Lie algebras of vector fields on the line and in the plane

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI, GPT-6.1 Sol (OpenAI), in Codex at Ultra. Public domain (CC0).

Finite-dimensionality places a strong restriction on local motions of a line. In a plane, the decisive extra question is whether the motions preserve a family of curves. We will prove the line classification and all three primitive complex plane normal forms, then state the remaining plane list with its historical locators. The proof also explains why first derivatives alone do not determine projective isotropy.

We use analytic germs at a specified regular point, holomorphic germs in the complex case. A Lie algebra consists of vector fields independent over the constant field, with bracket $[X,Y]=XY-YX$. Equivalence means pushforward by a local analytic diffeomorphism. Transitivity and invariant foliations have the local meanings established in Transitivity and primitivity. The projective fields come from Prolongation, differential invariants and the projective group.

1. The line: a dimension bound and actual coordinates

The regular-point hypothesis is essential. The one-dimensional algebra generated by $x^2\partial_x$ is nonzero as an algebra of germs at zero, but every field in it vanishes there. It cannot be changed into translations by a diffeomorphism fixing that base point. The theorem below assumes that evaluation at the base point is nonzero. Every nonzero analytic algebra on a connected line has such points on an open dense set.

Theorem 1.1 (line classification). Let $\mathfrak g$ be a finite-dimensional nonzero algebra of analytic vector-field germs on a line over $\mathbb K=\mathbb R$ or $\mathbb C$. At a point where some field is nonzero, it is analytically equivalent to exactly one of

In particular, its dimension is at most three at a regular point.

Proof. Straighten a nonzero field as in One-parameter groups, obtaining $\partial_x\in\mathfrak g$. Let $V$ be the finite-dimensional space of its coefficient functions. Since $[\partial_x,f\partial_x]=f'\partial_x$, differentiation preserves $V$.

For a nonzero germ $f$, let $\operatorname{ord}_0f$ be its vanishing order. There is a largest order $r$ among the nonzero elements of $V$. Indeed, the descending subspaces whose first $k$ Taylor coefficients vanish eventually stabilize; their intersection is zero by analyticity, so the stabilized subspace is zero. Each successive quotient has dimension at most one. A function of order $r$, together with its derivatives, supplies orders $r,r-1,\ldots,0$. Hence every quotient up to $r$ has dimension one and

For leading terms $a x^j$ and $b x^k$ with $j\ne k$,

If $r\ge3$, choose fields of orders two and $r$. Their bracket has order $r+1$ and nonzero leading coefficient, contradicting maximality. Thus $r\le2$.

If $r=0$, the straightened algebra is already the first form in (1.1). If $r\ge1$, there is a field $H=f(x)\partial_x$ with $f(0)=0$ and, after a constant rescaling, $f'(0)=1$. The function $1/f(x)-1/x$ has a removable singularity at zero. Define

The integrand is analytic, the integral is its local analytic primitive, and $u'(0)=1$. Differentiating gives $f(x)u'(x)=u$, so $H=u\partial_u$ in this coordinate.

The operator $\operatorname{ad}H$ on coefficients is $L=u\,d/du-1$. It preserves the transformed finite-dimensional coefficient space. Choose a nonzero polynomial $P$ that annihilates this finite-dimensional operator, for example its characteristic polynomial. If $g(u)=\sum_{k\ge0}a_k u^k$ belongs to the space, then

Only finitely many $k$ can have $a_k\ne0$, since $P$ has finitely many roots. Thus every coefficient is a polynomial. Polynomial spectral projectors for the distinct numbers $k-1$ extract each of its homogeneous monomials as another element of the space. Coordinate changes preserve vanishing orders, whose maximum is still $r$. Consequently the space is contained in $\langle1,u,\ldots,u^r\rangle$. Equation (1.2) gives equality. This proves (1.1). The dimensions distinguish its three forms. $\square$

The last form is $\mathfrak{sl}_2$ as an abstract algebra. For

equation (1.3) gives $[H,E]=2E$, $[H,F]=-2F$, and $[E,F]=H$. Its flows are projective transformations of a line. For example, $u^2\partial_u$ has flow $u\mapsto u/(1-tu)$ where the denominator is nonzero. Local coordinates do not imply a global projective action on the original domain.

