5
$\begingroup$

I am reading this great work here and I am trying to make sense of a specific derivation around the middle of the page. In particular, it seems they are claiming that:

$(\nabla_v w) (f) = (v^{\alpha} \nabla_{\alpha} w^{\beta})\nabla_{\beta}f$

where we work on a smooth manifold $M$, $v, w$ are smooth vector fields (i.e. sections of the tangent bundle $\mathcal{T}(M)$), $f$ is a scalar function and $\nabla$ is the connection on $\mathcal{T}(M)$.

My understanding is that they are using abstract index notation. But when I consider, for simplicity, a local set of coordinates with a local basis $e_i := \frac{\partial}{\partial x^i}$ I get a seemingly different answer. Specifically, if we write $v = v^i e_i$ and $w = w^j e_j$ one gets:

$(\nabla_v w) (f) = \left(\left(v^i \nabla_{e_i} w^k + v^iw^j \Gamma_{ij}^k \right)e_k\right)(f) = \left(v^i \nabla_{e_i} w^k + v^iw^j \Gamma_{ij}^k \right)\nabla_{e_k} f :=(\nabla_v w)^k \nabla_{e_k} f$

where $(\nabla_v w)^k$ means the $k$-th component of the vector $\nabla_v w$ in local coordinates. So my answer seems to be:

$(\nabla_v w) (f) = \left(v^i \nabla_{e_i} w^k\right) \nabla_{e_k} f + \left(v^iw^j \Gamma_{ij}^k \right)\nabla_{e_k} f$

But if their derivation is correct and I am interpreting abstract notation properly, it seems like I should instead be getting:

$ (\nabla_v w) (f) = \left(v^i \nabla_{e_i} w^k\right) \nabla_{e_k} f$

What am I missing?

Edit: I wish I could accept multiple answers. Huge Credit to @peek-a-boo and @Jackozee Hakkiuz for adding incredible insight to the problem. If anyone is reading this in the future, I highly recommend going over both answers.

Edit2: Also for those future readers, I highly suggest this high level debate here. One thing it illustrates clearly is that quantities like $\nabla_{\alpha} w^{\beta}$ are a priori ill-defined which can lead to understandable confusion. As a result, certain notational conventions are required, ones which are often author dependent. This makes it especially challenging for anyone who approaches the topic using different sources, since finding notational inconsistencies is almost inevitable.

Edit3: For my own clarity I would also like to give a formal answer that bridges the gap between the two notations, so here it goes:

$\underline{\textbf{(A posteriori) ANSWER:}}$

Let $\nabla_{\alpha} w^{\beta}:= (\nabla w)_{\alpha}^{\beta}$ be the placeholder (abstract) notation for the $(1,1)$-tensor field $\nabla w$ (said differently, in local coordinates $(\nabla w)_{\alpha}^{\beta}$ corresponds to $(\nabla w)_{k}^{i} (e_i \otimes \epsilon^k)$ or $(\nabla w)_{k}^{i} e_i \epsilon^k$ for short). We will see that this notation is consistent with the numerical one (i.e. the one in local coordinates).

The setup is the same as above but we also add a local co-basis $\epsilon^j := dx^j$. We observe the following:

  • $v = v^k e_k$ (as a $(1,0)$-tensor field, aka a vector field)
  • $\nabla w = (\nabla w)^i_j e_i \epsilon^j$ (as a $(1,1)$-tensor field)
  • $e_i (f) = \nabla_{e_i} f$ (by definition)
  • $\nabla_v w = (\nabla w)(v)$ (by definition)

One can then perform the following calculations:

$$(\nabla_v w)(f) = ((\nabla w)(v))(f) = ((\nabla w)^i_j e_i \epsilon^j v^k e_k)(f) = (v^k(\nabla w)^i_k) e_i(f) = (v^k(\nabla w)^i_k) \nabla_{e_i}f$$

which is "consistent" with the abstract index notation of:

$$(\nabla_v w)(f) = (v^{\alpha} \nabla_{\alpha} w^{\beta})\nabla_{\beta}f = (v^{\alpha} (\nabla w)_{\alpha}^{\beta}) \nabla_{\beta}f$$

