## Section5Examples

The proof of Theorem 4.1 of abnormality of ODE trajectories gives a constructive method to find any polynomial ODE trajectory as an abnormal curve. This section covers the method more concretely, showing how the correct Carnot group and covector is found for some specific examples. The practical version of the proof of Theorem 4.1 is the following algorithm.

An implementation of Algorithm 5.1 using the SageMath computer algebra system  is available in .

###### Remark5.2.

The naive upper bound computed in step 4 of Algorithm 5.1 for the nilpotency step $s$ as a function of the rank and degree of the polynomials is in general horribly inefficient, see the explosive growth already for the smallest ranks $r$ and degrees $d$ listed in Figure 5.3. In practice, solutions are found in much smaller nilpotency steps, see Subsection 5.4 for a particularly egregious example where the a priori upper bound is $s=724\text{,}$ but the system has nontrivial solutions already for $s=13\text{.}$ For practical computations it is more efficient to form the linear system in smaller steps $s'\leq s$ and increase $s'$ one by one until a nontrivial solution is found.

In fact, computations suggest that there exists a much more refined bound $s=s(d)$ independent from the rank $r\text{.}$ Linear ODEs with randomly chosen coefficients in ranks 2, 3, and 4 always had solutions in step $s=7\text{,}$ despite the a priori bounds for $s$ being 11, 89, and 386 respectively. Similarly the quadratic ODEs of Subsection 5.3 and Subsection 5.4 in ranks 2 and 3 both have solutions in step $s=13$ despite the increase of the a priori bound from $s=38$ to $s=724\text{.}$

### Subsection5.1A logarithmic spiral (r2s7) Figure 5.4. The logarithmic spiral $e^{-t/4}(\cos t,\sin t)$

The logarithmic spiral

\begin{equation*} \gamma\colon [0,\infty)\to\RR^2,\quad \gamma(t) = e^{-t}(\cos t,\sin t) \end{equation*}

is a trajectory of the linear ODE

\begin{align*} \dot{x}_1 \amp= P_1(x) = -x_1-x_2\\ \dot{x}_2 \amp= P_2(x) = x_1-x_2 \end{align*}

with $\lim\limits_{t\to\infty}\gamma(t)=0\text{.}$ The construction of Algorithm 5.1 will show that the logarithmic spiral $\gamma$ lifts to an abnormal curve in the free Carnot group of rank 2 and step 7.

Step 1 Define polynomials $Q_{1}:=x_1-x_2$ and $Q_{2}:=x_1+x_2$ so that $(Q_{1},Q_{2})$ is orthogonal to the ODE vector $(P_1,P_2)\text{.}$

Step 2 Suppose $Q\in\polyring{\hallset}$ is a polynomial such that $X_1Q = Q_{1}$ and $X_2Q = Q_{2}$ and compute the higher order derivatives. The restricted action $\freelie{2}\acts\polyring{x_1,x_2}$ is defined by $X_1=\partial_1$ and $X_2=\partial_2\text{,}$ as can be seen by considering the coordinate expressions of the left-invariant horizontal vector fields in any rank 2 Carnot group. Since $\deg(X_1Q)=\deg(X_2Q)=1\text{,}$ the only nontrivial derivative is

\begin{equation*} X_{12}Q = [X_1,X_2]Q = X_1(X_2Q)-X_2(X_1Q) = 2\text{,} \end{equation*}

and all the higher order derivatives are zero.

Step 3 In the action $\freelie{2}\acts\polyring{\hallset}\text{,}$ each Hall basis element acts by $X_{w}=\partial_{w}+\rho_{w}\text{,}$ with $\rho_{w}$ some derivation whose kernel contains all polynomials in the variables of degree equal or lower than $w\text{.}$ Hence the nonzero partial derivatives determine the variables of the polynomial $Q\text{,}$ which in this case means that $Q\in\polyring{x_1,x_2,x_{12}}\text{.}$ The action $\freelie{2}\acts\polyring{x_1,x_2,x_{12}}$ is then determined by exponential coordinates adapted to the deg-left-right Hall set on any rank 2 Carnot group of step at least $2\text{,}$ the prototypical example being the Heisenberg group. Explicitly, the derivations are

\begin{align*} X_1 \amp= \partial_1\amp X_2 \amp= \partial_2+x_1\partial_{12}\amp X_{12} \amp= \partial_{12}\text{.} \end{align*}

