text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
1 + tv1
y = p2 + tv2
and
z = p3 + tv3.
Example 1.7. If P = (1, −2, 3) and Q = (1, 0, −1), then �v = (0, 2, −4)
and a general point of the line containing P and Q is given parametri
cally by
(x, y, z) = (1, −2, 3) + t(0, 2, −4) = (1, −2 + 2t, 3 − 4t).
Example 1.8. Where do the two lines l1 and l2
(x, y, z) = (1, −2 + 2t, 3 − 4t)
and
(x, y, z) = (2t − 1, −3 + t, 3t),
intersect?
We are looking for a point (x, y, z) common to both lines. So we have
(1, −2 + 2s, 3 − 4s) = (2t − 1, −3 + t, 3t).
Looking at the first component, we must have t = 1. Looking at the
second component, we must have −2 + 2s = −2, so that s = 0. By
inspection, the third component comes out equal to 3 in both cases. So
the lines intersect at the point (1, −2, 3).
Example 1.9. Where does the line
(x, y, z) = (1 − t, 2 − 3t, 2t + 1)
intersect the plane
We must have
2x − 3y + z = 6?
2(1 − t) − 3(2 − 3t) + (2t + 1) = 6.
Solving for t we get
9t − 3 = 6,
so that t = 1. The line intersects the plane at the point
(x, y, | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
t − 3 = 6,
so that t = 1. The line intersects the plane at the point
(x, y, z) = (0, −1, 3).
3
Example 1.10. A cycloid is the path traced in the plane, by a point
on the circumference of a circle as the circle rolls along the ground.
Let’s find the parametric form of a cycloid. Let’s suppose that the
circle has radius a, the circle rolls along the x-axis and the point is at
the origin at time t = 0. We suppose that the cylinder rotates through
an angle of t radians in time t. So the circumference travels a distance
of at. It follows that the centre of the circle at time t is at the point
P = (at, a). Call the point on the circumference Q = (x, y) and let O
be the centre of coordinates. We have
(x, y) =
−→
OQ =
−→
OP +
−→
P Q.
Now relative to P , the point Q just goes around a circle of radius a.
Note that the circle rotates backwards and at time t = 0, the angle
3π/2. So we have
−→
P Q = (a cos(3π/2 − t), a sin(3π/2 − t)) = (−a sin t, −a cos t)
Putting all of this together, we have
(x, y) = (at − a sin t, a − a cos t).
4
MIT OpenCourseWare
http://ocw.mit.edu
18.022 Calculus of Several Variables
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
of Several Variables
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
LECTURE 4
Broken circuits, modular elements, and supersolvability
This lecture is concerned primarily with matroids and geometric lattices. Since
the intersection lattice of a central arrangement is a geometric lattice, all our results
can be applied to arrangements.
4.1. Broken circuits
y in L, we have seen (Theorem 3.10) that
For any geometric lattice L and x
1)rk(x,y)µ(x, y) is a positive integer. It is thus natural to ask whether this integer
(
−
has a direct combinatorial interpretation. To this end, let M be a matroid on the
< um.
set S =
Recall that a circuit of M is a minimal dependent subset of S.
. Linearly order the elements of S, say u1 < u2 <
u1, . . . , u
{
m}
· · ·
→
Definition 4.10. A broken circuit of M (with respect to the linear ordering O of
, where C is a circuit and u is the largest element of C (in the
S) is a set C
ordering O). The broken circuit complex BCO(M ) (or just BC(M ) if no confusion
will arise) is defined by
− {
u
}
BC(M ) =
S : T contains no broken circuit
T
{
∗
.
}
Figure 1 shows two linear orderings O and O� of the points of the affine matroid
M of Figure 1 (where the ordering of the points is 1 < 2 < 3 < 4 < 5). With respect
to the first ordering O the circuits are 123, 345, 1245, and the broken circuits are
12, 34, 124. With respect to the second ordering O� the circuits are 123, 145, 2345,
and the broken circuits are 12, 14, 234.
It is clear that the broken circuit complex | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
and the broken circuits are 12, 14, 234.
It is clear that the broken circuit complex BC(M ) is an abstract simplicial
BC(M ). In Figure 1 we
BC(M ) and U
T , then U
complex, i.e., if T
≤
∗
≤
1
5
3
5
2
4
2
4
3
1
Figure 1. Two linear orderings of the matroid M of Figure 1
41
42
R. STANLEY, HYPERPLANE ARRANGEMENTS
have BCO(M ) = 135, 145, 235, 245 , while BCO� (M ) = 135, 235, 245, 345 . These
simplicial complexes have geometric realizations as follows:
�
�
�
�
3
5
1
4
2
1
5
2
4
3
Note that the two simplicial complexes BCO(M ) and BCO� (M ) are not iso
morphic (as abstract simplicial complexes); in fact, their geometric realizations are
not even homeomorphic. On the other hand, if fi(�) denotes the number of i-
1) of the abstract simplicial complex
dimensional faces (or faces of cardinality i
�, then for � given by either BCO(M ) or BCO� (M ) we have
−
f−1(�) = 1, f0(�) = 5, f1(�) = 8, f2(�) = 4.
Note, moreover, that
ψM (t) = t3
5t2 + 8t
4.
−
In order to generalize this observation to arbitrary matroids, we need to introduce
a fair amount of machinery, much of it of interest for its own sake. First we give
a fundamental formula, | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
much of it of interest for its own sake. First we give
a fundamental formula, known as Philip Hall’s theorem, for the M¨obius function
value µ(ˆ ˆ
0, 1).
