# Flows that do not commute A vector field $V$ on the plane moves points. Starting at $P$ and following the integral curve of $V$ for parameter distance $s$ lands at a point we write $\phi^V_s(P)$. The map $\phi^V_s$ is the flow of $V$. This page is about what happens when you follow two fields in turn, first one order and then the other, and why the difference is measured by the Lie bracket $[V,W]$. The source is Schutz, *Geometrical Methods of Mathematical Physics*, §2.13–2.15. ## 1. Moving along a vector field Start in one dimension. For a function $f(\lambda)$, define the shift operator $T_\varepsilon$ by $(T_\varepsilon f)(\lambda)=f(\lambda+\varepsilon)$. Differentiating in $\varepsilon$, $ \frac{d}{d\varepsilon}(T_\varepsilon f)(\lambda) = f'(\lambda+\varepsilon) = \Big(\frac{d}{d\lambda}\,T_\varepsilon f\Big)(\lambda), \qquad T_0 = I . $ So $T_\varepsilon$ solves $\dfrac{dT}{d\varepsilon} = \dfrac{d}{d\lambda}\,T$ with $T_0=I$, the same equation as $y'=ay$, $y(0)=1$. Its solution is the exponential, $ T_\varepsilon = \exp\!\Big(\varepsilon\frac{d}{d\lambda}\Big) = I + \varepsilon\frac{d}{d\lambda} + \frac{\varepsilon^2}{2!}\frac{d^2}{d\lambda^2} + \cdots, $ and applying it to $f$ at $\lambda$ is the Taylor series of $f(\lambda+\varepsilon)$, for analytic $f$. On the plane a vector field $V=V^1\partial_x+V^2\partial_y$ acts on functions as a derivative, $Vf=V^i\partial_i f$. This is the derivative along the integral curves of $V$, with respect to their parameter. The same argument gives $ f\big(\phi^V_\varepsilon(P)\big) = \big(e^{\varepsilon V} f\big)(P) = \Big(f + \varepsilon\,Vf + \tfrac{\varepsilon^2}{2}\,V(Vf) + \cdots\Big)(P). $ Taking $f$ to be a coordinate function $x^j$ gives the coordinates of the point you reach. The series uses only $V$ and its derivatives at $P$, and it reproduces the finite motion. The applet's third mode shows this for a linear field, where $V(Vx)=A^2x$ and the series is $\sum (sA)^n P/n!$. ## 2. The incomplete parallelogram From $P$, go $\varepsilon$ along $V$ to $R$, then $\delta$ along $W$ to $A$. Separately, go $\delta$ along $W$ to $R'$, then $\varepsilon$ along $V$ to $B$: $ A = \phi^W_\delta\big(\phi^V_\varepsilon(P)\big), \qquad B = \phi^V_\varepsilon\big(\phi^W_\delta(P)\big). $ The operator order needs care. Let $g=e^{\delta W}f$, so $g(Q)=f(\phi^W_\delta(Q))$. Then $ \big(e^{\varepsilon V}e^{\delta W}f\big)(P) = \big(e^{\varepsilon V}g\big)(P) = g\big(\phi^V_\varepsilon(P)\big) = f\Big(\phi^W_\delta\big(\phi^V_\varepsilon(P)\big)\Big) = f(A). $ The motion taken first is the operator written first. So $x^j(A) = \big(e^{\varepsilon V}e^{\delta W}x^j\big)(P)$ and $x^j(B) = \big(e^{\delta W}e^{\varepsilon V}x^j\big)(P)$. Expand both products to second order: $ e^{\varepsilon V}e^{\delta W} = I + \varepsilon V + \delta W + \tfrac{\varepsilon^2}{2}V^2 + \varepsilon\delta\,VW + \tfrac{\delta^2}{2}W^2 + O(3), $ $ e^{\delta W}e^{\varepsilon V} = I + \varepsilon V + \delta W + \tfrac{\varepsilon^2}{2}V^2 + \varepsilon\delta\,WV + \tfrac{\delta^2}{2}W^2 + O(3). $ Everything cancels in the difference except the cross terms: $ x^j(A) - x^j(B) = \varepsilon\delta\,\big(V(Wx^j) - W(Vx^j)\big)(P) + O(3). $ Since $Wx^j = W^j$ and $Vx^j=V^j$, the bracket is $V(W^j)-W(V^j)=V^i\partial_iW^j - W^i\partial_iV^j$. This is the $j$th component of the Lie bracket, so $ \boxed{\;A - B = \varepsilon\delta\,[V,W](P) + O(3), \qquad [V,W]^j = V^i\partial_i W^j - W^i\partial_i V^j.