Notes on Fulton, Section 1.2
Convex Polyhedral Cones
Katherine Christianson
March 9, 2016
1
Fundamental Notions
First, some notation. We will follow the notation of Fulton (see the beginning of Section 1.1 for more information). Given a lattice N in Rn (i.e. a
subset isomorphic as an abelian group to Zn ), we define the vector space
NR to be the subspace of Rn spanned by vectors in N . (The more formal
but equivalent definition given in Fulton is: NR = N ⌦Z R.) We can also
define the dual lattice M = Hom(N, Z). (That is, M consists of homomorphisms from N to Z.) It turns out that M is isomorphic to Zn . The
isomorphism is as follows: given a homomorphism ' : N ! Z, use the
isomorphism N ⇠
= Zn to consider this as a homomorphism Zn ! Z. Then,
let zi be the image of the element of Zn with a 1 in the ith copy of Z and
a 0 in every other copy, and send ' 7! (z1 , . . . , zn ).
Exercise. (Optional.) Check that the map defined above is an isomorphism.
Since M is isomorphic to a lattice, we can also turn it into a vector space
MR just as we did for N . Essentially by its definition, MR is isomorphic to
the dual space (NR )⇤ .
Throughout, we will take V = NR , where N is some lattice. Note in
particular that this means V ⇤ = MR . We denote by h, i the dual pair
corresponding to V and V ⇤ : that is, for any v 2 V and u 2 V ⇤ , we have
hv, ui = u(v) 2 R.
Notice that for any v1 , v2 2 V , we may map v2 by the natural isomorphism
V ! V ⇤ . Suppose that u is the image of v2 under this map. Then, we will
abuse notation and write hv1 , v2 i for hv1 , ui. Under this definition, hv1 , v2 i
is simply the standard inner product on Rn (which V inherits).
Now, we are ready to begin our discussion of convex polyhedral cones.
We’ll start with some definitions.
1
Definition 1. We define a convex polyhedral cone to be a subset of Rn of
the form
= {r1 v1 + · · · + rs vs : ri 2 R, ri 0},
where v1 , . . . , vs 2 V . We say that the vi (or sometimes the rays R 0 vi )
are the generators of . Given a convex polyhedral cone , we define the
dimension dim to be dimension of the linear subspace R· = +( ) ⇢ Rn
and the codimension of to be dim V
dim . We define the dual cone
_
of by
_
= {u 2 V ⇤ : hu, vi 0 for all v 2 V }.
The following theorem is fundamental to the general theory of convex
polytopes (which is much broader than that of convex polyhedral cones),
but it is somewhat tedious to prove.
Theorem 2. Let C be a non-empty open convex set in Rn , and let x
be a point of Rn with x 62 C. Then, there exists a hyperplane H of Rn
which strictly separates x and C: that is, x is in one of the open halfplanes
bounded by H, and C is in the other open halfplane bounded by H.
Proof. See [1, Section2.2, Theorem 1].
This theorem has a corollary which is fundamental to the properties of
convex polyhedral cones.
Corollary 3. Let be a convex polyhedral cone, and let x 2 Rn such that
x 62 . Then, there exists an element u 2 _ such that hx, ui < 0.
Proof. By the Theorem, there exists some hyperplane H which strictly
separates x and . Let n be the vector normal to H which lies in the same
halfplane as . Then, hn, vi 0 for all v 2 (see the Lemma 6 for further
explanation of this fact), so n 2 _ by definition. However, since x lies in
the open halfplane in which n lies, we have that hx, ni < 0.
Corollary 4. Let
be a convex polyhedral cone. Then, (
_ _
) = .
Proof. By definition, every element of has positive inner product with
every element of _ , so ✓ ( _ )_ . Conversely, for any v 2 ( _ )_ , if v 62 ,
then the above corollary gives us a vector u 2 _ such that hv, ui < 0,
which by definition means that v 62 ( _ )_ , a contradiction. So, we must
have v 2 , which means that ( _ )_ ✓ .
We now wish to develop some terminology and machinary to help us
understand the boundaries of convex polyhedral cones.
Definition 5. Given a (convex) set S ⇢ Rn , we say that a hyperplane
H ⇢ Rn is a supporting hyperplane of S if all of S is contained in one of the
halfspaces bounded by H and if H contains at least one boundary point of
S.
2
The following lemma (which is not in Fulton) gives us some crucial
geometric intuition about the relationship between supporting hyperplanes
and dual cones.