Integrating in the maximal variable $x_{12}$ gives

\begin{equation*} Q = 2x_{12} + Q^{(2)} \end{equation*}

for some polynomial $Q^{(2)}\in\polyring{x_1,x_2}\text{.}$ The remainder $Q^{(2)}$ satisfies the PDE

\begin{align*} \partial_1Q^{(2)} \amp= X_1(Q-2x_{12}) = X_1Q = x_1-x_2\\ \partial_2Q^{(2)} \amp= X_2(Q-2x_{12}) = X_2Q - 2x_1 = -x_1+x_2\text{.} \end{align*}

Integrating in the variables $x_2,x_1\text{,}$ a solution is $Q^{(2)} = \frac{1}{2}x_1^2-x_1x_2+\frac{1}{2}x_2^2\text{,}$ so the full solution is

\begin{equation} Q = \frac{1}{2}x_1^2-x_1x_2+\frac{1}{2}x_2^2 + 2x_{12}\text{.}\label{eq-Q}\tag{5.1} \end{equation}

Step 4 The polynomial $Q$ contains variables of degrees $1$ and $2\text{,}$ so $m=3\text{.}$ The dimensions of the layers $1,2,3$ of the free Lie algebra of rank 2 are $2,1,2\text{,}$ respectively. To determine a sufficient nilpotency step, the generating function to consider is

\begin{equation*} \Delta(t) = \frac{(1-2(1-t^2))t^3}{(1-t)^2(1-t^2)}\text{.} \end{equation*}

The first few terms of its series expansion are

\begin{equation*} \Delta(t) = - t^{3} -2 t^{4} -2 t^{5} -2 t^{6} - t^{7} + 2 t^{9} + 4 t^{10} + 7 t^{11} + \cdots\text{.} \end{equation*}

Since $-1-2-2-2-1+2+4+7 = 5\geq 1\text{,}$ the logarithmic spiral will have an abnormal lift at least in step $11\text{.}$

Step 5 Let $\mathfrak{g}=\freelie{2,11}/[\lowercentralseriesterm{\freelie{2,11}}{3},\lowercentralseriesterm{\freelie{2,11}}{3}]$ be the quotient eliminating variables of degree $\geq 3$ from the abnormal polynomials $\abnormalpolynomial{112}{\covector}$ and $\abnormalpolynomial{212}{\covector}\text{,}$ which in nilpotency step 11 are polynomials of degree 8. By Lemma 3.5, the abnormal polynomial $\abnormalpolynomial{112}{\covector}$ has the simple expression

\begin{equation} \abnormalpolynomial{112}{\covector}(x_1,x_2,x_{12}) = \sum_{a+b+2c\leq 8}\frac{\covector_{(12)^c(2)^b(1)^{a+2}2}}{a!b!c!}x_1^ax_2^bx_{12}^c\text{.}\label{eq-P112}\tag{5.2} \end{equation}

The abnormal polynomial $\abnormalpolynomial{212}{\covector}$ is not as simple, since the Lie bracket $\ad{X_{12}}^c\ad{X_2}^b\ad{X_1}^aX_{212}$ is a deg-left-right Hall tree only when $a=0$ by the characterization of Lemma 2.9. For $a\gt 0\text{,}$ a normal form may be computed by considering the restricted adjoint representation $\ad{}\colon\mathfrak{g}\to\mathfrak{gl}(\lowercentralseriesterm{\mathfrak{g}}{3})\text{.}$ By the construction of the quotient $\mathfrak{g}\text{,}$ the only nontrivial commutator in $\mathfrak{gl}(\lowercentralseriesterm{\mathfrak{g}}{3})$ is $[\ad{X_1},\ad{X_2}] = \ad{X_{12}}\text{.}$ That is, a normal form may be computed using only the relation