2. Filtration and the isotropy representation

Let a finite-dimensional algebra $\mathfrak g$ be transitive at $p$ in an $n$-dimensional analytic manifold. Put $\mathfrak g_{-1}=\mathfrak g$ and, for $i\ge0$, let $\mathfrak g_i$ be the fields whose coefficients vanish to order at least $i+1$ at $p$. In particular, $\mathfrak g_0$ is the isotropy algebra.

Proposition 2.1. The filtration is intrinsic and eventually zero. Whenever $i,j\ge-1$ and $i+j\ge-1$,

Evaluation identifies $\mathfrak g/\mathfrak g_0$ with $T_pM$. The isotropy action on this quotient is

with kernel $\mathfrak g_1$.

Proof. In coordinates, $[X,Y]^a=X^b\partial_bY^a-Y^b\partial_bX^a$. Differentiation reduces vanishing order by at most one; multiplication adds orders. This gives (2.1), including the case of bracketing with a field of order zero. The chain rule shows that vanishing order is unchanged under an invertible analytic coordinate change. Finite-dimensionality and uniqueness of an analytic germ from its Taylor coefficients give eventual zero exactly as in the line proof.

Transitivity makes evaluation surjective, with kernel $\mathfrak g_0$. For $X(p)=0$,

This proves (2.2). Jacobi implies that $\rho$ is a representation; equivalently, direct differentiation shows $D[X,Y]_p=DY_pDX_p-DX_pDY_p$ for $X,Y$ vanishing at $p$. Its kernel is the fields with zero constant and linear terms, namely $\mathfrak g_1$. $\square$

The minus sign in (2.2) is required by our bracket convention. The subspace of coefficient matrices $DX_p$ is still closed under matrix commutators, since changing every bracket sign does not change the underlying subspace.

Write $V=T_pM$. The leading term of a field in $\mathfrak g_i$ embeds

Bracketing with a field whose value at $p$ is $v$ differentiates a leading homogeneous term in direction $v$. Thus higher homogeneous terms must have all their directional derivatives in the preceding graded space. This compatibility is often much stronger than a dimension estimate.

3. Why primitive complex plane isotropy contains $\mathfrak{sl}_2$

From now on the manifold is complex two-dimensional and the algebra is transitive and primitive. Primitivity means that no regular invariant one-dimensional foliation exists near the base point.

Lemma 3.1. The isotropy representation on $V\cong\mathbb C^2$ is irreducible. Its image is either $\mathfrak{sl}(V)$ or $\mathfrak{gl}(V)$.

Proof. An isotropy-invariant line $L\subset\mathfrak g/\mathfrak g_0$ has inverse image $\mathfrak k$ in $\mathfrak g$. The quotient $\mathfrak k/\mathfrak g_0$ is one-dimensional, $[\mathfrak g_0,\mathfrak k]\subset\mathfrak k$, and the bracket of a lift of a generator with itself is zero. Hence $\mathfrak k$ is a subalgebra strictly between $\mathfrak g_0$ and $\mathfrak g$. The subgroup–foliation correspondence proved in lesson 8 gives an invariant curve foliation, a contradiction. Conversely, such a foliation has an isotropy-invariant tangent line at $p$. This proves irreducibility and also its converse in this transitive plane setting.

Let $\mathfrak h\subset\mathfrak{gl}_2(\mathbb C)$ be the image. A solvable complex matrix algebra has a common eigenvector by the fully proved Lie theorem in The adjoint group, composition and isomorphism, so $\mathfrak h$ is not solvable. Project onto traceless matrices by $A\mapsto A-\tfrac12\operatorname{tr}(A)I$. This is a Lie algebra homomorphism. A proper subalgebra of $\mathfrak{sl}_2$ has dimension at most two, hence is solvable: a two-dimensional algebra has derived algebra of dimension at most one. The projection kernel is central, so a solvable projected algebra would make $\mathfrak h$ solvable. Therefore the projected image is all of $\mathfrak{sl}_2$.

