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EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 7, No. 4, 2014, 472-485
ISSN 1307-5543 – www.ejpam.com

The Linear Span of Four Points in the Plücker’s Quadric in P5

Jacqueline Rojas1,∗, Ramón Mendoza2

1 CCEN - Departamento de Matemática - UFPB Cidade Universitária, 58051-900,

João Pessoa - PB - Brasil
2 CCEN - Departamento de Matemática - UFPE Cidade Universitária, 50740-540, Recife - PE - Brasil

Abstract. Given four (distinct) lines ℓ1, ℓ2, ℓ3, ℓ4 in P3. Let Pi (i = 1, . . . , 4) be the image of ℓi in
the Plücker’s quadric Q ⊂ P5 under the Plücker embedding P (in (1)). Set Λ =




P1, . . . , P4

�

be the
linear span of those four points in P5. The purpose of this article is to write specifically what kind of
quadric Λ∩Q can be, taking under considerations all possible configurations of these four lines in P3.
In particular, having in mind the classical problem in Schubert Calculus: How many lines in 3-space meet

four given lines in general position? whose answer is 2 (see p. 272 in [3] or p. 746 in [4]). We verified
that four lines in P3 are in general position if and only if Λ is a 3-plane and Λ ∩Q is an irreducible
quadric surface. In fact, we prove that there are exactly two solutions if and only if Λ is a 3-plane and
Λ∩Q is a nonsingular quadric.

2010 Mathematics Subject Classifications: AMS 14N05, 14N20

Key Words and Phrases: Plücker’s quadric, linear span, 4-line problem.

1. Introduction

Plücker’s coordinates were introduced by the German geometer Julius Plücker (1801-
1868) in the 19th century, as a way to assign six homogenous coordinates to each line in
the complex projective 3-space P3. Since they satisfy a homogeneous quadratic equation, it
follows an embedding of the 4-dimensional space of lines in P3 (denoted by G1(P

3) and called
grassmannian of lines in P3) onto a nonsingular quadric hypersurface Q in P5 (see Proposi-
tion 3). Thus, reminding the Schubert’s classical enumerative problem: How many lines in

3-space meet four given lines in general position? Whose answer can be found in many texts
and is given by: “there are two lines in P3 which meet 4 given lines in general position” (see
Example 14.7.2 at p. 272 in [3]). In fact, if you want to know an algorithm to determine
explicit solutions, then see [7]. On the other hand, any beginner in the art of solving enumer-
ative problems will ask: what does general position means? In Algebraic Geometry, general
position is a notion of genericity for a set of points, or other geometric objects. It means the

∗Corresponding author.

Email addresses: jacq@mat.ufpb.br (J. Rojas), ramon@dmat.ufpe.br (R. Mendoza),

http://www.ejpam.com 472 c© 2014 EJPAM All rights reserved.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 473

general case situation, as opposite to some more special or coincident cases that are possible.
Its precise meaning differs in different settings. For example, in [8] the authors imposed the
condition ℓi ∩ ℓ j = ; (1≤ i < j ≤ 4) to the four given lines ℓ1, ℓ2, ℓ3, ℓ4 in P3 and, even under
this assumption they found (in one case) infinitely many solutions for the 4-lines problem (cf.
3.1 in the last subsection). So, this condition is not enough for the four given lines to be in
general position.

In this article, we use the identification between lines in P3 and points in the quadric
hypersurface Q to explain what is the precise meaning of general position for that problem.
Nevertheless, the emphasis in our work lies on to take under considerations all possible config-
urations of these four lines in P3 and write specifically what kind of quadric Λ∩Q can be. One
key ingredient in this work lies on the well known description of all linear subspaces contained
in a quadric hypersurface in P3 and P5 (see subsection 2.1 and 3.1).

Finally, we note that Plücker was a geometer that firmly believed in the importance of
the applications of Mathematics to the physical sciences. So, in 1847 he turned to Physics,
accepting the chair of Physics at Bonn and working on magnetism, electronics and atomic
physics. He anticipated Gustav Kirchhoff and Robert Wilhelm Bunsen in announcing that the
lines of the spectrum were characteristic of the chemical substance which emitted them, and
in indicating the value of this discovery in chemical analysis. According to Johann Hittorf he
was the first who saw the three lines of the hydrogen spectrum, which a few months after his
death were recognized in the spectrum of the solar protuberances.

2. Notations and Preliminary Results

We denote by C the field of complex numbers. Let V be an n-dimensional vector space
over C. Denote by [v1, . . . , vk] the subspace of V generated by the vectors v1, . . . , vk ∈ V .

The k-grassmannian associated to the vector space V . For each integer k,
0≤ k ≤ n= dim V , we denote by Gk(V ) the set of all k-dimensional linear subspaces of V and
call it the k-grassmannian associated to V . In the particular case k = 1, the 1-grassmannian
associated to V it is also called projective space associated to V and it is denoted by P(V ) (i.e.
P(V ) := G1(V )). We use the notation Pn instead of P(Cn+1) and p = [a0 : . . . : an] for
p = [(a0, . . . , an)] ∈ P

n, just for the sake of simplicity.
If W ∈ Gk+1(V ) then P(W ) ⊆ P(V ) will be called k-linear subspace of P(V ). The set of all k-

linear subspaces of P(V ) will be denoted by Gk(P(V )), the grassmannian of k-linear subspaces

of P(V ). Moreover, we shall call G1(P(V )), G2(P(V )) and Gn−1(P(V )) the grassmannian of
lines, planes and hyperplanes in P(V ), respectively. So, since a line ℓ in P3 is equal to P(W )
for some W ∈ G2(C

4), we have the correspondence

G2(C
4) −→ G1(P

3)

W 7−→ P(W ),

between the 2-grassmannian associated to C4 and the grassmannian of lines in P3. Thus, all
assertions involving G2(C

4) can be translated into G1(P
3).