Applying the above $a$ times, the resulting normal form for $a\gt 0$ is

Lemma 2.5 then gives the explicit expression

\begin{align} \abnormalpolynomial{212}{\covector}\amp(x_1,x_2,x_{12}) = \sum_{b+2c\leq 8}\frac{\covector_{(12)^c(2)^{b+1}12}}{b!c!}x_2^bx_{12}^c\label{eq-P212}\tag{5.3}\\ \amp+ \sum_{\substack{a+b+2c\leq 8\\a\geq 1}}\frac{\covector_{(12)^c(2)^{b+1}(1)^{a+1}2}+(a-1)\covector_{(12)^{c+1}(2)^b(1)^a2}}{a!b!c!}x_1^ax_2^bx_{12}^c\text{.}\notag \end{align}

For $i=1,2\text{,}$ define polynomials

\begin{equation} \factorpolynomial{\factorcoefficient{i}}(x_1,x_2,x_{12}) := \sum_{a+b+2c\leq 6}\factorcoefficient{i,a,b,c}x_1^ax_2^bx_{12}^c\label{eq-factorpoly}\tag{5.4} \end{equation}

and compute for $w_{1}=112$ and $w_{2}=212$ the difference polynomials

\begin{equation} R_i = \sum_{a+b+2c\leq 8}R_{i,a,b,c}(\covector,\factorcoefficient{i})x_1^ax_2^bx_{12}^c := \abnormalpolynomial{w_{i}}{\covector}-\factorpolynomial{\factorcoefficient{i}}Q\label{eq-Ri}\tag{5.5} \end{equation}

using the explicit expressions (5.1)(5.4).

Step 6 Consider the linear system

In the full set of $(\covector,\factorcoefficient{})$ variables, there are 190 equations and 220 variables: 120 variables $\covector_{112},\covector_{212},\ldots$ and 100 variables $\factorcoefficient{i,a,b,c}\text{.}$ Reducing to a system in only the $\factorcoefficient{}$ variables as in the proof of Proposition 3.6 leaves a system of 95 equations and 100 variables. The solution space is however much bigger than the difference: there is a 38 dimensional space of solutions $(\covector,\factorcoefficient{})\text{,}$ all with a nonzero $\factorcoefficient{}$ component.

One of the simplest solutions gives a degree 7 covector

\begin{align*} \covector\amp= 3 \covector_{1111112} - 3 \covector_{2211112} + 6 \covector_{2221112} - 9 \covector_{2222112} + 18 \covector_{2222212} \\ \amp + 2 \covector_{1211112} - 2 \covector_{1222112} + 8 \covector_{1222212} + 4 \covector_{1212112} + 8 \covector_{1212212} \end{align*}

Substituting the above solution $\covector$ into the formulas (5.2) and (5.3) and factoring gives the abnormal polynomials

\begin{align*} \abnormalpolynomial{112}{\covector} \amp = \frac{1}{4} (x_{1}^{2} + 2 x_{1} x_{2} - 3 x_{2}^{2} + 4 x_{12})Q\\ \abnormalpolynomial{212}{\covector} \amp = \frac{1}{2}(x_{1}^{2} + 3 x_{2}^{2} + 4 x_{12})Q \end{align*}

The conclusion is that the logarithmic spiral $\gamma(t)=e^{-t}(\cos t,\sin t)$ lifts to an abnormal curve in the free Carnot group of rank 2 and step 7.