−
Lemma 4.4. Let P be a finite poset with ˆ
Let ci denote the number of chains ˆ0 = y0 < y1 <
0 and ˆ
1, and with M¨
< y = ˆ
i
1 in P . Then
obius function µ.
µ(ˆ ˆ0, 1) =
c1 + c2
−
−
· · ·
c3 +
.
· · ·
Proof. We work in the incidence algebra I(P ). We have
µ(ˆ ˆ
0, 1) = α
0, 1)
−1(ˆ ˆ
= (ζ + (α
= ζ(ˆ ˆ
0, 1)
−
ζ))−1(ˆ0, 1)
ˆ
ζ)(ˆ ˆ
(α
0, 1) + (α
ζ)2(ˆ
0, ˆ1)
.
−
−
This expansion is easily justified since (α
length less than k. By definition of the product in I(P ) we have (α
and the proof follows.
− · · ·
0, 1) = 0 if the longest chain of P has
ˆ0, 1) = ci,
�
ˆ 1
0, ˆ . Define
}
�(P �) to be the set of chains of P �, so �(P �) is an abstract simplicial complex. The
reduced Euler characteristic of a simplicial complex � is defined by
Note. Let P be a finite poset with ˆ | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
ial complex � is defined by
Note. Let P be a finite poset with ˆ
1, and let P � = P
0 and ˆ
ζ)k (ˆ ˆ
ζ)i(ˆ
− {
−
−
−
−
where fi is the number of i-dimensional faces F
with Lemma 4.4 shows that
−
ψ˜(P ) =
f−1 + f0
f1 +
,
· · ·
� (or #F = i + 1). Comparing
≤
µ(ˆ ˆ
0, 1) = ˜(�(P �)).
ψ
Readers familiar with topology will know that ˜(�) has important topological sig-
nificance related to the homology of �. It is thus natural to ask whether results
ψ
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
43
3 2
1
312
1
1
3
2
2
3
1 2
3
1
1
1
2
1
3
2
1
1
2
1
2
(a)
(b)
(c)
Figure 2. Three examples of edge-labelings
concerning M¨obius functions can be generalized or refined topologically. Such re
sults are part of the subject of “topological combinatorics,” about which we will
say a little more later.
Now let P be a finite graded poset with ˆ
0 and ˆ
1. Let
(x, y) : x � y in P
,
}
{
the set of (directed) edges of the Hasse diagram of P .
E(P ) =
Definition 4.11. An E-labeling of P is a map ϕ : E(P )
P then | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
finition 4.11. An E-labeling of P is a map ϕ : E(P )
P then there exists a unique saturated chain
∃
P such that if x < y in
C : x = x0 � x1 � x1 �
satisfying
→
We call C the increasing chain from x to y.
→
· · ·
ϕ(x0, x1)
ϕ(x1, x2)
� xk = y
ϕ(xk−1 , xk ).
· · ·
→
∅
∃
) = i. (The one-element subsets
i
}
{
Figure 2 shows three examples of posets P with a labeling of their edges, i.e.
a map ϕ : E(P )
P. Figure 2(a) is the boolean algebra B3 with the labeling
ϕ(S, S
are also labelled with a small
i.) For any boolean algebra Bn, this labeling is the archetypal example of an E-
labeling. The unique increasing chain from S to T is obtained by adjoining to S
the elements of T
S one at a time in increasing order. Figures 2(b) and (c) show
two different E-labelings of the same poset P . These labelings have a number of
different properties, e.g., the first has a chain whose edge labels are not all different,
while every maximal chain label of Figure 2(c) is a permutation of
i
}
{
−
Theorem 4.11. Let ϕ be an E-labeling of P , and let x
the M¨obius function of P . Then (
decreasing saturated chains from x to y, i.e.,
−
.
} | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
ius function of P . Then (
decreasing saturated chains from x to y, i.e.,
−
.
}
y in P . Let µ denote
→
1)rk(x,y)µ(x, y) is equal to the number of strictly
1, 2
{
1)rk(x,y)µ(x, y) =
(
−
x = x0 � x1 �
{
#
� xk = y : ϕ(x0, x1) > ϕ(x1, x2) > > ϕ(xk−1 , xk )
}
· · ·
.
· · ·
Proof. Since ϕ restricted to [x, y] (i.e., to E([x, y])) is an E-labeling, we can assume
[x, y] = [ˆ ˆ0, 1] = P . Let S =
< aj−1.
1], with a1 < a2 <
[n
a1, a2, . . . , aj−1}
{
∗
−
· · ·
44
R. STANLEY, HYPERPLANE ARRANGEMENTS
that
(27)
Define κP (S) to be the number of chains ˆ0 < y1 <
rk(yi) = ai for 1
j
i
Claim. κP (S) is the number of maximal chains ˆ
· · ·
1. The function κP is called the flag f -vector of P .
0 = x0 � x1 � � xn = ˆ
< yj−1 < ˆ
1 in P such that
1 such
→
→
−
· · ·
ϕ(xi−1 , xi) > ϕ(xi , xi+1)
i
S, 1
i n.
To prove the claim, let ˆ = y0 < y1 <
1
j
i
0
· · ·
1. By the definition of E-labeling, there exists a | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
<
1
j
i
0
· · ·
1. By the definition of E-labeling, there exists a unique refinement
≤
⊆
< yj−1
→ →
ˆ
< yj = 1 with rk(yi) = ai for
→
→
−
ˆ
0 = y0 = x0 � x1 �
� xa1 = y1 � xa1 +1 �
· · ·
� xa2 = y2 �
· · ·
· · ·
� xn = yj = ˆ
1
satisfying
ϕ(x0, x1)
ϕ(xa1 , xa1 +1)
→
ϕ(x1, x2)
→
ϕ(xa1 +1, x
→
· · ·
a1 +2)
ϕ(xa1 −1, xa1 )
→
ϕ(xa2 −1, xa2 )
→
· · ·
→
S, so (27) is satisfied. Conversely, given
Thus if ϕ(xi−1 , xi) > ϕ(xi , xi+1), then i
a maximal chain ˆ
1 satisfying the above conditions on ϕ,
let yi = xai . Therefore we have a bijection between the chains counted by κP (S)
and the maximal chains satisfying (27), so the claim follows.