Commutators of lifts have zero trace, and their projections span $[\mathfrak{sl}_2,\mathfrak{sl}_2]=\mathfrak{sl}_2$. Thus $\mathfrak h$ itself contains every traceless matrix. It is $\mathfrak{sl}_2$ or, if it contains any nonzero trace, $\mathfrak{gl}_2$. $\square$

This argument uses complex eigenvectors. Over $\mathbb R$, a rotation algebra can be irreducible and solvable, so the same reduction is false. In particular, real rotations on a sphere and real conformal motions have additional primitive forms.

4. The quadratic terms and the end of the filtration

Use linear coordinates $x,y$ on $V$. Denote its six-dimensional space of quadratic vector fields by $\mathcal Q=S^2V^*\otimes V$. There are two invariant subspaces:

For $Q=\ell(x,y)(x\partial_x+y\partial_y)$, divergence is $3\ell$. Since divergence maps quadratic fields onto linear functions, $\dim\mathcal H=4$, $\dim\mathcal P=2$, and $\mathcal Q=\mathcal H\oplus\mathcal P$. Traceless linear fields preserve both subspaces. For $\mathcal H$, use $\operatorname{div}[L,Q]=L(\operatorname{div}Q)$ when $\operatorname{div}L=0$; for $\mathcal P$, use $[L,\ell E]=L(\ell)E$, where $E=x\partial_x+y\partial_y$ commutes with linear fields.

Lemma 4.1. Each subspace in (4.1) is an irreducible $\mathfrak{sl}_2$ module, and every invariant subspace of $\mathcal Q$ is a sum of some of these two subspaces.

Proof. Let $R=x\partial_y$, $S=y\partial_x$, and $H=x\partial_x-y\partial_y$. Their brackets are $[H,R]=2R$, $[H,S]=-2S$, and $[R,S]=H$. The space $\mathcal P$ has basis $xE,yE$, interchanged up to nonzero factors by $R,S$, with distinct $H$ weights $1,-1$. Any invariant subspace contains a weight vector, obtained by a polynomial projector in $H$, and hence both basis vectors.

The map

identifies homogeneous cubic polynomials with $\mathcal H$. Indeed, the coefficients in (4.2) have zero divergence. Conversely, the one-form $-Q^y dx+Q^xdy$ for a divergence-free quadratic field is closed and has a homogeneous cubic primitive, obtained by integration or the radial formula. The map is equivariant for traceless linear fields because these preserve $dx\wedge dy$. On the cubic basis $x^3,x^2y,xy^2,y^3$, the $H$ weights are $3,1,-1,-3$, and $R,S$ connect successive vectors with nonzero factors. The same weight-projector argument proves irreducibility.

To separate the two summands without importing complete reducibility, use the operator

on the module, where each letter denotes its action by bracket. Its commutators with $H,R,S$ vanish, by the three displayed bracket relations. A highest-weight vector of weight $\lambda$ satisfies $Rv=0$ and $RSv=\lambda v$, so $Cv=\lambda(\lambda+2)v$. Commutation then gives $C=15I$ on $\mathcal H$ and $C=3I$ on $\mathcal P$. The polynomials $(C-3I)/12$ and $(15I-C)/12$ project onto the two summands. An invariant subspace is stable under these projectors, and each resulting intersection is zero or the entire irreducible summand. $\square$

Lemma 4.2. In a finite-dimensional transitive plane algebra whose linear isotropy contains $\mathfrak{sl}_2$, the quadratic graded space is zero or $\mathcal P$. In the latter case the linear isotropy is $\mathfrak{gl}_2$. There are no cubic or higher graded terms.

Proof. The quadratic graded space $\mathfrak g_1/\mathfrak g_2$ is an invariant subspace of $\mathcal Q$. It cannot contain $\mathcal H$. Otherwise choose actual fields with leading terms

both in $\mathcal H$. For $A_k=x^k\partial_y$,

Starting with $k=2$, repeated bracketing of the actual fields has these nonzero successively higher leading terms; higher remainders cannot affect them. This contradicts termination of the finite filtration. Lemma 4.1 now leaves only zero or $\mathcal P$.