Lemma 6. Let
be a convex polyhedral cone. Then, for any x 2 Rn ,
_
x2
if and only if the vector x is orthogonal to a supporting hyperplane
H of Rn passing through the origin and x lies in the same halfspace bounded
by H as does.
Proof. Suppose x 2 _ . Then, notice that the hyperplane H = {y 2 V :
hy, xi = 0} ⇢ V defines two halfspaces: the halfspace R+ in which x lies
has the property that hy, xi 0 for all y in R+ ; and the other halfspace,
R , has the property that hy, xi  0 for all y in R . Since hx, vi 0 for
all v 2 , lies entirely in R+ . Moreover, the origin is a boundary point
of which is contained in H, so H is a supporting hyperplane of , and x
is orthogonal to H.
Conversely, suppose that x 2 Rn is orthogonal to a supporting hyperplane H of Rn . Let R be the halfspace bounded by H which contains x,
and suppose that ⇢ R. Then, as above, every element y in R has the
property that hy, xi
0; in particular, this holds for every element of ,
which implies that x 2 _ by definition.
Definition 7. Given a convex polyhedral cone , we say that ⌧ is a face
of if ⌧ = \ u? = {v 2 : hv, ui = 0} for some u 2 _ . Any face ⌧
of with ⌧ 6= is called a proper face of . If ⌧ is a face of which has
codimension 1 (in the linear subspace generated by ), we say that ⌧ is a
facet of .
Remark. 0 2
itself.
_
always, and
\ 0? = , so that
is always a cone of
Given the above Lemma, for any v 2 _ , v ? is a supporting hyperplane
of . Thus, we can understand the faces of as intersections of supporting
hyperplanes with itself. Notice that, for any linear subspace W of
and any v 2 _ , we must have W ✓ v ? : for any w 2 W , hw, vi
0 by
_
definition of ; but w 2 W as well, and we must also have h w, vi 0,
so this means that hw, vi = 0. Thus, any linear subspace of is contained
in every face of .
Example. Consider the following cone:
y
H
x
3
Here, H is a supporting hyperplane of ; its intersection with , which is
{0}, is a face of . Likewise, the lines generated by the vectors (0, 1) and
(1, 1) are supporting hyperplanes of ; their intersections with , which
are the rays R 0 (0, 1) and R 0 (1, 1) (respectively), are faces of with
codimension 1 and hence are facets of .
The following proposition gives us several important properties of the
faces of convex polyhedral cones.
Proposition 8. Let
generators of , u 2
be a convex polyhedral cone, {v1 , . . . , vr } be a set of
, and ⌧ = \ u? .
_
(1) ⌧ is a convex polyhedral cone.
(2) Any face of ⌧ is a face of .
(3) For any other face ⌧ 0 of , ⌧ \ ⌧ 0 is again a face of .
(4) If ⌧ is a proper face, then ⌧ is contained in some facet of . In fact,
⌧ is the intersection of all facets of which contain it.
(5) If spans V , then the (topological) boundary of is the union of the
proper faces (or equivalently, the proper factes) of .
Proof.
(1) ⌧ consists of linear combinations of the set {v1 , . . . , vr } \ u? with
non-negative coefficients, so that ⌧ is a convex polyhedral cone with
generating set {v1 , . . . , vr } \ u? .
(2) Suppose that = ⌧ \ v ? for some v 2 ⌧ _ . Now, for all vi 2 ⌧ ,
hv, vi i 0. So, define
p = 1 + max
vi 62⌧
|hvi , vi|
.
hvi , ui
By definition of p, hvi , v +pui 0 for all i, which implies that v +pu 2
_
. We show that = \ (v + pu)? , which implies that is a face
of . For any element w of = \ u? \ v ? , we have
hw, v + pui = hw, vi + phw, ui = 0,
so that w 2 \ (v + pu)? . To show the reverse inclusion, it suffices
to check that, for every vi 2 \ (v + pu)? , vi 2 . Notice that, by
definition of p, for any vi 62 ⌧ , we have
✓
◆
|hvi , vi|
hvi , v+pui = hvi , vi+phvi , ui hvi , vi+ 1 +
hvi , ui hvi , ui > 0,
hvi , ui
4
where the final inequality holds because hvi , ui
0 for all i, with
equality if and only if vi 2 ⌧ . So, any vi not in ⌧ cannot be in
(v + pu)? . Suppose vi 2 \ (v + pu)? . Then, vi 2 ⌧ by the arguments
we’ve just given, which in particular means that vi 2 u? . So, we have
0 = hvi , v + pui = hvi , vi + phvi , ui = hvi , vi,
which means that vi 2 v ? . So, vi 2
\ u? \ v ? = .
