I now realize my question can be stated very concisely.

In Chapter 11 of MTW, will the meaning be changed if in every instance we make the replacement


If so, how?

I will add that I have now discovered that MTW do in fact give the defintion $d\mathbf{v}/d\lambda\equiv\nabla_{\mathbf{u}}\mathbf{v}\equiv$ "covariant derivative of $\mathbf{v}$ along $\mathbf{u}$", where $\mathbf{u}=d/d\lambda.$

See Box 10.2 enter image description here

Original version of the question

It appears that the commutator of covariant derivatives $\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]$ and the commutator of the same two vectors $\left[\mathbf{a},\mathbf{b}\right]$ are actually the same thing. The only difference is in how they are applied to tensors of rank grater than 0.

Boldface type $\mathbf{a}$ represents a vector field. Fraktur type $\mathfrak{a}$ represents a member of $\mathbf{a}$. The non-standard use of angle brackets $\mathfrak{a}\left\langle f\right\rangle$ means to apply the vector $\mathfrak{a}$ to the scalar function $f$. The commutator of the vector fields $\mathbf{a},\mathbf{b}$ is defined in terms of an arbitrary scalar field as

\begin{align*} \mathfrak{a}\left\langle f\right\rangle & =\frac{df}{d\alpha}\\ \mathfrak{b}\left\langle f\right\rangle & =\frac{df}{d\beta}\\ \left[\mathbf{a},\mathbf{b}\right]\left( f\right) & \equiv\mathfrak{a}\left\langle \mathbf{b}\left\langle f\right\rangle \right\rangle -\mathfrak{b}\left\langle \mathbf{a}\left\langle f\right\rangle \right\rangle\\ & = f_{,\gamma}\left(b^{\gamma}{}_{,\alpha}\mathrm{a}^{\alpha}-a^{\gamma}{}_{,\beta}\mathrm{b}^{\beta}\right)\\ & = \mathfrak{e}_{\gamma}\left\langle f\right\rangle\left(b^{\gamma}{}_{,\alpha}\mathrm{a}^{\alpha}-a^{\gamma}{}_{,\beta}\mathrm{b}^{\beta}\right)\\ & = \mathfrak{e}_{\gamma}\left\langle f\right\rangle \mathrm{w}^{\gamma}\\ & = \mathfrak{w}\left\langle f\right\rangle . \end{align*}

Since $f$ is arbitrary, the commutator thus defined is a vector, $\left[\mathbf{a},\mathbf{b}\right] = \mathfrak{w}.$

Box 11.5 gives the curvature tensor as

\begin{align*} \mathfrak{R}\left(\dots,\mathbf{c},\mathbf{a},\mathbf{b}\right) &\equiv\mathscr{R}\left(\mathbf{a},\mathbf{b}\right)\mathbf{c},\\ \mathscr{R}\left(\mathbf{a},\mathbf{b}\right) &\equiv\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]-\nabla_{\left[\mathbf{a},\mathbf{b}\right]} \end{align*}

Suppose we define the following \begin{align*} \nabla_{\mathfrak{a}}\mathbf{c} &\equiv\mathfrak{a}\left\langle \mathbf{c}\right\rangle ,\\ \left[\mathbf{a},\mathbf{b}\right]\left[ \mathbf{c} \right] &\equiv\mathfrak{a}\left\langle \mathbf{b}\left\langle \mathbf{c} \right\rangle \right\rangle -\mathfrak{b}\left\langle \mathbf{a}\left\langle \mathbf{c} \right\rangle \right\rangle \text{, and }\\ \left[\mathbf{a},\mathbf{b}\right]\left\langle \mathbf{c} \right\rangle &\equiv \mathfrak{w}\left\langle \mathbf{c} \right\rangle = \nabla_{\left[\mathbf{a},\mathbf{b}\right]}\mathbf{c}. \end{align*}

Will this lead to

$$\mathscr{R}\left(\mathbf{a},\mathbf{b}\right)\mathbf{c} = \left[\mathbf{a},\mathbf{b}\right]\left[ \mathbf{c} \right] - \left[\mathbf{a},\mathbf{b}\right]\left\langle \mathbf{c} \right\rangle ? $$

Original incomplete draft of the question:

The original draft of this question was becoming very long because I wanted to define everything explicitly. I still have considerably more to add in order to make things rigorous. But it should at least provide some additional context.

All mappings are assumed to be $\mathscr{C}^{\infty}.$ In the following, the notational distinction between fixed vectors and corresponding vector fields is non-standard. I do not extend this practice to arbitrary one-forms, simply because I ran out of imagination. In this discussion the use of angled brackets $\left\langle \dots\right\rangle $ affixed to the right of a vector indicates the ``application'' of the vector to the expression contained in the brackets. I call this \emph{applicator notation}, and use it to remove ambiguity. Applying a vector to an expression means to differentiate the argument with respect to the parameter of the locally unique tangent path determined by the vector. This, of course, requires that all such vectors are specific to a manifold point (i.e., are tangent vectors).