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 474

Next we introduce the notion of algebraic projective set in Pn. We will see in Proposition 1
that k-linear subspaces of Pn are examples of algebraic sets.

Algebraic projective sets in Pn. Let C[X ] = C[X0, . . . , Xn] be the polynomial ring over C in
the variables X0, . . . , Xn. Now, for each integer d ≥ 0 consider the vector subspace C[X ]d ,
generated by all monomials in X0, . . . , Xn of degree d. Each element in C[X ]d will be called
an homogeneous polynomial of degree d. If F ∈ C[X ]d , then we define, Z (F), the zero set of
F in Pn by

Z (F) =
¦

[v] ∈ Pn | F(v) = 0
©

.

For example, if L ∈ C[X ]1, then Z (L) = P(W ) with W = {v ∈ Cn+1 | L(v) = 0}. Therefore,
Z (L) is a hyperplane of Pn, if L 6= 0 else Z (L) = Pn.

If d ≥ 1, then an element [F] in the projectivization of C[X ]d will be called hypersurface of

degree d in Pn. From here on, when we say: Let X ⊂ Pn be the reduced hypersurface defined
by F ∈ C[X ]d , that means that X = Z (F) ⊂ Pn and F is square-free.

A subset X of Pn will be called algebraic projective set, if there exist homogeneous polyno-
mials F1, . . . , Fk in C[X ] such that X = Z (F1)∩ · · · ∩Z (Fk).

Next we show that any r-linear subspace in Pn is the intersection of exactly n− r hyper-
planes.

Proposition 1. Let Λ be an r-linear subspace in Pn with n > r. Then, there are exactly n − r

linearly independent linear forms L1,. . . ,Ln−r in C[X ] such that

Λ = Z (L1)∩ · · · ∩Z (Ln−r).

Proof. Assume that Λ = P(W ) with W ∈ Gr+1(C
n+1). Let α = {e1, . . . , en+1} be the canon-

ical base of Cn+1 and let β = {e∗1, . . . , e∗n+1} be the associated dual base of (Cn+1)∗. Next we
consider the linear isomorphism

ϕ : (Cn+1)∗ → C[X ]1
e∗

i
7→ X i−1.

Now, let W 0 be the annihilator of W , then ϕ(W 0) is an (n− r)-dimensional linear subspace of
C[X ]1. Finally, an easy verification show that any base {L1, . . . , Ln−r} of ϕ(W 0) verified that
Λ = Z (L1)∩ · · · ∩Z (Ln−r).

Incidence of r-linear subspaces with reduced hypersurfaces in Pn. The following proposi-
tion will play an important role in our investigations.

Proposition 2. Let Z ⊂ Pn be the reduced hypersurface defined by the homogeneous polynomial

F of degree d and Λ be an r-linear subspace of Pn with r ≥ 1. Then the following conditions are

verified.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 475

(1) Z ∩Λ 6= ;;

(2) Z ∩Λ consists of infinitely many points, if Λ ⊂ Z or r ≥ 2, else Z ∩Λ consists of at most d

points.

Proof. To arrive at statements (1) and first part of (2) have in mind that dim Z = n − 1,
dimΛ = r and apply Theorem 7.2 at p. 48 in [5]. Already, the last part of statement (2)
follows easily from the fundamental theorem of algebra.

Projective Tangent Space and Nonsigular Reduced hypersurface. Let Z ⊂ Pn be the reduced
hypersurface defined by F ∈ C[X ]d and p = [v] ∈ Z. Let F ′v : Cn+1 −→ C be the differential of
F at v. We define the projective tangent space, TpZ , to the hypersurface Z at p by

TpZ = P(ker(F ′v)) =
¦

[u0 : . . . : un] ∈ P
n |

n
∑

i=0

∂ F

∂ X i

(v) · ui = 0
©

.

Now, note that:

• follows from the Euler relation
∑n

i=0
∂ F
∂ X i
· X i = dF that p ∈ TpZ .

• TpZ is a hyperplane in Pn if and only if p 6∈ ∩n
i=0Z (

∂ F
∂ X i
).

A point p ∈ Z satisfying the last condition above will be called a nonsingular point of Z , else p

will be called a singular point of Z . If all the points in Z are nonsingular, then Z will be called
a nonsingular hypersurface in Pn.

For example, after a linear change of coordinates (see Theorem 4 at p. 411 in [2]) we
concluded that Z (X 2

0+X 2
1 +X 2

2 +X 2
3) is the unique nonsingular quadric surface in P3, whereas

the singular reduced quadric surfaces in P3 correspond either to the union of two planes or a
quadric cone. Moreover, it is straightforward to show that the vertex of a quadric cone is its
unique singular point. In the case of the union of two (distinct) planes, it is verified that the
points in the line where the two planes meet are their singular points.

