha}{}_{\overline{\alpha}}\},$ which will also be a linear transformation. So we have

$$M^{\alpha}{}_{\overline{\alpha}}\mathfrak{e}_{\alpha}=\mathfrak{e}_{\alpha}M^{\alpha}{}_{\overline{\alpha}}=fL^{\alpha}{}_{\overline{\alpha}}\mathfrak{e}_{\alpha}=f\mathfrak{e}_{\overline{\alpha}}.$$

But I am told to let the basis vectors be defined as directional derivative operators along the coordinate curves. So that

$$\mathfrak{e}_{\alpha}\equiv\partial_{\alpha}=\frac{\partial}{\partial x^{\alpha}}.$$

Without any other guidelines on how to use these operators

$$M^{\alpha}{}_{\overline{\alpha}}\mathfrak{e}_{\alpha}=fL^{\alpha}{}_{\overline{\alpha}}\mathfrak{e}_{\alpha}=f\mathfrak{e}_{\overline{\alpha}},$$

as before.

But now, following the standard rules of calculus $$\mathfrak{e}_{\alpha}L^{\alpha}{}_{\overline{\alpha}}=\partial_{\alpha}fL^{\alpha}{}_{\overline{\alpha}}=\frac{\partial f}{\partial x^{\alpha}}L^{\alpha}{}_{\overline{\alpha}}+f\frac{\partial L^{\alpha}{}_{\overline{\alpha}}}{\partial x^{\alpha}}.$$

Even without introducing the scalar function $f$, if $\{L^{\alpha}{}_{\overline{\alpha}}\}$ is independent of position, a direct application of the rules of calculus leads to

$$\mathfrak{e}_{\alpha}L^{\alpha}{}_{\overline{\alpha}}=\mathfrak{\partial}_{\alpha}L^{\alpha}{}_{\overline{\alpha}}=\frac{\partial L^{\alpha}{}_{\overline{\alpha}}}{\partial x^{\alpha}}=0.$$

I am confident that this is not the intended result, but the authors who have introduced this alleged improvement never tell me how to correctly apply it.

When should $\mathfrak{e}_{\alpha}\equiv\partial_{\alpha}=\frac{\partial}{\partial x^{\alpha}}$ be treated as if it were a traditional basis vector, and when should it be applied as a derivative operator?

• I will be very surprised if there is anyone who can make any sense of this post. Commented Jul 19, 2018 at 21:10
• See Gravitation, by Misner, Thorne & Wheeler, Chapter 9 press.princeton.edu/titles/11169.html and Geometrical Methods of Mathematical Physics, by Schutz, Chapter 2 cambridge.org/us/academic/subjects/mathematics/… Commented Jul 20, 2018 at 0:42
• Among the 62 things that make no sense here are: 1) You have a single linear transformation. What's with the subscripts and superscripts? And why do you refer the singleton set containing that transformation as if it were the transformation itself? 2) Your transformation should act either on the left or on the right, not both, so your first displayed equation is nonsense (unless there's some additional structure you haven't told us about that allows operators to act on both sides. If so, you cannot possibly expect us to guess what that structure might be; you've got to tell us.) Commented Jul 20, 2018 at 1:50
• 3) It appears from your notation that you are assuming the transformation $L$ permutes the given basis. But the fact that you never explicitly say this leaves the reader wondering what you are assuming. 4) What on earth is a "scalar function of position"? This is the first time the word "position" has appeared in this post, and I haven't the foggiest notion what it means here. Neither will anyone else. 5) Things get even more incomprehensible from here..... Commented Jul 20, 2018 at 1:52
• Anybody qualified to answer my question wouldn't be asking your questions. Commented Jul 20, 2018 at 2:04

Apparently vector symbols such as $\mathbf{v}$, $\vec{v}$, $\mathbf{e}_i$, $\mathfrak{v}$, etc., are not to be understood as syntactically identical to the symbols for the "identical" entities they symbolize. That is:

Using the notation: $$\frac{\partial{}}{\partial{x^\alpha}}\equiv\partial{}_\alpha,$$ $$\mathfrak{v}\cdot\nabla\equiv\partial{}_{\mathfrak{v}},$$

And the definitions: $$f\text{ is a scalar field},$$ $$\mathfrak{v}\equiv\partial{}_{\mathfrak{v}},$$ $$\mathfrak{e}_{\alpha}\equiv\partial{}_{\alpha},$$

then $$\mathfrak{v}f\ne\partial{}_{\mathfrak{v}}f,$$ $$\mathfrak{e}_{\alpha}f\ne\partial{}_{\alpha}f.$$

In order to "apply" the directional derivative $\partial_{\mathfrak{v}}$ which is the vector $\mathfrak{v}$, when symbolized by $\mathfrak{v},$ we must affix additional notation to the symbol $\mathfrak{v}$.

Schutz symbolizes vectors by $\overline{v};$ and the directional derivative by $\overline{v}\left(f\right)\equiv\partial_{\overline{v}}f.$ Not to be confused with, for example, $x^{i}\left(P\right)$ which gives the $i^{th}$ coordinate as a function of abstract position.

Misner, Thorne and Wheeler write $\mathbf{v}\left[f\right]\equiv\partial_{\mathbf{v}}f$.

Now that I am more familiar with the context, it is fairly easy to separate the wheat from the chaff. Not so when I first encountered these ideas.

The place where the identification of a vector with the corresponding directional derivative seems most useful is the commutator; which MTW define as

$$\left[\mathbf{u},\mathbf{v}\right]\left[f\right]\equiv\mathbf{u}\{\mathbf{v}\left[f\right]\}-\mathbf{v}\{\mathbf{u}\left[f\right]\}.$$

Yes! They really did immediately throw in a new form of parentheses. And to keep things clear, they shamelessly write

$$\left[\mathbf{u},\mathbf{v}\right]f,$$

to mean the same thing. So, clearly one should assume

$$\mathfrak{v}f\ne\partial{}_{\mathfrak{v}}f.$$

Yes; I am being facetious. This confusion is what stopped me from continuing with MTW and Schutz. None of the "difficult" concepts were difficult for me.

Teacher: "Forget everything you know. Erase the entire blackboard."

Student: "OK. Done."

Teacher: "Now, draw a straight line."

Student: "Not only do I not know what you mean by straight line. I don't think housekeeping is going to like me drawing on the wall."