Initially we shall limit our definition of application $\mathbf{\mathfrak{u}}\left\langle \dots\right\rangle $ to scalar fields. To be explicit, given a parameterized path $\mathscr{P}\left[\lambda\right],$ we define a vector in the abstract using $\mathbf{\mathfrak{u}}\equiv d\mathscr{P}/d\lambda.$ This means that for every scalar field $f\left[\mathscr{P}\right],$ we have

\begin{align*} \mathscr{P}_{o} & =\mathscr{P}\left[\lambda_{o}\right]\\ \Delta\mathscr{P} & =\mathscr{P}\left[\lambda_{o}+\Delta\lambda\right]-\mathscr{P}_{o}\\ \frac{df}{d\lambda} & \equiv\lim_{\Delta\lambda\to0}\left\{ \frac{f\left[\mathscr{P}_{o}+\Delta\mathscr{P}\right]_{\text{transport agnostic}}-f\left[\mathscr{P}_{o}\right]}{\Delta\lambda}\right\} \\ \nabla f & \equiv\lim_{\Delta\mathscr{P}\to\mathfrak{0}}\left\{ \frac{f\left[\mathscr{P}_{o}+\Delta\mathscr{P}\right]_{\text{transport agnostic}}-f\left[\mathscr{P}_{o}\right]}{\Delta\mathscr{P}}\right\} \\ \mathfrak{u}\left\langle f\right\rangle & \equiv\frac{df}{d\lambda}\iff\mathfrak{u}\equiv\frac{d\mathscr{P}}{d\lambda} \end{align*}

We have also introduced an abstract definition of the gradient $\nabla f$ of a scalar field by omitting reference to any specific curve along which differentiation is performed.

The annotation ``transport agnostic'' pertains to more general covariant derivative definition, and means that the smooth path followed when taking the limit is irrelevant when differentiating a scalar field. The use of fraktur font $\mathfrak{u},\mathfrak{a},\mathfrak{b},$ etc., indicates fixed vector members of the corresponding vector fields $\mathbf{u},\mathbf{a},\mathbf{b},$ etc. Since every $\mathfrak{u}\left\langle f\right\rangle \in\mathbb{R}$ the corresponding $\mathbf{u}\left\langle f\right\rangle $ is a scalar field. The expression $\mathfrak{a}\left\langle \mathbf{u}\left\langle f\right\rangle \right\rangle \in\mathbb{R}$is the derivative of a scalar field.

We now define the commutator of two vector fields acting on a scalar field at $\mathscr{P}_{o}$ as follows

\begin{align*} \mathfrak{a}\left\langle f\right\rangle & =\frac{df}{d\alpha}\\ \mathfrak{b}\left\langle f\right\rangle & =\frac{df}{d\beta}\\ \left[\mathbf{a},\mathbf{b}\right]\left\langle f\right\rangle _{o} & \equiv\mathfrak{a}\left\langle \mathbf{b}\left\langle f\right\rangle \right\rangle -\mathfrak{b}\left\langle \mathbf{a}\left\langle f\right\rangle \right\rangle \end{align*}

In the case of a scalar field the expression $\left[\mathbf{a},\mathbf{b}\right]\left\langle f\right\rangle $ turns out to mean $\mathbf{\mathfrak{w}}\left\langle f\right\rangle $where $\mathbf{\mathfrak{w}}=\left[\mathbf{a},\mathbf{b}\right]_{o}.$ That is, we define the vector $\mathbf{\mathfrak{w}}$ in terms of its application to the arbitrary scalar field $f$ as

\begin{align*} \mathfrak{w}\left\langle f\right\rangle _{o} & \equiv\left[\mathbf{a},\mathbf{b}\right]\left\langle f\right\rangle _{o}\\ \iff\mathfrak{w} & \equiv\frac{d}{d\alpha}\left[\frac{d\mathscr{P}}{d\beta}\right]-\frac{d}{d\beta}\left[\frac{d\mathscr{P}}{d\alpha}\right] \end{align*}

To show that $\mathfrak{w}$ is, in fact a vector we appeal to the definition (with some notational liberties taken) of a tangent vector to a differentiable manifold given by Ciufolini and Wheeler in their Gravitation and Inertia.

Coordinate dependent definition: A tangent vector $\mathfrak{v}$ at a point $\mathscr{P}$ of a differentiable manifold is a mathematical object that, in a coordinate system, is represented by a set of $n$ numbers $\mathrm{v}^{i}$ at $\mathscr{P}$, components of $\mathfrak{v}$, that, under coordinate transformation $x^{\bar{i}}=\mathit{x}^{\bar{i}}\left[\left\{ x^{i}\right\} \right],$ change according to the transformation law; $$ v^{\bar{i}}=\left[\frac{\partial\mathit{x}^{\bar{i}}}{\partial x^{i}}\right]_{\mathscr{P}}v^{i} $$

Definition independent of coordinates: A tangent vector at a point $\mathscr{P}$ is a mapping $\mathfrak{v}_{\mathscr{P}}$that to each differentiable function defined in the neighborhood of $\mathscr{P}$ assigns one real number, and which is linear and satisfies the Leibniz rule. That is $$ \mathfrak{v}_{\mathscr{P}}\left\langle \mathrm{a}f+\mathrm{b}g\right\rangle =\mathrm{a}\mathfrak{v}_{\mathscr{P}}\left\langle f\right\rangle +\mathrm{b}\mathfrak{v}_{\mathscr{P}}\left\langle g\right\rangle , $$ linearity; and $$ \mathfrak{v}_{\mathscr{P}}\left\langle f\cdot g\right\rangle =\mathfrak{v}_{\mathscr{P}}\left\langle f\right\rangle g\left[\mathscr{P}\right]+f\left[\mathscr{P}\right]\mathfrak{v}_{\mathscr{P}}\left\langle g\right\rangle , $$ Leibniz rule; where $\mathrm{a},\mathrm{b}$ are real numbers and $f,g$ are differentiable functions.