\;} $ Here $O(3)$ means terms of total degree three in $\varepsilon$ and $\delta$. The bracket is the vector from $B$ to $A$. Schutz's (2.9) writes the operators for $A$ in the opposite order, which (acting on functions) describes the trip to $B$, so his (2.11)–(2.12) state the vector from $A$ to $B$ instead. With the definitions above the vector runs from $B$ to $A$. The shear example settles it. Take $V=\partial_x$ and $W=x\,\partial_y$, so $[V,W]=\partial_x(x)\,\partial_y = \partial_y$. From $(x,y)$, $ A = \big(x+\varepsilon,\; y+\delta(x+\varepsilon)\big), \qquad B = \big(x+\varepsilon,\; y+\delta x\big), \qquad A-B = (0,\ \varepsilon\delta), $ exactly. This is $+\varepsilon\delta\,\partial_y$. Preset 4 in the applet shows it at every point. ## 3. Exercise 2.1: the polar unit vectors The unit basis vectors of polar coordinates are $ \hat r = \cos\theta\,\partial_x + \sin\theta\,\partial_y = \Big(\frac{x}{r},\frac{y}{r}\Big), \qquad \hat\theta = -\sin\theta\,\partial_x + \cos\theta\,\partial_y = \Big(-\frac{y}{r},\frac{x}{r}\Big). $ In terms of the coordinate fields, $\hat r = \partial_r$ and $\hat\theta = \tfrac1r\,\partial_\theta$, since moving a distance $t$ along a circle of radius $r$ changes the angle by $t/r$. The coordinate fields commute, $[\partial_r,\partial_\theta]=0$, so the only contribution comes from differentiating the factor $1/r$: $ [\hat r,\hat\theta] = \Big[\partial_r,\ \tfrac1r\partial_\theta\Big] = \partial_r\Big(\frac1r\Big)\,\partial_\theta = -\frac{1}{r^2}\,\partial_\theta = -\frac{\hat\theta}{r}. $ The same answer from components: the components of $\hat\theta$, $(-\sin\theta,\cos\theta)$, do not change along $\hat r$, so $V^i\partial_iW^j=0$. Along $\hat\theta$ the components of $\hat r$ turn at rate $1/r$, so $W^i\partial_iV^j = \tfrac1r(-\sin\theta,\cos\theta)$. The bracket is $0-\hat\theta/r = (y/r^2,\,-x/r^2)$. Since it is not zero, $\hat r,\hat\theta$ is not a coordinate basis, which is what the exercise asks. Presets 2 and 3 have the same integral curves, rays and circles. They differ only in how fast the parameter runs along the circles: $\partial_\theta$ turns every point through angle $t$, while $\hat\theta$ moves every point a distance $t$, which is a smaller angle farther out. On the trip to $A$ the turn happens at radius $r+\varepsilon$, so it covers a smaller angle than the turn at radius $r$ on the trip to $B$. $A$ lags $B$ around the circle, which is the $-\hat\theta$ direction. At $P=(1,0)$ with $\varepsilon=\delta=0.1$ the applet gives $A-B=(0.00095,\,-0.00995)$ against $\varepsilon\delta[V,W](P)=(0,\,-0.01)$. ## 4. When the bracket vanishes, the flow parameters are coordinates Two facts from §2.15 (and Schutz's Exercise 2.2(b)): - If $[V,W]=0$ in a region, the flows commute exactly, not only to second order: $\phi^W_t\circ\phi^V_s = \phi^V_s\circ\phi^W_t$ for all $s,t$ small enough that the paths stay in the region. - Then the map $\Phi(s,t)=\phi^W_t(\phi^V_s(P))$ has $\partial\Phi/\partial t = W$ (the last step is along $W$) and, writing it the other way round as $\phi^V_s(\phi^W_t(P))$, $\partial\Phi/\partial s=V$. So $(s,t)$ are coordinates near $P$ with $V=\partial_s$ and $W=\partial_t$. Two qualifiers. - **Only locally.