I am totally new to the subject differential geometry, and that probably reflects itself in the naive question that I'm trying to formulate. I hope this question does not get closed because of this.

Let $M$ be a smooth manifold, equipped with an affine connection $\nabla$. Given two vector fields $X$ and $Y$, there are three ways we can "differentiate" $X$. The first one is to take Lie derivative $\mathscr{L}_Y X$, the second one is to do $\nabla_Y X$ and the third one, the most naive one, is just to take $DX: TM \to TTM$, which is just the total differential of $X$, viewed as a map $X: M\to TM$.

Note that the most naive way is very similar to taking exterior derivative of a function, and so, philosophically, it should be thought of as the "total derivative" and all others are somehow just partial derivatives.

This makes me believe (naively) that there is conceptually clean way to work from the most naive way of differentiating and deduce everything else (exterior derivatives (of higher wedges as well), Lie derivatives and connections) from it. I hope someone can clarify this for me.

Thank you.

  • 1
    $\begingroup$ You don't bother with $TTM.$ I remember Jeff Cheeger gong over this from his own book, Cheeger and Ebin, Comparison Theorems in Riemannian Geometry, in this case pages 2-3. He gives enough information there to justify writing expressions such as $\nabla_{c'}c'$ $\endgroup$
    – Will Jagy
    Apr 22, 2012 at 20:03

2 Answers 2


Your $DX$ generalizes to the setup where $X$ is a section $M\to E$ of a vector bundle $E$ on $M$, giving a (fiberwise linear) map $DX:TM\to TE$. At each point $p\in M$ this gives a linear map $D_pX:T_pM\to T_{X(p)}E$. There is a canonical short exact sequence $0\to E_p\to T_{X(p)}E\to T_pM$. The composed map $T_pM\to T_{X(p)}E\to T_pM$ is the identity, so the "TM part" of $DX$ does not depend on $X$.

A connection on $E$ gives a splitting of that exact sequence, so in the presence of a connection you can speak of the "E part" of your total derivative. That's $\nabla X$.

So you can sort of say that $\nabla X$ knows no more and no less than your $DX$ (but it doesn't appear until you choose a connection).

Note that $\nabla_YX$ is linear over the functions: we have $\nabla_{fX}Y=f\nabla_YX$ for a smooth function $f$ on $M$.

The Lie derivative is rather different. It's not linear over the functions in the $Y$ variable, and it's only defined when $E=TM$. Maybe someone else can say something more positive about where it fits in.


The language of Ehresmann connections is probably what you're looking for?

$$\nabla_X Y = c \circ DY \circ X$$

where $c : T(TM) \to TM$ is the Ehresmann connection. Here I'm using bundle conventions for derivatives, if $f : M \to N$ is smooth, $Df : TM \to TN$ is the tangent bundle map.

For Lie brackets, one way I like to formulate them is as an obstruction to certain types of 2nd order derivatives existing.

Let $\pi_M : TM \to M$ be tangent bundle projection, $\pi_{TM} : T^2 M \equiv T(TM) \to TM$ be double tangent bundle projection.

$$ \mathcal L(M) = \{ (V,W) \in (T^2M)^2 : D\pi_M(V) = \pi_{TM}(W), \pi_{TM}(V) = D\pi_M(W)\}$$

There is a map $[\cdot, \cdot] : \mathcal L(M) \to TM$ that you might as well call the `Lie Bracket' and it satisfies $[V,W]=0$ if and only if there exists a function of two variables

$f : (-\epsilon,\epsilon)^2 \to M$, $f(x,y)$ such that

$$\frac{\partial^2 f}{\partial x\partial y}(0,0) =V$$ $$\frac{\partial^2 f}{\partial y\partial x}(0,0) =W$$

where $\frac{\partial f}{\partial x} : (-\epsilon,\epsilon)^2 \to TM$ is the bundle theoretic partial derivative, i.e. $$\frac{\partial f}{\partial x}(x,y) = Df_{(x,y)}(1,0)$$

So while $\nabla_X Y$ and $DY \circ X$ are rather closely related, the Lie bracket's role is as an obstruction to 2nd order data being derivatives of multi-variable functions -- it's a very different object.

Oh, and of course, $[DY \circ X, DX \circ Y] = [X,Y]$, i.e. this Lie bracket specializes the the regular Lie bracket when you stick in vector fields.

  • $\begingroup$ Thanks a lot for the very instructive answer. I need to read more about Ehresmann connections to fully understand what you wrote. $\endgroup$
    – QcH
    Apr 29, 2012 at 2:10

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