Equivalent definition independent of coordinates: Given a differentiable curve $c\left[t\right],$ that is, a differentiable mapping from an interval of the real numbers into $\mathcal{M},$ and given a function $f$ on $\mathcal{M}$ differentiable at $\mathscr{P},$ the tangent vector to the curve at $\mathscr{P}=c\left[t\right]$ is defined by $$ \mathbf{\mathfrak{v}}_{\mathscr{P}}^{c}\left\langle f\right\rangle =\left[\frac{df\left[c\left[t\right]\right]}{dt}\right]_{t_{\mathscr{P}}}, $$ and one may write in a local coordinate system $\left\{ x^{i}\right\} _{n},$ $$ \mathbf{\mathfrak{v}}_{\mathscr{P}}^{c}\left\langle f\right\rangle =\left[\frac{\partial f}{\partial x^{i}}\right]_{\mathscr{P}}\left[\frac{dx^{i}\left[\vec{c}\left[t\right]\right]}{dt}\right]_{t_{\mathscr{P}}} $$ generalization of ordinary definition of a tangent vector to a curve in $\mathbb{R}^{n}.$ These definitions of tangent vector are equivalent.

We shall distinguish between components of a vector $\mathfrak{a}=\left\{ \mathrm{a}^{i}\right\} $ and those of the corresponding vector field $\mathbf{a}=\left\{ a^{i}\right\} ,$ by using Roman font for the former, and default Latin font for the latter. We also abbreviate Leibniz derivative notation by $\partial_{i}f\equiv df/d\mathrm{x}^{i}.$ There is a subtle, but important distinction between this and the subscript comma notation $f_{,i}\equiv\mathfrak{e}_{i}\left\langle f\right\rangle ,$ which only has the meaning $\partial_{i}=f_{,i}$ when $\mathfrak{e}_{i}=d\mathscr{P}/d\mathrm{x}^{i},$ that is when $\left\{ \mathfrak{e}_{i}\right\} $ is a coordinate induced basis. Finally, we introduce the coordinate-based definition of the gradient of a scalar field as the set of partial derivatives $\nabla f\equiv\left\{ \partial_{i}f\right\} ;$ (not per MTW) and the use of infix dot notation to express the contraction of the gradient with a vector $\partial_{i}f\mathrm{a}^{i}=\nabla f\cdot\mathfrak{a}$ (MTW write $\left\langle \nabla f,\mathfrak{a}\right\rangle $ instead). In more complicated expressions, the post-fix of the contraction vector is significant, in that it means to contract on the index of $\nabla$ rather than on one of the arguments to $\nabla.$

We now show $\mathfrak{w}\equiv\left[\mathbf{a},\mathbf{b}\right],$ may be expressed in component form independent of $f,$ and is therefor a vector. Since proving this for coordinate bases does the same for general bases, we shall assume we are working in coordinate bases where $\partial_{i}f=f_{,i}.$ Since we are differentiating scalar fields (even for components $a^{i},$etc.) we know that mixed partials commute: $\partial_{j}\partial_{i}f=f_{,ij}=f_{,ji}=\partial_{i}\partial_{j}f.$

\begin{align*} \mathfrak{a}\left\langle f\right\rangle = & \frac{df}{d\alpha}=\frac{\partial f}{\partial\mathrm{x}^{i}}\frac{dx^{i}}{d\alpha}=\partial_{i}f\mathrm{a}^{i}=\nabla f\cdot\mathfrak{a}=f_{,i}\mathrm{a}^{i}\\ \mathfrak{b}\left\langle f\right\rangle = & \frac{df}{d\beta}=\frac{\partial f}{\partial\mathrm{x}^{i}}\frac{dx^{i}}{d\beta}=\partial_{i}f\mathrm{b}^{i}=\nabla f\cdot\mathfrak{b}=f_{,i}\mathrm{b}^{i}\\ \mathfrak{w}\left\langle f\right\rangle = & \left[\mathbf{a},\mathbf{b}\right]\left\langle f\right\rangle =\frac{d}{d\alpha}\left[\frac{df}{d\beta}\right]-\frac{d}{d\beta}\left[\frac{df}{d\alpha}\right]\\ = & \mathfrak{a}\left\langle \mathbf{b}\left\langle f\right\rangle \right\rangle -\mathfrak{b}\left\langle \mathbf{a}\left\langle f\right\rangle \right\rangle \\ = & \nabla\left[\nabla f\cdot\mathbf{b}\right]\cdot\mathfrak{a}-\nabla\left[\nabla f\cdot\mathbf{a}\right]\cdot\mathfrak{b}\\ = & \left[f_{,\beta}b^{\beta}\right]_{,\alpha}\mathrm{a}^{\alpha}-\left[f_{,\alpha}a^{\alpha}\right]_{,\beta}\mathrm{b}^{\beta}\\ = & \left(f_{,\beta\alpha}\mathrm{b}^{\beta}+f_{,\beta}b^{\beta}{}_{,\alpha}\right)\mathrm{a}^{\alpha}-\left(f_{,\alpha\beta`}\mathrm{a}^{\alpha`}+f_{,\alpha}a^{\alpha}{}_{,\beta}\right)\mathrm{b}^{\beta}\\ = & \left(f_{,\beta\alpha}-f_{,\alpha\beta`}\right)\mathrm{b}^{\beta}\mathrm{a}^{\alpha}+\left(f_{,\beta}b^{\beta}{}_{,\alpha}\mathrm{a}^{\alpha}-f_{,\alpha}a^{\alpha}{}_{,\beta}\mathrm{b}^{\beta}\right)\\ = & f_{,\gamma}\left(b^{\gamma}{}_{,\alpha}\mathrm{a}^{\alpha}-a^{\gamma}{}_{,\beta}\mathrm{b}^{\beta}\right)\\ \equiv & \nabla f\cdot\left[\mathbf{a},\mathbf{b}\right]\\ \mathrm{w}^{\gamma}= & b^{\gamma}{}_{,\alpha}\mathrm{a}^{\alpha}-a^{\gamma}{}_{,\beta}\mathrm{b}^{\beta}\\ \mathfrak{w}= & \frac{d\mathscr{P}}{d\omega}=\mathfrak{e}_{\gamma}\mathrm{w}^{\gamma} \end{align*}