### Subsection5.2Planar linear ODEs (r2s7)

The technique of Algorithm 5.1 can be applied to ODEs depending on free parameters $\freeparam_1,\ldots,\freeparam_k\in\RR$ by replacing all the considerations over the field $\RR$ with the polynomial ring $\polyring{\freeparam_1,\ldots,\freeparam_k}$ or the fraction field $\fractionfield{\freeparam_1,\ldots,\freeparam_k}$ where necessary. Solving the linear system in step 6 is where the only difference appears. The difference is that solving a linear system with coefficients in $\fractionfield{\freeparam_1,\ldots,\freeparam_k}$ does not yield a universal solution, since the resulting nonzero covector may vanish for specific choices of parameters. Accounting for the vanishing leads to a semialgebraic description of the covector with finitely many different expressions depending on the parameters $\freeparam_1,\ldots,\freeparam_k\text{.}$

Consider a generic homogeneous planar linear ODE

\begin{align} \dot{x}_1 \amp = \freeparam_{11}x_1 + \freeparam_{12}x_2\label{eq-linear-ode}\tag{5.6}\\ \dot{x}_2 \amp = \freeparam_{21}x_1 + \freeparam_{22}x_2\notag \end{align}

with parameters $\freeparam_{11},\freeparam_{12},\freeparam_{21},\freeparam_{22}\in\RR\text{.}$ As in Subsection 5.1 for a logarithmic spiral, following Algorithm 5.1 will show that for any such ODE all trajectories whose closures meet the origin lift to abnormals in the free Carnot group of rank 2 and step 7.

Step 1 Let $\fractionfield{\vectorparam{\freeparam}}:=\fractionfield{\freeparam_{11},\freeparam_{12},\freeparam_{21},\freeparam_{22}}$ be the fraction field with the parameters as indeterminates. Suppose $Q\in\polyringextended{\vectorparam{\freeparam}}{\hallset}$ is a polynomial with coefficients in the fraction field such that $X_1Q=\freeparam_{21}x_1 + \freeparam_{22}x_2$ and $X_2Q=-\freeparam_{11}x_1 - \freeparam_{12}x_2\text{.}$

Steps 25 proceed exactly as in Subsection 5.1, since the action $\freelie{2}\acts\polyring{\hallset}\text{,}$ the Poincaré series, and the abnormal polynomials are all essentially uneffected by the field extension $\RR\into \fractionfield{\vectorparam{\freeparam}}\text{,}$ and $\deg(X_1Q)=\deg(X_2Q)=1$ as before. The solution of the PDE for $Q$ is

\begin{equation*} Q = \frac{1}{2}\freeparam_{21}x_1^2 + \freeparam_{22}x_1x_2 - \frac{1}{2}\freeparam_{12}x_2^2 -(\freeparam_{11}+\freeparam_{22})x_{12}\text{.} \end{equation*}

The abnormal polynomials $\abnormalpolynomial{112}{\covector},\abnormalpolynomial{212}{\covector}$ are exactly the same as in (5.2) and (5.3), and the difference polynomial coefficients $R_{i,a,b,c}(\covector,\factorcoefficient{i})$ are again defined by (5.5).

Step 6 The solution space in the Lie algebra $\freelie{2,11}/[\lowercentralseriesterm{\freelie{2,11}}{3},\lowercentralseriesterm{\freelie{2,11}}{3}]$ of nilpotency step 11 is a 31 dimensional space over $\fractionfield{\vectorparam{\freeparam}}\text{.}$ The simplest solutions are found already in step $s'=7\text{.}$ In $\freelie{2,7}/[\lowercentralseriesterm{\freelie{2,7}}{3},\lowercentralseriesterm{\freelie{2,7}}{3}]\text{,}$ the solution space is 2 dimensional and an example solution has the nonzero coefficients

