# Section 4.2 in Loring Tu's Differential Geometry

Section 4.2 in Loring Tu's Differential Geometry: 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.

• 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. Mar 14, 2021 at 5:12
• 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/… Mar 14, 2021 at 6:47
• @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
– BCLC
Apr 30, 2021 at 11:17
• @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)
– BCLC
Apr 30, 2021 at 11:19
• @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 $\nabla$.
– BCLC
Apr 30, 2021 at 13:37

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.

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 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.

• 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 $T$. Apr 30, 2021 at 13:18
• 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? Apr 30, 2021 at 13:30
• @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.
– BCLC
Apr 30, 2021 at 13:33
• @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?
– BCLC
Apr 30, 2021 at 13:44
• 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. Apr 30, 2021 at 15:50