0 = x0 � x1 �
≤
� xn = ˆ
· · ·
· · ·
Now for S
[n
∗
−
1] define
(28)
λP (S) =
(
T →S
�
1)#(S−T )κP (T ).
−
The function λP is called the flag h-vector of P . A simple Inclusion-Exclusion
argument gives
(29)
κP (S) =
λP (T ),
T →S
�
−
∗
[n
for all S
the number of | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
(S) =
λP (T ),
T →S
�
−
∗
[n
for all S
the number of maximal chains ˆ
if and only if i
decreasing maximal chains ˆ
1]. It follows from the claim and equation (29) that λP (T ) is equal to
1 such that ϕ(xi ) > ϕ(xi+1 )
· · ·
1]) is equal to the number of strictly
� xn = ˆ
T . In particular, λP ([n
0 = x0 � x1 �
0 = x0 � x1 � � xn = ˆ
1 of P , i.e.,
≤
−
· · ·
ϕ(x0, x1) > ϕ(x1, x2) > > ϕ(xn−1, xn).
· · ·
Now by (28) we have
λP ([n
−
1]) =
(
T →[n−1]
�
=
1)n−1−#T κP (T )
−
(
1)n−k
−
= (
−
k∗1 ˆ
�
0=y0 <y1 <···<yk =ˆ
1
�
1)k ck,
1)n
(
k∗1
�
−
where ci is the number of chains ˆ0 = y0 < y1 <
follows from Philip Hall’s theorem (Lemma 4.4).
1 in P . The proof now
�
We come to the main result of this subsection, a combinatorial interpretation
< yi = ˆ
· · ·
of the coefficients of the characteristic polynomial ψM (t) for any matroid M .
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
45
5
5
4
5
4 2
3
1
4
1
5
1
4
2
2
3
2
1
3
2 3 4 | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
1
4
1
5
1
4
2
2
3
2
1
3
2 3 4
5
2
5 5
4
5
4
5
Figure 3. The edge labeling ˜� of a geometric lattice L(M )
Theorem 4.12. Let M be a matroid of rank n with a linear ordering x1 < x2 <
< xm of its points (so the broken circuit complex BC(M ) is defined), and let
i
→ →
n. Then
· · ·
0
1)i[tn−i]ψM (t) = fi−1(BC(M )).
(
−
Proof. We may assume M is simple since the “simplification”
M has the same
lattice of flats and same broken circuit complex as M (Exercise 1). The atoms xi of
L(M ) can then be identified with the points of M . Define a labeling ˜ϕ : E(L(M ))
P as follows. Let x � y in L(M ). Then set
∃
�
(30)
Note that ˜ϕ(x, y) is defined since L(M ) is atomic.
˜
i : x
ϕ(x, y) = max
{
⇒
xi = y
.
}
As an example, Figure 3 shows the lattice of flats of the matroid M of Figure 1
with the edge labeling (30).
Claim 1. Define ϕ : E(L(M ))
P by
∃
ϕ(x, y) = m + 1
−
Then ϕ is an E-labeling.
˜ϕ(x, y).
To prove this claim, we need to show that for all x < y in L(M ) there is a | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
we need to show that for all x < y in L(M ) there is a
unique saturated chain x = y0 � y1 �
· · ·
˜
ϕ(y1, y2)
� yk = y satisfying
˜
ϕ(yk−1, yk).
˜ϕ(y0, y1)
⊂
The proof is by induction on k. There is nothing to prove for k = 1. Let k > 1 and
assume the assertion for k
1. Let
· · ·
⊂
⊂
−
i : xi
j = max
{
→
y, xi
x
.
}
⇔→
zi and xj
For any saturated chain x = z0 � z1 �
zi+1. Hence ˜ϕ(zi, zi+1) = j. Thus if ˜
xj
then ˜ϕ(z0, z1) = j. Moreover, there is a unique y1 satisfying x = x0 � y1
˜ϕ(x0, y1) = j, viz., y1 = x0
xj . (Note that y1 � x0 by semimodularity.)
� zk = y, there is some i for which
˜
ϕ(zk−1, zk),
y and
ϕ(z0, z1)
· · ·
· · ·
→
⊂
⊂
→
⇔→
⇒
46
R. STANLEY, HYPERPLANE ARRANGEMENTS
By the induction hypothesis there exists a unique saturated chain y1 � y2 �
� yk = y satisfying ˜
ϕ(y1, y2),
˜
ϕ(yk−1, yk). Since ˜
ϕ(y0, y1) = j > ˜
ϕ(y1, y2) | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
k). Since ˜
ϕ(y0, y1) = j > ˜
ϕ(y1, y2)
· · ·
the proof of Claim 1 follows by induction.
· · ·
⊂
⊂
Claim 2. The broken circuit complex BC(M ) consists of all chain labels ϕ(C),
where C is a saturated increasing chain (with respect to ˜
L(M ). Moreover, all such ϕ(C) are distinct.