If the linear isotropy is traceless and $Q\in\mathcal P$ is a leading quadratic term, its derivatives in every constant direction must be traceless linear fields, by transitivity and (2.3). Their traces are the corresponding derivatives of $\operatorname{div}Q=3\ell$. All vanish only when $\ell=0$. Thus nonzero $\mathcal P$ requires $\mathfrak{gl}_2$.

If the quadratic graded space is zero, a putative nonzero higher leading term can be differentiated repeatedly in constant directions to a nonzero quadratic term. This is impossible. Hence $\mathfrak g_1=0$ in that case.

If the quadratic space is $\mathcal P$, let $Z$ be a possible homogeneous cubic leading term. Its two partial derivatives have the form $\partial_jZ=\ell_j(x,y)(x,y)$, regarded as coefficient vectors. Euler's identity gives

so $Z=f(x,y)(x,y)$ with quadratic scalar $f$. Then $\partial_jZ=f_j(x,y)(x,y)+f e_j$. Since both $\partial_jZ$ must be proportional to $(x,y)$, the wedges of $f e_1$ and $f e_2$ with $(x,y)$ vanish. Thus $fy=fx=0$ as polynomials, and $f=0$. There are no cubic terms. Repeated differentiation rules out all higher terms. Therefore $\mathfrak g_2=0$. $\square$

This is the missing qualification in the assertion that isotropy is determined by its first-order part. In the projective case, $\mathfrak g_0$ has dimension six, its linear image has dimension four, and its kernel $\mathfrak g_1=\mathcal P$ has dimension two. First-order determination holds for the two affine cases, not for all primitive cases.

5. Analytic normalization when the linear isotropy is $\mathfrak{gl}_2$

We need to turn graded information into actual analytic coordinates. The scalar matrix in $\mathfrak{gl}_2$ supplies an isotropy field with linear part $E=(x,y)$. The following special linearization has a direct proof.

Lemma 5.1 (Euler linearization). An analytic field $Z(w)=w+O(|w|^2)$ near zero is analytically conjugate to the Euler field $w\cdot\partial_w$.

Proof. In the real case first extend the coefficients holomorphically to a complex ball, using their power series. Shrink the ball so that $Z(w)=w+R(w)$ with $|R(w)|\le\varepsilon|w|$ there, where $0<\varepsilon<1/2$. For real $t\ge0$, the backward flow $w_t=\Phi_Z^{-t}(w)$ solves $w_t'=-w_t-R(w_t)$. Differentiating its squared Hermitian norm gives

It remains inside the ball, exists for all $t\ge0$, and satisfies $|w_t|\le e^{-(1-\varepsilon)t}|w|$. The continuation assertion follows from the local ODE theorem on smaller closed balls. Since also $|R(w)|\le C|w|^2$,

The right side is integrable uniformly on a smaller ball. Hence

exists uniformly there and is $w+O(|w|^2)$. It is holomorphic: on each smaller polydisc the componentwise Cauchy integrals for the holomorphic approximants pass to the uniform limit and give Cauchy representations for the limit itself. It has an analytic inverse near zero by the analytic inverse theorem stated in lesson 1. The flow identity yields $h(\Phi_Z^s(w))=e^s h(w)$ for sufficiently small real $s$; differentiating at zero gives $Dh(w)Z(w)=h(w)$. In the holomorphic case this identity is holomorphic and gives the conjugacy as well. Real coefficients make $h$ real on real points. $\square$

After this normalization, the algebra contains the actual Euler field. On homogeneous coefficient fields of degree $d$, its adjoint action is multiplication by $d-1$. As in the line proof, a polynomial annihilating its action on the finite-dimensional algebra forces every field to have only finitely many homogeneous terms, and polynomial spectral projectors put each homogeneous term in the algebra. Lemma 4.2 bounds the degrees by two. Transitivity supplies the full two-dimensional space of constants; the linear image supplies all four linear fields. The quadratic space is either zero or the two projective fields $xE,yE$. The resulting algebras are exactly

The second is the eight-dimensional projective algebra $\mathfrak{sl}_3$, with the matrix-to-field sign identification proved in lesson 9.