2.1. Lines on Reduced Quadrics Surfaces in P3

Next, we present some observations about lines contained on reduced quadrics surfaces in
P

3.

• Union of two planes. Of course any line in this surface will be contained in one of the
planes. Thus this surface could have at most two disjoint lines.

• Quadric cone. Any line in this surface passes through its vertex. Thus this surface does
not contain disjoint lines.

• The nonsingular quadric surface. As we shall describe in the next Lemma, a nonsin-
gular quadric surface in P3 contains exactly two families of lines parametrize by P1. In
fact, this Lemma is part of exercise 2.15 in Hartshorne’s book [5]. See the proof at p.
478-479 in [4] or [8].



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 476

Lemma 1. Let Q be a nonsingular quadric surface in P3. Then there exist two families of lines

L = {Lp}p∈P1 andM = {Mp}p∈P1 in Q such that

(1) Lp ∩ Lq = ; and Mp ∩Mq = ; for all Lp, Lq ∈ L , Mp, Mq ∈M and p 6= q ∈ P1.

(2) Lp ∩Mq 6= ; for all Lp ∈ L , Mq ∈M and p,q ∈ P1.

(3) If ℓ is a line contained in Q then ℓ ∈ L or ℓ ∈M .

(4) Given x ∈Q there exist unique lines Lp(x) ∈ L and Mq(x) ∈M such that {x} = Lp(x)∩Mq(x).

One other simple but important fact which will help us to prove our main result (Theorem 1)
is the next Lemma.

Lemma 2. Given the lines ℓ1, ℓ2 and ℓ3 in P3 such that ℓi ∩ ℓ j = ; for 1≤ i < j ≤ 3, there exists

a nonsingular quadric surface Q in P3 containing ℓ1, ℓ2 and ℓ3.

Proof. Take 3 points p1i , p2i , p3i on each line ℓi (i = 1,2,3) and note that

W = {G ∈ C[X ]2 | G(pi j) = 0 for all 1≤ i, j ≤ 3} (here n= 3)

is a subspace of C[X ]2 of dimension greater than or equal to 1. So we can choose G 6= 0 in
W and take Q = Z (G). Now follows from Proposition 2 that each line ℓi is contained in Q.
Finally, since Q contains three pairwise disjoint lines, then Q must be a nonsingular surface in
P

3.

3. The Plücker’s Quadric Q in P5

The Plücker embedding P : Gk(P(V )) −→ P(Λ
k+1V ) is the map that enable us to identify

the grassmannian Gk(P(V )) with a projective variety in P(Λk+1V ) (Λk+1V denotes the (k+1)-
th exterior power of V ). It is defined by P(W ) 7→ [u0 ∧ . . .∧ uk], if W = [u0, . . . , uk].

The Plücker’s quadric Q in P5. In what follows we will consider V = C4. Now, fix the base
{ei ∧ e j}1≤i< j≤4 of Λ2

C
4 where {ei}

4
i=1 is the canonical basis of C4. Let us consider

W = [u, v] ∈ G2(C
4) with u= (u0,u1,u2,u3) and v= (v0, v1, v2, v3), then we see that

u∧ v=
∑

1≤k<l≤4

wk−1,l−1ek ∧ el where wi j = ui v j − u j vi for 0≤ i < j ≤ 3.

Thus the Plücker embedding P above in coordinates is given by

P :G1(P
3) −→P5

P([u, v]) 7−→[w01 : w02 : w03 : w12 : w13 : w23].
(1)

Proposition 3. The Plücker map P : G1(P
3) −→ P5 defined in (1) is an embedding of the

grassmannian G1(P
3) over the nonsingular quadric hypersurface Q = Z (F) ⊂ P5 where

F = X0X5 − X1X4 + X2X3 ∈ C[X ] (n= 5).



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 477

Proof. See Theorem 11 at p. 409 in [2] or p. 209-211 in [4].

We will refer to Q in Proposition 3 as the Plücker’s quadric Q (in P5).

Next, we will make a remark that tells us who is the image of the families of lines L and
M of a nonsingular quadric surface in P3, under the Plücker embedding P (cf. Lemma 1).

Remark 1. It can be shown that P (L ) = Z (X0 + X5, X1 − X4, X2 + X3, F) and

P (M ) = Z (X0− X5, X1+ X4, X2− X3, F), so the image of the families of lines L andM , under

the Plücker embeddingP (in (1)) are nonsingular conics lying in complementary planes (also see

p. 196 in [6]).

3.1. Linear Subspaces in the Plücker’s quadric Q

We begin by proving a simple result, which allows us to conclude that the Plücker’s quadric
Q does not contain 3-planes or hyperplanes. In fact, it is verified that Q only contains lines
and planes, and this will be described in Propositions 5 and 6, respectively (see [1]).

Proposition 4. Let Z ⊂ Pn be a nonsingular quadric hypersurface defined by G ∈ C[X ]2 and Λ

a r-linear subspaces of Pn . If Λ ⊂ Z, then we have 2r < n.