** In preset 6 the rotation and dilation commute, and $(s,t)$ are log-polar coordinates: $s$ is the angle and $t=\ln r$ plus a constant. But $s$ and $s+2\pi$ give the same point, so the coordinates hold on a patch around $P$, not on the whole plane. - **The fields must be independent.** If $V$ and $W$ are parallel at $P$, the grid collapses onto a curve and $\Phi$ is not one-to-one. In preset 6 both fields vanish at the origin, and nothing near there can be a coordinate patch. When the bracket is not zero, the two grids $\phi^W_t(\phi^V_s(P))$ and $\phi^V_s(\phi^W_t(P))$ separate, more so far from $P$. For the shear, the second-mode readout gives exactly $S^2$: the two grids differ by $(0,\,st)$ at the node $(s,t)$. ## 5. The applet <div class="applet" data-applet="lie-bracket"></div> **Parallelogram.** Click or drag to set $P$. The solid path goes $\varepsilon$ along $V$ (blue) then $\delta$ along $W$ (green) to $A$. The dashed path goes the other way to $B$. The black arrow runs from $B$ to $A$, and the wide pink arrow from $B$ is the prediction $\varepsilon\delta[V,W](P)$. When the prediction is good the two arrows lie on top of each other. The box in the corner magnifies the neighbourhood of $A$ and $B$ by the factor it prints. Without it the gap is invisible for small $\varepsilon$. **Play** sends a dot along each path at the same time. The sliders are logarithmic, from 0.001 to 1, and **δ = ε** ties them together. **Scaling panel** (under the readout). For the current $P$, with $\delta=\varepsilon$, the dots are $|A-B|$ for $\varepsilon$ from $10^{-3}$ to $1$. The line is $\varepsilon^2\,|[V,W](P)|$. **Flow grids.** The two grids of §4, solid for $V$ first and dashed for $W$ first, with $s,t$ from $-S$ to $S$. **Series.** For a single linear field $V=Ax$, the dashed curve is the exact orbit $e^{sA}P$. The solid curve is the series through the $(sA)^N P/N!$ term. The presets: | | $V$ | $W$ | $[V,W]$ | |---|---|---|---| | 1 | $\partial_x$ | $\partial_y$ | $0$ | | 2 | $\partial_r=(x/r,\,y/r)$ | $\partial_\theta=(-y,\,x)$ | $0$ | | 3 | $\hat r=(x/r,\,y/r)$ | $\hat\theta=(-y/r,\,x/r)$ | $-\hat\theta/r=(y/r^2,\,-x/r^2)$ | | 4 | $\partial_x$ | $x\,\partial_y$ | $\partial_y$ | | 5 | $\partial_x$ | $-y\,\partial_x + x\,\partial_y$ | $\partial_y$ | | 6 | $-y\,\partial_x + x\,\partial_y$ | $x\,\partial_x+y\,\partial_y$ | $0$ | | 7 | $A_1x$ | $A_2x$ | $(A_2A_1-A_1A_2)\,x$ | | 8 | typed | typed | computed numerically | Fields with $1/r$ (presets 2 and 3, and typed fields that blow up at the origin) are not defined at the origin. The applet stops and says so when $P$ or a path comes within $0.05$ of it. Things to try: - In preset 4, make $\varepsilon$ small and watch the magnifier. The black and pink arrows stay on top of each other for every $\varepsilon$. - In preset 3, drag $P$ toward the origin with $\varepsilon$ fixed. The gap grows like $1/r$. - Switch between presets 2 and 3 at the same $P$. Same curves, but only one of them closes. - In preset 7, press **a pair that commutes**, then