$\underline{\textbf{The Subtleties:}}$

There are some subtleties here, I feel they need to be addressed:

  1. The first one is that in some sense we got "lucky" that the two expressions (abstract and numerical) seem to match each other one to one (symbol-wise). This is due to the fact that both results above, correspond to scalars so there are no implicit vectors/covectors "leftover" as implied from the abstract notation. To see this explicitly, observe the following correspondences between the two notations:

$v^{\alpha} \leftrightarrow v^k e_k$

$(\nabla w)_{\alpha}^{\beta} \leftrightarrow (\nabla w)_{l}^{i}e_i \epsilon^l$

$\nabla_{\beta} f \leftrightarrow (\nabla_{e_m} f) \epsilon^m $

This means that when we put everything together we get:

$$ (v^{\alpha} (\nabla w)_{\alpha}^{\beta}) \nabla_{\beta}f \leftrightarrow (v^k e_k (\nabla w)_{l}^{i}e_i \epsilon^l) ((\nabla_{e_m} f) \epsilon^m) = (v^k (\nabla w)_{l}^{i} \nabla_{e_m} f) e_k e_i \epsilon^l \epsilon^m = (v^k (\nabla w)_{k}^{i}) \nabla_{e_i} f $$

Thus, it is only because we have the perfect amount of contractions (i.e. the final result is a $(0,0)$-tensor (aka. a scalar)) that the two notations "match" each other. In the general case, the two notations will not match one another as (in some sense) the abstract notation "implies" the basis vectors while the regular one does not. (For example, even though $v^{\alpha} \leftrightarrow v^k e_k$, the two expressions are not in a "perfect" one to one notational correspondence, symbol-wise).

  1. Ambiguity can be eliminated. If one uses correspondences analogous to the ones seen above, ambiguities can be eliminated. This is best seen by performing calculations on different parentheses placements. For example, we know that in local coordinates:

$$\nabla_v w = v^i \nabla_{e_i} (w^k e_k) = v^i (\nabla_{e_i} w)^k e_k $$ where $ \nabla_{e_i} w = \left(\frac{\partial w^k}{\partial x^i} + w^j \Gamma_{ij}^k \right) e_k$ so that $ (\nabla_{e_i} w)^k = \frac{\partial w^k}{\partial x^i} + w^j \Gamma_{ij}^k$. But then, one can calculate $v^{\alpha}\nabla_{\alpha} w^{\beta}$ in three different ways:

$$\boxed{v^{\alpha} (\nabla_{\alpha} w)^{\beta} \leftrightarrow v^i (\nabla_{e_i} w)^k e_k = \nabla_v w}$$

$$\boxed{ v^{\alpha} (\nabla w)_{\alpha}^{\beta} \leftrightarrow v^i (\nabla w)^k_i e_k = \nabla_v w }$$

$$\boxed{ v^{\alpha} (\nabla_{\alpha}w^{\beta}) \leftrightarrow v^i (\nabla_{e_i}( w^k e_k))= v^i(\nabla_{e_i} w)^k e_k = \nabla_v w }$$

This means that the notation $v^{\alpha}\nabla_{\alpha} w^{\beta}$ is (a posteriori) unambiguous.

  1. Some ambiguity can still arise (if one is not careful). For example, take the notation $\nabla_{\alpha} w^{\beta}$ with one covariant and one contravariant index. In local coordinates, does it correspond to $(\nabla_{e_i} (w^k e_k)) \epsilon^i$ or $(\nabla_{e_i} w^k) \epsilon^i e_k$? In other words, where should the "implied" basis vectors go? Unless we create some notion of priority of operation, we will have an unbridgeable ambiguity, since the two expressions above are different. However, loosely speaking, the operator $\nabla$ acts on the vector $w$ after it is made "whole" (i.e. after we make the correspondence $w^{\beta} \leftrightarrow w^k e_k$), so we need to put the implied basis vectors $e_k$ first:

$$\nabla_{\alpha} w^{\beta} \leftrightarrow (\nabla_{e_i} (w^k e_k)) \epsilon^i = (\nabla_{e_i} w)^k e_k \epsilon^i$$

