## Section4Proofs of the main theorems

### Subsection4.1Abnormality of ODE trajectories

In Theorem 1.1 the Carnot group $F$ is only concerned with the degree of the polynomial ODE and not the particular ODE itself. Making this dependency explicit leads to the following stronger statement.

By the expansion of the vector field $P$ in the left-invariant frame, if $x\colon(a,b)\to G$ is a horizontal trajectory of the vector field $P\text{,}$ then the derivative of the trajectory is

$$\frac{d}{dt}x(t) = P(x(t)) = \sum_{i=1}^r P_i(x(t))\tilde{X}_i(x(t))\text{.}\label{eq-horizontal-trajectory}\tag{4.1}$$

Since the statement of the theorem only concerns horizontal trajectories there is no loss of generality in assuming that the other coefficients $P_{r+1},\ldots,P_n$ are all zero. Then the vector field $P$ has a well defined lift from $G$ to a horizontal vector field $\tilde{P}$ in the free Carnot group of rank $r$ and step equal to the step of $G\text{,}$ and all horizontal trajectories of $P$ lift to trajectories of the vector field $\tilde{P}\text{.}$ Hence the claim holds if and only if the claim holds when $G$ is a free Carnot group. In particular, there is no loss of generality in assuming that $G$ is a Carnot group that is compatible with the deg-left-right Hall set $\hallset\text{.}$ By considering coordinates adapted to the Hall set $\hallset\text{,}$ the polynomials $P_1,\ldots,P_r$ are identified with elements of $\polyring{\hallset}$.

Let $Q_{1},\ldots,Q_{r}\in\polyring{\hallset}$ be polynomials such that not all $Q_{i}$ are zero, $\sum P_iQ_{i} = 0\text{,}$ and $\max_i\deg(Q_{i})\leq \max_i\deg(P_i)\leq d\text{.}$ By Lemma 3.1, there exists a polynomial $Q\in\polyring{\hallset}$ such that $X_iQ=Q_{i}\text{.}$ Let $x(t)$ be a horizontal trajectory of $P$ whose closure contains the identity. Since $\sum P_i(X_iQ)=0\text{,}$ the identity (4.1) implies that the polynomial $Q$ is constant along the horizontal trajectory $x(t)\text{.}$ Adding in a constant if necessary, there is no loss of generality in assuming that $Q(x(t))\equiv 0\text{.}$

From the construction of Lemma 3.1, it follows that the polynomial $Q$ only contains variables of degrees at most $\max_i\deg(Q_{i})+1\leq d+1$ and the degree $\deg(Q)$ is bounded by the same quantity. Let $m:=d+2\text{.}$ By Proposition 3.6, there exists some high enough step $s\geq m$ and a covector $\covector\in\freelie{r,s}^*$ such that the polynomial $Q$ is a common factor for all abnormal polynomials $\abnormalpolynomial{X}{\covector}$ for $X\in\freelielayer{r,s}{m}\text{.}$ Since the closure of the trajectory $x(t)$ by assumption contains the identity, Lemma 3.2 implies that the trajectory $x(t)$ has an abnormal lift in the free Carnot group $\freecarnot{r,s}\text{.}$

For an arbitrary horizontal trajectory $x(t)$ of the vector field $P$ in $G\text{,}$ let $z\in G$ be any point in the closure of the trajectory, and let $y(t):=L_{z^-1}x(t)$ be the left translation of the trajectory $x(t)\text{.}$ The derivative of the translated trajectory is

\begin{equation*} \frac{d}{dt}y(t) = \sum_{i=1}^r P_i(x(t))\Big((L_{z^{-1}})_*\tilde{X}_i(x(t))\Big) = \sum_{i=1}^r P_i(L_{z}y(t))\tilde{X}_i(y(t))\text{.} \end{equation*}

Since left translations preserve the degrees of polynomials, the earlier argument implies that there exists a polynomial $Q_z\in\polyring{\hallset}$ such that $Q_z(y(t))\equiv 0$ and the degree of $Q_z$ and maximum degree of variables of $Q_z$ is again strictly less than $m\text{.}$ Repeating the rest of the argument, Proposition 3.6 implies that the trajectory $y(t)$ also has an abnormal lift in the free Carnot group $\freecarnot{r,s}$ for the same step $s\geq m$ as before. Since abnormality is preserved under left translation, it follows that the original horizontal trajectory $x(t)$ of $P\colon \freecarnot{r,s}\to T\freecarnot{r,s}$ lifts to an abnormal curve in $F\text{,}$ proving the claim.

