I've come accross this physical interpretation for $ [X,Y] $ which I don't understand :

  • Follow $X$ for some time $\epsilon$;
  • Follow $Y$ for $\epsilon$;
  • Follow -X for $\epsilon$;
  • Follow -Y for $\epsilon$;

In the limit as $\epsilon$ approaches 0, the result of the above motion approaches the Lie Bracket $[X,Y]$.

Maybe someone can elucidate this for me?

  • $\begingroup$ This might be related.. $\endgroup$ – anon Jul 8 '12 at 16:40
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    $\begingroup$ A note on magnitudes might be in order: When $\epsilon$ is small, the above procedure approximates following $[X,Y]$ for $\epsilon^2$. There is a hilarious example in one of Edward Nelsons lecture notes from around 1970 (I think), using double commutators to show that you can drive a car sideways, at least to within any desired tolerance. But the $\epsilon^2$ factor makes this method of parallel parking exceedingly laborious. $\endgroup$ – Harald Hanche-Olsen Jul 8 '12 at 17:03
  • $\begingroup$ As Harald mentions, the Lie Bracket is the 2nd derivative of the function of $\epsilon$ you describe in your question, evaluated at $\epsilon = 0$. $\endgroup$ – Ryan Budney Aug 7 '12 at 19:04
  • $\begingroup$ @HaraldHanche-Olsen I think you mean the example starting on page 33 in Nelson's 1967 notes on tensor analysis. Fantastic! I only knew a simplified version of this example from lectures by Salamon, see also example 1.68 on page 34 (42 of the pdf) in these notes by Salamon-Robbin. $\endgroup$ – t.b. Sep 6 '12 at 22:15
  • $\begingroup$ @t.b. That's the one! $\endgroup$ – Harald Hanche-Olsen Sep 7 '12 at 9:32

The phrase 'flow $g$ along $Y$ a small distance $\epsilon$' is very nice, but all it means ultimately is just 'take the directional derivative of the function (e.g. surface) $g$ along $Y$'. Thus $X(Y)g$ means 'take the directional derivative of $Yg$ along $X$'. But $Yg$ is itself a directional derivative. Thus, $X(Y)g$ means 'take the directional derivative along $X$ of the directional derivative of $g$ along $Y$'.

Intuitively this is a generalisation of $\frac{\partial^2 g}{\partial x \partial y}$, since in the Lie bracket the two vector fields $X$ and $Y$ do not have to be orthogonal.

The second half of the Lie bracket then subtracts the same derivations in reverse order. If the two derivations commute, the Lie bracket is zero.

The vector flow terminology has definite aesthetic appeal (my original background is in fluid mechanics), but it remains difficult for me to visualise intuitively how the four arcs shrinking to a point as $\epsilon\rightarrow 0$ end up with this double derivation. The slope of a surface $g(x,y)$ in a particular direction at a given point $(x,y)$, on the other hand, is immediately obvious; and how that slope then may change in the direction of the other vector field is also intuitively clear. Once this has been established, the fact that these two double derivations may differ holds no mysteries.

Of course, the picture of the four arcs and whether they close or not remains the best visualisation of the Lie bracket once what it means has been understood.


To apply a vector field $V$ to a function $g$ (at a point $p$), take the directional derivative of $g$ along $V$. This is to say flow $g$ along $V$ some small distance $\epsilon$, take the difference quotient, and let $\epsilon \to 0$.

The Lie bracket $[X,Y]$ is defined as the vector field given by $[X,Y]f = X(Yf) - Y(Xf)$. So, loosely speaking, we are infinitesimally flowing along $Y$, then $X$ and also along $X$, then $Y$, and taking the difference. Since subtracting is adding the opposite, we're flowing infinitesimally along $Y$, $X$, then $-X$, and finally $-Y$.

I'm assuming you want an informal description, not a formal reason to think that $[X,Y]$ is the flow along $X$, then $Y$, then back along $X$, then back along $Y$. If you want the latter, I'll have to rewrite this.

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    $\begingroup$ Isn't the order wrong though? You have $Y,X,-X,-Y$ whereas it should be $X,Y,-X,-Y$. $\endgroup$ – user18063 Jul 8 '12 at 17:34

Write $\mathrm{Fl}^X_t(m)$ for the flow of a vector field $X$, that is $\forall m\in M,\forall t\in I_m$ where $I_m$ is the maximal interval on which the flow through $m$ is defined $$\frac{d}{dt}\mathrm{Fl}^X_t(m)=X_{\mathrm{Fl}^X_t(m)}$$ $$\mathrm{Fl}^X_0(m)=m.$$ Fix a point $m$ on your manifold $M$, and consider the curve $$c:(-\epsilon,+\epsilon)\rightarrow M,~t\mapsto c(t)=\mathrm{Fl}^Y_{-t}(\mathrm{Fl}^X_{-t}(\mathrm{Fl}^Y_t(\mathrm{Fl}^X_t(m))))$$ defined on a small neighborhood of $0$. This curve is what you get if you first follow the flow of $X$ for $t$ seconds, then the flow of $Y$ for $t$ seconds, the go back $t$ seconds along the flowlines of $X$ and go back again $t$ seconds along the flow lines of $Y$. This curve is drawn on the manifold, and passes through $m$ at $t=0$. It is also smooth because the flows of $X$ and $Y$ are smooth.

You can ask for the question "what is its derivative?", in particular what is $$\frac{d}{dt}\bigg|_{t=0}c(t)?$$ Since we are differentiating a curve, the result will be a tangent vector at $c(0)=m$. It turns out, but this requires some calculations, that the answer is $$\frac{d}{dt}\bigg|_{t=0}c(t)=[X,Y]_m.$$

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    $\begingroup$ This contradicts my remark on the question itself. Which of us is wrong? I think it is you, for if you replace $t$ by $-t$ in your construction, this would correspond to replacing $[X,Y]$ by $[-X,-Y]$ – i.e., by itself. So the curve you constructed must have a zero derivative at $t=0$, by symmetry. This is fixable, however, by taking a suitable square root. $\endgroup$ – Harald Hanche-Olsen Jul 8 '12 at 17:19
  • $\begingroup$ @Harald Hanche-Olsen There is indeed something amiss in what I wrote, maybe we should replace $t$ by $\sqrt t$. I thought this might work, but I probably need to write something like $$\frac{d}{dt}\frac{d}{ds}\mathrm{Fl}^Y_{-s}(\mathrm{Fl}^X_{-t}(\mathrm{Fl}^Y_s(\mathrm{Fl}^X_t(m))))$$ $\endgroup$ – Olivier Bégassat Jul 8 '12 at 17:26
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    $\begingroup$ No, that's too complicated. I think you just need to differentiate $c(\sqrt{t})$ at $t=0$ to get what you want. (But I don't have the time to check it. I should be packing for my vacation.) $\endgroup$ – Harald Hanche-Olsen Jul 8 '12 at 17:29
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    $\begingroup$ @HaraldHanche-Olsen Well this approach using two variables does work, this is how I learnt it initially. But I'll see if I can make it work with one variable. $\endgroup$ – Olivier Bégassat Jul 8 '12 at 17:33

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