(3) Suppose ⌧ 0 = \ (u0 )? , where u0 2 _ . We prove that ⌧ \ ⌧ 0 =
\ (u + u0 )? ; then, by linearity of the inner product, u + u0 2 _ , so
we will have that ⌧ \ ⌧ 0 is a face of . Now, ⌧ \ ⌧ 0 = \ u? \ (u0 )? .
For any v 2 \ u? \ (u0 )? , we have
hv, (u + u0 )i = hv, ui + hv, u0 i = 0 + 0 = 0.
This proves that ⌧ \ ⌧ 0 ✓ \ (u + u0 )? . On the other hand, given
any v 2 \ (u + u0 )? , we have
hv, (u + u0 )i = hv, ui + hv, u0 i = 0.
But since v 2 and u, u0 2 _ , hv, ui, hv, u0 i
possible if hv, ui = hv, u0 i = 0, in which case v 2
proves that \ (u + u0 )? ✓ ⌧ \ ⌧ 0 .
0, so this is only
\ u? \ (u0 )? . This
(4) By replacing V with the span of if necessary, we may assume that
spans V . Then, let W be the span of ⌧ , and for every i, let vi denote
the image of vi in V /W . Define H = u? , which is a hyperplane
through the origin, and let H be the image of H in V /W . Then, H is
again a hyperplane through the origin, and H does not contain any
point of the image of except for 0 (if it did, such a point would
correspond to an element of ⌧ in V and hence would be sent to 0 in
V /W ). This means that, given any line L through the origin in V /W ,
we can rotate H about L through some small enough (but nonzero)
angle and again obtain a supporting hyperplane of the image of .
Some such rotation will result in a hyperplane which still supports
the image of but now contains a nonzero vi for some i (intuitively,
we are just rotating H onto the boundary of the image of ); let n
denote the normal vector to this hyperplane which lies in the same
halfplane as the image of . Then, n corresponds to a vector n̂ 2 V
such that n̂ 2 _ , u? ⇢ n̂? , and vi 2 n̂? . So, \ n̂? is a face of
which contains ⌧ ; moreover, since vi 6= 0, vi 62 ⌧ , but vi 2 \ n̂? ,
so this new face has a strictly smaller codimension than that of ⌧ .
Repeating this process, we can eventually obtain a face containing ⌧
with codimension equal to 1, which is then a facet of .
5
Notice that in the above argument, when the codimension of ⌧ is
2, the image of in V /W will be a cone in R2 , so that there are exactly
two rotations of H which will result in the desired rotated hyperplane.
(Intuitively, these correspond to aligning H with either outside edge
of the cone that is the image of .) Following the above argument
with each of these rotated hyperplanes then gives rise to precisely two
facets which contain ⌧ . The intersection of the hyperplanes in V /W
is precisely 0, which means that the intersection of the hyperplanes in
Rn is precisely ⌧ . We then procede by induction on the codimension
of ⌧ . By the above, we can find some facet containing ⌧ . Then, ⌧ is
a face of the cone , so by induction hypothesis, ⌧ is the intersection
of all facets in containing it. These facets are faces of by (2)
above, and they have codimension 2, so by our base case they are
each the intersection of two facets of . Then, the intersection of all
these facets of is precisely ⌧ . By our construction, these are precise
the facets of which contain ⌧ .
(5) Let W be the union of the proper faces of . Recall that the topological boundary of a set S consists of all points of S which are limit
points of some set disjoint from S and are also limit points of S. Now,
for any w 2 W , w lies on a supporting hyperplane of , so it contains
points of which are arbitrarily close to it but are not in (just take
points on the other side of the supporting hyperplane from ). On
the other hand, we can fix any point v in the interior of and take
the line from v to w. By convexity of , this line is entirely contained
in , so it contains points in which are arbitrarily close to w. This
proves that w is in the boundary of .
Conversely, let v be a point in the boundary of . Then, we can
find a sequence {wi } converging to v with wi 62 for all i. For all i,
we can use Corollary 3 to produce some ui 2 _ with hui , wi i < 0.