6. The special affine case without a splitting theorem

Suppose the linear isotropy is $\mathfrak{sl}_2$. Lemma 4.2 gives $\mathfrak g_1=0$. Thus $\mathfrak g_0$ itself is isomorphic to $\mathfrak{sl}_2$, and $\dim\mathfrak g=5$. We will produce two commuting translations inside it, rather than assume a general extension theorem.

Choose $H,R,S$ in $\mathfrak g_0$ satisfying the same three Chevalley relations as above. Let $C$ be (4.3), now with each letter acting by its adjoint action on $\mathfrak g$. It commutes with $\mathfrak g_0$. On the adjoint module $\mathfrak g_0$, its value is $8I$; this follows from highest weight two and the calculation in Lemma 4.1. On the quotient $\mathfrak g/\mathfrak g_0\cong V$, it is $3I$, since this is the standard two-dimensional module with highest weight one. Therefore

The distinct roots give the invariant decomposition

Indeed, $C-3I$ maps the algebra into $\mathfrak g_0$, and the $8$ eigenspace is exactly $\mathfrak g_0$; spectral projection onto the $3$ eigenspace identifies it with the quotient.

The module $\Lambda^2A$ is trivial, because the standard $\mathfrak{sl}_2$ representation has trace zero. The bracket is an equivariant map from $\Lambda^2A$ into $\mathfrak g$. Its image would consist of vectors fixed by all of $\mathfrak g_0$. Neither the adjoint module nor the standard module has such a nonzero vector: commuting with all of $\mathfrak{sl}_2$ in its adjoint module means being in its zero centre; the standard module has no common zero vector for its matrices. Thus $[A,A]=0$.

Evaluation sends $A$ isomorphically to $T_pM$. Choose its two generators, whose commuting analytic flows give a coordinate chart by the inverse theorem. In that chart they are $\partial_x,\partial_y$. Each isotropy field $X$ normalizes this constant space, so $[X,\partial_j]=-\partial_jX$ has constant coefficients. Hence $X$ has linear coefficients plus constants. Its value at the origin is zero, so the constants vanish, and its linear part is traceless. We have obtained exactly

Theorem 6.1 (primitive complex plane classification). Every transitive primitive finite-dimensional algebra of holomorphic vector-field germs on a complex surface is locally analytically equivalent to the special affine algebra (6.2), the affine algebra in (5.2), or the projective algebra in (5.2). All three are primitive and have respective dimensions five, six, and eight.

Proof. Lemma 3.1 gives the two possible linear images. Lemmas 4.1–4.2 and Sections 5–6 produce exactly the three forms. Conversely, their isotropy images contain $\mathfrak{sl}_2$, which has no invariant line in $\mathbb C^2$. The first paragraph of Lemma 3.1 then excludes invariant curve foliations. Their displayed independent coefficient fields give the dimensions. $\square$

7. The remaining complex plane list

Here is the rest of Lie's list, stated rather than proved exhaustive. The precise source is [Lie–Engel III, Chapter 4, §18, Theorem 6, pp. 71–73], with its derivation in Chapter 3, §§6–14, pp. 28–58, and the invariant-family refinements in Chapter 4, §§15–17, pp. 58–71. We have grouped the fields using coefficient spaces to make the parameter families readable. This is a classification of germs at generic regular points; exceptional singular base points are outside the statement. The three primitive entries are already proved above.

Put $p=\partial_x$, $q=\partial_y$, $Y=yq$, and

In $A_F$, all displayed coefficient functions are analytic and independent over $\mathbb C$; $r=0$ is allowed. They are arbitrary subject to that condition. Define also

Here $P_j=\langle1,x,\ldots,x^j\rangle$. For $U$, take $m,m_k\ge0$, distinct nonzero $\alpha_k$, $l\ge0$, $l+m+\sum m_k>0$, and normalize $\alpha_1=1$ if $l>0$. For $V$, take distinct $\alpha_k$, which may include zero, $l>0$, $m_k\ge0$, $l+\sum m_k>1$, and normalize the first exponent to $\alpha_1\in\{0,1\}$. This last normalization restricts only the first chosen exponent, not every exponent. It follows by scaling $x$ when a nonzero exponent is chosen first. All rows mean the constant linear span of the displayed fields, not a module over arbitrary functions.

The nine families in the first part of Lie's one-invariant-foliation list are:

The six families in its second part are:

Lie describes the unique invariant family in these two parts as counted simply or doubly. This historical multiplicity concerns invariant tangent directions in his characteristic equation; it does not mean that the second part has two different foliations. The separate two-foliation group appears next.

The nine entries with two invariant curve foliations are:

The three entries with a one-parameter family of invariant foliations are $\langle p,q,xp+Y\rangle$, $\langle q,xp+Y\rangle$, and $\langle p,q\rangle$. The last entry, with a two-parameter family, is $\langle q\rangle$. Include the zero algebra separately if trivial actions are allowed. These tables, together with the three primitive forms, reproduce the entire finite continuous complex plane list in Theorem 6. They include intransitive entries; they are not all transitive algebras.

Example 7.1 (why exponentials appear). If $\langle p,Uq\rangle$ is finite-dimensional, closure under $[p,f(x)q]=f'(x)q$ makes $U$ invariant under differentiation. Its minimal polynomial gives a constant-coefficient linear ODE for every element of $U$. The generalized eigenspaces of differentiation have bases $e^{\alpha x},xe^{\alpha x},\ldots,x^m e^{\alpha x}$: solving $(D-\alpha)^{m+1}f=0$ amounts to writing $f=e^{\alpha x}g$ with $g^{(m+1)}=0$. This proves the shape of the spaces in (7.2), although it does not prove the whole classification table.

Example 7.2 (arbitrary functions in an intransitive algebra). For any independent analytic $f_1(x),\ldots,f_s(x)$,

is abelian and has one-dimensional regular orbits. It can have arbitrarily large finite dimension: the line dimension bound does not apply to coefficients varying in a transverse parameter $x$. After shrinking where $f_1\ne0$, set $y_1=y/f_1(x)$ to normalize one coefficient to one. If $s\ge2$, a ratio with nonzero derivative at a generic point can be used as the new $x$ coordinate, normalizing a second coefficient to $x$. This explains the $A_Fq$ row. If all ratios are constant, the original fields were not independent. Adding $Y$ preserves this arbitrary coefficient space because $[Y,fq]=-fq$.

Example 7.3 (a transitive imprimitive algebra). For $\langle p,q,xq\rangle$, the foliation $x=\text{constant}$ is invariant. Each generator takes a function of $x$ to another function of $x$. The bracket $[p,xq]=q$ makes this algebra the three-dimensional Heisenberg algebra. It acts transitively in a plane even though its centre has a nontrivial action and each point has one-dimensional isotropy. It is the $Vq,p$ row with $V=P_1$.

The equivalence relation matters. Chapter 4, §19, pp. 74–78, examines which distinct point-action entries become equivalent under contact transformations. Our tables and Theorem 6.1 use point diffeomorphisms; allowing contact transformations changes the classification problem.

Closure of the coefficient families can be checked without the classification proof. For analytic functions $f,g$ of $x$,

and $[x^2p+rxY,fq]=(x^2f'-rxf)q$. The last expression sends $x^j$ to $(j-r)x^{j+1}$, so its apparent highest degree cancels on $P_r$. Differentiation preserves every exponential-polynomial summand in (7.2), as Example 7.1 proves. In the exceptional polynomial row put $H=xp+(ry+x^r)q$. Then $[p,H]=p+rx^{r-1}q$ and $[H,x^jq]=(j-r)x^jq$; these stay in its displayed span. The remaining rows involve only the line projective fields $p,xp,x^2p$ and $q,Y,y^2q$, their direct sums, their diagonal copy, or the displayed weighted linear combinations. Formula (1.3) and the four identities above check their brackets directly. These calculations establish closure for the stated analytic coefficient spaces; they do not establish the table's exhaustiveness or distinguish all equivalence parameters.