≤
To prove the distinctness of the labels ϕ(C), suppose that C is given by ˆ =0
� yk, with ˜ϕ(C) = (a1, a2, . . . , ak ). Then yi = yi−1 xai , so C is the
0 to some x
ϕ) from ˆ
y0 � y1 �
only chain with its label.
· · ·
Now let C and ˜ϕ(C) be as in the previous paragraph. We claim that the
set
contains no broken circuit. (We don’t even require that C is
xa1 , . . . , xak }
{
increasing for this part of the proof.) Write zi = xai , and suppose to the contrary
that B =
zi1 , . . . , zij }
{
be a circuit with r > ait for 1
1
→
j. Now for any circuit
is a broken circuit, with 1
i1 < < ij
· · ·
u1, . . . , uh}
{
r
}
∅ {
and any
h we have
k. Let B
t
→
→
→
x
⇒
u1
⇒
u2
⇒ · · · ⇒
uh = u1
⇒ · · · ⇒
ui−1
⇒
ui+1
⇒ · · · ⇒ | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
= u1
⇒ · · · ⇒
ui−1
⇒
ui+1
⇒ · · · ⇒
uh.
i
→ →
Thus
⇒
zi1
zij−1
zi2 ⇒ · · · ⇒
z = zi1 ⇒
xr = yij , contradicting the maximality of the label aij . Hence
zi2 ⇒ · · · ⇒
xr =
z⊆B
�
zij .
⇒
Then yij −1
x
⇒
BC(M ).
a1 , . . . , xak }
Conversely, suppose that T :=
≤
{
contains no broken circuit, with
xa1 , . . . , xak }
{
xai , and let C be the chain ˆ
0 := y0 � y1 � � yk.
a1 < < ak . Let yi = x
(Note that C is saturated by semimodularity.) We claim that ˜ϕ(C) = (a1, . . . , ak ).
If not, then yi−1
a1 ⇒ · · · ⇒
· · ·
· · ·
⇒
xj = yi for some j > ai. Thus
rk(T ) = rk(T
) = i.
xj
}
∅ {
Since T is independent, T
contains a broken circuit. This contradiction completes the proof of Claim 2.
contains a circuit Q satisfying xj
xj
}
{
≤
∅
Q, so T
To complete the proof of the theorem, note that we have shown that fi−1(BC(M ))
0 = y0 � y1 � � yi such that ˜
· · ·
is the number of chains C : ˆ
ϕ(C) is strictly increas-
ing, or equivalently, ϕ(C) is strictly decreasing. Since ϕ is an E-labeling, the proof
�
follows from Theorem 4.11.
Corollary 4.6. The broken circuit complex BC(M ) | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
s from Theorem 4.11.
Corollary 4.6. The broken circuit complex BC(M ) is pure, i.e., every maximal
face has the same dimension.
to be inserted.
�
Note (for readers with some knowledge of topology). (a) Let M be a matroid
BC(M ) if and only
on the linearly ordered set u1 < u2 < < um. Note that F
if F
BC(M ). Define the reduced broken circuit complex BCr (M ) by
BC(M ) : um
BCr (M ) =
m}
∅ {
· · ·
F
F
≤
≤
u
.
{
≤
⇔≤
}
Thus
BC(M ) = BCr(M )
∼
um,
the join of BCr(M ) and the vertex um. Equivalently, BC(M ) is a cone over BCr (M )
with apex um. As a consequence, BC(M ) is contractible and therefore has the ho
motopy type of a point. A more interesting problem is to determine the topological
nature of BCr(M ). It can be shown that BCr (M ) has the homotopy type of a wedge
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
47
M (1)
of λ(M ) spheres of dimension rank(M )
−
(the derivative of ψM (t) at t = 1). See Exercise 21 for more information on λ(M ).
1)rank(M )−1λ(M ) = ψ�
2, where (
−
(b) [to be inserted]
As an example of the applicability of our results on matroids and geometric
lattices to arrangements, we have the following purely combinatorial description of
the number of regions of a real central arrangement.
Corollary 4.7. Let A be a central arrangement in Rn, and let M | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
ollary 4.7. Let A be a central arrangement in Rn, and let M be the matroid
A, i.e., the independent sets of M are the linearly
defined by the normals to H
independent normals. Then with respect to any linear ordering of the points of M ,
r(A) is the total number of subsets of M that don’t contain a broken circuit.
≤
Proof. Immediate from Theorems 2.5 and 4.12.
�
4.2. Modular elements
We next discuss a situation in which the characteristic polynomial ψM (t) factors in
a nice way.
Definition 4.12. An element x of a geometric lattice L is modular if for all y
we have
≤
L
(31)
rk(x) + rk(y) = rk(x
y) + rk(x
y).
∈
Example 4.9. Let L be a geometric lattice.
⇒
0 and ˆ
(a) ˆ
(b) We claim that atoms a are modular.
1 are clearly modular (in any finite lattice).
Proof. Suppose that a
= y, so equation
(31) holds. (We don’t need that a is an atom for this case.) Now suppose
y) = 1 + rk(y), while rk(a) = 1 and
a
⇔→
�
rk(a
y) = rk(ˆ0) = 0, so again (31) holds.
By semimodularity, rk(a
y = a and a
y. Then a
y.
y
→
⇒
∈
⇒
∈
(c) Suppose that rk(L) = 3. All elements of rank 0, 1, or 3 are modular by
(a) and (b). Suppose that rk(x) | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
rank 0, 1, or 3 are modular by
(a) and (b). Suppose that rk(x) = 2. Then x is modular if and only if for
all elements y = x and rk(y) = 2, we have that rk(x
y) = 1.
