数理解析研究所講究録
第 729 巻 1990 年 32-49

Classification of projectively homogeneous surfaces
TAKESHI SASAKI
Department of Mathematics
Faculty of Integrated Arts and Sciences
Hiroshima University(広島大・総合科学 )Introduction
A surface inis called projectively homogeneous (homogeneous in short)
if a subgroup of $PGL_{4}$ acts transitively on this surface. Such surfaces areclassified by S. Lie relying on his group theory when the coefficient field is. In this $note^{*}$ we show how to classify such surfaces over the real fieldrelying on the projective differential geometry of surfaces. A surface inis locally seen as an immersion of a two-manifold with coordinates $(x, y)$into the affine space with coordinates $z=$
, which is transversal
to rays through the origin. The coordinates , as functions of $(x, y)$ , satisfy
a completely integrable system of linear differential equations of the form$P^{3}$$\mathbb{C}[L]$$P^{3}$$(z^{1}, \cdots , z^{4})$$z^{i}$$\{_{z^{xx}=mz_{xy}^{xy}+cz^{x}}^{z_{yy}=lz+az_{x}}:_{dz_{y}^{y}}^{bz}I_{qz}^{pz}$.

Our method of classification is first to find projective scalar invariants ofthis system and then to integrate the corresponding systems using the factthat such invariants are automatically constant on homogeneous surfaces.
Examination of homogeneity turns out to be easy and we get a table ofsuch surfaces. A classification of unimodular-aflSnely homogeneous surfaces
has been given by this method in [G].

In \S 1 we will review the projective differential geometric treatment of
surfaces and relate it to the above system. In this respect, the Fubini-Pickinvariant denoted by $F$ plays a fundamental role. Surfaces with $F\neq 0$ isclassified in \S 2. If $F=0$ , then surfaces are known to be ruled and its
classification is given in \S 3. In \S 4 we give some remarks.
*this note is based on [S] with some modification

33\S 1. Projective differential geometry of surfaces.
Let $M$ be a 2-manifold immersed in a real projective space. Such an
immersion has locally a lift to the affine space $fl^{4}-\{0\}$ . Let
be oneof lifts. Then we associate to each point of $M$ a set of four independentvectors
as the first vector so that this set moves
withsmoothly on $M$ . Such a set , that is only locally defined, is called aprojective frame field. Certainly we can assume is unimodular in thesense that $\det(e_{0}, e_{1}, e_{2}, e_{3})=1$ with respect to a fixed volume form of .
The dependence of on $M$ is denoted by a Pfaff syst $em$
$P^{3}$$e_{0}$$\{e_{0}, e_{1}, e_{2}, e_{3}\}$$e_{0}$$e$$e$$fl^{4}$

$e$

(1.1)

i.e.

$de=\omega e$$de_{\alpha}= \sum_{\beta}\omega_{\alpha}^{\beta}e_{\beta}$,

which is our starting setting. The integrability condition is(1.2)$d \omega_{\alpha}^{\beta}=\sum_{\gamma}\omega_{\alpha}^{\gamma}$A$\omega_{\gamma}^{\beta}$.

Now we restrict freedom of choice of projective frame fields. First, we may$e_{2}>becom$ es the tangent
assumeso that the vector space $<e_{0},$space of the cone over $e_{0}(M)$ with the origin as the vertex. This choice
implies$\omega_{0}^{3}=0$(1.3)$e_{1},$$0=d\omega_{0}^{3}=\omega_{0}^{\alpha}$A$\omega_{\alpha}^{3}=\omega_{0}^{i}\wedge\omega_{i}^{3}$.

Here we use the summation convention; repeated indices are summed up in
their range and the range forisis to 3 and the range for $i,j,$
1 to 3. We also use the notation
for . From (1.3) we see the existenceof $s\grave{y}mmetric$ functions
so that$\alpha,$$\beta,$$\cdots$$\omega^{i}$$0$$\cdots$$\omega_{0}^{i}$$h_{ij}$(1.4)$\omega_{i}^{3}=h_{ij}\omega^{j}$.

Put. Then we can see the conformal class of is independentof frames chosen (see PROPOSITION 1.1). From now on we assume that thisform is non-degenerate; an immersion or a surface with this assumption iscalled non-degenerate. We next take the exterior derivation of (1.4) to get
the identity$\varphi_{2}=h_{fj}\omega^{i}\omega^{j}$$\varphi_{2}$$(dh_{ij}-h_{ik}\omega_{j^{k}}-h_{kj}\omega_{i}^{k}+h_{ij}(\omega_{0}^{0}+\omega_{3}^{3}))\wedge\omega^{j}=0$

2

.

34
Hence we can put$h_{\mathfrak{i}jk}\omega^{k}=dh_{\dot{x}j}-h_{ik}\omega_{j}^{k}-h_{kj}\omega_{i}^{k}+h_{ij}(\omega_{0}^{0}+\omega_{3}^{3})$and we get a cubic form$\varphi_{3}=h_{ijk}\omega^{i}\omega^{j}\omega^{k}$.

DEFINITION. The calar function$s$$F=h_{ijk}h_{pqr}h^{ip}h^{jq}h^{kr}$,$(h^{ij})=(h_{ij})^{-1}$is called th Fubini-Pick invariant.
$e$

We can see moreover that our consideration can be restricted to the case$|\det(h_{ij})|=1$ , which enables us to choose a frame with the condition(1.5)$\omega_{0}^{0}+\omega_{3}^{3}=0$.