Since normalizing does not change the sign of inner products, we
may take all of the ui to have norm 1. Then, the ui lie in the sphere
S n , which is compact. Since compact implies sequentially compact
in the metric space Rn , there is a converging subsequence of the ui ;
so, throwing out some of the wi if necessary (note that this will still
produce a subsequence which converges to v), we may take the ui to
converge to some u0 . Since _ is closed and the ui are in _ , their
limit u0 must be in _ . In particular, hv, u0 i 0. If hv, u0 i > 0, then
hx, yi > 0 for all (x, y) in some neighborhood U ⇥ U 0 ⇢ Rn ⇥ Rn of
v ⇥ u0 (since the inner product is continuous). But since the wi and
the ui converge to v and u0 , respectively, there is some i such that
wi 2 U and ui 2 U 0 , so that hwi , ui i > 0, a contradiction. Thus,
hv, ui = 0, and v is in the face \ u?
spans V , the only
0 . Since
6
element orthogonal to all of is 0; the ui are in S n , so u0 is too,
which means u0 6= 0 and \ u?
\ u?
0 is a proper face. Thus, v 2
0
implies v 2 W .
So, we have shown that W is precisely the boundary of . We
finally note that by (4) above, W is also the union of all the facets of
.
Remark.
1. Notice that the proof of (1) in this proposition implies that a convex
polyhedral cone has a finite number of faces (since each is generated
by a subset of the generators of , and there are finitely many such
subsets).
2. A standard induction argument on point 3 of the proposition gives
that any intersection of finitely many faces of is again a face of .
2
Supporting Hyperplanes and Dual Cones
We now wish to give an alternate description of a convex polyhedral cone
in terms of its supporting hyperplanes. Let u 2 _ , and suppose that
⌧ = \ u? is a facet of . If spans V , then ⌧ has codimension 1 in V
itself, which means that there is only one dimension of V in ⌧ ? and hence
that u is unique up to scalar multiplication. In this case, we will write
u⌧ = u for ease of notation. Notice that since spans V , ⌧ spans the
supporting hyperplane u?
⌧ .
Theorem 9. Let be a convex polyhedral cone which spans V . Then, is
the intersection of the halfspaces H⌧ = {v 2 V : hv, u⌧ i 0} as ⌧ ranges
over all of the faces of .
Proof. Lemma 6 tells us that lies in each of the H⌧ and hence in their
intersection. On the other hand, suppose that v is an element of every H⌧
but v 62 . Fix some v 0 in the interior of , and let w be the point on the
line from v 0 to v that is closest to v while still lying in . Then w is on the
boundary of and hence is in some facet ⌧ of by point 5 of Proposition
0
8. So, w lies on the hyperplane u?
⌧ , and v and v lie on either side of this
hyperplane. Since u 2 _ , hu⌧ , v 0 i 0; moreover, v 0 is in the interior of ,
so hu⌧ , v 0 i > 0 (this inner product is only equal to 0 on the boundary of ,
by point 5 of Proposition 8). This implies that hu⌧ , vi < 0, so that v 62 H⌧ ,
a contradiction. So, v must be in , which proves the desired result.
Corollary 10 (Farkas’ Theorem). Let
_
is a convex polyhedral cone.
7
be a convex polyhedral cone. Then,
Proof. If spans V , then we claim that the cone generated by the u⌧ (for
every facet ⌧ of ) is precisely _ . Clearly this cone is contained in _ ,
since all the u⌧ are in _ . Conversely, if u 2 _ is not in the cone generated
by the u⌧ , then Corollary 3 applied to the cone generated by the u⌧ gives
us a vector v such that hv, u⌧ i
0 for all ⌧ (so that v 2 by the above
Theorem) but such that hv, ui < 0. However, this contradicts the fact that
u is in _ . So, the cone generated by the u⌧ must in fact be all of the _ .
If now does not span V , then it spans some linear subspace W = R· .
Then, consider the image _ of _ in V ⇤ /W ? . _ is the dual cone of the
image of in V /W ? , and since spans V /W ? by definition of W , the
above argument tells us that _ is generated by the u⌧ of the cone . So,
_
is generated by vectors whose images are these u⌧ along with {±wi },
where {wi } is a basis of W ? .
This theorem has a couple important consequences. First, we can write
conveniently as the intersection of a finite number of halfspaces:
= {v 2 V : hv, ui i
0 for all i},
where {ui } is a generating set of _ . Secondly, the arguments of the theorem and its corollary give us a practical algorithm for finding the generators of the cone _ . After finding a basis of W ? as required in the
proof of the corollary, we reduce the problem to finding the u⌧ of a convex
polyhedral cone which spans V . To do this, we take a generating set
S = {v1 , . . . , vr } of , and for each linearly independent subset of n 1
vectors T = {vi1 , . . . , vin 1 } ⇢ S, we consider the hyperplane HT generated
by the elements of T . If HT is a supporting hyperplane of , i.e. if there
is a vector uT normal to HT such that huT , vi i 0 for all i, then HT = ⌧
is a facet of , and uT is the u⌧ corresponding to this facet; otherwise, we
throw out uT and move on to the next choice of T .