(d) Let L = Bn. If x
have x
have
∈
y = x
⊕
Bn then rk(x) = #x. Moreover, for any x, y
Bn we
y. Since for any finite sets x and y we
≤
y and x
≤
y = x
∅
⇒
∈
#x + #y = #(x
y) + #(x
y),
⊕
∅
it follows that every element of Bn is modular. In other words, Bn is a
modular lattice.
(e) Let q be a prime power and Fq the finite field with q elements. Define
Bn(q) to be the lattice of subspaces, ordered by inclusion, of the vector
space Fn . Note that Bn(q) is also isomorphic to the intersection lattice
of the arrangement of all linear hyperplanes in the vector space Fn(q).
Figure 4 shows the Hasse diagrams of B2(3) and B3(2).
q
≤
Note that for x, y
Bn(q) we have x
= x + y
(subspace sum). Clearly Bn(q) is atomic: every vector space is the join
(sum) of its one-dimensional subspaces. Moreover, Bn(q) is graded of rank
n, with rank function given by rk(x) = dim(x). Since for any subspaces
x and y we have
y and x
y = x
y | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
(x). Since for any subspaces
x and y we have
y and x
y = x
y
⊕
∈
⇒
dim(x) + dim(y) = dim(x
⊕
y) + dim(x + y),
⇔
48
R. STANLEY, HYPERPLANE ARRANGEMENTS
010
100
110
001
101
011
111
B (3)
2
B3(2)
Figure 4. The lattices B2 (3) and B3 (2)
it follows that L is a modular geometric lattice. Thus every x
modular.
L is
≤
Note. A projective plane R consists of a set (also denoted R) of
points, and a collection of subsets of R, called lines, such that: (a) every
two points lie on a unique line, (b) every two lines intersect in exactly one
point, and (c) (non-degeneracy) there exist four points, no three of which
are on a line. The incidence lattice L(R) of R is the set of all points
and lines of R, ordered by p < L if p
1 adjoined. It
is an immediate consequence of the axioms that when R is finite, L(R)
is a modular geometric lattice of rank 3. It is an open (and probably
intractable) problem to classify all finite projective planes. Now let P and
Q be posets and define their direct product (or cartesian product ) to be
the set
L, with ˆ
0 and ˆ
≤
P
Q =
(x, y) : x
{
P, y
,
Q
≤
}
(x | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
�
P
Q =
(x, y) : x
{
P, y
,
Q
≤
}
(x , y�) if x
�
≤
→
x� and y
×
y . It is easy
ordered componentwise, i.e., (x, y)
to see that if P and Q are geometric (respectively, atomic, semimodular,
Q (Exercise 7). It is a consequence of the
modular) lattices, then so is P
“fundamental theorem of projective geometry” that every finite modular
geometric lattice is a direct product of boolean algebras Bn, subspace
3, lattices of rank 2 with at least five elements
lattices Bn(q) for n
⊂
(which may be regarded as B2(q) for any q
2) and incidence lattices of
finite projective planes.
→
×
→
⊂
�
(f) The following result characterizes the modular elements of Γn, which is
the lattice of partitions of [n] or the intersection lattice of the braid ar
rangement Bn.
Proposition 4.9. A partition β
Γn is a modular element of Γn if
≤
and only if β has at most one nonsingleton block. Hence the number of
modular elements of Γn is 2n
n.
−
Proof. If all blocks of β are singletons, then β = ˆ0, which is modular by
(a). Assume that β has the block A with r > 1 elements, and all other
blocks are singletons. Hence the number β of blocks of β is given by
|
|
LECTURE 4. BROKEN CIR | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
β of blocks of β is given by
|
|
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
49
r + 1. For any π
Γn, we have rk(π) = n
n
−
π . Let k = π and
|
|
|
|
−
≤
B
j = #
{
≤
r) and β
π : A
−
π) = r
B =
.
�}
⊕
π = k
| ⇒ |
j, and rk(β
−
⇒
−
|
Then β
∈
rk(π) = n
modular.
π = j + (n
|
k, rk(β
−
∈
B1 and b
π =
≤
(B1
{
Let a
≤
Then
Conversely, let β =
B1, B2, . . . , Bk}
{
B2, and set
b)
−
∅
a, (B2
a)
−
.
b, B3, . . . , Bk
}
∅
j + 1. Hence rk(β) = r
π) = n
1,
1, so β is
k + j
−
−
−
with #B1 > 1 and #B2 > 1.
β
|
|
= π = k
|
|
b, . . . , B3, . . . , Bk}
β
β
∈
⇒
−
π =
π =
a, b, B1
{
B1
{
a, B2
B2, B3, . . . , Bl}
∅
Hence rk(β) + rk(π) = rk(β
∈
−
β
⊆ |
∈
π = k + 2
|
β
|
π = k
⇒ |
1.
−
⊆
π) + rk(β
π), so β is not modular.
⇒
�
� | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
1.
−
⊆
π) + rk(β
π), so β is not modular.
⇒
�
∈
y = ˆ
0 and x
In a finite lattice L, a complement of x
y = ˆ
L such that
1. For instance, in the boolean algebra Bn every element has
x
a unique complement. (See Exercise 3 for the converse.) The following proposition
collects some useful properties of modular elements. The proof is left as an exercise
(Exercises 4–5).
L is an element y
≤
⇒
≤
Proposition 4.10. Let L be a geometric lattice of rank n.
(a) Let x
L. The following four conditions are equivalent.
≤
(i) x is a modular element of L.
(ii) If x
(iii) If x and y are complements, then rk(x) + rk(y) = n.
(iv) All complements of x are incomparable.
y = ˆ0, then rk(x) + rk(y) = rk(x
y).