Then we see

PROPOSITION 1.1. (1) Let and be $tvvo$ proje ctive frame fields. Thenth $ey$ are related by a fram change$\tilde{e}$

$e$

$e$

$\tilde{e}=ge$

where$g=(\begin{array}{lll}\lambda 0 0b a 0\mu c t/\end{array})$,

being a $2\cross 2- m$ atrix, with the property $|\det a|=|\lambda\nu|=1$ .
are related as
and functions(2) The forms$a$$\varphi_{2},\tilde{\varphi}_{2},$$F,\tilde{F}$$\varphi_{3},\tilde{\varphi}_{3}$$\tilde{\varphi}_{2}=\lambda\nu^{-1}\varphi_{2}$,$\tilde{\varphi}_{3}=\lambda\nu^{-1}\varphi_{3}$PROPOSITION 1.2. The coefficients$h_{ijk}$.

,$\tilde{F}=\lambda^{-1}\nu F$

satisfies th $ec$ on dition$h^{ij}h_{ijk}=0$.

This condition is called the apolarity condition.
We next take the exterior derivation of (1.5) to get the identity$(\omega_{i}^{0}-h_{ij}\omega_{3}^{j})$A $\omega^{i}=0$ .

Hence we may put$\omega_{i}^{0}-h;_{j}\omega_{3}^{j}=L_{ij}\omega^{j}$Without proof we cite a

3

.

35
PROPOSITION 1.3. $Ass$ um th immersion is non-degenerat \’e. Then th ere
exists locally a proje ctive frame field with th property
$e$

$e$

$e$

$\omega_{0}^{3}=\omega_{0}^{0}+\omega_{3}^{3}=0$,$|\det(h_{ij})|=1$,$h^{ij}L_{ij}=0$.

Such frames are related by a ransformation $earrow\tilde{e}=ge$ where components
enoted as in Proposition 1.1, satisfy $|\det a|=|\lambda\nu|=1,$ $b=\nu^{-1}ah^{t}c$ ,
of
.
and
$t$

$d$

$g,$

$\mu=\frac{1}{2}\nu^{-1}ch^{t}c$

We finally introduce the last invariant$\gamma_{i}$$\omega_{3}^{0}=-\gamma_{i}\omega^{i}$by.

The covariance of the invariants are seen to be as follows:(1.6)$\lambda^{2}\nu\tilde{h}_{ijk}=h_{pqr}a_{i}^{p}a_{j}^{q}a_{k}^{r}$(1.7)$\lambda^{2}\tilde{L}_{ij}=(L_{k1}-\lambda h_{k1m}c^{m})a_{i}^{k}a_{j}^{1}$(1.8)$\lambda^{2}\tilde{\gamma}_{j}=(\nu\gamma_{k}+c^{i}L_{ik}-\frac{1}{2}\nu^{-1}c^{i}c^{\ell}h_{ilk})a_{j}^{k}$.

For later use we cite a
THEOREM 1.4. A non-degenera $tes$ urface withis a quadra $tics$ urface. A non-degenerate urface is ruled if and only if th invariant $F$ vanishes.
$\varphi_{3}=0$

$e$

$s$

For a precise treatment of the above materials with proofs, refer the notes[S].

\S 2. Homogeneous surfaces with $F\neq 0$ .
2.1 Let us consider homogeneous surfaces whose Fubini-Pick invariant $F$
does not vanish. To such a surface we as sociate a distinguished frame thatwill now be constructed. Take a normalized frame with
(2.1)$\omega_{0}^{0}+\omega_{3}^{3}=0$and

4$h^{ij}L_{ij}=0$

36
and satisfying$h=(\begin{array}{l}l00\epsilon\end{array}),$(2.2)$\epsilon=\pm 1$; i.e.

$\omega_{1}^{3}=\omega^{1}$,$\omega_{2}^{3}=\epsilon\omega^{2}$.

The apolarity condition is $h_{111}+\epsilon h_{122}=h_{112}+\epsilon h_{222}=0$ . By the formula(1.6), we find a frame with
(2.3)

and

$h_{111}=-\epsilon h_{122}=-2$$h_{112}=h_{222}=0$.

Then, by definition of the cubic form,(2.4)$\omega_{1}^{1}=\omega^{1}$,$\omega_{2}^{2}=-\omega^{1}$and$\omega_{2}^{1}+\epsilon\omega_{1}^{2}=-2\epsilon\omega^{2}$

Note here thatby a frame change whenin the notation of \S 1. Hence we can assume$\tilde{\omega}_{0}^{0}=\omega_{0}^{0}-b_{i}\omega^{i}$(2.5)$\omega_{0}^{0}=\omega_{3}^{3}=0$.

$\lambda=1$

and $a=I_{2}$.

Now it is important to see that there exist only finite frames satisfying(2.1) to (2.5). Therefore, if the surface is homogeneous and $F\neq 0$ , thenthe coefficients of remaining components of must be constant. We define
coefficients as follows:$\omega$(2.6)$\omega_{1}^{2}=a\omega^{1}+(b-1)\omega^{2}$$\omega_{1}^{0}=\mu 0^{1}+q\omega^{2}$$\epsilon\omega_{2}^{1}=-a\omega^{1}-(b+1)\omega^{2}$,$\omega_{3}^{1}=p’\omega^{1}+q’\omega^{2}$$\omega_{3}^{0}=u\omega^{1}+v\omega^{2}$,$\omega_{2}^{0}=q\omega^{1}+r\omega^{2}$

,

,

,

$\omega_{3}^{2}=\epsilon q^{l}\omega^{1}+\epsilon r^{J}\omega^{2}$

,

.

