On page 159 of "A Comprehensive Introduction To Differential Geometry Vol.1" by Spivak has written:

We thus see that the bracket $[X,Y]$ measures, in some sense, the extent to which the integral curves of $X$ and $Y$ can be used to form the "coordinate lines" of a coordinate system. If $X$ and $Y$ are two vector fields in a neighborhood of p, then for sufficiently small $h$ we can

(1) follow the integral curve of $X$ through $p$ for time $h$ ;

(2) starting from that point, follow the integral curve of $Y$ for time $h$;

(3) then follow the integral curve of $X$ backwards for time $h$ ;

(4) then follow the integral curve of $Y$ backwards for time $h$.

enter image description here


Before reading this book I thought that $\mathcal{L}_{X}Y=[X,Y]$ calculates changes of $Y$ along Integral curve of $X$.But in this Figure, the integral curves of both vector fields are used. I'm confused. Can someone help me?


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    $\begingroup$ A lot of beginners confuse $[X,Y]=-[Y,X]$ (which is defined for any pair of smooth vector fields) with $\nabla_X Y$ (which is only defined if you have a connection $\nabla$). However, you can see that they can't, generally, be the same just by taking a few examples. $X = y\partial_x$ and $Y = \partial_y$ in the $xy$-plane, for example. With the usual flat connection on the plane, $Y$ is parallel, so $\nabla_X Y=0$. But $\nabla_Y X$ is not zero, it's $[Y,X]$. When $\nabla$ is torsion-free, $[X,Y]=\nabla_X Y - \nabla_Y X$, so you can see how the flows of each vector field contribute to $[X,Y]$. $\endgroup$ – Robert Bryant Apr 17 '13 at 12:38
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    $\begingroup$ I'd say "along the flow of X", rather than "along integral curve of X" $\endgroup$ – Peter Dalakov May 3 '16 at 9:52
  • $\begingroup$ @RobertBryant The OP does not mention connections here. The question is how does one intuitively understand that the two characterizations of $[X,Y]$, one in terms of moving $X$ along the flow of $Y$, and the other more symmetrical in terms of the flows of both $X$ and $Y$, are consistent and should be expected to give the same result. $\endgroup$ – Arnaud Mortier Jun 25 '20 at 12:25
  • $\begingroup$ @ArnaudMortier: The OP didn't explain the meaning of the term "changes of $Y$ along integral curve of $X$", and I was pointing out that it should NOT be interpreted in the usual mistaken way, i.e., as essentially the covariant derivative of $Y$ w.r.t. $X$. When one defines $[X,Y]$ ( as Lie did) as the commutator, i.e., $[X,Y]f$ is set to be $X(Yf)-Y(Xf)$, it is obvious that $[X,Y]=-[Y,X]$. It was later that the operator $\mathcal{L}_X$ on tensors was defined, and when one looks at the definition, there is no reason to believe that $\mathcal{L}_{fX} = f\mathcal{L}_X$ for all $f$. $\endgroup$ – Robert Bryant Jun 25 '20 at 12:49

It is correct "that $\mathcal{L}_{X}Y=[X,Y]$ calculates changes of $Y$ along integral curve of $X$".

Edit: Namely, if $Fl^X_t$ is the flow of $X$, then $\mathcal L_XY = [X,Y] = \frac{d}{dt}|_{t=0} (Fl^X_t)^*Y$, the derivative of a smooth curve in the space of vector fields on the manifold (or on open subsets if $X$ does not have a global flow).

Spivak's description is another view. A general version of this view that "infinitesimal versions of group commutators are Lie brackets" is here:

  • Markus Mauhart, Peter W. Michor: Commutators of flows and fields. Archivum Mathematicum (Brno) 28,3-4 (1992), 228--236, arXiv:math.DG/9204221 (pdf).

See also 3.16 of here.


Arnold liked to call the Lie derivative the "fisherman derivative": you sit on the banks of a river and measure the change in the objects flowing in front of your eyes.

More concretelly, denote by $\Phi^t$ the local flow generated by the vector field $X$. Fix a point $p_0$ in the manifold $M$ and set $p_t:=\Phi^t(p_0)$. Then

$$L_XY(p_0)=[X,Y]_{p_0}= \lim_{t\to 0} \frac{1}{t}\Bigl(\;\Phi^{-t}_* Y_{p_t}- Y_{p_0}\;\Bigr)\in T_{p_0}M, $$

where $\Phi^{-t}_*: T_{p_t}M\to T_{p_0} M$ denotes the differential of $\Phi^{-t}$. For a proof I refer to Section 3.1.2 of these lectures.