\begin{align*} \covector_{(1)^{4}(112)} \amp= 3 \freeparam_{21}^2 (6\freeparam_{11}^2 - 5\freeparam_{12}\freeparam_{21} + 13\freeparam_{11}\freeparam_{22} + 2\freeparam_{22}^2)\\ \covector_{(2)(1)^{3}(112)} \amp= 3 \freeparam_{21} (\freeparam_{11}\freeparam_{12}\freeparam_{21} + 5\freeparam_{11}^2\freeparam_{22} - 3\freeparam_{12}\freeparam_{21}\freeparam_{22} + 11\freeparam_{11}\freeparam_{22}^2 + 2\freeparam_{22}^3)\\ \covector_{(2)^{2}(1)^{2}(112)} \amp= -4\freeparam_{11}^2\freeparam_{12}\freeparam_{21} + 3\freeparam_{12}^2\freeparam_{21}^2 - 3\freeparam_{11}\freeparam_{12}\freeparam_{21}\freeparam_{22}\\ \amp+ 8\freeparam_{11}^2\freeparam_{22}^2 - 2\freeparam_{12}\freeparam_{21}\freeparam_{22}^2 + 18\freeparam_{11}\freeparam_{22}^3 + 4\freeparam_{22}^4\\ \covector_{(2)^{3}(1)(112)} \amp= -3 \freeparam_{12} (\freeparam_{11}\freeparam_{12}\freeparam_{21} + 3\freeparam_{11}^2\freeparam_{22} - \freeparam_{12}\freeparam_{21}\freeparam_{22} + 5\freeparam_{11}\freeparam_{22}^2)\\ \covector_{(2)^{4}(112)} \amp= -3 \freeparam_{12}^2 (-2\freeparam_{11}^2 + \freeparam_{12}\freeparam_{21} - \freeparam_{11}\freeparam_{22} + 2\freeparam_{22}^2)\\ \covector_{(12)(1)^{2}(112)} \amp= \freeparam_{21} (\freeparam_{11} + \freeparam_{22}) (-4\freeparam_{11}^2 + 3\freeparam_{12}\freeparam_{21} - 9\freeparam_{11}\freeparam_{22} - 2\freeparam_{22}^2)\\ \covector_{(12)(2)(1)(112)} \amp= - (\freeparam_{11} + \freeparam_{22}) (\freeparam_{11}\freeparam_{12}\freeparam_{21} + 3\freeparam_{11}^2\freeparam_{22} - \freeparam_{12}\freeparam_{21}\freeparam_{22} + 7\freeparam_{11}\freeparam_{22}^2 + 2\freeparam_{22}^3)\\ \covector_{(12)(2)^{2}(112)} \amp= \freeparam_{12} (\freeparam_{11} + \freeparam_{22}) (2\freeparam_{11}^2 - \freeparam_{12}\freeparam_{21} + 3\freeparam_{11}\freeparam_{22})\\ \covector_{(12)^{2}(112)} \amp= -(\freeparam_{11} + \freeparam_{22})^2 (-2\freeparam_{11}^2 + \freeparam_{12}\freeparam_{21} - 5\freeparam_{11}\freeparam_{22} - 2\freeparam_{22}^2)\\ \covector_{(2)^{4}(212)} \amp= 12 (\freeparam_{11} + \freeparam_{22}) \freeparam_{12}^3\\ \covector_{(12)(2)^{2}(212)} \amp= 2 \freeparam_{12}^2 (\freeparam_{11} + \freeparam_{22})^2 \end{align*}

For some specializations $\vectorparam{\freeparam}\in\RR^4$ such as $\freeparam_{11}=-2\text{,}$ $\freeparam_{12}=\freeparam_{21}=0\text{,}$ $\freeparam_{21}=1\text{,}$ the above covector vanishes and hence is not a valid abnormal covector. Nonetheless the existence of a generic solution implies that solutions exist also for every other choice of the parameters by the following brief argument:

The kernel of a matrix $A$ is nontrivial if and only if the matrix $A^TA$ has zero determinant. Let $A$ be a matrix with coefficients in $\polyring{\freeparam_1,\ldots,\freeparam_k}\text{.}$ If $A$ has a nontrivial kernel over $\fractionfield{\freeparam_1,\ldots,\freeparam_k}\text{,}$ then

\begin{equation*} \det(A^TA) = 0\in\polyring{\freeparam_1,\ldots,\freeparam_k}. \end{equation*}

Consequently for any specialization $\vectorparam{\freeparam}\in\RR^k$

\begin{equation*} \det(A(\vectorparam{\freeparam})^TA(\vectorparam{\freeparam})) = \det(A^TA)(\vectorparam{\freeparam}) = 0\in\RR, \end{equation*}

so the specialized matrix $A(\vectorparam{\freeparam})$ also has a nontrivial kernel.