,
and
Here we have used the fact that, if we putthenandare symmetric matrices in view of (2.5). Assume nowisthe coefficients are constant. The integrability condition$\omega_{i}^{0}=p_{ij}\omega^{j}$$(p_{ij})$$h_{ij}\omega_{3}^{j}=q_{ij}\omega^{j}$$(q_{ij})$$d\omega=\omega\wedge\omega$5

37
written in terms of these coefficients as follows:(2.7)$d\omega^{1}=\epsilon$(2.8.1)a$\omega^{1}\wedge\omega^{2}$,$p’+\epsilon r’=p+\epsilon r$$d\omega^{2}=b\omega^{1}\wedge\omega^{2}$,$q’-q=3\epsilon a$(2.8.2) $p+\epsilon r’=\epsilon a^{2}+(b-1)(b-2),$ $p’+\epsilon r=\epsilon a^{2}+(b+1)(b+2)$(2.8.3)$\epsilon a(p-\epsilon r)+2(b-1)q=v,$ $\epsilon a(p’-\epsilon r’)+2(b+1)q’=-v$(2.8.4) $(b-1)(p’-\epsilon r’)-2aq’=-u$ ,(2.8.5)$(b+1)(p-\epsilon r)-2aq=u$$q(p’-\epsilon r’)-q’(p-\epsilon r)=\epsilon$

a $u+bv$ .

In order to solve this system (2.8) put $2t=p’+\epsilon r’=p+\epsilon r,$ $2k=p-\epsilon r$
and $p’-\epsilon r’=2k’$ . Then from (2.8.2),
and $k-k’=-3b$ . From
(2.8.3), $\epsilon a(k+k’+3)+b(q+q’)=0$ and from (2.8.4), $a(q+q’)-b(k+k‘ -3)=$. Hence $6ab+(a^{2}+\epsilon b^{2})(q+q’)=0$ . Assume here $a\neq 0$ and $b\neq 0$ . Then
. Combining with (2.8.1), we have $2q=-3\epsilon a-6ab/(a^{2}+\epsilon b^{2})$ ,$2t=\epsilon a^{2}$$0$$a^{2}+\epsilon b^{2}\neq 0$$2q’=3\epsilon a-6ab/(a^{2}+\epsilon b^{2}))2k=-3b+3(\epsilon b^{2}-a^{2})/(a^{2}+\epsilon b^{2}),$$2k’=$: and consequently, $v=-6ab(\epsilon a^{2}+b^{2}-$$1)/(a^{2}+\epsilon b^{2})$ and $u=3(a^{2}-\epsilon b^{2})(\epsilon a^{2}+b^{2}-1)/(a^{2}+\epsilon b^{2})$ .
Now (2.8.5)
shows $(3b^{2}-\epsilon a^{2})(\epsilon a^{2}+b^{2}-4)=0$ . So we have two cases $3b^{2}=a^{2}$
or $\epsilon a^{2}=4-b^{2}$ . However the first case is reduced to the case where$3b+3(\epsilon b^{2}-a^{2})/(a^{2}+\epsilon b^{2})$638

2.2 The surface corresponding to each solution is obtained by solving a
Pfaffian equation $de=\omega e$ . We first carry out the integration for the case[2]. Since $d\omega^{1}=0$ andthere is a local coordinate system$\omega^{1}=dx$$\omega^{2}=e^{bx}dy$$(x, y)$ so that
. Let $z=e_{0}$ . Thenandshows $e_{1}=z_{x}$ and $e_{2}=z_{y}e^{-bx}$ . Since $de_{1}=z_{xx}dx+z_{xy}dy=\omega_{1}^{\alpha}e_{\alpha}$ and$d\omega^{2}=b\omega^{1}\wedge\omega^{2_{3}}$$de_{0}=\omega^{1}e_{1}+\omega^{2}e_{2}$739$de_{2}=(e^{-bx}z_{y})_{x}dx+(e^{-bx}z_{y})_{y}dy=\omega_{2}^{\alpha}e_{\alpha}$, we have$z_{xx}= \frac{3}{2}(1-b)z+z_{x}+e_{3}$$z_{xy}=(b-1)z_{y}$$z_{yy}e^{-2bx}= \frac{3}{2}\epsilon(b-1)z-\epsilon(b+1)z_{x}+\epsilon e_{3}$.

Hence, putting $z=e^{(b-1)x}w$ and cancelling , we have$e$$w_{xy}=0$(2.9)$w_{xx}-\epsilon e^{-2bx}w_{yy}=(4-b)w_{x}$.

It is easy to integrate this system and to obtain solutions:$\{1,y,$ $e^{(4-b)x}$$\{1,y,x$ ,$\{1,y,$ $e^{8x}$,$\frac{y^{2}}{2}+\frac{\epsilon}{2b(b+4)}e^{-2bx}\}$$\frac{y^{2}}{2}+\frac{\epsilon}{64}e^{-8x}\}$

,

,

,

$\frac{y^{2}}{2}+\frac{\epsilon}{8}xe^{8x}\}$,when $|b|\neq 4,$ $b=4$ and $b=-4$ respectively. Then the corresponding
surfaces are respectively

1o$Z=Y^{2}+\epsilon X^{k}$$2^{o}$$Z=Y^{2}+\epsilon e^{X}$$3^{o}$$Z=Y^{2}+\epsilon X\log X$(X, $Y,$$Z$$k\neq 0,1,2$,) being certain affine coordinates. These are all homogeneous.