8. Linear and projective groups: what classification is being made

On a projective line the full algebra is the three-dimensional algebra in Section 1. Over $\mathbb C$, a nonzero matrix in $\mathfrak{sl}_2$ is either nilpotent or has two distinct eigenvalues. Thus a one-dimensional projective subalgebra is conjugate to translations or to dilations. A two-dimensional subalgebra is solvable and preserves a common line in $\mathbb C^2$ by Lie's theorem; after choosing that line as the point at infinity, it is the affine projective subalgebra. The three-dimensional case is the full projective algebra. The one-dimensional translation and dilation subgroups are distinct under projective conjugation, although Theorem 1.1 makes their regular local actions equivalent under arbitrary analytic coordinate changes. This distinction is treated in [Lie–Engel III, Chapter 2, §4, pp. 12–18].

For linear homogeneous plane groups, the origin is fixed. Hence their actions at the origin fall outside the regular transitive classification above. Their coefficient matrices form subalgebras of $\mathfrak{gl}_2$. Over $\mathbb C$, Lemma 3.1 shows that an irreducible such subalgebra is $\mathfrak{sl}_2$ or $\mathfrak{gl}_2$. Reducible subalgebras can be made upper triangular. The full normal-form catalogue, including its continuous parameters, is [Chapter 2, §5, pp. 18–28]; we do not claim that the two irreducible entries exhaust the reducible list.

The full plane projective algebra is $\mathfrak{sl}_3$. Its proper subalgebras include primitive affine and special affine actions, many imprimitive polynomial entries in Section 7, and intransitive stabilizer actions. Chapter 5, §§20–28, pp. 78–109, classifies its subgroups using changes restricted to projective coordinates and projective duality. An invariant line in the projective plane is different from an invariant foliation on a two-dimensional affine chart. For example, the affine group preserves the line at infinity but is primitive on its affine open orbit. Chapter 6, §§29–31, pp. 109–121, treats linear homogeneous groups in three variables; their relation to projective plane actions is projectivization, whose kernel consists of scalar matrices. These two additional catalogues are cited with their scope, not proved or replaced by the table for unrestricted local equivalence.

9. Exercises

Exercise 1 (introductory). Compute all brackets among $\partial_x,x\partial_x,x^2\partial_x$ and identify a Chevalley basis for $\mathfrak{sl}_2$.

Exercise 2 (intermediate). Show that $\langle\partial_x,x\partial_x,x^3\partial_x\rangle$ is not a Lie algebra. Explain both the immediate failure and the later obstruction that would remain after adjoining its first missing bracket.

Exercise 3 (intermediate). Prove directly that the special affine algebra (6.2) is primitive, using invariant tangent lines and its isotropy.

Exercise 4 (intermediate). Find an invariant foliation for $\langle\partial_x,\partial_y,x\partial_y\rangle$. Compute its isotropy at the origin and the associated intermediate subalgebra.

Exercise 5 (advanced). Give a complete proof of the line classification, including the analytic coordinate normalization. Explain why singular base points and projective-only coordinate changes require different statements.

10. Solutions

Solution 1. Formula (1.3) gives $[\partial_x,x\partial_x]=\partial_x$, $[\partial_x,x^2\partial_x]=2x\partial_x$, and $[x\partial_x,x^2\partial_x]=x^2\partial_x$. The reverse brackets are their negatives, and self-brackets are zero. The choices $E=\partial_x$, $H=-2x\partial_x$, $F=-x^2\partial_x$ give $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$.

Solution 2. The bracket $[\partial_x,x^3\partial_x]=3x^2\partial_x$ is absent from the proposed span; independence follows by comparing polynomial coefficients. Adjoining it does not close the algebra: $[x^2\partial_x,x^3\partial_x]=x^4\partial_x$. More generally, $[x^2\partial_x,x^k\partial_x]=(k-2)x^{k+1}\partial_x$ for $k\ge3$, producing arbitrarily high orders. This is exactly the maximal-order contradiction used in Theorem 1.1.