Similarly, the hessian of a smooth scalar function has a unique correspondence in local coordinates, given by:

$$\color{red}{\nabla_{\alpha} \nabla_{\beta} f \leftrightarrow (\nabla_{e_i} [(\nabla_{e_j} f) \epsilon^j])\epsilon^i} = \dots = [(\partial_i \partial_j f) - (\partial_k f)\Gamma_{ij}^k]\epsilon^j\epsilon^i $$

From there, we can derive the expression of the Torsion (defined by $T(f) = (\nabla_{\alpha} \nabla_{\beta} - \nabla_{\beta} \nabla_{\alpha}) (f)$) when acting on the scalar :

$$T(f) = [(\Gamma_{ij}^k - \Gamma_{ji}^k) - [e_i, e_j]^k](\partial_k f)\epsilon^i \epsilon^j $$

And also the expression for Torsion itself as a (1,2) vector field:

$$\boxed{T = [(\Gamma_{ij}^k - \Gamma_{ji}^k) - [e_i, e_j]^k]\epsilon^i \epsilon^j e_k} $$

$\endgroup$
12
  • 2
    $\begingroup$ Interesting that I worked 40 years in differential geometry and never before heard of an “abstract index.” Dare I guess this is a physics thing — again? $\endgroup$ Sep 10, 2022 at 1:18
  • 1
    $\begingroup$ @Ted Shifrin Yes it is. Introduced by Sir Roger Penrose, my understanding is that it is used to describe invariant objects like tensors when a particular basis is not present, using "placeholder" labels/indices in a consistent way with where the indices would be (if such a basis was present). You can find more here. $\endgroup$
    – Pellenthor
    Sep 10, 2022 at 4:43
  • 1
    $\begingroup$ Going through your last edit I can't avoid the feeling that there is some misunderstanding. To use abstract indices you don't need to think about basis vectors. To write $\nabla_vw$ with abstract indices, write it as the $(1,1)$-tensor $\nabla w$ acting on a vector $v$ $$(\nabla w)(v).$$ Then assign the indices based on the variance. $$(\nabla w)_\alpha{}^\beta v^\alpha.$$ The repeated $\alpha$ means application of a linear map $\endgroup$ Sep 13, 2022 at 11:33
  • 1
    $\begingroup$ I don't think there is a conceptual error, but I feel the idea of the "implied" basis vectors is a bit mislead. You can write basis-dependent expressions that involve abstract indices, for example $v^\alpha = v^ke_k{}^\alpha$. In that equality, the upper $\alpha$ marks what object is a vector, while the indices $k$ are just numbers. $\endgroup$ Sep 13, 2022 at 21:25
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Sep 13, 2022 at 21:41

2 Answers 2

4
$\begingroup$

What you’re missing is an understanding of where the implicit brackets lie.

It is indeed correct that (by definition of what it means for a vector field to act on a function) $(\nabla_vw)(f)=(\nabla_vw)^k\nabla_{e_k}f$. But this is also what the abstract index notation is saying. Let’s stick to greek for abstract indices. Then:

  • $\nabla_{\beta}f$ stands for the $(0,1)$ tensor field $\nabla f\equiv df$.
  • $\nabla_{\alpha}w^{\beta}$ stands for the $(1,1)$ tensor field $\nabla w$ (recall that if $T$ is an $(r,s)$ tensor field then $\nabla T$ is an $(r,s+1)$ tensor field).
  • Now for contractions: $v^{\alpha}\nabla_{\alpha}w^{\beta}$ stands for the contraction of the $(1,1)$ tensor field $\nabla w$ with the $(1,0)$ tensor field, i.e vector field, $v$ to get $(\nabla w)(v)=\nabla_vw$. Finally, $v^{\alpha}\nabla_{\alpha}w^{\beta}\nabla_{\beta}f$ thus stands for the contraction of $\nabla_vw$ with $\nabla f=df$, which by definition is also what it means for a vector field to act on a smooth function, i.e $(\nabla_vw)(f)$.