Proof. Set Λ = P(W ) where W = [w0, . . . ,wr]. Assume that G =
∑n

i=0 X 2
i

(otherwise make
a linear change of coordinates). So G induz the bilinear form B : Cn+1 ×Cn+1 → C given by
B(v,w) = v0w0 + · · ·+ vnwn, if v= (v0, . . . , vn) and w= (w0, . . . , wn). Let

W⊥ = {v ∈ Cn+1 | B(v,w) = 0 for all w ∈W}

be the the orthogonal subspace associated to W .
Note that W⊥ = ker(T0) ∩ · · · ∩ ker(Tr) where Ti is the linear functional over Cn+1 given

by Ti(v) = B(v,wi), i ∈ {0, . . . , r}. Since these linear functionals are linearly independent we
conclude that dim W⊥ = n+ 1− dim W = n− r.

On the other hand, Λ ⊂ Z if and only if G(w) = B(w,w) = 0 for all w ∈ W . Thus Λ ⊂ Z

if and only if W ⊆ W⊥. So, if Λ ⊂ Z then dim W ≤ dim W⊥ = n+ 1 − dim W . Therefore,
2 dim W ≤ n+ 1 which implies 2r < n.

In what follows, we denoted by



ℓ1, . . . ,ℓk

�

the smallest linear subspace of Pn containing
the lines ℓ1, . . . ,ℓk of Pn. For example, if ℓ1 and ℓ2 are two distinct lines in Pn having a com-
mon point, then




ℓ1,ℓ2
�

is a plane, else



ℓ1,ℓ2
�

is a 3-plane.

Next we give the description of lines and planes in Q.

Lines in the Plücker’s quadric Q. The lines in the Plücker’s quadric Q are parametrized by
the incidence variety Γ = {(p,Π) | p ∈ Π} ⊂ P3 ×G2(P

3). In fact, let p ∈ P3 and Π ⊂ P3 be a
plane through p. Set

Ωp(Π) =
¦

ℓ ∈ G1(P
3) | p ∈ ℓ ⊂ Π
©

.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 478

Proposition 5. Let P : G1(P
3) → P5 be the Plücker embedding in (1). If L is a line in P5

contained in Q, then we have:

(i) P (Ωp(Π)) is a line in P5 contained in Q for all (p,Π) ∈ Γ.

(ii) If P0, P ∈ L are two different points such that P0 =P (ℓ0) and P =P (ℓ), then ℓ0∩ℓ= {p}
and they determine the plane Π =




ℓ0,ℓ
�

. In fact, for every point Q =P (m) ∈ L holds that

m ∈ Ωp(Π).

Planes in the Plücker’s quadric Q. There are two families of planes in the Plücker’s

quadric Q parametrized by P3 and its dual
∨

P
3= G2(P

3), respectively. In fact, let p ∈ P3 and
Π ⊂ P3 be a plane. Set

Ωp =
¦

ℓ ∈ G1(P
3) | p ∈ ℓ
©

and Ω(Π) =
¦

ℓ ∈ G1(P
3) | ℓ ⊂ Π
©

.

Proposition 6. Let P :G1(P
3) → P5 be the Plücker embedding in (1). If Λ is a plane in P5

contained in Q, then we have:

(i) P (Ωp) and P (Ω(Π)) are planes in P5 contained in Q.

(ii) The pre-image of Λ under the Plücker embedding P is either: Ωp for some p ∈ P3 or Ω(Π)

for some Π ∈ G2(P
3). In fact, if Pi = P (ℓi), i = 0,1,2 are three points in Q whose linear

span it is equal to Λ, then Λ = Ωp if p ∈ ∩2
i=0ℓi (and these lines are non-coplanar) else

Λ = Ω(Π) with Π = 〈ℓ0,ℓ1,ℓ2〉.

4. Description of Λ∩Q According to the Relative Position of the Four Given

Lines in P3

We begin this section by introducing some more notation. As we did for lines we denote by
〈p1, . . . , pk〉 the linear span of p1,. . . ,pk ∈ P

n (i.e. the smallest linear subspace of Pn containing
these points). Moreover, if p is a point and ℓ is a line in Pn, then 〈p,ℓ〉 also denote the linear
span of p and ℓ in Pn. So 〈p,ℓ〉= ℓ, if p ∈ ℓ else 〈p,ℓ〉 is a 3-plane.

Let ℓ1,ℓ2,ℓ3 and ℓ4 be four distinct lines in P3. Let Pi ∈ Q (i = 1, . . . , 4) be the image of
ℓi under the Plücker embedding P (in (1)) and Λ = 〈P1, P2, P3, P4〉 ⊂ P

5. Set S be the set of
solutions of the 4-lines problem in Schubert calculus, i.e.

S =
�

ℓ ∈ G1(P
3) | ℓ∩ ℓi 6= ; for 1≤ i ≤ 4

	

.

We have three major cases to be considered (cf. subsections 4.1, 4.2 and 4.3). In each
of these cases, the linear space Λ and Λ ∩Q are reported in tables, whose rows are labeled
according to the position that the lines can have in P3. Next, we determine Λ and Λ ∩Q in
some of these cases, which we believe are sufficient to illustrate the technique used in this
computations, we left to the reader the other cases

In what follows, for simplicity, we will use the notation Ωp(Π), Ωp and Ω(Π) for lines and
planes in Q, according to the identification given by the Plücker embedding in Propositions 5
and 6, respectively.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 479

4.1. The Four Lines are Concurrent, Say p ∈ ∩4
i=1ℓi.

Table 1: Four Lines are Concurrent.

