M231: Exercise sheet 8
Subspaces in quadrics
1. Let S be a non-singular 2n-dimensional quadric in a complex projective space. Show
that S contains a projective subspace of dimension n.
Hint: Show that there is a basis v0 , . . . , v2n+1 with respect to which the quadric is
given by λ0 λ1 + λ2 λ3 + · · · + λ2n λ2n+1 = 0 and contemplate the span of the v2i .
2. Let S ⊂ P(R4 ) be a real, non-singular quadric surface. For any A ∈ S, we know that
A⊥ ∩ S is a line-pair or a point. Suppose that A⊥ ∩ S is a line-pair.
Show that, for all B ∈ S, B ⊥ ∩ S is a line-pair also.
Deduce that if A⊥ ∩ S is a point, so is each B ⊥ ∩ S.
Hint: Think about how S intersects the line A⊥ ∩ B ⊥ .
On pencils
3. Let A1 , A2 , A3 , A4 be four points in general position in a projective plane and let X be
a distinct fifth point.
(a) Show that there is a unique conic containing A1 , A2 , A3 , A4 , X.
(b) Show that conic is non-singular if and only if no three of these five points are
collinear.
4. Find all of the conics which pass through the points [1, 0, 0], [1, 0, 1], [1, 1, 0], and [1, 1, 1]
in P (R3 ). What are the singular conics in this pencil?
Hint: Answer the second part first!
5. Show that the set of conics tangent to the line λ0 − λ1 + λ2 = 0 at [1, 1, 0] and to the
line λ1 + λ2 = 0 at [1, 0, 0] is a pencil.
What are the singular conics in this pencil?
What is the base locus of the pencil?
Further questions
6. Let Q0 , Q1 be quadratic forms on a complex vector space V . Show that the corresponding quadrics in P(V ) coincide if and only if Q1 = λQ0 , for some λ ∈ C \ {0}.
Hint: Deal with the case dim P(V ) = 1 first.
7. Let P(V ) be a complex projective line.
(a) Show that the space P(S 2 V ∗ ) of quadrics in P(V ) is a projective plane.
(b) Show that the set of singular quadrics is a non-singular conic in P(S 2 V ∗ ).
(c) Deduce the existence of a bijection from P(V ) to a non-singular conic.
[This is yet another proof of the result proved in lectures that a non-singular conic
is bijective with a projective line.]
April 25, 2017
Home page: http://go.bath.ac.uk/ma30231
M231: Exercise sheet 8—Solutions
1. Since the quadric is non-singular, there is a basis w0 , . . . , w2n+1 of the underlying 2n + 2dimensional vector space with respect to which the symmetric bilinear form B defining the
quadric is given by
X
X
X
B(
ai wi ,
bi wi ) =
ai bi
√
√
Now set v2i = w2i − −1w2i+1 , v2i+1 = w2i + −1w2i+1 , 0 ≤ i ≤ n and note that B(v2i , v2j ) =
0 for 0 ≤ i, j ≤ n so that span{v2i } is an n-dimensional subspace of the quadric.
2. Let B 6= A ∈ S,A⊥ ∩ S = L1 ∪ L2 and L = A⊥ ∩ B ⊥ which is a line. Since the Li and L lie
in the plane A⊥ , they must intersect at points that lie in A⊥ ∩ B ⊥ ∩ S and there are two
possibilities. First, if L intersects the Li at separate points X1 , X2 , then the Xi are two points
in B ⊥ ∩ S so that the latter contains more than one point and so must be a line-pair. The
only other possibility is for the Li to intersect L at A. Then A ∈ B ⊥ ∩ S and now B ⊥ ∩ S
contains the two distinct points A, B and that again B ⊥ ∩ S is a line-pair.
We have shown that as soon as one tangent plane intersects S in a line-pair, all tangent planes
intersect S that way. So, as soon as one tangent plane intersects in a point, they all do.
3. We know from lectures that the conics through the Ai form a pencil {δ0 Q0 + δ1 Q1 } and that
the intersection of any two of these conics is exactly the Ai . With this in mind:
(a) If X = [v] is distinct from the Ai , at least one of Q0 (v), Q1 (v) 6= 0 (else X is in the
intersection of the corresponding C0 and C1 and so is one of the Ai ). Now X lies in the
conic with quadratic form Q1 (v)Q0 − Q0 (v)Q1 and this conic is unique since another
would intersect this one at X, forcing X to be one of the Ai .
(b) If the conic is non-singular then no three points on it can be collinear since no line intersects
such a conic in more than two points. Since A1 , . . . , A4 , X all lie on the conic, no three
are collinear.
Conversely, from lectures, we know that the singular conics in the pencil are the line pairs
through the Ai and if X lies on such a line-pair, then X is collinear with two of the Ai .