2.3 The case [3] is similarly treated. Since $d\omega^{1}=d\omega^{2}=0$ , we may put$\omega^{1}=dx$
and $\omega^{2}=dy$ . The system for $z=e_{0}$ is $z_{xx}-\epsilon z_{yy}=2pz+2z_{x}$and $z_{xy}=qz-z_{y}$ . Introducing a new variable by $z=e^{-x}w$ and writing$(x, y)$ for} ), we have$w$$( \frac{1}{2}x,$(2.10)$y$$\{\begin{array}{l}w_{xx}-\epsilon w_{yy}=pw+2w_{x}w_{xy}=qw\end{array}$8

40
where and are new constants. This system is equivalent to the system
considered by Wilczynski in [W] when $\epsilon=-1$ . We follow his argumentwith a little modification.
$p$

$q$

First assume $q=0$ . Then the surface is a translation surface $w=f(x)+$$g(y)$ , where $f$ andare determined by $f_{xx}-2f_{x}-pf=\epsilon g_{yy}+pq=const$ .
,Since the characteristic equations are $(\rho-1)^{2}=p+1$ and
we must consider five cases separately: $p<-1,$ $p=-1,$ $-1<p<0$ ,$p=0$ and $p>0$ . Let $p<-1$ and $\epsilon=-1$ . Then the solutions are$\{e^{x}\cos\lambda x, e^{x}\sin\lambda x , \cos\mu y, \sin\mu y\}$ where $\lambda^{2}=-p-1$ and $\mu^{2}=-p$ .
This yields $the_{-x}surface(Y^{2}+Z^{2})/(1+X^{2})=.exp(k\tan^{-1}X)-x$ for $X=$
$Y=e$ $\cos\mu y/\cos\lambda x$ and $Z=e$ $S\ln\mu y/\cos\lambda x$ . $k=-2/\lambda$ .
, solutions are, which yields
When$(Y^{2}+Z^{2})/X=exp(k\tan^{-1}Z/Y)$ where $k=2/\lambda$ . This is equivalent to$YZ/(1+X^{2})=exp(ktan^{-1}X)$ by a projective transformation. So bothare together written as$g$$\rho^{2}=-\epsilon p$$\tan\lambda x,$$\epsilon=1$$\{e^{x}\cos\lambda x, e^{x}\sin\lambda x, e^{\mu y}, e^{-\mu y}\}$$\tan^{-1}Z=k\log(\frac{X^{2}-\epsilon Y^{2}}{1+Z^{2}})$$4^{o}$$k\neq 0$

by a trivial change of coordinates. Remaining four cases are similarly integrated and we have$5^{o}$$e^{Z}=X^{2}+\epsilon Y^{2}$$(p=-1)$$6^{o}$$Z^{k}=X^{2}-Y^{2}$$k>2$ when$\epsilon=$$k>2$ when $\epsilon=-1$ and $2>k>1$ when$\epsilon=$$(-1<p<0)$

$-1$

$1$

and $2>k>1$ when$\epsilon=1$$7^{o}$$Z=Y^{2}-\epsilon\log X$$8^{o}$$Z^{k}=X^{2}+Y^{2}$$(p>0)$ .

$(p=0)$

has two sets ofThese are all homogeneous; for example, the surface
rotations when $\epsilon=-1$ and spiral motions in and translations in when$4^{o}$$x$$\epsilon=1$$y$.

We next consider the case $q\neq 0$ . The system has $e^{ax+by}$ as a solution
where $a^{2}-\epsilon b^{2}=p+2a$ and $ab=q$ ; hence$a^{4}-2a^{3}-pa^{2}-\epsilon q^{2}=0$

$g$

.

41
Types of solutions of this algebraic equations are divided into six cases:(i) $(r, r,r, s)$ , (ii)
, (iii) $(r, r, s, t)$ , (iv)and (vi) $(r, s, t, u)$ whereare different reals andare non-realdifferent complex numbers. Other combinations do not occur because of$q\neq 0$ . The case (i) occurfs only when $p=-9/8,$ $q=3\sqrt{3}/16$ and.
The solutions are $\{1, 3x-\sqrt{3}y, 8\sqrt{3}y+(3x-\sqrt{3}y)^{2}, e^{-x-\sqrt{y}}\}$ . However
this yields a surface equivalent to ; so excluded. The solutions of (ii) are$(\alpha,\overline{\alpha}, \beta,\overline{\beta}),$$(r,r, \alpha,\overline{\alpha})$$r,$ $s,$ $t,$ $u$$\alpha,$$(v)(r, s, \alpha,\overline{\alpha})$$\beta$$\epsilon=1$$7^{o}$$\{e^{rx+by}, (rx-by)e^{rx+by}, e^{-\lambda^{2}x} \cos(\lambda x-\frac{b}{\lambda}y), e^{-\lambda^{2}x} \sin(\lambda x-\frac{b}{\lambda}y)\}$where $b=q/r$ and $\lambda^{2}=a-1>0$ . This represents$Z+\log(X^{2}+Y^{2})=k\tan^{-1}\underline{Y}$$9^{o}$

$X$

$k>0$

$(\epsilon=-1)$.