Subcase Position of the lines S Λ Λ∩Q

1.1
ℓ1,ℓ2,ℓ3 and ℓ4 are contained
in the plane Π

Ωp ∪Ω(Π) Line Ωp(Π)

1.2
ℓ1,ℓ2,ℓ3 and ℓ4 are non
coplanar lines

Ωp Plane Ωp

1.1 Naturally any line passing trough p is a solution. On the other hand, if ℓ ∈ S is a
solution such that p 6∈ ℓ, it certainly meets the lines ℓ1 and ℓ2 at two different points, which
implies that ℓ ⊂ Π. Therefore S = Ωp ∪ Ω(Π). On the other hand, since Pi ∈ Ωp(Π) for
i = 1, . . . , 4, then Λ ≡ Ωp(Π) is a line contained in Q.

Figure 1: Case 1.1

4.2. The Four Lines have no Common Point and at Least Two are Coplanar.

Of course in all the subcases listed below we consider an index reordering if necessary.

Table 2: Four Lines have no Common Point and at Least Two are Coplanar.

Subcase Position of the lines S Λ Λ∩Q

2.1
ℓ1,ℓ2,ℓ3 and ℓ4 are contained
in the plane Π

Ω(Π) Plane Ω(Π)

2.1 Of course any line contained in the plane Π is a solution. Now, since the four given
lines have not common point, then any solution will meets at least two of these at different
points. Therefore, S = Ω(Π). Now, since the four lines are not concurrent, then three of these
four points (the Pi ’s) determine the plane Ω(Π). In fact, Λ ≡ Ω(Π), so it is a plane contained
in Q.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 480

Table 3: 2.2 Exactly Three Lines are Coplanar. Assume that ℓ1,ℓ2 and ℓ3 are contained in the plane Π and
Π∩ ℓ4 = {q}.

Subcase Position of the lines S Λ Λ∩Q

2.2.1
⋂3

i=1 ℓi = {p} Ωp(〈p,ℓ4〉)∪Ωq(Π) Plane
Union of
two lines

2.2.2
⋂3

i=1 ℓi = ; Ωq(Π) 3-plane
Union of

two planes

2.2.1 Let ℓ ∈ S be a solution which meets the lines ℓ1 and ℓ2 at two different points, then
ℓ ∈ Ωq(Π) (since Π ∩ ℓ4 = {q}). Else ℓ meets ℓ1 and ℓ2 at p and ℓ4 at some point different
of p, then ℓ ⊂ 〈p,ℓ4〉. So ℓ ∈ Ωp(〈p,ℓ4〉). Therefore, S = Ωq(Π) ∪Ωp(〈p,ℓ4〉) and it can be
identified with two projective lines having a common point (just a cross!). On the other hand,
since P1, P2 and Pi lies on the line Ωp(Π) and P4 6∈ Ωp(Π) (p 6∈ ℓ4) we conclude that Λ is a plane
in P5 not contained in Q. So Λ∩Q is a conic. Note that the line L1,2 = 〈P1, P2〉 ⊂ Λ∩Q and
P4 6∈ L1,2. Therefore, Λ∩Q is the union two lines. In fact, Λ∩Q = L1,2∪ L where L = 〈P4, M〉
with M =P (〈p,q〉).

Table 4: 2.3 Any Three Lines are non Coplanar and at Least Two Pair of Lines are Coplanar with
ℓi 6= Π1 ∩Π2∀ i. Assume that ℓ1 ∩ ℓ2 = {p} and ℓ3 ∩ ℓ4 = {q}. Set Π1 = 〈ℓ1,ℓ2〉 and Π2 = 〈ℓ3,ℓ4〉.

Subcase Position of the lines S Λ Λ∩Q

2.3.1 p,q ∈ Π1 ∩Π2 Ωp(Π2)∪Ωq(Π1) Plane
Union of
two lines

2.3.2 p ∈ Π1 ∩Π2 6∋ q Ωp(Π2) 3-plane
Union of

two planes

2.3.3 p 6∈ Π1 ∩Π2 6∋ q
¦

Π1 ∩Π2,ℓp,q

©

3-plane
Nonsingular

quadric

Figure 2: Case 2.3.3

2.3.3 Let ℓ be a solution. Next we consider the following two cases.

(i) Assume that p ∈ ℓ. Suppose that ℓ meets the lines ℓ3 and ℓ4 at two different points,
then ℓ ⊂ Π2 which implies that p ∈ Π1 ∩Π2. But this is impossible, so q ∈ ℓ. Therefore
ℓ= ℓp,q. Similarly, we conclude that ℓ= ℓp,q, if q ∈ ℓ.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 481

(ii) Assume that p 6∈ ℓ and q 6∈ ℓ. Then ℓ meets the lines ℓ1 and ℓ2 (respectively, ℓ3 and
ℓ4) at two different points which implies that ℓ ⊂ Π1 (respectively, ℓ ⊂ Π2). Therefore,
ℓ= Π1 ∩Π2.

