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Section 4.2 in Loring Tu's Differential Geometry:

enter image description here

My Question: Since $D_XY −D_YX = [X,Y]$, then why define the quantity $T(X,Y)=D_XY −D_YX - [X,Y]$? Isn't $T$ always equal to $0$? I got very confused, and I want to know whether I have got anything wrong.

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    $\begingroup$ The wording is unfortunate. In general we are interested in derivative $D$ different from the directional derivative. In that case the torsion might be non-zero. $\endgroup$ Mar 14, 2021 at 5:12
  • $\begingroup$ I think you might like the answers to a similar-in-spirit question over on MathOverflow, in particular Tom Boardman's answer: mathoverflow.net/questions/20493/… $\endgroup$ Mar 14, 2021 at 6:47
  • $\begingroup$ @ArcticChar what's unfortunate with the wording? If $[X,Y]$ is defined as the Lie bracket in '(A.2)', then $T(X,Y)=0$. If $[X,Y]$ is not defined as the Lie bracket in '(A.2)', then we might not have $T(X,Y)=0$. Am I wrong? I posted answer $\endgroup$
    – BCLC
    Apr 30, 2021 at 11:17
  • $\begingroup$ @SammyBlack What's the relevance? That question is asking for intuition on torsion. This question is asking how we don't always have $T=0$. I think it's pretty easy/simple/shallow based on an oversight: If $[X,Y]$ is defined as the Lie bracket in '(A.2)', then $T(X,Y)=0$. If $[X,Y]$ is not defined as the Lie bracket in '(A.2)', then we might not have $T(X,Y)=0$. Am I wrong? Or is there some hard/complicated/deep thing that I missed? (Note: 'easy' here is not meant as an attack against the OP. It's meant to wonder why 2 commenters here are talking as if there's some deeper meaning or something) $\endgroup$
    – BCLC
    Apr 30, 2021 at 11:19
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    $\begingroup$ @ArcticChar oh wait never mind. you're right. forgot this book already apparently. Lol hehe. I got confused with the $D$ vs $\nabla$ and the lie bracket thing. I thought it was that $[,]$ means something else later on. Actually it's that $D$ means something else later on in the sense that $D$ is generalised to $\nabla$. $\endgroup$
    – BCLC
    Apr 30, 2021 at 13:37

2 Answers 2

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You're completely correct. This is a slightly unfortunate sentence. Later in the book, Tu will introduce the more general notion of an affine connection $\nabla$ on $TM$. This is a gadget quite similar to $D$, in that it is a map $\nabla:\mathfrak{X}(M)\times \mathfrak{X}(M)\to \mathfrak{X}(M)$ which is written $\nabla(X,Y)=\nabla_X Y$ and "differentiates" $Y$ with respect to $X$.

It satisfies moreover the properties of being $C^\infty(M)$ linear in $X$ and $\Bbb{R}-$linear in $Y$. The point of saying all of this is that for a general affine connection $\nabla$, we define the quantity $T(X,Y)=\nabla_X Y-\nabla_Y X-[X,Y]$ to be the torsion of $\nabla$, which is a tensor that eats a pair of vector fields and returns a vector field.

The reason we want to introduce this terminology is that a Riemannian manifold $(M,g)$ has a unique torsion free connection $\nabla$ compatible with the metric $g$. Compatibility here means that for all $X,Y,Z\in \mathfrak{X}(M)$, we have $$ X g(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_X Z)\:\:\:\:\text{(a version of the product rule)}. $$ We call this the Levi-Civita connection and it shows us that a Riemannian manifold comes for free with a "canonical" choice of connection. This is in turn useful, because it gives us a notion of parallel transport of vector fields. Given a parametrized curve $\gamma:I\to M$, we say that a vector field $V$ along $\gamma$ is parallel with respect to $\nabla$ if $$ \nabla_{\gamma'(t)}V=0\:\:\:\text{(parallel transport equation)}. $$ If you look here: https://mathoverflow.net/questions/20493/what-is-torsion-in-differential-geometry-intuitively at Anonymous's answer, they provide an example of a connection on $\Bbb{R}^3$ which is not the Levi-Civita connection (because it has nonzero torsion) and with respect to which the parallel translation rotates a vector as it "moves" along a curve. This perhaps explains the reason why it is called torsion. $T(X,Y)\equiv 0$ means (roughly) that there is no twisting in the translation in some sense.

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Edit 1: Arctic Char is right. I forgot this book already apparently. Lol hehe. I got confused with the $D$ vs $\nabla$ and the lie bracket thing. I thought it was that $[,]$ means something else later on. Actually it's that $D$ means something else later on in the sense that $D$ is generalised to $\nabla$.

For $T_\nabla(X,Y) := \nabla_X Y - \nabla_Y X- [X,Y]$, we have $T_\nabla(X,Y)=0$ for $\nabla=D$ because of the way the Lie bracket is defined. Actually the way the Lie bracket is defined gives us

$T_\nabla(X,Y) := \nabla_X Y - \nabla_Y X- (D_XY - D_YX)$


Edit 2: Wait I think it need not be Lie bracket in general. This book often refers to Tu's other book An Introduction to Manifolds. In Section 14, we have brackets

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It's not explicitly mentioned, but I think we can apply for $\mathfrak X(M)$ the zero bracket. But yeah it's probably just the $D$ vs $\nabla$ thing. if it's lie bracket, then check if $\nabla = D$. if it's not lie bracket, then depends on what the bracket is.

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    $\begingroup$ I think it's more what ArcticChar was saying in the comments: in my experience, when talking about vector fields $[X,Y]$ always indicates the usual Lie bracket. On the other hand, we often consider different notions of derivative (for example, the Levi-Civita connection associated to a Riemannian metric), and these can possibly lead to non-zero $T$. $\endgroup$ Apr 30, 2021 at 13:18
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    $\begingroup$ I'm sorry, I'm not following your second sentence. You're saying that in Tu's book, he eventually uses the notation $[X,Y]$ for vector fields to mean something other than the Lie brakcet? $\endgroup$ Apr 30, 2021 at 13:30
  • $\begingroup$ @JasonDeVito ok i checked the book again. I think I was mixing up $D$ and $\nabla$ vs the lie bracket thing. no need to apologise. thanks! editing answer now. $\endgroup$
    – BCLC
    Apr 30, 2021 at 13:33
  • $\begingroup$ @JasonDeVito wait. please see my 2nd edit. i think we can do like $[X,Y]=0$ ? i mean, is the zero bracket a lie bracket? $\endgroup$
    – BCLC
    Apr 30, 2021 at 13:44
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    $\begingroup$ There are certainly common situations where the notation $[X,Y]$ can refer to things other than Lie bracket of vector fields. This most prominently occurs in the study of Lie algebras, where $X$ and $Y$ need not be vector fields at all (just vectors in some random vector space). The claim I am making is that, in my own experience, when $X$ and $Y$ are vector fields on a manifold, the notation $[X,Y]$ has always meant the usual Lie bracket of vector fields. $\endgroup$ Apr 30, 2021 at 15:50

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