∈
⇒
(b) (transitivity of modularity) If x is a modular element of L and y is modular
in the interval [ˆ0, x], then y is a modular element of L.
(c) If x and y are modular elements of L, then x
y is also modular.
∈
The next result, known as the modular element factorization theorem [16], is
our primary reason for defining modular elements — such an element induces a
factorization of the characteristic polynomial.
Theorem 4.13. Let z be a modular element of the geometric lattice L of rank n. | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
13. Let z be a modular element of the geometric lattice L of rank n.
Write ψz (t) = ψ[ˆ
0,z](t).
Then
(32)
ψL(t) = ψz (t)
�
µL(y)tn−rk(y)−rk(z)
y : y≥z=ˆ0
�
.
⎟
⎠
⎞
Example 4.10. Before proceeding to the proof of Theorem 4.13, let us consider
an example. The illustration below is the affine diagram of a matroid M of rank
3, together with its lattice of flats. The two lines (flats of rank 2) labelled x and y
are modular by Example 4.9(c).
⇔
⇔
50
R. STANLEY, HYPERPLANE ARRANGEMENTS
y
x
x
y
Hence by equation (32) ψM (t) is divisible by ψx(t). Moreover, any atom a of
the interval [ˆ0, x] is modular, so ψx(t) is divisible by ψa(t) = t
1. From this it
is immediate (e.g., because the characteristic polynomial ψG(t) of any geometric
lattice G of rank n begins xn
, where a is the number of atoms of G) that
ψx(t) = (t
5). On the other hand, since y is
1)(t
modular, ψM (t) is divisible by ψy(t), and we get as before ψy(t) = (t
3) and
ψM (t) = (t
5). Geometric lattices whose characteristic polynomial
factors into linear factors in a similar way due to a maximal chain of modular
elements are discussed further beginning with Definition 4.13.
5) and ψM (t) = (t
axn−1+
· · ·
1)(t
3 | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
4.13.
5) and ψM (t) = (t
axn−1+
· · ·
1)(t
3)(t
1)(t
3)(t
1)(t
−
−
−
−
−
−
−
−
−
−
−
−
Our proof of Theorem 4.13 will depend on the following lemma of Greene [11].
We give a somewhat simpler proof than Greene.
Lemma 4.5. Let L be a finite lattice with M¨obius function µ, and let z
following identity is valid in the M¨obius algebra A(L) of L:
≤
L. The
(33)
πˆ0 :=
µ(x)x =
�x⊆L
⎤
�
�v⊇z
µ(v)v
�
⎢
⎤
�
�y≥z=ˆ0
µ(y)y
.
�
⎢
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
51
Proof. Let πs for s
then given by
≤
L be given by (8). The right-hand side of equation (33) is
µ(v)µ(y)(v
y) =
⇒
µ(v)µ(y)
πs
v⊇z
�
y≥z=ˆ
0
s∗v∞y
�
v⊇z
�
y≥z=ˆ
0
=
πs
µ(v)µ(y)
s
�
v⊇s,v⊇z
�
y⊇s,y≥z=ˆ0
=
πs
s
�
µ(v)
v⊇s≥z
�
�
ˆ0,s�z
��
�
�
⎡
⎡
⎡
⎡
⎡
⎢
⎤
⎥
⎥
⎥
⎥
⎥
�
�
⎤
⎤
µ(y)
� | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
⎥
⎥
⎥
⎥
�
�
⎤
⎤
µ(y)
�
y⊇s
�
y≥z=ˆ0
⎥
⎥
�
⎡
⎡
⎢
=
πs
s≥z=ˆ0
�
= πˆ0.
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
�
µ(y)
y⊇s
�
y≥z=ˆ0 (redundant)
�
ˆ0,s
��
�
�
�
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎡
⎢
�
Proof of Theorem 4.13. We are assuming that z is a modular element of
the geometric lattice L.
Claim 1. Let v
z and y
illustrated below).
→
ˆ
z = 0 (so v
∈
∈
y = 0). Then z
ˆ
(v
∈
⇒
y) = v (as
z y
v
z
v yv
v
y
0
R. STANLEY, HYPERPLANE ARRANGEMENTS
52
y))
→
rk(z
Proof of Claim 1. Clearly z
y)
rk(v). Since z is modular we have
(v
⇒
∈
v, so it suffices to show that rk(z
(v
∈
⇒
⊂
(v
∈
⇒
y)) = rk(z) + rk(v
= rk(z) + rk(v
y)
y)
⇒
⇒
−
−
rk(z
y)
(rk(z) + rk(y) | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
y)
y)
⇒
⇒
−
−
rk(z
y)
(rk(z) + rk(y)
⇒
−
y))
rk(z
∈
0
= rk(v
y)
rk(y)
⇒
(rk(v) + rk(y)
−
→
−
rk(v
∈
0
)
y)
−
�
��
�
= rk(v),
proving Claim 1.
��
rk(y) by semimodularity
�
�
Claim 2. With v and y as above, we have rk(v
Proof of Claim 2. By the modularity of z we have
⇒
y) = rk(v) + rk(y).
⇒
By Claim 1 we have rk(z
z we have
∈
(v
⇒
∈
rk(z
(v
y)) + rk(z
(v
y)) = rk(z) + rk(v
y).
⇒
⇒
⇒
y)) = rk(v). Moreover, again by the modularity of
rk(z
(v
y)) = rk(z
y) = rk(z) + rk(y)
rk(z
−
y) = rk(z) + rk(y).
∈
⇒
⇒
It follows that rk(v) + rk(y) = rk(v
y), as claimed.