For the case (iii) it is seen that we may put $r=1-\lambda^{2},$and $t=-\lambda-\lambda^{2}$ , when
. Since solutions are { $e^{rx+by},$ ( $rx$ –by)$e^{rx}+,$ $e\iota Ie^{tx+y}tA$ } $(b=r/q)$ , the surface is$s=\lambda-\lambda^{2}$$|\lambda|\neq 1$$10^{o}$$Z= \frac{1}{\lambda-1}\log X-\frac{1}{\lambda+1}\log Y,$$\lambda^{2}<1)$($\epsilon=1$

when $\lambda^{2}>1;\epsilon=-1$ when.

In the case (iv) the independent solutions over

$C$

are

$\{\exp((a\pm i\lambda)x+\frac{q}{a\pm i\lambda}y), \exp((1-a\pm i\mu)x+\frac{1}{1-a\pm i\mu}y)\}$for some parameterssolutions we see$11^{o}$$\lambda$

and

$\mu$

(a is a function of$\log\frac{X^{2}+Y^{2}}{1+.Z^{2}}=k\tan^{-1}\frac{Y}{X}+l\tan^{-1}Z$$\lambda$

and

$\mu$

). Taking real,

and being real parameters determined by and . In the case (v)$exp(rx+z_{y),exp(sx}’+z_{y)\};r}s$ andsolutions are {are functions of and . The corresponding surface is$k$$\ell$$\lambda$$\mu$$exp((a\pm i\lambda)x+\overline{a}\pm^{l}\overline{i\lambda}y),$$\lambda$

$a$

$s$

$12^{o}$$\log Z=k\tan^{-1}\frac{Y}{X}+\frac{p}{2}\log(X^{2}+Y^{2})$.

The last case (vi) yields$13^{o}$$Z=X^{k}Y^{\ell}$.

We here do not try to determine $ex$ act ranges of

10

$k$

and

$\ell$

in$11^{o},$$12^{o}$and42$13^{o}$$12^{o}$. For non-degeneracy it is necessary to assume $kl\neq 0$ and $p\neq 1$ for
and $kP(k+P-1)\neq 0$ for. Above surfaces are also homogeneous.
$13^{o}$2.4 We integrate the final case [1]. Choose coordinates $(x, y)$ such that$e^{abx}dy$ anddy–b $dx$ . Put $z=e_{0}$ . Then $e_{1}=e^{-abx}z_{y}+ \frac{1}{a}z_{x}$and. Hence we have$\omega^{1}=$$\epsilon\omega^{2}=\frac{b}{a}e^{abx}$$e_{2}=- \frac{\epsilon}{b}z_{x}$$\{\begin{array}{l}uz_{xy}+\frac{1}{a}z_{xx}=(b-1)z_{x}+abuz_{y}+\frac{3}{4}ab(b+2)zu^{2}z_{yy}+\frac{4}{ab^{2}}uz_{xy}=\frac{4}{ab^{2}}(b^{2}-l)z_{x}+\frac{2}{b}(b+2)uz_{y}+\frac{6}{b}(1-b^{2})z\end{array}$where $u=e^{-abx}$ . Referring thatfor $v=y$ , we see(2.11)$e^{\frac{3}{2}abx}$

is a solution, put$z=e^{\frac{3}{2}abx}w$. Then,$\{\begin{array}{l}u(w_{uu}-w_{vv})=(b+1)w_{u}-w_{v}bu(w_{uu}-2w_{vv})=(b+1)w_{u}-w_{v}\end{array}$

The independent solutions are$w_{0}=1$,$w_{1}=u+(b+1)v$ ,$w_{2}=u^{2}-2uv-(b+1)v^{2}$$w_{3}=u^{3}+(b-1)u^{2}v-(b-1)uv^{2}+(1-b^{2})v^{3}/3$ .

The corresponding equation of the surface is$((b+1)w_{2}+w_{1}^{2})^{3}= \frac{g}{16}(b+2)((b+1)^{2}w_{3}-\frac{2}{3}(b-1)w_{1}^{3}-(b^{2}-1)w_{1}w_{2})^{2}$

when $|b|\neq 1,2$ and called Enriques surface. By a projective change of
coordinates this is written$14^{o}$$( Y-X^{2})^{3}=k(Z-\frac{2}{3}X^{3}-XY)^{2},$ $k\neq 0,$$\frac{9}{4}$.

The case $|b|=1$ is included in . The case $|b|=2$ is excluded because$a\neq 0$ . However the case $b=2$ is interesting because it yields a surface ofdegree 4: $27Z^{2}-X^{2}Y^{2}-42X^{4}-4Y^{3}-18XYZ=0$ , which is known to
be a homogeneous developable surface.
$1^{o}$

11

43\S 3. Homogeneous surfaces with $F=0$ .
3.1 This section treats the case $F=0$ . Since the surface is quadratic
when the cubic form vanishes (THEOREM 1.4), we here consider surfaces
with $F=0$ but with non-vanishing cubic form. Such a surface is ruled.
Since it must be of indefinite type, we start with a normalized frame with$\hslash=(l l)$ . We first see the existence of a frame with the following property:(3.1)$\omega_{1}^{3}=\omega^{2}$,$\omega_{2}^{3}=\omega^{1}$,$\omega_{1}^{2}=\omega^{1}$,$\omega_{2}^{1}=0$$h_{111}=-2$ ,$h_{112}=h_{122}=h_{222}=0$ ,$\omega_{0}^{0}-3\omega_{1}^{1}=0$,$L_{ij}=0$

and$\gamma_{2}=0$.