Now, note that the line L1,2 = 〈P1, P2〉 ⊂ Λ ∩ Q and Pi 6∈ L1,2, i = 3,4. So 〈P1, P2, P3〉 is a
plane not contained in Q (since p 6∈ ℓi and ℓi 6⊂ Π1 for i = 3,4). On the other hand, the line
L3,4 = 〈P3, P4〉 is also contained in Λ∩Q and we observe that L1,2 and L3,4 are disjoint (since
L1,2 = Ωp(Π1) and L3,4 = Ωq(Π2)), so P4 does not belong to the plane 〈P1, P2, P3〉. Therefore,
Λ is a 3-plane. Keeping in mind that Λ∩Q is a quadric surface containing four non coplanar
points and two disjoint lines, we conclude that Λ∩Q is a union of two planes or a nonsingular
quadric. Suppose that Λ ∩Q is a union of two planes, say Λ ∩Q = Λ1 ∪ Λ2. Thus we can
assume that L1,2 ⊂ Λ1 and L3,4 ⊂ Λ2. Then necessarily Λ1 it is either Ωp or Ω(Π1), and in
the same form Λ2 it is either Ωq or Ω(Π2). But in any case, Λ1 ∩Λ2 will be empty or a point.
Therefore, Λ∩Q is a nonsingular quadric surface.

Table 5: 2.4 Any Three Lines are non Coplanar and at Least Two Pair of Lines are Coplanar with Π1∩Π2 = ℓ1.
Assume that ℓ1 ∩ ℓ2 = {p} and ℓ1 ∩ ℓ3 = {q}. Set Π1 = 〈ℓ1,ℓ2〉 and Π2 = 〈ℓ1,ℓ3〉. Let {ri}= Πi ∩ ℓ4 for i = 1, 2.

Subcase Position of the lines S Λ Λ∩Q

2.4.1 p = q Ωp(〈p,ℓ4〉) 3-plane
Union of

two planes

2.4.2
p 6= q and p ∈ ℓ4

(or p 6= q and q ∈ ℓ4)
Ωp(〈p,ℓ3〉)

(or Ωq(〈q,ℓ2〉))
3-plane

Union of
two planes

2.4.3
r1 = r2 and

#{p,q, r1}= 3

¦

ℓ1

©

3-plane
Quadric

cone

2.4.4 p 6= q and r1 6= r2

¦

ℓq,r1
,ℓp,r2

©

3-plane
Nonsingular

quadric

2.4.4 Let ℓ ∈ S . Here we have two possibilities:

(i) p 6∈ ℓ. Then ℓ ∩ ℓ1 6= ℓ ∩ ℓ2 and we have that ℓ ⊂ Π1. Thus ℓ ∩ ℓ3 ⊂ Π1 ∩ ℓ3 = {q} and
ℓ∩ ℓ4 ⊂ Π1 ∩ ℓ4 = {r1}. Therefore, ℓ= ℓq,r1

(q 6= r1 since ℓ4 6⊂ Π1).

(ii) p ∈ ℓ. Note that ℓ meets the line ℓ3 at a point p1 (p1 ∈ Π2) different from p. Thus
ℓ ⊂ Π2. On the other hand ℓ∩ ℓ4 ⊂ Π2 ∩ ℓ4 = {r2}. Therefore, ℓ= ℓp,r2

.

As in case 2.3.3 we have that 〈P1, P2, P3〉 is a plane not contained in Q. Since the lines
L1,2 = 〈P1, P2〉 and L1,3 = 〈P1, P3〉 are contained in 〈P1, P2, P3〉 ∩Q we conclude that
〈P1, P2, P3〉 ∩Q = L1,2 ∪ L1,3. So P4 6∈ 〈P1, P2, P3〉. Therefore, Λ is a 3-plane. Keeping in mind
that Λ ∩ Q is a quadric surface containing four non coplanar points, {P1} = L1,2 ∩ L1,3 and
L1,4 = 〈P1, P4〉 6⊂ Λ∩Q (since L1,4 6⊂ Q), we conclude that Λ∩Q is a union of two planes or a
nonsingular quadric. Suppose that Λ∩Q is a union of two planes, say Λ∩Q = Λ1∪Λ2. Thus
we can assume that L1,2 ⊂ Λ1. Now, note that L2,3 6⊂ Λ ∩Q (since L2,3 6⊂ Q). So P3 6∈ Λ1,
which implies L1,3 ⊂ Λ2. Finally, observe that L2,4 and L3,4 are not contained in Λ∩Q. Thus
P4 6∈ Λ1 ∪Λ2. Therefore, Λ∩Q is a nonsingular quadric.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 482

Figure 3: Case 2.4.4

Table 6: 2.5 Exactly Two Lines are Coplanar. Assume that ℓ1 and ℓ2 are contained in the plane Π,
ℓ1 ∩ ℓ2 = {p}, Π∩ ℓi = {pi} for i = 3, 4 and p3 6= p4. Let C = {p, p3, p4}.

Subcase Position of the lines S Λ Λ∩Q

2.5.1
p = p3

(or p = p4)
Ωp(〈p,ℓ4〉)

(or Ωp(〈p,ℓ3〉))
3-plane

Union of
two planes

2.5.2
#C = 3 and

p ∈ ℓp3,p4

¦

ℓp3,p4

©

3-plane
Quadric

cone

2.5.3
p 6∈ ℓp3,p4

(so #C = 3)

¦

〈p,ℓ3〉 ∩ 〈p,ℓ4〉,ℓp3,p4

©

3-plane
Nonsingular

quadric

2.5.3 Let ℓ be a solution and consider the following two possibilities.

(i) ℓ ⊂ Π. Since ℓ∩ ℓi ⊂ Π∩ ℓi = {pi} then pi ∈ ℓ for i = 3,4. Therefore ℓ= ℓp3,p4
.