The above shows that for any choice of parameters $\freeparam_{11},\freeparam_{12},\freeparam_{21},\freeparam_{22}\in\RR\text{,}$ all trajectories to the origin for the linear ODE (5.6) lift to abnormals in the free Carnot group of rank 2 and step 7.

Finding an explicit covector also in the cases where the generic one vanishes would require a more careful look at the computation to solve the linear system over the fraction field $\fractionfield{\vectorparam{\freeparam}}\text{.}$ In the standard Gauss-Jordan algorithm, each attempted division by a nonconstant polynomial $P$ splits the consideration into two cases: the points within the variety $P=0$ and those outside. Inside the variety, the polynomial is replaced by zero and the Gauss-Jordan procedure continues. Outside the variety, the polynomial $P$ can freely be used as a denominator. Expanding away the denominators in the end leads to finitely many different reduced echelon forms in distinct semialgebraic varieties, with each echelon form consisting of coefficients in the polynomial ring $\polyring{\vectorparam{\freeparam}}\text{.}$ Nontrivial elements of the solution space can then be read off from the echelon form, again splitting into cases based on the vanishing of the coefficients. This procedure defines an admissible abnormal covector $\covector$ as a semialgebraic function $\covector(\vectorparam{\freeparam})\text{.}$

However even in the relatively simple case of the linear ODE (5.6), solving the resulting $44\times 45$ system in nilpotency step 7 as described above leads to a rather cumbersome expression, and this will not be pursued here.

### Subsection5.3The Hawaiian earring (r2s13)

The Hawaiian earring is a countable union $E=\bigcup_{n\in\NN}E_n$ of circles $E_n = S^1((0,1/n),1/n)$ all with a common tangency point at 0. The punctured circles $E_n\setminus\{(0,0)\}$ have parametrizations that are trajectories of the complex ODE $\dot{z}=z^2\text{,}$ i.e., the planar quadratic ODE

\begin{align*} \dot{x}_1 \amp= x_1^2-x_2^2,\qquad\\ \dot{x}_2 \amp= 2x_1x_2\text{.} \end{align*}

Since all the punctured circles are trajectories of the same ODE and have a common point in their closure, the construction of Algorithm 5.1 implies they all have the same covector. The concatenation resulting in the full Hawaiian earring will be shown to lift to an abnormal curve in the free Carnot group of rank 2 and step 13.

Step 1 Set $X_1Q:=2x_1x_2$ and $X_2Q:=-x_1^2+x_2^2\text{.}$

Step 2 The nonzero commutators are

\begin{align*} X_{12}Q \amp = [X_1,X_2]Q = X_1(X_2Q)-X_2(X_1Q) = -4x_1\\ X_{112}Q \amp = [X_1,X_{12}]Q = X_1(X_{12}Q)-X_{12}(X_1Q) = -4\text{.} \end{align*}

Step 3 The action $\freelie{2}\acts \polyring{x_1,x_2,x_{12},x_{112}}$ is

\begin{align*} X_1 \amp= \partial_1,\amp X_2 \amp= \partial_2+x_1\partial_{12}+\frac{1}{2}x_1^2\partial_{112},\amp X_{12} \amp= \partial_{12}+x_1\partial_{112},\amp X_{112} \amp= \partial_{112}\text{.} \end{align*}

Integrating in $x_{112}$ and then $x_2$ and $x_1$ gives the solution

\begin{equation*} Q = x_1^2x_2+\frac{1}{3}x_2^3 - 4x_{112}\text{.} \end{equation*}

Step 4 Set $m=4\text{.}$ The dimensions of the first four layers $\freelielayer{r}{k}$ are $2,1,2,3\text{.}$ The generating function determining the step upper bound is

\begin{align*} \Delta(t) \amp = \frac{(1-3(1-t^3))t^3}{(1-t)^2(1-t^2)(1-t^3)^2}\\ \amp=-2 t^{4} - \cdots -20 t^{30} +45 t^{31} + \cdots +930 t^{38} + \cdots\text{.} \end{align*}