In fact, by the identity $F=2h_{111}h_{222}$ we may assume $h_{222}=0$ . The
identity $h_{112}=h_{122}=0$ follows from the form of . Consider a frame,change by a transformation whose components are$h$$\lambda\nu=1,$$g$$a=(\begin{array}{ll}\alpha 00 \alpha^{-1}\end{array})$.
and. Thentransforms as
andso that $h_{111}=-2$ ; and this condition restricts
So there exist. Then computation shows
frame changes to the case.
. Thus we can find a frame withandfollow from the form of the cubic form.
The identities, we see, which shows
Taking exterior derivation of$L_{22}=0$ . Similarly $L_{11}=0$ is seen by derivation of. Also theshows. Notice here that frame
exterior derivation ofchanges which keep the condition (3.1) are now restricted to$b_{1}=\lambda c^{2},$$b_{2}=\lambda c^{1}$$\mu=\lambda c^{1}c^{2}$$\lambda\tilde{h}_{111}=h_{111}\alpha^{3}$$h_{111}$$\lambda$$\alpha$$\lambda=\alpha^{3}$$\tilde{\omega}_{0}^{0}-3\tilde{\omega}_{1}^{1}=$$\omega_{0}^{0}-3\omega_{1}^{1}+\lambda(-4c^{2}\omega^{1}+2c^{1}\omega^{2})$$\omega_{1}^{2}=\omega^{1}$$\omega_{0}^{0}-3\omega_{1}^{1}=0$$\omega_{2}^{1}=0$$0=L_{22}\omega^{1}\wedge\omega^{2}$$\omega_{2}^{1}=0$$\omega_{1}^{2}=\omega^{1}$$\gamma_{2}=0$$\omega_{2}^{0}-\omega_{3}^{1}=0$(3.2)$g(\alpha)=(\begin{array}{llll}\alpha^{3} \alpha \alpha^{-1} \alpha^{-3}\end{array})$

We put(3.3)$\omega_{1}^{1}=a\omega^{1}+L^{2}$,$\omega_{1}^{0}=\omega_{3}^{2}=\mu\prime^{1}+q\omega^{2}$$\omega_{2}^{0}=\omega_{3}^{1}=r\omega^{1}+s\omega^{2}$

Coefficients change under(3.4)$\tilde{p}=\alpha^{-4}p$,$g(\alpha)$$\tilde{q}=\alpha^{-6}q$,$\omega_{3}^{0}=t\omega^{1}$.

as follows,$\tilde{r}=\alpha^{-6}r$12,$\tilde{s}=\alpha^{-8}s$,$t=\alpha^{-8}t\sim$.

44
The integrability condition(3.5)$d\omega=\omega$$d\omega^{1}=-2b\omega^{1}\wedge\omega^{2}$A,$\omega$

induces the following identities$d\omega^{2}=4a\omega^{1}$A$\omega^{2}$,

4 $bp-6aq=q_{1}-p_{2}$ ,$r=-2q,$ $q-r=3(b_{1}-a_{2})+6ab$ ,(3.6)$6rb-8as=s_{1}-r_{2}$ ,$s=t$ ,

6 $bt=-t_{2}$ ,where indices imply the derivation with respect to$a_{2}\omega^{2}$$\omega^{1}$.

and$\omega^{2}$: $da=a_{1}\omega^{1}+$3.2 Now assume $t\neq 0$ . Then by (3.4) we may assume further $|t|=1$ .
So, if the surface is homogeneous, remaining coefficients are assumed to beconstant. Then from (3.6), $a=b=q=r=0$ . Next assume $s=t=0$ .
and are constant. But thenIf $r\neq 0$ , then by a similar reasoning$b=0$ and following $r=0$ by (3.6), which is a contradiction. Thus $r=0$ .
Whqn $b\neq 0$ , we have $|b|=1$ and $a=p=0$ . When $b=0$ , we have either$a=0$ and $p=0,1,$ $-1$ or $|a|=1$ and is arbitrary. Therefore homogeneous
ruled surfaces are among four cases:
$a,$

$b,$

$p$

$q$

$p$

[4]

$|t|=1$ ,$a=b=q=r=0$[5]$p=0$ or$\pm 1$[6]$|a|=1$ ,$b=q=r=s=t=0$ .

[7]$|b|=1$ ,$a=p=b=q=r=s=t=0$ .

,$a=b=q=r=s=t=0$

We first examine the case [7]. In this case $\phi^{2}=0$ by (3.5) and consequently all diagonal forms are closed. Hence there is a frame change
which transforms the coframe into$g(\alpha)$$(\begin{array}{llll}0 \omega^{1} \omega^{2} 0 0 \omega^{l} \omega^{2} 0 \omega^{l}0 0\end{array})$.

Moreover the same reasoning applies to the case [6] when $p=0$ . Thus thecase [7] is a subcase of [6].

13

45
3.3 We next integrate the case [5]. Since $\omega^{1}=dx$ and $\omega^{2}=dy$ for somelocal coordinates $(x, y)$ , the associate system of differential quations is
$e$

(3.7)

$\{\begin{array}{l}z_{xx}=pz+z_{y}p=0,\pm 1z_{yy}=0\end{array}$

When $p=1$ , the independent solutions are$2y)e^{-x}\}$ and the surface is$\{e^{-x}$,$e^{x}$, $(x+2y)e^{x},$ $(x-$$Z=XY+X\log X$ .

$15^{o}$When $p=-1$ , the independent solutions are $\{\cos x$ , $\sin x,$ $x\sin x+$$2y\cos x$ , $x\cos x-2y\sin x$ } and the surface is$Z=XY+(1+X^{2})\tan^{-1}X$ .