Example. Consider the following cone:
y
_
x
Since has two linearly independent generators, it spans all of R2 = V .
As discussed in the previous example, the two hyperplanes corresponding
to facets of are the lines generated by (0, 1) and (1, 1). We then find
that (1, 0) and (1, 1) are vectors orthogonal to these lines which dot nonnegatively with all of . So, these two vectors are the u⌧ , and they form a
generating set for _ , as shown in the diagram.
8
Notice that even when spans V , the number of generators of the dual
cone is based on the number of facets of , not its minimal number of generators. This is unlike the behavior of vector spaces, where we always have
an isomorphism between a space and its dual. (For an explicit construction
of a cone whose dual has a di↵erent number of minimal generators, see (7)
in the notes for chapter 1 at the end of Fulton.)
Now that we know that dual cones are, in fact, convex polyhedral cones,
we can relate the faces of a cone to the faces of its dual. To do this, we
require a preliminary definition.
Definition 11. Given a convex polyhedral cone , we say that the relative
interior of is the topological interior of when considered as a subset of
R · (which is given the subspace topology).
Remark. If {v1 , . . . , vr } is a generating
Pset for a convex polyhedral cone
and v is in , then we can write v = ri=1 ai vi , where ai 0 for all i. If
v is in the relative interior of , then all of the ai must be positive.
Theorem 12. Given a convex polyhedral cone
order-reversing correspondence
{faces of }
⌧
! {faces of
! ⌧⇤ =
_
, there is a one-to-one,
_
}.
\ ⌧?
In particular, for any face ⌧ of , dim ⌧ + dim ⌧ 0 = dim V , and the smallest
face of is \ ( ).
Proof. We first have to check that this correspondence makes sense, i.e.
that ⌧ ⇤ is a face of _ . Given any face ⌧ of , ⌧ contains some point v 2
in its relative interior. Notice that v 2 = ( _ )_ , so that F = _ \ v ? is
a face of _ . We claim that F is equal to _ \ ⌧ ? = ⌧ ⇤ . Given an element
u 2 _ \ ⌧ ? , we have that hu, ti = 0 for all t 2 ⌧ . In particular, hu, vi = 0,
which proves that u 2 F . Conversely, take any u
P2 F . Now, if {v1 , . . . , vr }
is a generating set of ⌧ , then we can write v = ri=1 ai vi , where ai > 0 for
all i (since v is in the relative interior of ⌧ ). Then, we have
hu, vi =
r
X
i=1
ai hu, vi i = 0.
Now, u 2 _ , so hu, vi i 0 for all i. Since the ai are all strictly positive,
these inner products must be 0, which means that hu, xi = 0 for all x 2 ⌧
and hence that u 2 ⌧ ? . Thus, u 2 _ \ ⌧ ? . This proves that ⌧ ⇤ = F is
a face of _ . So, the correspondence given in the problem is well-defined.
9
One easily sees that it is order-reversing: if ⌧ ⇢ ⌧ 0 are two faces of , then
⌧ ? (⌧ 0 )? , which implies that ⌧ ⇤ (⌧ 0 )⇤ .
Now, let {v1 , . . . , vr } be a generating set of . Then, our above algorithm for creating a generating set of _ tells us that we can take a
generating set of the form U [ W , where U = {u1 , . . . , us } is the set of all
the u⌧ for some facet ⌧ of and W = {w1 , . . . , wt } is a basis for the orthogonal complement of the linear subspace R · . Suppose that ⌧ and ⌧ 0 are
two faces of with ⌧ ⇤ = (⌧ 0 )⇤ . Then, ⌧ and ⌧ 0 are intersections ofPfacets of
, so by the proof
of (3) in Proposition 8, we can write ⌧ = \ ( i2I ui )?
P
and ⌧ 0 = \ ( i2I 0 ui )? for some subsets I, I 0 ⇢ {1, . . . , s}. Now, consider
⌧ ⇤ = _ \ ⌧ ? . ⌧ ⇤ is generated by (U [ W ) \ ⌧ ? ; we claim that this generating set is precisely {ui }i2I [ W . The elements of W are in _ and are
orthogonal to all of and hence are in particular in ⌧ ? ; moreover, for any
i 2 I, ui is in ⌧ ? , and ui is an element of _ by definition. This proves
that {ui }i2I [ W ⇢ (U [ W ) \ ⌧ ? . On the other hand, given any generator
uj of _ such that uj 2 ⌧ ? , uj is orthogonal to all the elements of ⌧ , so
the facet of to which uj is normal contains ⌧ . Since ⌧ is an intersection
of the facets that contain it, this means that uj 2 {ui }i2I by definition.