(ii) ℓ 6⊂ Π. Now having in mind that ℓ∩ℓi 6= ; for i = 1,2 and ℓ 6⊂ Πwe concluded that p ∈ ℓ.
On the other hand ℓ∩ ℓi = {qi} ⊂ ℓi ⊂ 〈p,ℓi〉 for i = 3,4. Note that qi 6= p (since p 6∈ ℓi

for i = 3,4) which implies ℓ= ℓp,qi
⊂ 〈p,ℓi〉 for i = 3,4. Therefore, ℓ= 〈p,ℓ3〉 ∩ 〈p,ℓ4〉.

Note that Pi 6∈ L1,2 = 〈P1, P2〉 ⊂ Q for i = 3,4. Thus 〈P1, P2, P3〉 is a plane not contained in
Q since p 6∈ ℓ3 6⊂ Π. Since L1,2 ∪ {P3} ⊂ 〈P1, P2, P3〉 ∩Q, if P4 ∈ 〈P1, P2, P3〉 then P4 ∈ L1,2 or
P4 belong to the component L passing through P3 in the conic 〈P1, P2, P3〉 ∩Q = L1,2 ∪ L. But
P4 6∈ L1,2 because ℓ4 6⊂ Π. Moreover, P4 6∈ L since ℓ4 ∩ ℓ3 = ;. Thus Λ is a 3-plane. Keeping
in mind that Λ ∩Q is a quadric surface containing four non coplanar points and L1,3 is not
contained in Λ∩Q, we conclude that Λ∩Q is a union of two planes or a nonsingular quadric.
Suppose that Λ∩Q is a union of two planes, say Λ∩Q = Λ1 ∪Λ2. Thus we can assume that
L1,2 ⊂ Λ1. Next, note that L2,3 6⊂ Λ∩Q (since L2,3 6⊂ Q). So P3 6∈ Λ1, which implies L1,3 ⊂ Λ2.
Now, since P4 ∈ Λ1 ∪ Λ2 we conclude that L2,4 and L3,4 are contained in Λ ∩ Q which it is
impossible since ℓ4 ∩ ℓi = ;, i = 2,3. Therefore, Λ∩Q is a nonsingular quadric surface.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 483

Figure 4: Case 2.5.3

4.3. No Pair of Lines is Coplanar

Let Q be a nonsingular quadric surface in P3 containing ℓ1, ℓ2 and ℓ3 (as stated in Lemma 2).
Since ℓi ∩ ℓ j = ; for 1 ≤ i < j ≤ 3 then they belong to the same family of lines in Q (cf.
Lemma 1). So we will assume that ℓ1, ℓ2 and ℓ3 belong to the family L .

Note that any solution ℓ ∈ S meets the line ℓi at a point pi ∈ ℓi ⊂ Q for i = 1,2,3. Since
this three points p1, p2 and p3 are different and belong to ℓ ∩Q. Then follows from Proposi-
tion 2 that ℓ ⊂Q, which implies that ℓ ∈M . Therefore any solution belong to the familyM .

In order to determine the set of solutions S and the linear space Λ, we will consider the
following two subcases.

3.1 ℓ4 ⊂Q . In this case ℓ1,ℓ2,ℓ3 and ℓ4 belong to the same family L (since ℓ4 if disjoint to

the others three). Therefore S =M .

Now, since the line Li, j =



Pi , Pj

�

6⊂ Q for 1 ≤ i < j ≤ 3, we conclude from Proposition 6
that



P1, P2, P3

�

is a plane not contained in Q. But, from Remark 1 we conclude that



P1, P2, P3

�

∩Q = P (L ). Consequently, Λ =



P1, P2, P3

�

(keep in mind that P4 ∈ P (L )
too) and Λ∩Q is a nonsingular conic.

3.2 ℓ4 6⊂Q . In this case follows from Proposition 2 that ℓ4 ∩Q is non empty and

ℓ4 ∩Q = {x , y} (with x and y do not necessarily distinct). Let Mx and My be the unique
lines in the familyM passing through x and y , respectively (cf. (4) in Lemma 1). Indeed
Mx and My are solutions (in fact, if p ∈ {x , y} then Mp ∩ ℓi is non empty for i = 1,2,3
because ℓi ∈ L and Mp ∩ ℓ4 = {p}). Now, we will show that Mx and My are the unique
solutions. Let ℓ ∈M be a solution, then ℓ∩ ℓ4 ⊂Q∩ ℓ4 = {x , y} which implies that x ∈ ℓ
or y ∈ ℓ. Therefore follows from (4) in the Lemma 1 that ℓ= Mx or ℓ= My .

Again, since the line Li, j =



Pi, Pj

�

6⊂ Q for 1 ≤ i < j ≤ 3, we conclude that



P1, P2, P3

�

is a plane not contained in Q and



P1, P2, P3

�

∩Q is a nonsingular conic. On the other hand,
the condition ℓ4 6⊂ Q implies that ℓ4 6∈ L . So, we are left to conclude that P4 6∈




P1, P2, P3

�

.
Thus Λ is a 3-plane in P5. One more time, since Λ ∩Q is a quadric surface containing four
non coplanar points and, the line Li, j 6⊂ Q for 1 ≤ i < j ≤ 3, we conclude that Λ ∩ Q is a
quadric cone or a nonsingular quadric surface. In fact, since ℓ4 6⊂Q, it must be either: tangent
or secant to the surface Q.