Let us use latin indices for ‘actual’ indices. An expression like $\nabla_iw^j$ should NOT be interpreted as $\nabla_{e_i}(w^j)$, instead it should be interpreted as $(\nabla_{e_i}w)^j$. You’ll often see brackets omitted, so when in doubt, do the covariant derivatives first, and only then extract the components.


For instance, you may see the definition of the hessian of a smooth function written as $\nabla_{\alpha}\nabla_{\beta}f$. This stands for the $(0,2)$ tensor field $\nabla(\nabla f)$. The $(i,j)$ components of this tensor field are NOT $\nabla_{e_i}(\nabla_{e_j}(f))$. They’re more complicated.

$\endgroup$
2
  • $\begingroup$ Excellent answer thanks! Basically, $\nabla_{\beta} f$ is a tensor field (equal to $\nabla f := \nabla_{[\cdot ]} f$) but $\nabla_{e_i} f$ is clearly not (it is the i-th component of the tensor $\nabla f$). This is very unfortunate because there seems to be an unbridgeable "inter-inconsistency" between notations despite their usefulness and "intra-consistency" at each individual level. Differential Geometry is a wonderful notational mess. $\endgroup$
    – Pellenthor
    Sep 9, 2022 at 20:54
  • 5
    $\begingroup$ @Pellenthor indeed. Perhaps you may have heard that differential geometry is the study of things invariant under change of notation $\endgroup$
    – peek-a-boo
    Sep 9, 2022 at 21:11
2
$\begingroup$

To add to peek-a-boo's answer, I want to emphasize what I think is the key point: in abstract index notation you have $$\nabla_\alpha(w^\beta)=(\nabla w)_\alpha{}^\beta$$ and hence there is no ambiguity in what $\nabla_\alpha w^\beta$ means.

This is false for numerical indices, where you have $$\nabla_i(w^j)=\partial_i(w^j)$$ but $$(\nabla w)_i{}^j= \partial_i(w^j) + \Gamma_{ik}^j w^k$$ so it remains to say what $\nabla_iw^j$ means. The traditional choice is to set $$\nabla_iw^j=(\nabla w)_i{}^j.$$

$\endgroup$
5
  • $\begingroup$ Yes, that makes perfect sense thanks. This is exactly the type of "bridge" I was looking for, between the two types of notations. I may be in the minority here, but I am really not a fan of what abstract notation is meant to represent. If I have to make this translation in my head every time I do calculations, then it sort of defeats the purpose. Is there anything wrong with using $(\nabla_w)_\alpha^{\beta}$ instead of $\nabla_{\alpha} w^{\beta}$ when the former is so much more descriptive than the latter? $\endgroup$
    – Pellenthor
    Sep 9, 2022 at 21:55
  • $\begingroup$ @Pellenthor What translation are you referring to? I prefer to think entirely in abstract index notation precisely because there is no ambiguity: $\nabla_\alpha w^\beta = (\nabla w)_\alpha{}^\beta= \nabla_\alpha(w^\beta)$ so you can use either of the three without danger of confusion. $\endgroup$ Sep 9, 2022 at 22:17
  • $\begingroup$ In my opinion, the choice of defining $\nabla_i w^j = (\nabla w)_i{}^j$ when using numerical indices is the unnatural one. For me it would be much more intuitive to have $\nabla_iw^j=\nabla_i(w^j)=\partial_i(w^j)$, (and judging by the calculations in your post, it looks that it is what you tried to use) but this is not the usual convention for numerical indices. $\endgroup$ Sep 9, 2022 at 22:21
  • $\begingroup$ I think we were in agreement from the beginning. The point I was trying to make is that - for me - it makes little sense to ever use $\nabla_{\alpha} w^{\beta}$ when I always have to "translate" it into $(\nabla_w)_{\alpha}^{\beta}$ which is the only unambiguous version which is also "compatible" with the "regular" one (even if both abstract versions mean the same thing). In any case, I added an addendum to the question to reflect that. Thank you again for your insight! $\endgroup$
    – Pellenthor
    Sep 9, 2022 at 22:28
  • 1
    $\begingroup$ Oh, I think I see what you mean. I'm glad the confusion is now cleared up. :) $\endgroup$ Sep 9, 2022 at 22:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .