The Sard problem in step 2 and in filiform Carnot groups

We study the Sard problem for the endpoint map in some well-known classes of Carnot groups. Our first main result deals with step 2 Carnot groups, where we provide lower bounds (depending only on the algebra of the group) on the codimension of the abnormal set; it turns out that our bound is always at least 3, which improves the result proved in arXiv:1503.03610 and settles a question emerged in arXiv:1709.02854. In our second main result we characterize the abnormal set in filiform groups and show that it is either a horizontal line, or a 3-dimensional algebraic variety.


Introduction
One of the main open questions in sub-Riemannian geometry is arguably the Sard problem for the endpoint map, cf.[2] or Section 10.2 of [14]: in fact, the problem is ubiquitous in sub-Riemannian geometry, as it has implications on the regularity of geodesics, the regularity of the distance and of its spheres, the heat diffusion, the analytic-hypoellipticity of sub-Laplacians, etc.The Sard problem asks whether the abnormal set, i.e. the set of critical values of the endpoint map, is negligible or not; we refer to Section 2 for precise definitions.Despite such a simple formulation, only very partial results are known [1,5,6,16,18] even in settings with a rich structure such as Carnot groups [3,7,[9][10][11][12]15].The goal of this note is to provide a contribution in two meaningful classes of Carnot groups: those with nilpotency step 2, and filiform ones.
The Sard problem in step 2 Carnot groups has already been answered affirmatively in [12]; however, the question of getting finer estimates on the size of the abnormal set was left open.In fact, in [15] it was conjectured that the abnormal set Abn G in a Carnot group G of step 2 has codimension at least 3.The conjecture is true in free Carnot groups of step 2, as proved in [12], and in some classes of step 2 Carnot groups [15] including all groups of topological dimension up to 7. Our first main result, Theorem 1.1 below, is a full, positive answer to this question: we prove in fact that Abn G has codimension at least 3 in every step 2 Carnot group G. Actually, ¦ L.N. is partially supported by the Swiss National Science Foundation (grant 200021-204501 Regularity of sub-Riemannian exponential map.With this notation we can state our first main result.
Theorem 1.1.Let G be a Carnot group of step 2 given by G V 2 V W for some W ¤ 2 V .Let W u2 be the orthogonal to W in 2 V with respect to an adequate scalar product and let k : min t rankpωq | ω W u2 z t 0 u u.Then Abn G is contained in an algebraic variety of codimension 2 k 1.
In particular, the abnormal set Abn G has codimension at least 3.
We refer to Proposition 3.1 for the definition of adequate scalar product.The bound 2 k 1 on the codimension of Abn G is not always optimal, see Examples 3.6 and 3.7; however, in some cases it provides the exact estimate, as for instance in the well-known case of Heisenberg groups (see Ex. 3.5) where the abnormal set is the singleton t0u and the optimal bound is indeed greater than 3.
The idea behind the proof of Theorem 1.1 can be described as follows.There is a natural projection map π : F r Ñ G from the free Carnot group F r associated with f r onto G.Each horizontal curve from the identity in G can be uniquely lifted to a horizontal curve from the identity in F r .Abnormal curves in G are projections of abnormal curves in F r , but the converse is not always true: in step 2 Carnot groups one is able to characterize those abnormal curves in F r that project to abnormal curves in G and this leads to the precise description of Abn G contained in formula (3.3) (see also [15], Prop.2.5).In Proposition 3.4 we use this description to study Abn G and eventually prove Theorem 1.1.
Our second main result settles the Sard problem in another well-studied class of Carnot groups, the one of filiform groups, where the question was left open.Filiform groups are Carnot groups whose stratified Lie algebra In particular, dim G s 1.As proved in [17], filiform groups fall into two subclasses: type I filiform groups and type II filiform groups.While type I filiform groups can be of any step s ¥ 2, type II ones always have an odd nilpotency step; since type I and type II filiform groups are isomorphic when s 3 (both coinciding with the well-known Engel groups), we adopt the convention that s ¥ 5 for type II filiform groups.We refer to Section 4.1 for precise definitions and we now pass to the statement of our second main result.Theorem 1.2.Let G be a filiform group of step s ¥ 3.
Theorem 1.2 follows from Propositions 4.2 and 4.4 together with Remark 4.5; in these results the singular controls and the associated abnormal curves are also characterized.The statement of Theorem 1.2 does not include the non-interesting cases s 1, when G R 2 and Abn G r, and s 2, when G is the first Heisenberg group.
Theorems 1.1 and 1.2 show that the codimension of the abnormal set is at least 3 both in step 2 and in filiform Carnot groups; actually, to our best knowledge there is currently no example of a Carnot group (nor of an equiregular sub-Riemannian manifold) where the abnormal set has codimension less than 3.This might lead to formulate the following "strong" Sard conjecture: is it true that, in Carnot groups and/or in equiregular manifolds, the codimension of the abnormal set is at least 3? Observe that there are counterexamples when the equiregularity assumption is not in force, the most noticeable being the Martinet structure in R 3 [13], where the abnormal set is contained in a plane and has therefore codimension 1.At any rate, the answer to this question seems for the moment out of reach.

Preliminaries
A Carnot group G of rank r and step s is a connected, simply connected and nilpotent Lie group whose Lie algebra g, here identified with the tangent at the group identity e, admits a stratification of the form: Denoting by L g the left-translation on G by an element g G, we consider the endpoint map where we denoted by γ u : r0, 1s Ñ G the absolutely continuous curve issuing from e, whose derivative is given by pdL γuptq q e uptq for a.e.t r0, 1s.Any such curve γ u is called horizontal.
The following Proposition 2.1 was proved in Proposition 11 of [7].Let us state some notation.We fix a basis X 1 , . . . ,X r of g 1 , so that we can identify g 1 R r by R r u Ô X u g 1 where As customary, for X, Y g we write ad X pY q : rX, Y s, while for p G we denote by R p : G Ñ G the associated right-translation R p pqq qp.Eventually, given t ¥ 0 and an integer j ¥ 1 we introduce the j-dimensional simplex Σ j ptq of side t by With this notation one has the following.
Proposition 2.1.Let u L 1 pr0, 1s, g 1 q be a control; then In particular d e R γup1q pg 1 q Im pd u F q.
In particular, a point g G belongs to Abn G if and only if there exists an abnormal curve joining e and g.
In the setting of Carnot groups of step 2, Proposition 2.1 allows to characterize abnormal curves in a particularly simple way.Let G be a fixed Carnot group of step 2 with Lie algebra g g 1 g 2 and denote by π 1 : g 1 g 2 Ñ g 1 the canonical projection onto the first layer.Given an horizontal curve γ in G we define Starting from Proposition 2.1 it is not difficult to realize that Im pd u F q d e R γup1q pI γ q; in particular, (2.4) For more details see Section 2 of [15].The characterization (2.4) of abnormal curves, in turn, allows to find an explicit, purely algebraic formula (see (3.3) below) for the abnormal set Abn G : we however postpone its proof in order to fist settle some notation and preliminary material about step 2 Carnot groups.

The Sard problem in step 2 Carnot groups
In this section we consider Carnot groups of step s 2 and a fixed rank r ¥ 2. Recall that a Carnot group is free if the only relations imposed on its Lie algebra are those generated by the skew-symmetry and Jacobi's identity.Let us denote by F r the free Carnot group of step 2 and rank r and by V pf r q 1 the first layer of the associated Lie algebra f r .Clearly, f r V rV, V s can be identified with V 2 V .
The following proposition is standard and we omit its proof.Proposition 3.1.Given a free Lie algebra f r V 2 V and a scalar product on V , there is a unique way to extend it to a scalar product on f r such that for every orthonormal basis t e 1 , . . ., e r u of V , the basis t e 1 , . . ., e r u t e i e j | i j u is orthonormal.We will refer to such a scalar product as an adequate scalar product on f r .
Let now G be a fixed Carnot group of step 2 and rank r; then its Lie algebra g g 1 g 2 can be seen as the quotient of f r through a linear subspace W ¤ 2 V , so that Identifying G and g through the exponential map, we have G ! πpf r q where π is the quotient map.
The necessity of using the language of multi-vectors motivates the following subsection, where we fix some terminology and state some preliminary facts.Most of them are well-known, see however Section I.1 of [8].

Some tools from multi-linear algebra
Let V be a real vector space of dimension r.We will refer to as the rank k Grassmanian of V .It has a structure of real projective algebraic variety, with no singular points.Let us consider the following open subset of V k : then Grpk, V q also has a structure of smooth manifold such that the map span : The topology of such manifold is equivalent to the topology of Grpk, V q as an algebraic variety (with the topology given by the inclusion).
Given a basis e 1 , . . ., e r of V , the second exterior power in this case we write rankpωq k and define the support (or span, see also [4], Sect.5) of ω sptpωq : spantv 1 , . . ., v 2k u.
As a matter of fact, sptpωq is the smallest subspace Once a basis for V is fixed, the space 2 V can be canonically identified with the space Skewpr, Rq of skew-symmetric r ¢ r matrices by Observe that the rank of A (as a matrix) is twice the rank of ωpAq (as a 2-vector).We define and we write down the following identifications: We observe that A ¤k is an affine algebraic variety, as we impose every minor of order 2k 1 to be zero, while A k is an affine semi-algebraic variety, as we impose every minor of order 2k 1 to be zero and the sum of all squared minors of order 2k to be positive, so that at least one is non-zero.
We will now prove that A k , endowed with the topology induced by the inclusion in Skewpr, Rq, has a structure of smooth manifold of dimension kp2r ¡ 2k ¡ 1q.We will also provide a useful system of local coordinates.The identification Skewpr, Rq R rpr¡1q{2 is understood.
Lemma 3.2.Let ω e 1 e 2 ¤ ¤ ¤ e 2k¡1 e 2k be a 2-vector of rank k; let t e 1 , . . ., e 2k u be completed to a basis t e 1 , . . ., e r u of V and consider the associated identifications (3.2).Then there exists an open cone U A k containing ω and rational functions In particular, the function ρ : pλ t , a s t , b s t q 1¤t¤k,2t 1¤s¤r : U Ñ R kp2r¡2k¡1q is an homeomorphism into its image and A k has the structure of kp2r ¡ 2k ¡ 1q-dimensional smooth manifold.Proof.Let A ω be the matrix associated to ω with respect to the basis t e 1 , . . ., e r u.For A A k we consider the map pe i e j q .
We define the rational functions , for every s 3, . . ., r and let pe i e j q.
We now notice that sptpω 1 pAqq t e 3 , . . ., e r u, moreover °i A i2 e i and e 2 °j¥3 A1j A12 e j are jointly linear independent from t e 3 , . . ., e n u: therefore, ω 0 pAq is the sum of ω 1 pAq and a simple (i.e., with rank 1) vector with support in direct sum with sptpω 1 pAqq, and this implies that rank ω 1 pAq rank ω 0 pAq ¡ 1 k ¡ 1.We now consider the open cone which is also a neighbourhood of ω, and we recall that rational functions are closed under composition.
Iterating this process k times, we obtain an open cone U : U k A k containing ω and rational functions The vector function ρ : pλ t , a s t , b s t q 1¤t¤k,2t 1¤s¤r is clearly smooth on U , since its components are rational functions.The function η : R kp2r¡2k¡1q Ñ A k defined by coincides with ρ ¡1 on ρpU q, and its coefficients η ij are evidently polynomial functions in pλ t , a s t , b s t q.Thus ρ is an homeomorphism onto its image and A k is a topological manifold.Finally, transition maps are compositions of rational function with a polynomial map, hence smooth.Therefore A k is also a smooth manifold.
Remark 3.3.In the previous construction, λ t and a s t are homogeneous rational functions of degree one while b s t are homogeneous rational functions of degree zero.

Estimate of the abnormal set in Carnot groups of step 2
Hereafter we will always consider a free Lie algebra f r V 2 V of step 2 and rank r ¥ 2 equipped with an adequate scalar product, as in Proposition 3.1; the symbol u will indicate orthogonality with respect to such scalar product.Recall that the Lie algebra g g 1 g 2 of a Carnot group G of step 2 and rank r can always be written as in (3.1) for some subspace W ¤ 2 V , that is fixed from now on.Let us introduce the notation: if A ¤ V and B ¤ 2 V (so that A, B are subspaces of f r too), then Let us identify F r f r and G g by the associated exponential maps and let π : F r f r Ñ g G the quotient map.Given a horizontal curve γ 0 : r0, 1s Ñ G such that γ 0 p0q 0, there exists a unique horizontal curve γ : r0, 1s Ñ F r such that γ 0 π ¥γ and γp0q 0. The curves γ 0 and γ are associated with the same control uptq; recalling the notation in (2.3), one has π 1 pγ 0 q π 1 pγq, thus P γ0 P γ and I γ0 πpI γ q.In particular, an abnormal curve γ : r0, 1s Ñ F r is the lift of an abnormal curve on G if and only if πpI γ q $ g, i.e., if and only if We can compute I u γ pV rP γ , V sq u rP γ , V s u2 rP u1 γ , P u1 γ s 2 P u1 γ ¨, where we used the fact that the subspaces rP γ , V s and rP u1 γ , P u1 γ s are orthogonal and complementary in 2 V .Since γ P γ rP γ , P γ s, we deduce that This formula for Abn G is equivalent to Proposition 2.5 of [15].If g Abn G , then g π The following estimate is the most important result of the present section.
Proposition 3.4.The set A is contained in an algebraic variety of codimension 2k 1.
Proof.We define We notice that A k,W and A m k,W are cones and semi-algebraic subvarieties of A k .Moreover Bpλωq Bpωq for λ Rz t 0 u and dim Bpωq pr ¡ 2kqpr ¡ 2k 1q 2 where Rpm, k, W q is the smallest dimension of a smooth semi-algebraic variety containing A m k,W .Indeed, let , that is a semi-algebraic variety (since it is image of a semi-algebraic variety through an algebraic map) whose dimension is not greater than dim A m k,W ¡1 (since each non-empty fibre of t Bpωq | ω A m k,W u contains a one-dimensional set as Bpλωq Bpωq).Finally we recall that the described local parametrization of A m k,W also provides an algebraic map A m k,W Ñ V 2k that maps ω to a basis of sptpωq, thus we can locally find a basis to Bpωq that depends algebraically on ω, using this basis we can parametrize Bpωq with dim Bpωq parameters.As a result, for every x E m k,W there is a local algebraic submersion from a neighbourhood of R Rpm,k,W q to (possibly a superset of) a neighbourhood of x.
The final step of the proof is to estimate Rpm, k, W q: setting n : dim W , we will prove that where we used Lemma 3.2.Let ω 0 A m k,W be fixed, then ω 0 e 1 e 2 ¤ ¤ ¤ e 2k¡1 e 2k for linearly independent vectors e 1 , . . ., e 2k V , non necessarily orthogonal.Let us consider an orthonormal basis t e 2k 1 , . . ., e r u of sptpω 0 q u1 so that t e 1 , . . ., e 2k , e 2k 1 , . . ., e r u is a basis for V .Considering this basis, Lemma 3.2 provides the local (around ω 0 ) parametrization of A k p 2 V q k (which is a superset of A m k,W ) given by ωpξ, a, bq ω 0 where Q ts pξ, a, bq are homogeneous polynomial of degree 2 in pξ, a, bq.In the previous parametrization we used Lemma 3.2 making the useful substitution λ i 1 ξ i , so that ω 0 ωp0, 0, 0q.Let us now consider the basis v 1 , . . ., v r of V defined by Let θ h ij be the coordinates of θ h with respect to the basis t v i v j | i j u, i.e., θ h i j θ h ij pv i v j q, h 1, . . ., n .

By (3.6)
θ h ij 0 whenever n ¡ m 1 ¤ h ¤ n, i ¤ 2k and i j. (3.7) The orthogonality between ωpξ, a, bq and W can be expressed by the system xωpξ, a, bq, θ The latter system defines the algebraic variety A k,W that contains A m k,W .We can estimate the dimension of A k,W (and therefore the dimension of A m k,W ) computing the rank of dP p0, 0, 0q were P is the polynomial function P pξ, a, bq ¤ ¦ ¥ P 1 pξ, a, bq . . .

Now we finally compute
Moreover we know that rank where we used the fact that t θ 1 , . . ., θ n u is a basis for W and (3.6).Then and rank pdPp0, 0, 0qq rank Thus we proved that at any point k,W , we can conclude that also E k,W is contained in an algebraic variety of codimension 2k 1.
We can now prove one of our main results.
One may ask whether the estimate on the dimension of the abnormal set provided by Theorem 1.1 is optimal.Heisenberg groups provide a family of Carnot groups where this estimate is optimal.
Example 3.5.The k-th Heisenberg group is the stratified group H k whose Lie algebra stratification g g 1 g 2 is given by and the only non-zero Lie brackets between generators are rX i , Y i s T for i 1, . . ., k .

It can be easily checked that g
hence W u is a one dimensional subspace spanned by a 2-vector of rank k.This implies that k k, and it is well known (see e.g.[15], Ex. 2.4) that Abn H k t0u has codimension 2k 1.
On the other hand, this is not always the case.The estimate on dim E k,W we found in Proposition 3.4 may be loose due to two possible issues: First issue: we may have sptpω 1 q sptpω 2 q for two linearly independent 2-vectors ω 1 , ω 2 W u2 z t 0 u, as in Example 3.6 below.Second issue: we computed the dimension of the algebraic variety defined by P 1 pξ, a, bq ¤ ¤ ¤ P n¡m pξ, a, bq 0, while the variety defined by P 1 pξ, a, bq ¤ ¤ ¤ P n pξ, a, bq 0 may have a lower dimension, despite not being a smooth manifold around ω 0 .Observe that this does not constitute a problem when A 0 k,W is non-empty, while it is the reason behind the following Example 3.7.We now provide two examples of Carnot groups whose abnormal sets has even codimension: therefore, in both cases the estimate provided by Theorem 1.1 is not optimal.
Example 3.6.Let G be the 6-dimensional Carnot group of step 2 whose stratified algebra g g 1 g 2 is such that and the only non-vanishing commutation relations between the generators are given by rX 1 , X 2 s rX 3 , X 4 s T 1 , rX 1 , X 4 s rX 2 , X 3 s T 2 .
We can see g as the quotient f 4 {W of the free algebra f 4 V 2 V (where V : g 1 ) by the subspace W ¤ 2 V defined by The adequate scalar product on f 4 we consider is the one such that X 1 , X 2 , X 3 , X 4 is orthonormal.We claim that W u2 does not contain simple 2-vectors.Indeed, the determinant of the skew-symmetric matrix A tω1 sω2 associated with the linear combination tω 1 sω 2 is detpA tω1 sω2 q 0 t 0 s ¡t 0 s 0 0 ¡s 0 t ¡s 0 ¡t 0 pt 2 s 2 q 2 , which is not null as long as t and s are not both zero.It follows that every non trivial linear combination ω of ω 1 and ω 2 has rank 2 and, in particular, spt ω V .Recalling (3.4) hence Abn G t0u has codimension 6.
In the previous example, the codimension is less than 5 (as estimated by Thm.1.1) because of the first issue, as ω 1 and ω 2 are two linearly independent 2-vectors in W u2 with the same support.In order to construct an example where the codimension is 4, we will instead build on the second issue.Namely, we will provide an example where A 1,W $ r, but A 0 1,W r, taking advantage of the geometry of simple 2-vectors.Example 3.7.Let G be the 6-dimensional Carnot group of step 2 whose stratified algebra g g 1 g 2 is such that and the only non-vanishing commutation relations between the generators are given by We can see g as the quotient f 4 {W of the free algebra f 4 V 2 V (where V : g 1 ) by the subspace W ¤ 2 V defined by The adequate scalar product on f 4 we consider is the one such that X 1 , X 2 , X 3 , X 4 is orthonormal.The determinant of the skew-symmetric matrix A tω1 sω2 associated with the linear combination tω 1 sω 2 is detpA tω1 sω2 q 0 t 0 s ¡t 0 s 0 0 ¡s 0 0 ¡s 0 0 0 which is zero if and only if s 0; in particular, a linear combination ω tω 1 sω 2 tω 1 ha rank 1 if and only if it is a (non-zero) multiple of X 1 X 2 .Recalling (3.4) where we used the fact that

Filiform groups
A filiform group is a Carnot group associated with a stratified Lie algebra g g 1 ¤ ¤ ¤ g s such that dim g 1 2 and dim g 2 ¤ ¤ ¤ dim g s 1.
We fix a basis X 1 , . . ., X s 1 of g such that g 1 spantX 1 , X 2 u and g j spantX j 1 u d j 2, . . ., s.
There are two non-isomorphic classes of filiform groups only (see [17]), which we list according to their non-trivial bracket relations: Type I filiform groups, where the only non-trivial relations are given by Remark 4.1.We observe that the filiform groups of type I and II with nilpotency step s 3 are isomorphic (Engel group); we can therefore adopt the convention that the step of a type-II filiform is an odd integer s ¥ 5.

Type I filiform groups
Let us characterize singular curves in type I filiform groups of step s ¥ 3.For the well-known case s 2 (i.e., the first Heisenberg group) see Remark 4.3.Proposition 4.2.Let G be a type I filiform group of step s ¥ 3 with generators X 1 , . . ., X s 1 , as in Section 4.1.Then singular controls u L 1 pr0, 1s, g 1 q are exactly those for which u 1 0 a.e. on r0, 1s.In particular, abnormal curves are the absolutely continuous curves contained in the line t Þ Ñ expptX 2 q and Abn G texpptX 2 q | t Ru has codimension s.
Proof.Let u L 1 pr0, 1s, g 1 q be a singular control.Using (2.1) with Y X 2 and taking the bracket relations into account, Proposition 2.1 implies that the subspace span tr0,1s is contained in pd e R γup1q q ¡1 pIm d u F q. Since u is singular, there exists λ g ¦ such that λ $ 0 and λ u pd e R γup1q q ¡1 pIm d u F q.By (2.2) we have Since λ is orthogonal to all the elements in (4.1) we deduce that for every t r0, 1s For i 2, . . ., s ¡ 1 we introduce A i : r0, 1s Ñ R by Assume that C 0 : tt r0, 1s | u 1 ptq $ 0u is such that L 1 pC 0 q ¡ 0; then  for all h sufficiently small, implying that t h C 1 for all such values of h, contradicting the fact that C 1 does not contain any isolated point.
We have thus proved the following: almost every t C 1 is a common zero for the functions A 1 , . . ., A s .But this readily implies that λ s 1 ¤ ¤ ¤ λ 3 0, which is impossible since the covector λ has to be nonzero.
We conclude that a singular control u satisfies u 1 ptq 0 a.e. on r0, 1s.Conversely, every control u such that u 1 0 is indeed singular: in fact, in this case Proposition 2.1 gives Im pd u F q d e R γup1q pg 1 g 2 q unless u 2 0 on r0, 1s as well, in which case Im pd u F q d e R γup1q pg 1 q.In both cases, we deduce that u is singular because s ¥ 3, and this concludes the proof.Remark 4.3.When s 2 the filiform group G is the 3-dimensional Heisenberg group H 1 .In this setting one can follow the previous proof to find that the only singular control is the null one.In particular, Abn H 1 teu.

Type II filiform groups
We now study singular curve in type II filiform groups; recall that the nilpotency step of such groups is an odd integer not smaller than 5, see Remark 4.1.Proposition 4.4.If G is a type II filiform group of step s ¥ 5, then u L 1 pr0, 1s, g 1 q is a singular control if and only if there exists a R such that both the following statements (i) either u 1 ptq 0 or u 2 ptq 0 (ii) if u 1 ptq $ 0, then ³ t 0 u 2 pτq dτ a hold for a.e.t r0, 1s.
Before proving Proposition 4.4, let us discuss its implications about the geometry of abnormal curves in type II filiform groups.
Remark 4.5.Let us briefly discuss the geometry of the abnormal curves associated with the singular controls described in Proposition 4.4.Let u L 1 pr0, 1s, g 1 q be a control as in Proposition 4.4 and let X : r0, 1s Ñ g 1 be a primitive of u.Condition (i) implies that X is a concatenation of segments parallel to the coordinate axis in g 1 R 2 , while condition (ii) requires that the segments that are parallel to the first axis are all contained in the line X 2 a. See Figure 1.The abnormal curve γ u is uniquely determined by Xptq.
It is a standard task to deduce that Abn G texppaX 2 q exppbX 1 q exppcX 2 q | a, b, c Ru and, in particular, Abn G is an algebraic variety of dimension 3.In particular, Abn G has codimension s ¡ 2 ¥ 3.
We now prove Proposition 4.4.
Proof of Proposition 4.4.Step 1.Let u L 1 pr0, 1s, g 1 q be a singular control.Using (2.1) with Y X 2 and taking the bracket relations into account, Proposition 2.1 implies that the subspace span tr0,1s is contained in pd e R γup1q q ¡1 pIm d u F q. Since u is singular, there exists λ g ¦ such that λ $ 0 and λ u pd e R γup1q q ¡1 pIm d u F q.By (2.2) we have λ 1 λ 2 0.    We first prove statement (i), i.e., that u 1 u 2 0 a.e. on r0, 1s.Assume by contradiction that C 0 : tt r0, 1s | u 1 ptqu 2 ptq $ 0u is such that L 1 pC 0 q ¡ 0; then also C 1 : tt C 0 | t is a Lebesgue point of uu satisfies L 1 pC 1 q ¡ 0. Again, C 1 does not contain any isolated point and, arguing as in the proof of Proposition 4.2, we deduce that A λ 1 ptq ¤ ¤ ¤ A λ s ptq 0 for almost every t C 1 .This gives λ s 1 λ s ¤ ¤ ¤ λ 3 0, contradicting the fact that λ is non-zero.
Next, we prove that, if λ s 1 0, then necessarily u 1 0 a.e. on r0, 1s, so that also statement (ii) holds.Assume on the contrary that the set D 0 : tt r0, 1s | u 1 ptq $ 0u has positive measure; then also the set D 1 D 0 of Lebesgue points of u has positive measure and (using in a crucial way the assumption λ s 1 0) one can reason as before to deduce that λ 0, contradiction.
Eventually, we consider the case λ s 1 $ 0; we can also assume that the sets D 0 and D 1 introduced in the previous paragraph have positive measure, otherwise we have again u 1 0 a.e. on r0, 1s.Differentiating (4.7) and using (4.9) we deduce that there exists D 2 D 1 such that L 1 pD 1 zD 2 q L 1 pD 0 zD 2 q 0 and (4.10) Therefore 0 A λ s¡1 ptq λ s λ s 1 and we get that necessarily Since L 1 pD 0 zD 2 q 0, statement (ii) is proved for a.e.t.