This series has the partial sums $\sum_{k=0}^{37}\differenceinteger{k} = -205$ and $\sum_{k=0}^{38}\differenceinteger{k} = 725\text{,}$ so the naive upper bound for the abnormality step is $s=38\text{.}$

Step 5 Since the upper bound $s=38$ is so large, it is more practical to solve the system in step $s'\lt s$ and keep increasing $s'$ until a nontrivial solution is found. Let $\mathfrak{g}$ be the quotient $\mathfrak{g}=\freelie{2,s'}/[\lowercentralseriesterm{\freelie{2,s'}}{4},\lowercentralseriesterm{\freelie{2,s'}}{4}]\text{.}$ As before, the abnormal polynomials are computed by computing normal forms for the brackets

where $w_{1}=1112\text{,}$ $w_{2}=2112$ and $w_{3}=2212$ are the deg-left-right Hall words of degree 4. The normal forms are computed using the represention $\ad{}\colon\mathfrak{g}\to\mathfrak{gl}(\lowercentralseriesterm{\mathfrak{g}}{4})\text{,}$ where the only nontrivial commutators come from $[X_1,X_2]=X_{12}\text{,}$ $[X_1,X_{12}]=X_{112}$ and $[X_2,X_{12}]=X_{212}\text{.}$

Step 6 The first nontrivial solution is found in step $s'=13\text{.}$ The result is that the Hawaiian earring has an abnormal lift in the free Carnot group of rank 2 and step 13 with the covector $\covector$ whose nonzero components are

\begin{align*} \covector_{(2)^4(1)^5(1112)} \amp = 210\amp \covector_{(112)^2(2)^2(1)(1112)} \amp = 2\\ \covector_{(2)^6(1)^3(1112)} \amp = 150\amp \covector_{(2)^9(2112)} \amp = 280\\ \covector_{(2)^8(1)(1112)} \amp = 140\amp \covector_{(112)(2)^6(2112)} \amp = -15\\ \covector_{(112)(2)^3(1)^3(1112)} \amp = -15\amp \covector_{(112)^2(2)^3(2112)} \amp = 2\\ \covector_{(112)(2)^5(1)(1112)} \amp = -10\amp \covector_{(112)^3(2112)} \amp = -2\text{.} \end{align*}

The resulting abnormal polynomials have the factorizations

\begin{align*} \abnormalpolynomial{1112}{\covector} \amp = -\frac{1}{96} x_{1} x_{2}^{2} (-7 x_{1}^{2} x_{2} - x_{2}^{3} + 12 x_{112})Q\\ \abnormalpolynomial{2112}{\covector} \amp = \frac{1}{432} (-x_{2}^{3} + 3 x_{112}) (-15 x_{1}^{2} x_{2} - x_{2}^{3} + 12 x_{112})Q\\ \abnormalpolynomial{2212}{\covector} \amp = -\frac{1}{288} x_{1} (3 x_{1}^{4} x_{2} - x_{1}^{2} x_{2}^{3} - 6 x_{2}^{5} - 24 x_{1}^{2} x_{112} + 36 x_{2}^{2} x_{112})Q\text{.} \end{align*}

### Subsection5.4The Lorenz butterfly (r3s13) Figure 5.8. The trajectory of the Lorenz system starting from $(1,0,0)$ on the time interval $[0,T]$ for $T=10\text{,}$ $T=30\text{,}$ and $T=100\text{.}$

The Lorenz system

\begin{align*} \dot{x}_1 \amp= 10(x_2-x_1)\\ \dot{x}_2 \amp= 28x_1-x_2-x_1x_3\\ \dot{x}_3 \amp= x_1x_2-\frac{8}{3}x_3 \end{align*}

is a classical example of a polynomial ODE system that exhibits chaotic behavior. Consider the trajectory starting from the point $(x_1,x_2,x_3)=(1,0,0)\text{,}$ see Figure 5.8 for a visualization. Following Algorithm 5.1 will show that this trajectory lifts to an abnormal curve in the free Carnot group of rank 3 and step 13. Translating the initial point of the trajectory to the origin means that the ODE to study is

\begin{align*} \dot{x}_1 \amp= -10x_1+10x_2+10\\ \dot{x}_2 \amp= -x_1x_3+28x_1-x_2+x_3-28\\ \dot{x}_3 \amp= x_1x_2-x_2-\frac{8}{3}x_3\text{.} \end{align*}