So, we have that {ui }i2I [ W is a generating set for the face ⌧ ⇤ . Likewise,
{ui }i2I 0 [ W is a generating set for (⌧ 0 )⇤ . But by assumption, ⌧ = ⌧ ⇤ , so
they mustP
have the same setP
of generators; thus, I = I 0 , which implies that
?
⌧ = \ ( i2I ui ) = \ ( i2I 0 ui )? = ⌧ 0 . This proves injectivity of the
map ⌧ 7! ⌧ ⇤ .
For any face ⌧ of , we have (⌧ ⇤ )⇤ = \ ( _ \ ⌧ ? )? . Every element
of ⌧ is orthogonal to every element of _ \ ⌧ ? ⇢ ⌧ ? , and ⌧ ⇢ , so
we have ⌧ ⇢ (⌧ ⇤ )⇤ . Since our map is order-reversing, this implies that
⌧⇤
((⌧ ⇤ )⇤ )⇤ . On the other hand, ((⌧ ⇤ )⇤ )⇤ = _ \ ( \ (⌧ ⇤ )? )? . Now,
⌧ ⇤ ⇢ _ , and it is clearly orthogonal to every element of \ (⌧ ⇤ )? ⇢ (⌧ ⇤ )? ,
so ⌧ ⇤ ⇢ ((⌧ ⇤ )⇤ )⇤ , which implies that ⌧ ⇤ = ((⌧ ⇤ )⇤ )⇤ . By injectivity, we then
have that ⌧ = (⌧ ⇤ )⇤ . If we now take ⌧ to be a face of _ , then this statement
still holds (this amounts to swapping and _ , which does not a↵ect the
above arguments). So, ⌧ ⇤ is a face of which maps to the face ⌧ under
our correspondence, which proves that the map F 7! F ⇤ is surjective and
hence a bijection.
Because the correspondence is bijective and order-preserving, the smallest face of must correspond to the largest face of _ , namely _ itself.
Thus, the smallest face of is ( _ )⇤ = ( _ )_ \ ( _ )? = ( _ )? . We claim
that ( _ )? = \ ( ). Any v 2 ( _ )? is by definition in ( _ )_ =
and also in (( )_ )_ =
, so that v 2 \ ( ). Conversely, given any
v 2 \ ( ), we have for any element u of _ that hu, vi 0 (since v 2 )
and hu, vi = hu, vi 0 (since v 2 ). Thus, hu, vi = 0 for all u 2 _ ,
which means that v 2 ( _ )? .
For the smallest face of , ( _ )⇤ = ( _ )? , we have by the above that
10
dim(( _ )? ) + dim( _ ) = dim V . For a general face ⌧ of
some maximal chain
⌧0 * ⌧1 * · · · * ⌧` = ⌧
, there exists
of faces of . Taking dual faces, we get a chain
⌧ ⇤ = ⌧`⇤ * · · · * ⌧1⇤ * ⌧0⇤ ,
which must also be maximal (if we could add any face to this chain somewhere, its dual could be added to the original chain, which would contradict
maximality of that chain). Recall from our definition of the dual face that
⌧i⇤ is formed out of ⌧i? . Thus, for every linearly independent generator we
⇤
?
add to ⌧i⇤ to get ⌧i+1
, that generator is in ⌧i? but not in ⌧i+1
; so, for each
dimension we gain between ⌧0 and ⌧ in the chain, we lose precisely one
dimension from ⌧ ⇤ to ⌧0⇤ . Moreover, notice that ⌧0⇤ must be _ , or else we
could add this cone to the chain and violate maximality. So, we have that
dim(⌧ ) + dim(⌧ ⇤ ) = dim(⌧0 ) + dim(⌧0⇤ ) = dim(
_
) + dim((
_ ?
) ) = dim V.
Using the correspondence given in the above theorem, we can prove a
stronger version of Corollary 3.
Lemma 13 (Separation Lemma). Let and 0 be convex polyhedral cones,
and suppose that ⌧ = \ 0 is a face of both and 0 . Then, there is some
u 2 _ \ ( 0 )_ such that
⌧=
\ u? =
0
\ u? .
Proof. Define = + ( 0 ). Notice that is a convex polyhedral cone.
Fix any u in the relative interior of _ ; then, by Theorem 12, ( _ )⇤ = \u?
is the smallest face of , and
\ u? =
\(
\ u? ⇢
\ u? =
)=(
0
)\(
0
).