Proof of Theorem 1 . 1 .
Recalling (3.4), we have Abn G k E k,W and, by Proposition 3.4, each E k,W has codimension at least 2k 1 in G.The statement follows by noticing that k min

Figure 1 .
Figure 1.Example of an abnormal curve in type II filiform groups.
Let G be a Carnot group; we say that u L 1 pr0, 1s, g 1 q is a singular control if the differential d u F of the endpoint map at u is not surjective.A horizontal curve γ u from the identity e is called singular (or abnormal) curve if the associated control u is singular.The abnormal set Abn G G is the set of critical values of F , i.e., 1y ¤ ¤ ¤ xωpξ, a, bq, θ n y 0 which, due to xv i v j , e h e y xv i , e h y xv j , e y ¡ xv i , e y xv j , e h y δ ih δ j ¡ δ i δ jh δ ih δ j P 1 pξ, a, bq : °k i1 1 , X 2 ss, ...X s 1 rX 1 , X s s rX 1 , r. .., rX 1Type II filiform groups, where s is odd and the only non-trivial relations are given byX 3 rX 1 , X 2 s, X 4 rX 1 , X 3 s rX 1 , rX 1 , X 2 ss, . ..X s rX 1 , X s¡1 s rX 1 , r. .., rX 1 looooooomooooooon ps ¡ 2q-times , X 2 s . ..ss,X s 1 p¡1q i rX i , X s 2¡i s, for every i 2, . . ., s.
that C 1 does not contain any isolated point.Differentiating (4.3) and using (4.4) we find u 1 ptqA λ 2 ptq 0 for a.e.t r0, 1s, therefore In fact, by (4.4) one has that at every differentiability point t C 1 of A λ pt hq h ¤ u 1 ptqA λ 3 ptq ophq $ 0 for all h sufficiently small, implying that t h C 1 for all sufficiently small values of h, i.e. that t is isolated in C 1 .This would be a contradiction.Inductively, suppose that we have proved that A λ i ptq is zero for a.e.t C 1 and let us then show that A λ i 1 ptq 0 for a.e.t C 1 .If, by contradiction, this were not true, then at differentiability points of A λ i we would have 4.7)For i 2, . . ., s ¡ 2 we introduce A λ pτ s¡1 q . . .u 1 pτ i qdτ s¡1 . . .dτ i , A λ s¡1 ptq : λ s λ s 1 1, . . ., s ¡ 2, u 2 ptqA λ s ptq u 2 ptqλ s 1 , i s ¡ 1 for a.e.t r0, 1s.