The final equation is the assertion that since $\mathfrak{w}$ is a (tangent) vector it determines a locally unique manifold path, with a parameter which we have named $\omega$. From the preceding we have all the pieces needed to justify our definition of coordinate induced basis vectors, and show how they are used to represent vectors.

\begin{align*} \mathfrak{e}_{\iota}\equiv & \frac{d\mathscr{P}}{d\mathrm{x}^{\iota}}\\ \mathfrak{e}_{\iota}\left\langle x^{\kappa}\right\rangle = & \delta_{\iota}^{\kappa}\\ \mathfrak{e}_{\iota}\left\langle f\right\rangle = & \partial_{\iota}f\\ \mathfrak{v}\left\langle f\right\rangle = & \frac{df}{d\nu}=\partial_{\iota}f\mathrm{v}^{\iota}=\mathfrak{e}_{\iota}\left\langle f\right\rangle \mathrm{v}^{\iota}\\ \mathfrak{v}= & \mathfrak{e}_{\iota}\mathrm{v}^{\iota} \end{align*}

We now define the covariant derivative of a (contravariant) vector $\mathbf{v}$ along a curve tangent to the vector $\mathfrak{u}=d\mathscr{P}/d\lambda,$ with the assumption that ``parallel transport'' is clearly defined. We also generalize the definition of the gradient to vector fields. This definition of the gradient, and the use of dot product notation do not come from MTW.

\begin{align*} \nabla_{\mathfrak{u}}\mathbf{v}\equiv & \lim_{\Delta\lambda\to0}\left\{ \frac{\mathbf{v}\left[\mathscr{P}_{0}+\Delta\mathscr{P}\right]_{\text{parallel transported to}\mathscr{P}_{0}}-\mathbf{v}\left[\mathscr{P}_{0}\right]}{\Delta\lambda}\right\} \\ = & \lim_{\Delta\lambda\to0}\left\{ \frac{\mathbf{v}\left[\mathscr{P}_{0}+\Delta\lambda\mathfrak{u}\right]_{\text{parallel transported to}\mathscr{P}_{0}}-\mathbf{v}\left[\mathscr{P}_{0}\right]}{\Delta\lambda}\right\} \\ \nabla\mathbf{v}\equiv & \lim_{\Delta\mathscr{P}\to0}\left\{ \frac{\mathbf{v}\left[\mathscr{P}_{0}+\Delta\mathscr{P}\right]_{\text{parallel transported to}\mathscr{P}_{0}}-\mathbf{v}\left[\mathscr{P}_{0}\right]}{\Delta\mathscr{P}}\right\} \\ \nabla_{\mathfrak{e}_{\delta}}\mathbf{v}\equiv & \nabla_{\delta}\mathbf{v}\\ \nabla_{\mathfrak{u}}\mathbf{v}= & \nabla\left[\mathbf{v}\right]\cdot\mathfrak{u}=\nabla_{\delta}\mathbf{v}\mathrm{u}^{\delta}\\ \mathfrak{e}_{\sigma}\Gamma_{\iota\delta}^{\sigma}\equiv & \nabla_{\delta}\mathbf{e}_{\iota}\\ \nabla_{\mathfrak{u}}\mathbf{v}= & \nabla\left[\mathbf{v}\right]\cdot\mathfrak{u}=\nabla_{\delta}\left[\mathbf{e}_{\iota}v^{\iota}\right]\mathrm{u}^{\delta}\\ = & \left(\nabla_{\delta}\mathbf{e}_{\iota}v^{\iota}+\mathfrak{e}_{\iota}\nabla_{\delta}v^{\iota}\right)\mathrm{u}^{\delta}\\ = & \mathbf{\mathfrak{e}}_{\iota}\left(\Gamma_{\nu\delta}^{\iota}v^{\nu}+v^{\iota}{}_{,\delta}\right)\mathrm{u}^{\delta}\\ = & \mathfrak{e}_{\iota}v^{\iota}{}_{;\delta}\mathrm{u}^{\delta} \end{align*}

On an $n$--dimensional manifold, analogous to the component representation of a vector field, we introduce a system of $n$ scalar fields $\overset{\sim}{\sigma}\left[\mathscr{P}\right]=\left\{ \sigma_{\iota}\left[\mathscr{P}\right]\right\} ,$ which we shall call a one-form field. Thus, for every vector $\mathfrak{v}=\mathbf{v}\left[\mathscr{P}\right]$ we have $\overset{\sim}{\sigma}\cdot\mathfrak{v}=\sigma_{\iota}\mathrm{v}^{\iota}\in\mathbb{R}.$ In the case of the coordinate induced basis field, we define a special one-form field $\left\{ \mathbf{e}^{\iota}\right\} $ such that at every $\mathscr{P}$ we have $\mathbf{e}^{\iota}\cdot\mathbf{e}_{\kappa}=\delta_{\kappa}^{\iota}.$

Let us consider the general case of $f\left[\mathscr{P}\right]=\overset{\sim}{\sigma}\cdot\mathbf{v}.$

\begin{align*} \mathfrak{v}= & \mathfrak{e}_{\nu}\mathrm{v}^{\nu}\\ \mathrm{v}^{\iota}= & \mathfrak{e}^{\iota}\cdot\mathfrak{e}_{\nu}\mathrm{v}^{\nu}\\ f\left[\mathscr{P}\right]= & \overset{\sim}{\sigma}\left[\mathscr{P}\right]\cdot\mathbf{v}\left[\mathscr{P}\right]\\ \iff f= & \overset{\sim}{\sigma}\cdot\mathbf{v}=\sigma_{\iota}v^{\iota}=\sigma_{\iota}\mathfrak{e}^{\iota}\cdot\mathfrak{e}_{\nu}\mathrm{v}^{\nu} \end{align*}