Solution 3. The two translations give transitivity. The isotropy at the origin contains all traceless linear fields. An invariant foliation would have a tangent line preserved by all their linear parts. The diagonal field $x\partial_x-y\partial_y$ forces such a line to be one of the two coordinate axes. The field $x\partial_y$ moves the first axis to the second direction, and $y\partial_x$ moves the second to the first. Neither is invariant under all three. Thus no invariant tangent line and hence no invariant regular curve foliation exists. This proof is over $\mathbb C$ and also proves that this particular real special affine action is primitive.

Solution 4. All flows preserve the family $x=c$: translations in $x$ change $c$, while the other two fields stay on a leaf. At zero the only vanishing generator is $x\partial_y$, so $\mathfrak g_0=\langle x\partial_y\rangle$. The leaf tangent line is $\langle\partial_y\rangle$ modulo isotropy. Its inverse image is $\mathfrak k=\langle\partial_y,x\partial_y\rangle$, an abelian subalgebra properly between $\mathfrak g_0$ and the full three-dimensional algebra. This gives the foliation through the subgroup–foliation correspondence as well.

Solution 5. At a regular base point, straighten a nonzero field to $\partial_x$. The coefficient space is finite-dimensional and differentiation-invariant. Its vanishing-order filtration terminates by analyticity. If its maximal order is $r$, successive derivatives of a field of that order give all orders from zero to $r$, while each graded piece has dimension at most one. Hence its dimension is $r+1$. If $r\ge3$, bracketing orders two and $r$ gives order $r+1$ with nonzero leading coefficient, a contradiction. Thus $r\le2$.

For $r=0$ we have translations. Otherwise choose a field with a simple zero, rescale its derivative to one, and use the analytic coordinate (1.4) to make it $u\partial_u$. A polynomial annihilating its adjoint action shows that every coefficient is a polynomial, because each Taylor coefficient has the eigenvalue $k-1$ and only finitely many can be roots of that polynomial. Its spectral projectors extract each monomial. No degree exceeds $r$, since vanishing order is coordinate invariant, and dimension $r+1$ forces the full span $1,u,\ldots,u^r$. These are exactly (1.1).

If all fields vanish at the base point, straightening the initial field is impossible; $\langle x^2\partial_x\rangle$ is already a counterexample to regular translations there. If coordinate changes are restricted to projective transformations, a translation field and a dilation field are distinct: the former has one double zero on the projective line, the latter two simple zeros. Arbitrary local analytic changes away from their zeros can straighten either. Thus neither qualification can be omitted.

What this lesson does not prove

The entire complex imprimitive and intransitive table is stated from [Lie–Engel III, Chapter 4, §18, Theorem 6, pp. 71–73]. Closure in the explanatory families and examples is checked, but exhaustiveness and uniqueness of those table entries are not proved here. The line and primitive complex plane classifications, the filtration, the small-module argument, the required Euler linearization and the special affine splitting are proved in full. No Levi decomposition or Whitehead lemma is imported.

The finer classifications under projective conjugation, of reducible linear homogeneous plane groups, of projective plane subgroups, and of linear groups in three variables are cited at the exact chapter locators in Section 8. We do not give the classification at singular base points or the additional real primitive plane forms. Uniform limits of holomorphic maps on smaller domains are holomorphic, by the Cauchy formula from the prerequisite Complex Analysis; analytic flows and the inverse theorem have been established or located in lessons 3 and 1.

References

• Sophus Lie, with Friedrich Engel, Theorie der Transformationsgruppen, Volume III (1893), first division, Chapter 1, §§1–3, pp. 1–11; Chapter 2, §§4–5, pp. 12–28; Chapter 3, §§6–14, pp. 28–58; Chapter 4, §§15–19, pp. 58–78, especially Theorem 6, pp. 71–73. The printed page numbers identify the normal-form table and its parameter conditions.
• Sophus Lie, with Friedrich Engel, Theorie der Transformationsgruppen, Volume III, Chapter 5, §§20–28, pp. 78–109, and Chapter 6, §§29–31, pp. 109–121, for the separate projective and linear classifications.
• Sophus Lie, with Friedrich Engel, Theorie der Transformationsgruppen, Volume I (1888), Chapter 29, pp. 619 onwards, for characteristic properties of actions locally equivalent to projective groups. The line proof here uses its own finite filtration and coordinate argument.