change one entry of $A_2$ a little. - Type your own fields in preset 8, for example $V=(1,0)$ and $W=(0,\sin x)$, where the gap changes sign as $P$ crosses $x=\pi/2$. ## 6. Questions 1. Before trying them, predict which of presets 1–6 close ($A=B$). > [!question]- Answer > Presets 1, 2 and 6, the ones with $[V,W]=0$ in the table. For those the applet reports $|A-B|$ at roundoff, around $10^{-16}$, for every $\varepsilon$ up to 1, not just small ones. Presets 3, 4 and 5 do not close. 2. In preset 6 neither field is constant, and both integral curves are curved (circles and rays). Why does it close anyway? > [!question]- Answer > The bracket is zero. $V=(-y,x)$ and $W=(x,y)$ have $V^i\partial_iW^j = V^j$ and $W^i\partial_iV^j = V^j$, which cancel. Geometrically, the flow of $V$ is rotation $R(s)$ and the flow of $W$ is scaling by $e^t$, and rotating a scaled vector gives the same result as scaling a rotated one: $e^t R(s)P = R(s)\,e^tP$. Being curved has nothing to do with it. What matters is whether one flow carries the other field to itself, and rotation carries the radial field $(x,y)$ to itself. 3. In preset 5, which way does the gap point, and why does it not depend on $P$? > [!question]- Answer > Write $R(\delta)$ for the rotation by $\delta$ and $e_1=(1,0)$. Then $A = R(\delta)(P+\varepsilon e_1)$ and $B = R(\delta)P + \varepsilon e_1$, so > $A - B = \varepsilon\big(R(\delta)e_1 - e_1\big) = \varepsilon\,(\cos\delta - 1,\ \sin\delta).$ > $P$ cancels. The gap points mostly along $+y$, which is $[V,W]=\partial_y$, tipped toward $-x$ by the angle $\delta/2$. At $\varepsilon=\delta=0.3$ it is $(-0.0134,\ 0.0887)$ at every point. The length is $\varepsilon\cdot 2\sin(\delta/2)$, so the ratio to $\varepsilon\delta$ is $\sin(\delta/2)/(\delta/2)$, which is $0.9963$ at $\delta=0.3$. 4. What does the slope-2 line in the scaling panel tell you, and what happens to the dots for presets 1, 2 and 6? > [!question]- Answer > On log–log axes a power law $|A-B| = c\,\varepsilon^k$ is a straight line of slope $k$. With $\delta=\varepsilon$ the leading term is $\varepsilon^2|[V,W](P)|$, so the dots lie on a line of slope 2: halve $\varepsilon$ and the gap drops by a factor of four. The gap is second order, a product of the two step sizes, while each path has length of first order. Where the dots leave the line at large $\varepsilon$, the $O(3)$ terms are showing. For the shear they never leave, because the formula $A-B=(0,\varepsilon\delta)$ is exact. For commuting pairs the dots do not follow any line. They sit at the size of floating-point roundoff, about $10^{-16}$, across the whole range, and some are exactly zero (drawn hollow at $10^{-17}$). 5. In preset 3, set $\varepsilon=\delta=0.3$ with $P=(1,\,0.3)$. The length ratio $|A-B|/|\varepsilon\delta[V,W]|$ is $0.9998$, yet the two arrows differ in direction by about $15^\circ$. How can the size be so good while the direction is off? > [!question]- Answer > The $O(3)$ correction here is almost perpendicular to $[V,W]$: it points mostly along $\hat r$, while the bracket points along $-\hat\theta$. A perpendicular correction of size $c$ changes the angle by about $c$ but the length only by about $c^2/2$. Measured with $\varepsilon=\delta$ from $0.1$ down to $0.003$, the angle falls like $\varepsilon$ (fitted slope $0.99$) while the length error falls like $\varepsilon^4$ (slope $3.9$). Preset 5 splits the same way less extremely: angle $\delta/2$, length error $\delta^2/24$. There is no general rule. For linear fields (next question) the split is the other way round. 