  • $\begingroup$ This is of course correct but it misses the point of the question, which is how does one intuitively understand that the two characterizations of $[X,Y]$, one in terms of the flow of $Y$, and the other more symmetrical in terms of the flows of both $X$ and $Y$, are consistent and should be expected to give the same result. $\endgroup$ – Arnaud Mortier Jun 25 '20 at 12:22

The statement "$L_X Y=[X,Y]$ calculates changes of $Y$ along the integral curve of $X$" is not quite correct. Let me explain why.

Let $\phi_t^X$ denote the flow of $X$.

The key formula to understand the bracket (already mentioned above) is $$[X, Y] = \left. \frac{d}{dt} \right|_{t=0} \left( (\phi_t^X)^* Y \right).$$ (Pay attention to the location of the star: $\psi^* X$ and $\psi_* X$ denote respectively the pullback and the pushforward of $X$ by $\psi$; one of them corresponds to a "passive" change of coordinates while the other corresponds to an "active" transformation. Confusing them would lead to a sign error.)

Here is what this formula says. Imagine you push the vector $Y$ along the flow of $X$ for some time $\Delta t$, and you compare it to the vector that is already sticking out of the point you have reached. You divide the difference by $\Delta t$, and you make $\Delta t$ tend to $0$; this gives you $[X, Y]$.

The crucial thing to understand here is that when you do this, two things happen:

  • Obviously as you move along the integral curve of $X$, the value of $Y$ changes. The rate of this change is one of the terms that comprise $[X, Y]$. This is what you must have thought about when you said that "$L_X Y=[X,Y]$ calculates changes of $Y$ along the integral curve of $X$". But this is only part of the story, because...

  • The flow of $X$ is NOT a translation, because $X$ need NOT be locally constant. (In fact, on a general manifold, neither the notion of "translation" nor that of a "locally constant vector field" make sense, because these notions do depend on the coordinate system you choose.) So the flow of $X$ can stretch, squeeze or rotate the manifold, and then it stretches, squeezes or rotates $Y$ correspondingly. This means that even if $Y$ is "locally constant" (in some coordinate system), the bracket can still be nonzero.

These two contributions account respectively for the two terms in the right-hand side of the formula $$[X, Y] = \nabla_X Y - \nabla_Y X.$$ Check this! This is obvious for the first term, and requires some thinking for the second term.

If you do not know what the covariant derivative $\nabla$ is, you can think of vector fields on $\mathbb{R}^n$, and interpret $\nabla_X Y$ simply as "the directional derivative of $Y$ along $X$". This makes sense on $\mathbb{R}^n$, but not on an abstract manifold (if you try to define it with coordinates, you will get different values depending on what coordinate system you choose - unless you have an additional structure such as a Riemannian metric.)

The left-hand side, on the other hand, always makes sense, which is why it is introduced. The advantage is that it is invariant by diffeomorphisms (or, if you prefer, by change of coordinates). The drawback is that $(L_X Y)_x$ does not only depend on the value of $X$ at $x$, but on the value of $X$ on a whole neighborhood of $x$.

  • $\begingroup$ Disclaimer: software engineer. ;) Amazing explanations but the end let me wondering. The lie bracket is not linked to a given covariant derivative since you compute its torsion by subtracting your "second term" with the lie bracket. (Not confident at all) I think you meant its the pushforward of: the local derivative of Y along X with the manifold "flowing along" X minus of the local derivative of X along Y with the manifold "flowing along" Y this time. Although the antisymmetry is trivial with this formulation, the independence upon local coordinates is not. Again not confident at all. :S $\endgroup$ – matovitch Aug 28 '16 at 18:57
  • $\begingroup$ I do not understand what your are talking about; could you reformulate your question more clearly? For example when you say "it's the pushforward of...", what does "it" refer to? $\endgroup$ – Ilia Smilga Aug 29 '16 at 19:35
  • $\begingroup$ "It" was referring to [X,Y]. And by "push-forward", well I was thinking of the map from the local coordinate to the manifold (although I realize it does not mean a thing since one do not speak of a manifold embedded in a larger euclidean space a priori). (I am on the chat for a few minutes as this may be more an informal discussion than an interesting comment thread) $\endgroup$ – matovitch Aug 29 '16 at 20:00
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    $\begingroup$ @matovitch The Lie bracket can be defined in the absence of any connection. If you have a connection (i.e. a notion of parallel transport), its torsion measures the extent to which the parallel transport commutator $\nabla_XY-\nabla_YX$ differs from the Lie bracket. If the torsion is zero, they are the same. $\endgroup$ – S.Surace Jun 29 '18 at 7:14

Let me attempt to reconcile the two views on the Lie bracket.