Now substitute µ(v)v
µ(v)trk(z)−rk(v) and µ(y)y
µ(y)tn−rk(y)−rk(z) in the
right-hand side of equation (33). Then by Claim 2 we have
∃
⇒
vy
tn−rk(v)−rk(y)
= t
n−rk(v∞y)
.
⇒
∃
∃
Now v
tute µ(x)x
∃
preserved. We thus obtain
⇒
y is just vy in the M¨obius algebra A(L). Hence if we further substi
µ(x)tn−rk(x) | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
¨obius algebra A(L). Hence if we further substi
µ(x)tn−rk(x) in the left-hand side of (33), then the product will be
µ(x)t
n−rk(x)
=
x⊆L
�
�L (t)
as desired.
�
��
�
⎤
⎥
⎥
⎥
⎥
⎥
�
v⊇z
�
�
µ(v)trk(z)−rk(v)
�
µ(y)t
y≥z=ˆ0
⎤
⎡
⎡
�
⎡
�
⎡
⎡
⎢
�
�z (t)
��
n−rk(y)−rk(z)
�
⎢
,
�
Corollary 4.8. Let L be a geometric lattice of rank n and a an atom of L. Then
ψL(t) = (t
−
1)
µ(y)tn−1−rk(y).
y≥a=ˆ0
�
Proof. The atom a is modular (Example 4.9(b)), and ψa(t) = t
�
Corollary 4.8 provides a nice context for understanding the operation of coning
defined in Chapter 1, in particular, Exercise 2.1. Recall that if A is an affine
arrangement in K n given by the equations
1.
−
L1(x) = a1, . . . , Lm(x) = am,
then the cone xA is the arrangement in K n
with equations
×
K (where y denotes the last coordinate)
L1(x) = a1y, . . . , Lm(x) = amy, y = 0.
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
53 | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
53
Let H0 denote the hyperplane y = 0. It is easy to see by elementary linear algebra
that
L(A) ∪= L(cA)
− {
≤
x
L(A) : x
= L(A)
L(AH0 ).
H0}
⊂
−
Now H0 is a modular element of L(A) (since it’s an atom), so Corollary 4.8 yields
ψcA(t) = (t
−
1)
µ(y)t(n+1)−1−rk(y)
y∈∗H0
�
= (t
−
1)ψA(t).
There is a left inverse to the operation of coning. Let A be a nonempty linear
arrangement in K n+1 . Let H0
A. Choose coordinates (x0, x1, . . . , xn) in K n+1
so that H0 = ker(x0). Let A be defined by the equations
≤
x0 = 0, L1(x0, . . . , xn) = 0, . . . , Lm(x0, . . . , xn) = 0.
Define the deconing c−1A (with respect to H0) in K n by the equations
L1(1, x1, . . . , xn) = 0, . . . Lm(1, x1, . . . , xn) = 0.
Clearly c(c−1A) = A and L(c−1A) ∪= L(A)
x
L(A) : x
.
H0}
⊂
≤
− {
4.3. Supersolvable lattices
For some geometric lattices L, there are “enough” modular elements to give a
factorization of ψL(t) into linear factors.
Definition | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
enough” modular elements to give a
factorization of ψL(t) into linear factors.
Definition 4.13. A geometric lattice L is supersolvable if there exists a modular
maximal chain, i.e., a maximal chain ˆ
1 such that each xi
is modular. A central arrangement A is supersolvable if its intersection lattice LA
is supersolvable.
0 = x0 � x1 �
� xn = ˆ
· · ·
ˆ
· · ·
Note. Let ˆ = x0 � x1 �
0
� xn = 1 be a modular maximal chain of the
geometric lattice L. Clearly then each xi−1 is a modular element of the interval
0 = x0 � x1 � � xn = ˆ
0, xi]. The converse follows from Proposition 4.10(b): if ˆ
[ˆ
1
is a maximal chain for which each xi−1 is modular in [ˆ0, xi], then each xi is modular
in L.
Note. The term “supersolvable” comes from group theory. A finite group �
is supersolvable if and only if its subgroup lattice contains a maximal chain all of
whose elements are normal subgroups of �. Normal subgroups are “nice” analogues
of modular elements; see [17, Example 2.5] for further details.
· · ·
Corollary 4.9. Let L be a supersolvable geometric lattice of rank n, with modular
maximal chain ˆ
1. Let T denote the set of atoms of L, and
set
0 = x0 � x1 � � xn = ˆ
· · ·
(34)
Then ψL(t) = (t
e1)(t
e2)
· · ·
−
−
a
ei = #
{
≤
(t
T | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
t
e1)(t
e2)
· · ·
−
−
a
ei = #
{
≤
(t
T : a
xi, a
xi−1}
.
⇔→
→
en).
−
Proof. Since xn−1 is modular, we have
xn−1 = ˆ
0
y
∈
y
≤
√
T and y
⇔→
xn−1, or y = ˆ
0.
54
R. STANLEY, HYPERPLANE ARRANGEMENTS
By Theorem 4.13 we therefore have
ψL(t) = ψxn−1 (t)
�
µ(a)t
n−rk(a)−rk(xn−1 ) + µ(ˆ
0)tn−rk(ˆ
0)−rk(xn−1)
⎟
.
a⊆T
⎝ �
a∈⊇xn−1
⎝
⎞
1, µ(ˆ
0) = 1, rk(a) = 1, rk(ˆ
⎣
⎣
⎠
1, the
−
en. Now continue this with L replaced by [ˆ0, xn−1]
�
Note. The positive integers e1, . . . , en of Corollary 4.9 are called the exponents
Since µ(a) =
−
expression in brackets is just t
(or use induction on n).