An explicit value for the abnormality step $s$ of Theorem 4.1 can be computed from the rank $r$ and degree bound $d$ as described in Algorithm 5.1. The resulting values of $s$ for some of the smallest values of $r$ and $d$ are collected in Figure 5.3.

### Subsection4.2Concatenation of abnormals

Let $r$ be the rank of $G$ and let $s$ be the nilpotency step of $G\text{.}$ Since the abnormal curves in $G$ lift to the free Carnot group $\freecarnot{r,s}$ of rank $r$ and step $s\text{,}$ replacing $G$ with $\freecarnot{r,s}$ if necessary, there is no loss of generality in assuming that $G$ is compatible with the deg-left-right Hall set $\hallset\text{.}$

Let $\alpha\colon[a,b]\to G$ and $\beta\colon[c,d]\to G$ be two abnormal curves. The concatenation $\alpha\star\beta\colon[a,b-c+d]\to G$ in the group $G$ is defined by translating $\beta(c)$ to $\alpha(b)$ and then concatenating curves in the usual way. Since abnormality is preserved by left translation and reparametrization, it suffices to consider the case $[a,b]=[-1,0]\text{,}$ $[c,d]=[0,1]\text{,}$ and $\alpha(0)=\beta(0)=\identity{G}\text{.}$ Then the concatenation $\alpha\star\beta\colon[-1,1]\to G$ is defined by

\begin{equation*} \alpha\star\beta(t) = \begin{cases}\alpha(t),\amp t\leq 0\\\beta(t),\amp t\geq 0,\end{cases}\text{.} \end{equation*}

By the characterization of Lemma 2.2, abnormality means that there exist nonzero covectors $\covector_\alpha,\covector_\beta\in\mathfrak{g}^*$ such that $\abnormalpolynomial{X}{\covector_\alpha}\circ\alpha\equiv 0$ and $\abnormalpolynomial{X}{\covector_\beta}\circ\beta\equiv 0$ for all horizontal vectors $X\in\lielayer{\mathfrak{g}}{1}\text{.}$

Let $X,Y\in\lielayer{\mathfrak{g}}{1}$ be such that the abnormal polynomials $\abnormalpolynomial{X}{\covector_\alpha}$ and $\abnormalpolynomial{Y}{\covector_\beta}$ are nonzero, and let $Q:= \abnormalpolynomial{X}{\covector_\alpha}\abnormalpolynomial{Y}{\covector_\beta}$ be the product of the two. By Proposition 3.6 there exists a step $s_2\geq s$ and a covector $\covectorb\in\freelie{r,s_2}^*$ such that $Q$ is a common factor of all the abnormal polynomials $\abnormalpolynomial{Z}{\covectorb}$ for $Z\in\freelielayer{r,s_2}{s}\text{.}$ By construction $Q\circ(\alpha\star\beta)\equiv 0\text{,}$ so Lemma 3.2 then implies that the concatenation $\alpha\star\beta$ has an abnormal lift in the free Carnot group $F=\freecarnot{r,s_2}\text{.}$

By Proposition 3.6 the step $s_2$ depends only on the variables and degree of $Q\in\polyring{\hallset}\text{.}$ Since the step of the Carnot group $G$ is $s\text{,}$ the polynomials $\abnormalpolynomial{X}{\covector_\alpha}$ and $\abnormalpolynomial{X}{\covector_\beta}$ have (weighted) degree at most $s-1$ and hence $Q$ has degree at most $2s-2$ regardless of the specific covectors $\covector_\alpha$ and $\covector_\beta\text{.}$ Since the variables are necessarily some subset of the variables of the group $G\text{,}$ it follows that the step $s_2$ has an upper bound that depends only on the group $G\text{,}$ and not on the individual curves $\alpha$ and $\beta\text{.}$