We show that u has the desired properties. Since ⇢ , we have u 2 _ ⇢
_
. Moreover, ⌧ is in , hence in , and likewise ⌧ is in 0 and hence in
. So, ⌧ ⇢ \ ( ) = \ u? ⇢ u? , which means that ⌧ ⇢ \ u? .
Conversely, given any element v 2 \ u? , we have
v2
\(
)⇢
=
0
,
so that v = w0 w for some w0 2 0 and w 2 . So, v + w = w0 2 0 , and
since v 2 , v + w 2 , which implies that v + w 2 \ 0 = ⌧ . However,
one can prove that the sum of two elements of a cone are in a face if and
only if the summands are in a face (this is part of one of the exercises at
the end of these notes), so in fact v is in ⌧ . This proves that ⌧ = \ u? .
An identical argument but replacing u with u shows that u 2 ( 0 )_ (so
that u 2 ( 0 )_ ) and that ⌧ = 0 \ ( u)? = 0 \ u? .
11
3
Rational and Strongly Convex Cones
For our purposes, we will generally be interested in convex polyhedral cones
which are generated by lattice elements. To this end, we have the following
definition:
Definition 14. Let be a convex polyhedral cone. Then, we say that
is a rational convex polyhedral cone if it has a generating set {v1 , . . . , vr }
with vi 2 N for all i.
Notice that if the vi are all in N in the above algorithm to find a
generating set of _ , then the generating set of _ will also be in M . This
shows that _ is rational if is.
The following proposition is very important for the application of convex
polyhedral cones to toric varieties.
Proposition 15 (Gordan’s Lemma). Let be a rational convex polyhedral
cone. Then, S = _ \ M is a finitely generated semigroup.
Proof. Recall that a semigroup is a set with an associative binary operation.
Clearly S is a semigroup, with the operation given by addition. Since
is rational, we can take a generatingPset {u1 , . . . , ur } for _ such that
ui 2 _ \ M for all i. Define K = { ri=1 ti ui : 0  ti  1}. Since K is
a closed and bounded subset of V ⇤ ⇠
= Rn , it is compact, and since M is
discrete, we have that K \M is finite. (To see this, notice that M and K are
closed, hence so is K \ M . So, K \ M is compact. If this intersection were
infinite, then by compactness there would be a limit point x of any infinite
sequence of elements. But then x is an element of the discrete set K \ M
all of whose neighborhoods contain other points of K \ M , contradicting
discreteness.)
u =
Pr We claim that K \ M generates S . Let u 2 S , and write
_
0 for all i (we can do this because u 2 ). Then,
i=1 ai ui , where ai
write ai = zi + ti , where zi 2 Z 0 and 0  ti  1. We now have
u=
r
X
z i ui +
i=1
r
X
t i ui .
i=1
The former sum here is in K \M since each ui is. The latter sum is in K by
definition and in M since u and the former sum are (and since M is closed
under subtraction). Thus, we have written u as a sum of two elements of
K \ M , which proves that this set generates S .
We would like to apply what we know about the faces of a convex
polyhedral cone
to the rational case and in particular to understand
the relationships of S and S⌧ for some face ⌧ of . First, we require a
straightforward lemma.
12
Lemma 16. Let be a convex polyhedral cone, u 2
Then, ⌧ _ = _ + R 0 ( u).
_
, and ⌧ =
\ u? .
Proof. Since both ⌧ _ and _ + R 0 ( u) are convex polyhedral cones, it
suffices to check that their duals are equal. (⌧ _ )_ = ⌧ ; meanwhile,
(
_
+ R 0 ( u))_ =
\ ( u)_ .
(Clearly the right-hand side is a subset of the left-hand side. On the other
hand, anything in the left-hand side must in particular dot positively with
every element of _ and every element of R 0 ( u).) Now, any element v 2
has hv, ui 0, since u 2 _ . Thus, v 2 ( u)_ if and only if hv, ui 0 if
and only if hv, ui = 0. This proves that in fact \ ( u)_ = \ u? = ⌧ .
Proposition 17. Let be a rational convex polyhedral cone, and let u 2
S = _ \ M . Then, ⌧ = \ u? is a rational convex polyhedral cone. ⌧ is
clearly also a face of ; in fact, all faces of are of the form of ⌧ , and
S⌧ ⇢ S + Z 0 ( u).