$16^{o}$When $p=0$ , we get the Cayley surface$17^{o}$$Z=XY+ \frac{1}{3}X^{3}$.

3.4 For the case [6] we choose coordinates $(x, y)$ such that $\omega^{1}=dx$ and$\omega^{2}=e^{4ax}dy$ ; see (3.5). The system of differential equations for $z=e_{0}$ is(3.8)$\{\begin{array}{l}z_{xx}=(p-3)z+4az_{x}+e^{-4ax}z_{y}z_{yy}=0\end{array}$Putting $z=e^{2ax}w$ , we have $w_{xx}=(p+1)w+e^{-}w_{y}$ and $w_{yy}=0$ .
Hence $w=u(x)+.yv(x)$ . By a simple change of variables we see$\{\begin{array}{l}u_{xx}=pu+e^{x}vv_{xx}=pv.\cdot\end{array}$is a new constant. We have four subcases according as $p>0$ and $p\neq 1/4$ ,$p=1/4,$ $p<0$ , or $p=0$ . The space of independent solutions is respectively
wherein thi\’{s} order$p$$\{e^{-\lambda x} , e^{\lambda x} , \frac{1}{2\lambda+1}e^{(\lambda+1)x}+ye^{\lambda x} , \frac{1}{1-2\lambda}e^{(1-\lambda)x}+ye^{-\lambda x}\}$$p=\lambda^{2}$;$\{e^{-\frac{1}{2}x}, e^{\frac{1}{2}x}, \frac{1}{2}e^{\frac{3}{2}x}+ye^{\frac{1}{2}x}, xe^{\frac{1}{2}x}+ye^{-\frac{1}{2}x}\};\{1,$$x,$ $e^{x}+y,$$(x-2)e^{x}+$, a $e^{x}(\cos\lambda x+2\lambda\sin\lambda x)+y\cos\lambda x$ , awhere $p=-\lambda^{2}$ and $a=(1+4\lambda^{2})^{-1}$ . Therefore thecorresponding surfaces are respectively$xy\}$

and {$\cos\lambda x,$$e^{x}(\sin\lambda x-$$\sin\lambda x$$2\lambda\cos\lambda x)+y\sin\lambda x\}$14

46
$18^{o}$$Z=XY+X^{k}$ ,$19^{o}$$Z=XY+X^{2}\log X$$20^{o}$$Z=XY+e^{X}$$21^{o}$$\log\frac{X+YZ}{1+Z^{2}}=k\tan^{-1}Z$$k\neq 0,2$,

$k>0$ .

3.5 For the final case [4], the system is easily seen to be(3.9)$\{\begin{array}{l}z_{xx}=pz+z_{y}z_{yy}=tzt=\pm 1\end{array}$

Assume $t=1$ . Depending on the value of we have five subcases: $p<-1$ ,$p=-1,$ $-1<p<1,$ $p=1$ and $p>1$ . Respective solutions are given in thisorder: $\{e^{-y}\cos\lambda x , e^{-y}\sin\lambda x , e^{y}\cos\mu x , e^{y}\sin\mu x\}$ where $\lambda^{2}=1-p$ and$\mu^{2}=-1-p;\{e^{y}, xe^{y}, e^{-y}\cos\sqrt{2}x , e^{-y}\sin\sqrt{2}x\};\{e^{-y}\cos\lambda x$ ,,,} where $\lambda^{2}=1-p$ and $\mu^{2}=p+1;\{e^{-y}xe^{-y}$ ,where $\lambda^{2}=p-1$ andand$\mu^{2}=p+1$ . Surfaces are listed also in this order$p$$e^{-y}\sin\lambda x$$e^{\mu x+y},$$e^{\sqrt{2}x+y}$$e^{-\mu x+y}$)$e^{-\sqrt{2}x+y}\}$$\{e^{\lambda x-y}, e^{-\lambda x-y}, e^{\mu x+y}, e^{-\mu x+y}\}$$\tan^{-1}Z=k\tan^{-1}\underline{Y}$$22^{o}$

$k>1$

$X$

23

$o$

$Z= \tan^{-1}\frac{Y}{X}$$24^{o}$$\tan^{-1}Z=k\log\frac{Y}{X}$$25^{o}$$Z=Ye^{X}$$26^{o}$$Z=( \frac{Y}{X})^{k}$

,

,

$k>0$

$0<k<1$ .

When $t=-1$ , the solutions are$\cos(-\frac{x}{2\lambda}.+y),$surface$27^{o}$$e^{-\lambda x} \sin(-\frac{x}{2\lambda}+y)$$\{e^{\lambda x}\cos(\frac{x}{2\lambda}+y),$} where$e^{\lambda x} \sin(\frac{x}{2\lambda}+y),$$2\lambda^{2}=p+\sqrt{1+p^{2}}$. The resulting$1S$$\log\frac{X^{2}+Y^{2}}{1+Z^{2}}=k(\tan^{-1}\frac{Y}{X}-\tan^{-1}Z)$15,$e^{-\lambda x}$$k>0$ .

47

As is easily seen by looking at solutions, all surfaces are homogeneous.