4. These four points are in general position so the conics passing through them form a pencil and
the line pairs through these points are the singular ones. So we compute the lines through
these points: the line through [1, 0, 0] and [1, 0, 1] is λ1 = 0 while that through [1, 1, 0] and
[1, 1, 1] is λ0 − λ1 = 0 so that the line-pair is λ1 (λ0 − λ1 ) = 0. Again, the line through [1, 0, 0]
and [1, 1, 0] is λ2 = 0 and that through [1, 0, 1] and [1, 1, 1] λ0 − λ2 so that the line-pair is given
by λ2 (λ0 − λ2 ) = 0. We now know enough to describe the whole pencil: they are the quadrics
of the form
δ0 λ1 (λ0 − λ1 ) + δ1 λ2 (λ0 − λ2 ) = 0
and we readily check that the last singular quadratic is given at [δ0 , δ1 ] = [1, −1] by (λ1 −
λ2 )(λ0 − λ1 − λ2 ) = 0.
P
5. Write such a conic C as Q =
i,j βij λi λj = 0. Then [1, 0, 0] ∈ C means β00 = 0 while
[1, 1, 0] ∈ C forces β11 + 2β01 = 0. Thus our quadratic form has matrix


0
β01
β02
P = β01 −2β01 β12 
β02
β12
β22
so that the tangent line at [1, 0, 0] is given by (1, 0, 0)P (λ0 , λ1 , λ2 )T = 0, that is, β01 λ1 +β02 λ2 =
0. Since this must be the line λ1 + λ2 = 0, we get β01 = β02 . Similarly, the tangent at [1, 1, 0]
is given by β01 λ0 − β01 λ1 + (β01 + β12 )λ2 = 0. Since this is supposed to be the line given by
λ0 − λ1 + λ2 = 0 we conclude that β12 = 0. It follows that our conic is of the form
β01 (−2λ21 + 2λ0 λ1 + 2λ0 λ2 ) + β22 λ22 = 0,
for β01 , β22 not both zero. Conversely, any conic of this form is tangent at [1, 0, 0] and [1, 1, 0]
to the desired lines, as we can easily check. Thus the family of conics under consideration is a
pencil.
As to the singular quadrics in the pencil: at β01 = 0 we get the double line λ22 = 0. To find the
other, we contemplate the determinant for an arbitrary quadratic form in the family:
0
0
β01
β01 1
1 2 2
β01 −2β01
0 = β01
1 −2 0 = β01 (−2β01 + β22 ).
β01
β01 0 β22 0
β22 Thus there is one more singular conic when β22 = 2β01 : this has quadratic form 2λ22 − 2λ21 +
2λ0 λ1 + 2λ0 λ2 = (λ1 + λ2 )(λ0 − λ1 + λ2 ) which we recognise as the line-pair consisting of our
two tangent lines.
Finally the base locus is the intersection of our double line and line-pair and so consists of two
points which must be the original [1, 0, 0] and [1, 1, 0] that we started with.
6. First suppose that dim P(V ) = 1: in this case, a quadric either consists of two points [v] and
[w], say, or is a singular quadric [u]. In the first case, v, w are a basis of V and any symmetric
bilinear form giving rise to this quadric has B(v, v) = B(w, w) = 0 and B(v, w) 6= 0. These
form a 1-dimensional subspace in the space of such forms and we are done.
If the quadric is singular, extend u to a basis u0 . Then B(u, u) = B(u, u0 ) = 0 while B(u0 , u0 ) 6= 0
so again the symmetric bilinear forms giving this quadric form a 1-dimensional subspace.
Finally, let dim V be arbitrary and suppose that Q0 and Q1 are not scalar multiples of each
other. We show that the corresponding quadrics S0 , S1 are distinct. By hypothesis, there are
v, w ∈ V so that Q0 (v)Q1 (w) 6= Q0 (w)Q1 (v). Now intersect the Si with the line P(U ) through
[v] and [w]: we have that the Qi|U are not scalar multiples of each other so that, by the first
part, S1 ∩ L 6= S2 ∩ L, whence S1 6= S2 .
7.
(a) dim S 2 V ∗ = (1 + 1)(1 + 2)/2 = 3 so that P(S 2 V ∗ ) is a projective plane.
(b) Use a basis v0 , v1 of V to identify a symmetric bilinear form with the symmetric matrix
2
βij = B(vi , vj ). Then the degenerate B are the zeros of det B = β00 β11 − β01
which is a
2 ∗
non-degenerate quadratic form on S V . Thus the degenerate quadrics constitute a conic
in the space of all quadrics on a projective line!
(c) The singular conics are precisely the double-points so that the map that sends a point [v]
to the singular conic {[v]} is a bijection.