So long as we keep our wits about us, I see no reason not to extend the standard calculus notation to vector fields. That is

\begin{align*} \mathfrak{u}= & \frac{d\mathscr{P}}{d\upsilon}\\ \mathbf{v}\left[\mathscr{P}\right]= & \mathbf{e}_{\iota}v^{\iota}\\ \frac{d\mathbf{v}}{d\upsilon}\equiv & \nabla_{\mathfrak{u}}\mathbf{v}=\nabla\left[\mathbf{v}\right]\cdot\mathfrak{u} \end{align*}

  • 2
    $\begingroup$ In short, they are not the same thing. If you simply read off what you put into the commutator, one is the measure of the distance between two vectors in some space given by their basis vector $\textbf{e}_\alpha$. The other, is the commutator of of the covariant derivative along vectors a and b. This is much easier to see if you have the commutators "act" on some test function $f(x^\mu)$, or a vector $V^\alpha$. (In the first case, you get $ [u,v] = \left(u^\alpha v^\beta_{,\alpha} - v^\alpha u^\beta_{,\alpha}\right)\mathbf{e}_\beta$, second gives you the Ricci + Torsion tensor). $\endgroup$
    – MathZilla
    Commented Mar 14, 2022 at 20:06
  • $\begingroup$ My point is that MTW never write $\vec{v}\left[\vec{u}\right],$ but the expression has an obvious (to me) meaning. Its the same thing as $\nabla_{\vec{v}}\vec{u}.$ But when we do that, there are two different possible meanings for $\left[\vec{a},\vec{b}\right]\vec{c}.$ One is the covariant derivative of $\vec{c}$ along the vector $\left[\vec{a},\vec{b}\right].$ The other is equation 11.7. $\endgroup$ Commented Mar 14, 2022 at 20:29

2 Answers 2


Boldface type $\mathbf a$ represents a vector field. Fraktur type $\mathfrak a$ represents a member of $\mathbf a$.

Which one? Why wouldn't you use the standard notation $\mathbf a_p$ to refer to the vector from $\mathbf a$ attached to the point $p$?

The commutator of the vector fields $\mathbf{a},\mathbf{b}$ is defined in terms of an arbitrary scalar field as \begin{align*} \mathfrak{a}\left\langle f\right\rangle & =\frac{df}{d\alpha}\\ \mathfrak{b}\left\langle f\right\rangle & =\frac{df}{d\beta}\\ \left[\mathbf{a},\mathbf{b}\right]\left( f\right) & \equiv\mathfrak{a}\left\langle \mathbf{b}\left\langle f\right\rangle \right\rangle -\mathfrak{b}\left\langle \mathbf{a}\left\langle f\right\rangle \right\rangle\\ &\vdots \end{align*}

I have no idea what the first two lines mean. What are $\alpha$ and $\beta$? If $f$ is a function on a manifold $M$, its arguments are points $p\in M$. How are you defining these derivatives? The third line also doesn't make sense unless you specify what $\mathfrak a$ and $\mathfrak b$ are. As per my first comment, which element of $\mathbf a$ and $\mathbf b$ are they? Clearly it would matter in this calculation, but that would indicate that $[\mathbf a,\mathbf b]$ would depend on this mysterious choice of point which doesn't make any sense, since the expression $[\mathbf a,\mathbf b]$ doesn't reference one.

To be honest, I'm mostly just confused as to what you're asking here. A vector field like $\mathbf a$ or $\mathbf b$ is a linear map which eats a smooth function and spits out another smooth function. Being endomorphisms on the space of smooth functions, the composition of two vector fields is another such map, so we can naturally define the vector field commutator $$[\mathbf a,\mathbf b](f) := \mathbf a\big(\mathbf b(f)\big) - \mathbf b\big(\mathbf a(f)\big)$$

All of this is very natural,and none of it requires any additional structure. If we wish to introduce the covariant derivative, however, we need to make a choice of connection. Any notation which uses the connection should make some reference to this fact, so your definition $$\mathfrak a\langle\mathbf b\rangle := \nabla_\mathfrak a \mathbf b$$ is not good; the left-hand side suggests that this quantity is defined intrinsically from the vector $\mathfrak a$ and vector field $\mathbf b$, when in fact it is not.

The rest of your question utilized the ill-defined commutator expression you came up with in the beginning, so it doesn't make much sense to me either. $[\nabla_\mathbf a,\nabla_\mathbf b]$ and $[\mathbf a,\mathbf b]$ are emphatically not the same thing. To see this, it's sufficient to note that the former requires a connection to define while the latter does not.

The only difference is in how they are applied to tensors of rank grater than 0.

So in other words, the only similarity is in how they are applied to tensors of rank 0. That's a very strange criterion for calling them "the same thing."

With that being said, $[\mathbf a,\mathbf b]$ is a type of derivative - it is the Lie derivative of the vector field $\mathbf b$ with respect to the vector field $\mathbf a$. But I'm quite sure that's not what you're referring to here.