\S 4. Remarks
4.1 Summarizing calculations in previeous sections we have seen that nondegenerate projectively homogeneous surfaces are projectively congruent to
one of$0^{0}$$Z=X^{2}+Y^{2}$and surfaces in the list

$1^{o}\sim 27^{o}$:

1o$Z=Y^{2}+\epsilon X^{k}$$2^{o}$$Z=Y^{2}+\epsilon e^{X}$$3^{o}$$Z=Y^{2}+\epsilon X\log X$$4^{o}$$\tan^{-1}Z=k\log(\frac{X^{2}-\epsilon Y^{2}}{1+Z^{2}})$$5^{o}$$e^{Z}=X^{2}+\epsilon Y^{2}$$6^{o}$$Z^{k}=X^{2}-Y^{2}$$7^{o}$$Z=Y^{2}-\epsilon\log X$$8^{o}$$Z^{k}=X^{2}+Y^{2}$$9^{o}$$Z+ \log(X^{2}+Y^{2})=k\tan^{-1}\frac{Y}{X}$$k\neq 0,1,2$$k\neq 0$$k>2$ when $\epsilon=1$ and $2>k>1$ when $\epsilon=-1$$k>2$ when $\epsilon=-1$ and $2>k>1$ when $\epsilon=1$

,

$k>0$

$10^{o}$$Z= \frac{1}{\lambda-1}\log X-\frac{1}{\lambda+1}\log Y$$11^{o}$$\log\frac{X^{2}+Y^{2}}{1+Z^{2}}=k\tan^{-1}\frac{Y}{X}+\ell\tan^{-1}Z$$12^{o}$$\log Z=k\tan^{-1}\frac{Y}{X}+\frac{p}{2}\log(X^{2}+Y^{2})$$13^{o}$$Z=X^{k}Y^{\ell}$$14^{o}$$( Y-X^{2})^{3}=k(Z-\frac{2}{3}X^{3}-XY)^{2},$ $k\neq 0,$

16$\frac{9}{4}$48$15^{o}$$Z=XY+X\log X$$16^{o}$$Z=XY+(1+X^{2})\tan^{-1}X$$17^{o}$$Z=X Y+\frac{1}{3}X^{3}$$18^{o}$$Z=XY+X^{k}$ ,$19^{o}$$Z=XY+X^{2}\log X$$20^{o}$$Z=XY+e^{X}$$21^{o}$$\log\frac{X+YZ}{1+Z^{2}}=k\tan^{-1}Z$$22^{o}$$k\neq 0,2$$\tan^{-1}Z=k\tan^{-1}\underline{Y}$

,

$k>0$

$k>1$

$X$

$23^{o}$$Z= \tan^{-1}\frac{Y}{X}$$24^{o}$$\tan^{-1}Z=k\log\frac{Y}{X}$$25^{o}$$Z=Ye^{X}$$26^{o}$$Z=( \frac{Y}{X})^{k}$$27^{o}$$\log\frac{X^{2}+Y^{2}}{1+Z^{2}}=k(\tan^{-1}\frac{Y}{X}-\tan^{-1}Z)$

,

,

$k>0$

$0<k<1$,$k>0$ .

Remark that the above list does not contain $degene\tau atep$ rojectively homogeneous surfaces such as projective planes and a surface mentioned in
the end of \S 3.
is included
4.2 In the above list we find some duplications: The surface
$k=p$
when
. Similarly 24 $0=12^{o}(\ell=0),$ $26^{o}=13^{o}(k=-p)$ ,in$27^{o}=11^{o}(k^{2}=l^{2})$ and $17^{o}=18^{o}(k=3)$ . These are due to the nonuniqueness of the frames we have chosen (each surface could have several
but finite number of frames with the required property) and due to that theexact range of the parameters are not determined. For example the surface$Z=X^{k}Y^{1}$ is equivalent to $Z=X^{1-k-1}Y^{1}$ by a transformation (X, $Y,$)$(1/X,Y/X, Z/X)$ . However to understand surfaces geometrically, those$6^{o}$$13^{o}$$Z$

17$arrow$subcases mentioned above should be excluded from the case written in theright side of equalities.
4.3 In [L], Lie has given, other than quadratic surfaces and the Cayley$11$ surfaces with two-dimensional automorphism groupsurfaceover
C. They are denoted by I to VII and I’ to IV’. Those are identified withsurfaces considered over in the above list as follows:$17^{o},$$C$$I=14^{o},$ $\Pi=1^{o}$$VI=18^{o}$, III $=3^{o},$ $IV=20^{o},$ $V=19^{o}$, VII $=2^{o},$ $I’=7^{o},$ $\Pi’=15^{o}$ , III’ $=10^{o}+5^{o},$ $IV’=13^{o}$ .

4.4 In the projective study of surfaces we know several notions such as
projective minimality, Demoulin surfaces etc. In regard of these notions we
can examine following facts: The Enriques surface
is known to be anaffine sphere and, so, rojectively minimal. Surfaces in the classes [5] and[6] are Demoulin and rojectively minimal. We see surfaces in [3] and [4]are also projectively minimal. Surfaces in [4] are not Demoulin. Surfacesin [2] are rojectively minimal only when $|b|=2$ , i.e. $k=-2$ or in .
We omit the details.
$14^{o}$$p$$p$$\frac{2}{3}$$p$$1^{o}$REFERENCES[G] H. W. Guggenheimer, “Differential Geometry,” McGraw-Hill, 1963.
[L] S. Lie, Bestimmung aller Fl\"achen, die eine continuirliche Schar von projectiven Transformationen gestatten, Berichte Verh. Ges. Wiss. Leibzig 47 (1895), 209-260.

[S] T. Sasaki, Projective differential geometry and linear homogeneous differential equations, lecture notes at Brown Univ. 1988/89.
[W] E.J. Wilczynski, On a certain class of self-projective surfaces, Trans. Amer. Math. Soc.
14 (1913), 421-443.
(Feb. 1990)

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