Step 1 In rank bigger than 2 there is more freedom to find a polynomial $Q\in\polyring{\hallset}$ whose horizontal gradient is orthogonal to the ODE. One possible choice is

\begin{align*} X_1Q\amp:=-x_1x_3 + 28x_1 - x_2 + x_3 - 28\\ X_2Q\amp:=10x_1 - 10x_2 - 10\\ X_3Q\amp:=0\text{.} \end{align*}

Step 2 The action $\freelie{3}\acts\polyring{x_1,x_2,x_3}$ on the horizontal variables is again $X_1=\partial_1\text{,}$ $X_2=\partial_2\text{,}$ $X_3=\partial_3\text{,}$ and all other elements give the zero derivation. The nonzero higher order derivatives of $Q$ are

\begin{align*} X_{12}Q \amp = X_1(X_2Q)-X_2(X_1Q) = 11\\ X_{13}Q \amp = X_1(X_3Q)-X_3(X_1Q) = x_1-1\\ X_{113}Q \amp = [X_1,X_{13}]Q = X_1(X_{13}Q)-X_{13}(X_1Q) = 1\text{.} \end{align*}

Step 3 The action $\freelie{3}\acts \polyring{x_1,x_2,x_{12},x_{13},x_{113}}$ is

\begin{align*} X_1 \amp= \partial_1 \amp X_{12} \amp= \partial_{12}\\ X_2 \amp= \partial_2+x_1\partial_{12} \amp X_{13} \amp= \partial_{13}+x_1\partial_{113}\\ X_3 \amp= \partial_3+x_1\partial_{13}+\frac{1}{2}x_1^2\partial_{113} \amp X_{113} \amp= \partial_{113}\text{.} \end{align*}

Integrating the variables in the order $x_{113},x_{13},x_{12},x_3,x_2,x_1$ gives the solution

\begin{equation*} Q = -\frac{1}{2}x_1^2x_3 + x_{113} + 14x_1^2 - x_1x_2 - 5x_2^2 + x_1x_3 + 11x_{12} - x_{13} - 28x_1 - 10x_2\text{.} \end{equation*}

Step 4 The dimensions of the first four layers of the free Lie algebra $\freelie{3}$ are $3,3,8,18\text{.}$ The generating function

\begin{equation*} \Delta(t) = \frac{(1-18(1-t^3))t^4}{(1-t)^3(1-t^2)^3(1-t^3)^8} \end{equation*}

gives the a priori bound of step $s=724$ for when a lift becomes abnormal.

Step 5 With the unreasonably large nilpotency step $s=724\text{,}$ it is better to solve the system in step $s'\lt s$ and keep increasing $s'$ until a nontrivial solution is found. In the quotient $\mathfrak{g}=\freelie{3,s'}/[\lowercentralseriesterm{\freelie{3,s'}}{4},\lowercentralseriesterm{\freelie{3,s'}}{4}]\text{,}$ the abnormal polynomials can be computed via normal forms for Lie brackets using the smaller family of commutation rules of the restricted adjoint representation $\ad{}\colon\mathfrak{g}\to\mathfrak{gl}(\lowercentralseriesterm{\mathfrak{g}}{4})$ as in the previous examples.

Step 6 The first solution exists in step $s=13\text{,}$ so the trajectory starting from $(1,0,0)$ of the Lorenz system is abnormal in the free Carnot group of rank 3 and step 13. In rank 3 and step 13 the linear system consists of 81360 equations in 34465 variables $\covector$ and 9918 variables $\factorcoefficient{}\text{.}$ The solutions are however relatively sparse, with an example solution having 476 nonzero coefficients out of the possible 34465, suggesting that further simplifications using more refined quotient Lie algebras could be possible.