The question has now been heavily edited, and is now

In Chapter 11 of MTW, will the meaning be changed if in every instance we make the replacement $$[\mathbf a,\mathbf b]\mapsto [\nabla_{\mathbf a},\nabla_{\mathbf b}]$$

The essential content of this question, and your fairly extensive use of non-standard notation, is related to the difference between the vector field $\mathbf a$ and the differential operator $\nabla_\mathbf a$. On one hand, it's true that for any scalar field $f$, the two objects have the same effect, so $\nabla_\mathbf a f = \mathbf a(f)$. It's also true that $\mathbf a$ is defined to act on scalar fields only while $\nabla_\mathbf a$ is allowed to act on generic $(p,q)$-tensor fields (which includes scalar fields as a special case). You use this as justification for defining the action of $\mathbf a$ on tensor fields as $$\mathbf a\langle\mathbf T\rangle:= \nabla_{\mathbf a}\mathbf T$$ This is non-standard but perfectly well-defined, and in particular if you wish to replace every instance of $\mathbf a$ acting on a scalar field with $\nabla_{\mathbf a}$ acting on the same scalar field, then you are free to do so.

However, from a structural and conceptual standpoint, I think its extremely important to maintain the distinction between vector fields and covariant derivatives. Your proposed definition would essentially make $\mathbf a$ and $\nabla_\mathbf a$ exactly the same object, which I object to for the following reasons.

  1. Vector fields are defined at the level of a smooth manifold, and do not require the extra structure of a connection / parallel transport. As a result, things like the action of $\mathbf a$ on a scalar field and the commutator of two vector fields can be discussed in a context where no notion of parallel transport is available. Examples of such contexts are the symplectic manifolds which arise in Hamiltonian mechanics and Lie groups, defined to be smooth manifolds with some additional group structure. In both cases, the actions of vector fields and their commutators are critical components to the relevant formalism and simply cannot be framed in terms of a covariant derivative which specifically does not exist. Manifolds-with-connection are also important, especially in GR, but in the larger context of differential geometry they are a special case; your proposed notation makes this extremely unclear.

  2. Even on those manifolds on which we have a connection, the action of $\nabla_\mathbf a$ depends on which connection we choose. Using the special notation $\nabla$ provides a good reminder that the results we choose are predicated on a particular choice of connection; the Riemann curvature tensor is an example of a connection-dependent object which may vanish or not based on how we choose $\Gamma$. If a particular statement is made purely in terms of the action of vector fields, we can be assured that it is a connection-independent statement. By blurring (or in this case, completely eliminating) the distinction between $\mathbf a$ and $\nabla_\mathbf a$, you use notation which may or may not rely on some arbitrary connection, which is not a good idea.

In summary, it seems to me that you find it advantageous to overload notation and use as small a list of symbols as possible. You see that $\mathbf a$ and $\nabla_\mathbf a$ act the same way on scalar fields while the latter is defined on all tensor fields, and view this as a problem; but this is a feature, not a bug.

Vector fields and covariant derivatives are conceptually and structurally entirely different. Extending the intuitive idea of directional differentiation from scalar fields to tensor fields requires additional structure, and so the use of different notation to emphasize this is an unambiguously good idea.

Beyond that, your chosen notation is so non-standard that if you choose to make this the foundation of your differential geometry notation, you are going to need to extensively explain what you mean to any differential geometer you ever speak to. This will likely include a strenuous objection in line with the ones I've raised. In the best case scenario, this will add completely unnecessary layers of confusion.

  • $\begingroup$ $\mathfrak{a}\left\langle f\right\rangle =\frac{df}{d\alpha}$ is a vector applied to a scalar field. See Eq. 9.1 of MTW. The Lie derivative is recovered from my definition as shown in the original post. The Lie derivative is not the only use for a commutator. MTW never define the application of a vector to a vector field, nor do they define the application of $\left[\vec{a}, \vec{b}\right]$ to a vector field. So they have no need to reference the connection. My definition subsumes theirs, and explicitly mentions the connection. What else would $d\vec{v}/d\lambda$ mean? $\endgroup$ Commented Mar 15, 2022 at 3:15
  • $\begingroup$ @StevenThomasHatton Okay. So given a vector $\mathfrak a$, there is a curve $\gamma:\alpha\mapsto \gamma(\alpha)$ such that $\mathfrak a(f) := \frac{d}{d\alpha} f\big(\gamma(\alpha)\big)$. That still doesn't answer the most important question I raised - namely, which vector of $\mathbf a$ is $\mathfrak a$? You say that $\mathfrak a$ is a member of $\mathbf a$, but which one, and why is that one special? $\endgroup$
    – J. Murray
    Commented Mar 15, 2022 at 3:21
  • $\begingroup$ It's the vector that doesn't get differentiated. I only introduced that notation to keep track of which symbols needed to be vector fields. I was hoping it would be apparent by the context. $\endgroup$ Commented Mar 15, 2022 at 3:34
  • 1
    $\begingroup$ @StevenThomasHatton Given the amount of non-standard notation and terminology you have chosen to introduce in your post, it's a bit unreasonable to expect the reader to make contextual inferences. In any case, the answer to your titular question is no, for essentially the same reason that a vector $\mathfrak a$ and the covariant derivative $\nabla_\mathfrak a$ are not the same thing. Your entire question - including the very substantial expansion - reduce down to that. I fail to understand what you think this library of non-standard notation and terminology is actually getting you. $\endgroup$
    – J. Murray
    Commented Mar 15, 2022 at 3:58
  • 2
    $\begingroup$ @StevenThomasHatton I'll rephrase my question. What is it that your notation clarifies, which the standard notation leaves ambiguous or unclear? What would one gain by adopting it? What does it make more convenient? What are you aiming to do here, besides making a list of new symbols and definitions? $\endgroup$
    – J. Murray
    Commented Mar 15, 2022 at 4:36

My question is:

Suppose we define the following \begin{align*} \nabla_{\mathfrak{a}}\mathbf{c} &\equiv\mathfrak{a}\left\langle \mathbf{c}\right\rangle ,\\ \left[\mathbf{a},\mathbf{b}\right]\left[ \mathbf{c} \right] &\equiv\mathfrak{a}\left\langle \mathbf{b}\left\langle \mathbf{c} \right\rangle \right\rangle -\mathfrak{b}\left\langle \mathbf{a}\left\langle \mathbf{c} \right\rangle \right\rangle \text{, and }\\ \left[\mathbf{a},\mathbf{b}\right]\left\langle \mathbf{c} \right\rangle &\equiv \mathfrak{w}\left\langle \mathbf{c} \right\rangle = \nabla_{\left[\mathbf{a},\mathbf{b}\right]}\mathbf{c}. \end{align*} Will this lead to $$\mathscr{R}\left(\mathbf{a},\mathbf{b}\right)\mathbf{c} = \left[\mathbf{a},\mathbf{b}\right]\left[ \mathbf{c} \right] - \left[\mathbf{a},\mathbf{b}\right]\left\langle \mathbf{c} \right\rangle ?$$


\begin{align*} \mathfrak{R}\left(\dots,\mathbf{c},\mathbf{a},\mathbf{b}\right) &\equiv\mathscr{R}\left(\mathbf{a},\mathbf{b}\right)\mathbf{c},\\ \mathscr{R}\left(\mathbf{a},\mathbf{b}\right) &\equiv\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]-\nabla_{\left[\mathbf{a},\mathbf{b}\right]} \end{align*}

The answer is yes.

All derivatives in this context are covariant derivatives. Thus, we may write one expression for the commutator.

If that definition is assumed then, when operating on a scalar field, the commutator acts as the covariant derivative along a vector $\mathfrak{w}$. However, when operating on a tensor of rank 1 or greater, the commutator does not produce the same result as covariantly differentiating along the vector $\mathfrak{w}$. Thus we have to

\begin{align*} \mathfrak{v}= & \frac{d\mathscr{P}}{d\nu},\\ \mathfrak{v}\left\langle \mathbf{T}\right\rangle \equiv & \lim_{\nu\to0}\frac{\mathbf{T}_{\nu\parallel\to0}-\mathbf{T}_{0}}{\nu}\equiv\nabla_{\mathfrak{v}}\mathbf{T},\\ \nabla_{\mathfrak{a}}\mathbf{b} = & \mathfrak{a}\left\langle \mathbf{b}\right\rangle \\ \nabla_{\mathfrak{a}}\nabla_{\mathbf{b}}\mathbf{c} = & \mathfrak{a}\left\langle \mathbf{b}\left\langle \mathbf{c}\right\rangle \right\rangle \\ \left[\mathbf{a},\mathbf{b}\right]\left[\mathbf{c}\right] \equiv & \mathfrak{a}\left\langle \mathbf{b}\left\langle \mathbf{c}\right\rangle \right\rangle -\mathfrak{b}\left\langle \mathbf{a}\left\langle \mathbf{c}\right\rangle \right\rangle \\ = &\left(\nabla_{\mathfrak{a}}\nabla_{\mathbf{b}}-\nabla_{\mathfrak{b}}\nabla_{\mathbf{a}}\right)\mathbf{c}\\ = & \left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]\left[\mathbf{c}\right]\\ \left[\mathbf{a},\mathbf{b}\right]\left\langle \mathbf{c}\right\rangle \equiv & \nabla_{\left[\mathbf{a},\mathbf{b}\right]}\mathbf{c}\\ \mathscr{R}\left(\mathbf{a},\mathbf{b}\right)\mathbf{c}\equiv & \left(\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]-\nabla_{\left[\mathbf{a},\mathbf{b}\right]}\right)\mathbf{c}\\ = & \left[\mathbf{a},\mathbf{b}\right]\left[\mathbf{c}\right]-\left[\mathbf{a},\mathbf{b}\right]\left\langle \mathbf{c}\right\rangle \end{align*}

Explicitly addressing the the question as stated in the header:

Is MTW's covariant commutator $\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]$ really the same thing as their vector field commutator $\left[\mathbf{a},\mathbf{b}\right]?$

Consider the clarification in the body of the original question:

$\dots$ The only difference is in how they are applied to tensors of rank grater than $0$.

First what is meant by "the same thing"? In the sense that $\left[\mathbf{a},\mathbf{b}\right]$ is only defined as an operator on scalar fields, whereas $\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]$ is defined to operate on any rank and valence of tensor field, they are clearly different. However, for a scalar field $f$ we do have the identity


So as defined by MTW, they are identical over the intersection of their domains of definition. As such, the operator $\left[\mathbf{a},\mathbf{b}\right]\left[\_\right]$ is redundant.

Thus, every appearance of $\left[\mathbf{a},\mathbf{b}\right]$ could be replaced by $\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]$ without changing the logical meaning of the presentation. The only caveat is that when written without an argument, we have to assume the vector identity $$\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]\equiv\nabla_{\mathfrak{a}}\mathbf{b}-\nabla_{\mathfrak{b}}\mathbf{a},$$

which I believe MTW never explicitly define.

So, where I have written the covariant derivative along the vector $\left[\mathbf{a},\mathbf{b}\right]$ as $$\left[\mathbf{a},\mathbf{b}\right]\left\langle \mathbf{c}\right\rangle \equiv\nabla_{\left[\mathbf{a},\mathbf{b}\right]}\mathbf{c},$$

we would be obliged to write


But that is no more arduous a burden than the original identification of $\left[\mathbf{a},\mathbf{b}\right]$ with a vector.

The following has been moved from the question.