First, one has to wonder what it should mean that a vector field $Y$ is ``constant'' along $X$. This is ambiguous, as noticed by katz. One point is that it is not a property that depends solely on the values of $Y$ along $X$, contrary to its Riemannian counterpart: it should really depends on the (local) field $Y$. Another confusion not to make is that it cannot be simply defined in charts by looking whether $Y$ is constant in the Euclidean sense: this would certainly not be chart-independant (even if we ask the chart to be a flow box for $X$).

Since the model is when $X=\frac\partial{\partial x}$ and $Y=\frac\partial{\partial y}$ in the plane, the one thing we could ask to a ``constant along $X$'' field $Y$ would be that if one follows during a given time $h$ an integral curve of $Y$ starting from any point in a integral curve $\gamma$ of $X$, then one should end up in a given integral curve $\gamma'$ of $X$ that does not depend on the starting point (but only on $t$ and $\gamma$). In fact, one should even ask that the parametrization of $\gamma$ is respected. This is what you get if you can find some chart that is a flow box for both $X$ and $Y$, that is if are part of a coordinate system (up to minor cheating on colinearity).

But this is exactly the definition of Lie bracket given in Spivak, up to a little twist: one asks if following $X$ for some time $h$ then $Y$ for time $h$ gives you the same point than following $Y$ for time $h$ then $X$ for time $Y$.


It is not entirely correct to assert that "the Lie bracket measures the change in Y along integral curves of X", at least not in the context of Riemannian geometry which is of course Spivak's context. Note that the same phrase could be applied to $\nabla_X Y$ as well, so at best the phrase is ambiguous. Thanks to Peter for the interesting reference which I hope to study further. Note that interpretations of the Lie bracket in terms of actual infinitesimals have been worked out in various contexts, so that the 4-step procedure becomes literally correct without taking limits.

Note also that you can push forward the standard coordinate fields in the plane by an arbitrary diffeomorphism and obtain random-looking vector fields that Lie-commute by construction. From the Riemannian viewpoint, it is odd to insist that one of them "does not change" along the other.


One view of the Lie bracket that is intermediate between the other views expressed is in terms of the symmetries of the double tangent bundle. It may not be the most clean geometric picture, but it is very useful when you want to think about other problems -- for example it gives a very clean formulation of torsion, the Riemann curvature tensor, and the Frobenius theorem on integrability of sub-bundles (to foliations).

Given a function between manifolds $f : M \to N$ there is an induced map of tangent bundles

$$Tf : TM \to TN$$

where $TM$ is the space of pairs of points, one point in $M$, and the other in the tangent space.

$$TM = \{ (p,v) : p \in M, v \in T_p M \}$$

$$Tf(p,v) = (f(p), Df_p(v))$$

this gives a nice formulation of the chain rule: $T(f \circ g) = Tf \circ Tg$.

From this perspective, the double tangent bundle is the space

$$T^2 M = \{ (p,v,w,y) : p \in M, v \in T_p M, (w,y) \in T_{(p,v)} TM \}$$

The latter condition requires that $w \in T_p M$. Although it is maybe not obvious, there is a canonical involution of $T^2 M$ given by $\iota(p,v,w,y) = (p,w,v,y)$.

From this perspective, if $v,w : M \to TM$ are vector fields, you compute $Tv \circ w$ and $\iota \circ Tw \circ v$ and observe that these are taking values in the same fibers of $T^2 M \to TM$, so you can subtract them:

$$Tv \circ w - \iota \circ Tw \circ v$$

Moreover, one can check these are in the "vertical" fiber. Specifically, take the bundle projection map $\pi : TM \to M$, $\pi(p,v) = p$ then $Tv \circ w - \iota \circ Tw \circ v$ is taking values in $ker(T\pi)$, which one can identify with the tangent spaces of $M$. Call the identification $\alpha$, then you get the formula

$$[w,v] = \alpha \left( Tv \circ w - \iota \circ Tw \circ v \right).$$

I don't claim this is how you want to always think about the Lie bracket, but it is a very formally useful device, and this perspective can make computations of things like the Riemann curvature tensor a bit simpler. The integrability condition for the Frobenius theorem has a simple statement from this perspective, as well. If you have a sub-bundle of $TM$, it has a "tangent bundle" that is a sub-bundle of $T^2 M$. The Frobenius integrability statement has the relatively elegant formulation that this "tangent bundle" must be invariant under $\iota$. This is close to formally equivalent to the statement that tangent vector fields have Lie brackets that must be tangent, but I find it more pleasant -- and easier to compute, i.e. use as an obstruction.


refer to Wikipedia: https://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields in limit and flow section that hints what you want. also this PDF File below will be very useful And Directly discussing what you want: https://faculty.math.illinois.edu/~kapovich/481-14/commutator.pdf or check this one: http://pi.math.cornell.edu/~goldberg/Talks/Flows-Olivetti.pdf


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