6. In preset 7, choose any two matrices. The angle in the readout is $0.00^\circ$ even at $\varepsilon=1$, and the length ratio is the same at every $P$. Why? > [!question]- Answer > For a $2\times2$ matrix the Cayley–Hamilton theorem says $A^2 = (\operatorname{tr}A)A - (\det A)I$, so every power of $A$, and therefore $e^{sA}$, is a combination $\alpha I + \beta A$. Write $e^{\varepsilon A_1}=\alpha_1 I+\beta_1 A_1$ and $e^{\delta A_2}=\alpha_2 I+\beta_2A_2$. Multiples of $I$ commute with everything, so > $A - B = \big(e^{\delta A_2}e^{\varepsilon A_1} - e^{\varepsilon A_1}e^{\delta A_2}\big)P = \beta_1\beta_2\,(A_2A_1-A_1A_2)P = \frac{\beta_1\beta_2}{\varepsilon\delta}\;\varepsilon\delta\,[V,W](P),$ > exactly. The gap is parallel to the bracket at every scale, and only its length is off, by the factor $\beta_1\beta_2/(\varepsilon\delta)$, which does not involve $P$. For the default pair at $\varepsilon=\delta=0.1$ that factor is $0.99019$. This fails in three or more dimensions, where $e^{sA}$ needs $A^2$ as well. 7. In the series mode, the shear's series is exact from $N=1$ on. For the rotation from $P=(1,0)$ with $s$ up to $3$, the error at the end is $3.48$, $3.80$, $1.86$ and $0.052$ for $N=1,2,4,8$. Why exact in one case and slowly improving in the other? > [!question]- Answer > For the shear $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $A^2=0$, so every term after $sAP$ vanishes and $P+sAP$ is the whole orbit, a straight line. For the rotation, $A^2=-I$ and the terms never stop. The series is the Taylor series of $(\cos s,\sin s)$, and a polynomial of degree $N$ eventually runs off to infinity while the circle stays bounded. The error at $s$ is about the first omitted term, $s^{N+1}/(N+1)!$. On a fixed interval this goes to zero as $N$ grows ($N=20$ at $s=2$ is accurate to $4\times10^{-14}$), but for fixed $N$ it fails as $s$ grows. 8. In the flow-grids mode, what does it mean that the solid and dashed grids coincide for presets 1, 2 and 6, and why do they separate more the farther you go from $P$ in the other presets? > [!question]- Answer > If they coincide, the point reached does not depend on the order of the two motions, so each point near $P$ has a single label $(s,t)$, and these labels are coordinates (§4). Otherwise the mismatch at node $(s,t)$ is about $st\,|[V,W]|$ for small $s,t$, the parallelogram result with $\varepsilon=s$ and $\delta=t$. It grows with both distances. For the shear it is exactly $(0,\,st)$, so the largest mismatch is $S^2$ at the corners. > [!quote] Credit > The mathematics follows Bernard F. Schutz, *Geometrical Methods of Mathematical Physics* (Cambridge University Press, 1980), §2.13 (exponentiation of $d/d\lambda$), §2.14 (Lie brackets, Exercise 2.1, figure 2.21) and §2.15 (when a basis is a coordinate basis). The sign convention here differs from the book's printed (2.9)–(2.12), as explained in section 2.