Section 4.2 in Loring Tu's Differential Geometry

Question

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.

Answer 1

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.

Answer 2

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

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

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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.