Additional considerations:

When I originally posted this question, I didn't realize how "radical" my approach is. I have been doing this for a long time, and applying a vector to a vector field as covariant differentiation has always worked, and made sense to me. I only got tangled up when I started applying it to commutators.

I'm adding some screen-scrapes to illustrate what I mean by saying the two commutators are actually the same thing, using the notation I have defined in the original question. The three derivations are motivated by MTW Box 9.2. All three refer to the same diagram. In my personal notes the $\left[\left[\_,\_\right]\right]$ notation is what I use for a commutator, and my vector variables are written $\vec{\mathit{a}}.$ Some of the motivation for my "applicator notation" $\langle{\dots}\rangle$ is evident in the distinction between the value of a vector field at a point (approximately) determined by a vector displacement $\vec{\mathit{a}}\left[\mathfrak{b}\right],$ and the application of one vector field to another

$$\nabla_{\vec{\mathit{a}}}\vec{\mathit{b}} \equiv\nabla\left[\vec{\mathit{b}}\right]\cdot\vec{\mathit{a}} \equiv\vec{\mathit{a}}\langle\vec{\mathit{b}}\rangle.$$

Higher order error terms are omitted.

The commutator as closer of quadrilaterals

See MTW Box 9.2.

closed quadrilateral

Commutator operating "directly on the manifold" $\mathcal{P}$

Parallel transport the point $\mathcal{P}_{0}$ along the curve tangent to $d\mathcal{P}/d\alpha=\mathfrak{a}$ from $\alpha=0$ to $\alpha=1,$ and arrive at $\mathcal{P}_{1}.$ Parallel transport the point $\mathcal{P}_{1}$ along the curve tangent to $d\mathcal{P}/d\beta=\mathfrak{\vec{\mathit{b}}\left[a\right]}$ from $\beta=0$ to $\beta=1,$ and arrive at $\mathcal{P}_{4}.$

Go back to $\mathcal{P}_{0}$ and parallel transport it along $d\mathcal{P}/d\beta=\mathfrak{b},$ from $\beta=0$ to $\beta=1$ reaching $\mathcal{P}_{2},$ then along $d\mathcal{P}/d\alpha=\vec{\mathit{a}}\left[\mathfrak{b}\right]$ to $\mathcal{\mathcal{P}}_{3}.$ The geodesic connecting $\mathcal{P}_{3}$ with $\mathcal{P}_{4}$ has the tangent vector $\left[\vec{\mathit{a}},\vec{\mathit{b}}\right]=d\mathcal{P}/d\omega,$ such that $\mathcal{P}_{3}=\mathcal{P}\left[\omega=0\right]$ and $\mathcal{P}_{4}=\mathcal{P}\left[\omega=1\right].$

Of course "parallel transport the point" just means "transport the point", since points have no structure to keep parallel.

commutator acting on manifold

Commutator operating on a scalar field $f\left[\mathcal{P}\right]$

Do the same thing as in the previous example, except evaluate the scalar field $f$ at the points arrived at. The reason $\Delta{f}=\nabla_{\left[\vec{\mathit{a}},\vec{\mathit{b}}\right]}f$ is because $\nabla{f}$ is the rate of change of $f$ in every direction. So, treating $\mathfrak{w}=\left[\vec{\mathit{a}},\vec{\mathit{b}}\right]$ as a vector $\nabla{f}\cdot\mathfrak{w}=\nabla_{\mathfrak{w}}f$ is the first difference of $f$.

Again, parallel transport is indistinguishable from simple transport because $f\left[\mathcal{P}\right]$ is a number, and a number here is a number there, no matter how you carry it from place to place.

The important point is that the form of the derivation in this example of a scalar function is identical to that in the next example.

Commutator operating on a scalar field

Commutator operating on a vector field $\vec{\mathit{c}}\left[\mathcal{P}\right]$

As we see, the operation of $\left[\nabla_\vec{\mathit{a}},\nabla_\vec{\mathit{b}}\right]$ on a scalar field involves the same operations, performed in the same order as when operating on a vector field. The differences are in the results. For one, in the case of a scalar field the operation of the commutator as given in the definition turns out to be equivalent to differentiating along $\mathfrak{w}.$ As has already been indicated, differentiation of a scalar field along a vector is indistinguishable from covariant differentiation along a vector.

Differentiating along $\mathfrak{w}$ is meaningful in the case of a vector field, but it is not identical to performing the operations specified in the definition of the commutator. This is because the result of parallel transport is sensitive to the path taken when the manifold is not flat.

Commutator operating on a vector field

  • 1
    $\begingroup$ The answer to the title question - which you haven't referenced at all here - is an enthusiastic no. Perhaps you might clarify this. $\endgroup$
    – J. Murray
    Commented Mar 17, 2022 at 17:01
  • $\begingroup$ I have now explicitly addressed the question as stated in the header. The short answer is that we may make the replacement $\left[\mathbf{a},\mathbf{b}\right]\mapsto\left[\nabla_{\mathbf{a}},\nabla_{\mathbf{b}}\right]$ everywhere, without impacting the content of the theory. $\endgroup$ Commented Mar 18, 2022 at 9:30
  • $\begingroup$ It's clear that your mind is made up here, but I've added an elaboration to the end of my answer to clarify my objections in light of your edits. Best of luck. $\endgroup$
    – J. Murray
    Commented Mar 18, 2022 at 15:20
  • $\begingroup$ Argument by shibboleth is not mathematically valid. $\endgroup$ Commented Mar 19, 2022 at 22:01
  • 1
    $\begingroup$ I think I made two fairly strong points against your notational (re)definitions, and rather than engage with those criticisms, your response is to accuse me of mindlessly appealing to tradition? $\endgroup$
    – J. Murray
    Commented Mar 19, 2022 at 22:23

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