J. Rojas, R. Mendoza / Eur. J. Pure Appl. Math, 7 (2014), 472-485 484

Assume that ℓ4 is tangent to Q. Let x ∈ ℓ4 ∩Q be the tangent point. Let Mx ∈ M and
Lx ∈ L be the (only) lines in Q passing through x . It is verified that ℓ4 ⊂ TxQ = 〈Mx , Lx〉.
Therefore, ℓ4 ∈ Ωx(TxQ). Set M = P (Mx), L = P (Lx) ∈ Q. Thus L belong to the conic



P1, P2, P3

�

∩Q = P (L ). Moreover M ∈ 〈L, P4〉 ⊂ Λ ∩Q (since, x ∈ ℓ4 ⊂ 〈Mx , Lx〉). On the
other hand the lines Li = 〈M , Pi〉 ⊂ Λ∩Q, i = 1, . . . , 4. Therefore, Λ∩Q is a quadric cone.

Thus we have proved the following Theorem.

Theorem 1. Given four lines ℓ1, ℓ2, ℓ3, ℓ4 in P3. Set

S =
�

ℓ ∈ G1(P
3) | ℓ∩ ℓi 6= ; for 1≤ i ≤ 4

	

.

Let Pi (i = 1, . . . , 4) be the image of ℓi in the Plücker’s quadric Q ⊂ P5 under the Plücker em-

bedding P (in (1)). Set Λ =



P1, . . . , P4

�

be the linear span of those four points in P5. Then the

cardinal of the set S is either 1, 2 or infinite. Moreover we have.

(i) #S = 2 if and only if Λ is a 3-plane and Λ∩Q is a smooth quadric.

(ii) #S = 1 if and only if Λ is a 3-plane and Λ∩Q is a quadric cone.

(iii) If #S =∞ then

(a) S is a line if and only if Λ is a 3-plane and Λ∩Q is a union of two planes.

(b) S is a conic if and only if Λ is a plane and Λ∩Q is a nonsingular conic.

(c) S is a reduced and reducible conic if and only if Λ is a plane and Λ ∩Q is a reduced

and reducible conic.

(d) S is a plane if and only if Λ is a plane in Q.

(e) S is a union of two distinct planes if and only if Λ is a line in Q.

In terms of the lines position Theorem 1 tell us:

#S = 2⇐⇒







































• ∩4
i=1ℓi = ; and any three lines are non coplanar.

(i) ℓ1 ∩ ℓ2 = {p}, ℓ3 ∩ ℓ4 = {q} with p 6= q and p 6∈ Π1 ∩Π2 6∋ q,
if Π1 = 〈ℓ1,ℓ2〉, Π2 = 〈ℓ3,ℓ4〉.

(ii) ℓ1 ∩ ℓ2 = {p}, ℓ1 ∩ ℓ3 = {q} with p 6= q. If Πi = 〈ℓ1,ℓi+1〉
for i = 1,2, then it is verified that Πi ∩ ℓ4 = {ri} i = 1,2
and r1 6= r2.

(iii) ℓ1 ∩ ℓ2 = {p}. If Π = 〈ℓ1,ℓ2〉 then it is verified for i = 3,4
that Π∩ ℓi = {ri} , r3 6= r4 and p 6∈ 〈r3, r4〉.

• No pair of lines is coplanar.

(iv) The non singular quadric surface Q ⊂ P3 containing ℓ1, ℓ2
and ℓ3, meets ℓ4 exactly in two distinct points.



REFERENCES 485

Note that (i), (ii) and (iii) above correspond to subcases 2.3.3, 2.4.4 and 2.5.3 in subsec-
tion 4.2. Already (iv) corresponds to subcase 3.2 in subsection 4.3.

#S = 1⇐⇒

































• ∩4
i=1ℓi = ; and any three lines are non coplanar.

(α) ℓ1 ∩ ℓ2 = {p}, ℓ1 ∩ ℓ3 = {q} with p 6= q. If Πi = 〈ℓ1,ℓi+1〉
for i = 1,2, then it is verified that Πi ∩ ℓ4 = {ri} i = 1,2.
with r1 = r2 and #{p,q, r1}= 3.

(β) ℓ1 ∩ ℓ2 = {p}. If Π = 〈ℓ1,ℓ2〉 then it is verified for i = 3,4
that Π∩ ℓi = {ri}, r3 6= r4, p ∈ 〈r3, r4〉 and p 6= ri .

• No pair of lines is coplanar.

(γ) The non singular quadric surface Q ⊂ P3 containing ℓ1, ℓ2
and ℓ3, meets ℓ4 exactly in one point.

Note that (α) and (β) above correspond to subcases 2.4.3 and 2.5.2 in subsection 4.2.
Already (γ) correspond to subcase 3.2 in subsection 4.3.

ACKNOWLEDGEMENTS The first author wishes to express her gratitude to I. Vainsencher
(DM-UFMG) for a helpful conversation about the subject of this paper.

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[4] P. Griffiths and J. Harris. Principles of algebraic geometry. Wiley-Interscience. 1994.

[5] R. Hartshorne. Algebraic Geometry. Graduate Texts in Math. 52, Springer, New York. 1977.

[6] J. Harris and D. Eisenbud. The Geometry of Schemes. Graduate Texts in Math. 197, Springer,
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Tools, 4(3):11–22. 1999.

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