Proof. If ⌧ is any face of , then by the arguments in the proof of Theorem
12, ⌧ = \ u? for any u in the relative interior of ⌧ ⇤ = _ \ ⌧ ? . Now, _ is
rational, hence ⌧ ⇤ is, too (since it is generated by a subset of the generators
of _ ). So, we can take u to be in M (for instance, set u to be a sum of
some subset of the rational generators of ⌧ ⇤ ), which means that ⌧ is of the
desired form. Moreover, a subset of the rational generators of generate
⌧ , so ⌧ is rational.
Let v 2 S⌧ = ⌧ _ \ M . Using the same trick as in the proof of (2) in
Proposition 8, we have that v+pu is in _ for large enough p. Since v+pu 2
M as well, this implies that v + pu 2 S , whence v 2 S + Z 0 ( u).
Using this proposition, we can now understand what S⌧ looks like when
the face ⌧ is the intersection of two rational cones.
Theorem 18. Let and 0 be rational convex polyhedral cones, and suppose ⌧ = \ 0 is a face of both and 0 . Then,
S⌧ = S + S 0 .
_
Proof. Clearly S + S 0 ⇢ S⌧ : ⌧ _
, ( 0 )_ implies that S , S 0 ⇢ S⌧
(and hence so is any sum of elements from these). On the other hand,
by the proof of Lemma 13, we can find u 2 _ \ ( 0 )_ \ M such that
⌧ = \ u? = 0 \ u? . Then, using Proposition 17 along with the fact that
u 2 S 0 , we have
S⌧ ⇢ S + Z 0 ( u) ⇢ S + S 0 ,
as desired.
13
We end by discussing one more important type of convex polyhedral
cone.
Definition 19. Let
be a convex polyhedral cone. We say that
strongly convex if \ ( ) = {0}.
is
Using the above results, we can easily prove several equivalent conditions to strong convexity.
Proposition 20. Let
are equivalent:
be a convex polyhedral cone. Then, the following
(1)
is strongly convex;
(2)
contains no nonzero linear subspaces of V ;
(3) there exists some u 2
(4)
_
_
such that
\ u? = {0}; and
spans V ⇤ .
Proof. (1) and (2) are equivalent because \ ( ) is the largest linear
subspace of . (1) and (3) are equivalent because \ ( ) is the smallest
face of . (1) and (4) are equivalent because, by Proposition 12, dim( \
( )) + dim( _ ) = dim V = dim V ⇤ .
Any convex polyhedral cone is generated by a minimal set of generators S. If is strongly convex, then for any v 2 S, we can apply Corollary
3 to the cone generated by the set S \ {v} to get a vector w such that
hw, v 0 i 0 for all v 0 2 S \ {v} and hw, vi < 0. (Notice that v cannot be in
the cone generated by S \ {v}, since otherwise this cone would be all of ,
in which case we could remove v from S to get a strictly smaller generating
set for .) Then, by (3) of the above proposition, there exists some u 2 _
such that \ u? = {0}. Define
w0 = w
hw, vi
u.
hu, vi
Then, for any v 0 2 S \ {v}, hw0 , v 0 i > 0 (since hw, v 0 i 0 and hu, v 0 i > 0),
and
hw, vi
hw0 , vi = hw, vi
hu, vi = 0.
hu, vi
So, \ w? is a face of which is precisely the ray generated by v. Clearly
any one-dimensional face of is of this form, so the one-dimensional faces of
are precisely the rays generated by each element of the minimal generating
set S.
This fact allows us to identify the facets of many typical cones in 2 dimensions simply by inspection: we simply find a minimal set of generators,
and then each of these generates one facet of the cone. We can then easily
compute vectors normal to each of these generators and so obtain a basis
for _ (since these normal vectors are, up to sign, the u⌧ ).
14
4
Exercises
Both exercises are drawn from the end of Section 1.2 of Fulton.
Exercise. Let be a convex polyhedral cone and v be an element of .
Show that the following are equivalent:
(1) v is in the relative interior of ;
(2) hv, ui > 0 for all u 2
(3)
(4)
_
\ v? =
?
_
\
?
;
;
+ R 0 ( v) = R · ; and
(5) for all x 2 , there exists some p > 0 and some y in
pv = x + y.
such that
Exercise. Let ⌧ be a face of a convex polyhedral cone . Show that the
sum of two vectors in is in ⌧ if and only if both of the summands are in ⌧ .
Conversely, show that any convex subset of which satisfies this property
is a face of .
References
[1] Branko Grünbaum. Convex polytopes, volume 221 of Graduate Texts in
Mathematics. Springer-Verlag, New York, second edition, 2003. Prepared and with a preface by Volker Kaibel, Victor Klee and Günter M.
Ziegler.
15