0) = 0, and rk(xn−1) = n
−
of L.
Example 4.11.
(a) Let L = Bn, the boolean algebra of rank n. By Exam
ple 4.9(d) every element of Bn is modular. Hence Bn is supersolvable.
Clearly each ei = 1, so ψBn (t) = (t
1)n .
(b) Let L = Bn(q), the lattice of subspaces of F | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
= (t
1)n .
(b) Let L = Bn(q), the lattice of subspaces of Fq . By Example 4.9(e) every
element of Bn(q) is modular, so Bn(q) is supersolvable. If
denotes the
number of j-dimensional subspaces of a k-dimensional vector space over
Fq , then
k
j
�
�
n
−
−
ei = [i
1]
iq
q
=
[i−1]1
1
1 −
−
−
i−1
.
= q
Hence
1
qi−1
q
−
1
−
ψBn (q)(t) = (t
1)(t
q)(t
q 2
(t
)
· · ·
−
−
−
−
q n−1
).
In particular, setting t = 0 gives
µBn (q)(ˆ
Note. The expression
k
j
1) = (
1)n
q(n
2 ).
−
is called a q-binomial coefficient. It is a
polynomial in q with many interesting properties. For the most basic
properties, see e.g. [18, pp. 27–30].
�
�
−
(c) Let L = Γn, the lattice of partitions of the set [n] (a geometric lattice of
1). By Proposition 4.9, a maximal chain of Γn is modular if and
rank n
0 = β0 � β1 � � βn−1 = ˆ
only if it has the form ˆ
1, where βi for i > 0 has
· · ·
exactly one nonsingleton block Bi (necessarily with i + 1 elements), with
n−1 = [n]. In particular, Γn is supersolvable and has
B1
exactly n!/2 modular chains for n > 1. The atoms covered by βi are the
partitions with one nons | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
exactly n!/2 modular chains for n > 1. The atoms covered by βi are the
partitions with one nonsingleton block
Bi. Hence βi lies above
atoms, so
exactly
B2
j, k
B
· · ·
∗
⊇
⊇
}
{
i+1
2
⎜
�
ei =
�
i + 1
2
i
2
= i.
−
� �
1)(t
�
2)
n + 1) and µ�n (ˆ1) =
It follows that ψ�n (t) = (t
· · ·
−
1)!. Compare Corollary 2.2. The polynomials ψBn (t) and
(
−
ψ�n (t) differ by a factor of t because Bn(t) is an arrangement in K n of
1)n−1(n
(t
−
−
−
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
55
rank n
tion, then
−
1. In general, if A is an arrangement and ess(A) its essentializa
(35)
trk(ess(A))ψA(t) = trk(A)ψess(A)(t).
(See Lecture 1, Exercise 2.)
Note. It is natural to ask whether there is a more general class of geometric
lattices L than the supersolvable ones for which ψL(t) factors into linear factors
(over Z). There is a profound such generalization due to Terao [22] when L is an
intersection poset of a linear arrangement A in K n . Write K[x] = K[x1, . . . , xn]
and define
T(A) =
(p | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
. Write K[x] = K[x1, . . . , xn]
and define
T(A) =
(p1, . . . , pn)
{
Here we are regarding (p1, . . . , pn) : K n
≤
K[x]n : pi(H)
H for all H
∗
≤
K n, viz., if (a1, . . . , an)
A
.
}
K n, then
∃
(p1, . . . , pn)(a1, . . . , an) = (p1(a1, . . . , an), . . . , pn(a1, . . . , an)).
≤
The K[x]-module structure K[x]
T(A)
T(A) is given explicitly by
×
∃
(p1, . . . , pn) = (qp1, . . . , qpn).
q
≤
·
T(A). Since A is a linear
Note, for instance, that we always have (x1, . . . , xn)
arrangement, T(A) is indeed a K[x]-module. (We have given the most intuitive
definition of the module T(A), though it isn’t the most useful definition for proofs.)
It is easy to see that T(A) has rank n as a K[x]-module, i.e., T(A) contains n,
but not n + 1, elements that are linearly independent over K[x]. We say that A
is a free arrangement if T(A) is a free K[x]-module, i.e., there exist Q1, . . . , Qn
≤
T(A) can be uniquely written in the form
T(A) such that every element Q
K[x]. It is easy to see that if T(A) is free,
+ qnQ | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
K[x]. It is easy to see that if T(A) is free,
+ qnQn, where qi
Q = q1Q1 +
· · ·
then the basis
can be chosen to be homogeneous, i.e., all coordinates
Q1, . . . , Qn}
{
of each Qi are homogeneous polynomials of the same degree di. We then write
di = deg Qi. It can be shown that supersolvable arrangements are free, but there
are also nonsupersolvable free arrangements. The property of freeness seems quite
subtle; indeed, it is unknown whether freeness is a matroidal property, i.e., depends
only on the intersection lattice LA (regarding the ground field K as fixed). The
remarkable “factorization theorem” of Terao is the following.
≤
≤
Theorem 4.14. Suppose that T(A) is free with homogeneous basis Q1, . . . , Qn. If
deg Qi = di then
−
We will not prove Theorem 4.14 here. A good reference for this subject is [13,
· · ·
−
−
ψA(t) = (t
d1)(t
d2)
(t
dn).
Ch. 4].
Returning to supersolvability, we can try to characterize the supersolvable prop
erty for various classes of geometric lattices. Let us consider the case of the bond
lattice LG of the graph G. A graph H with at least one edge is doubly connected if
it is connected and remains connected upon the removal of any vertex (and all in
cident edges). A maximal doubly connected subgraph of a graph G is called a block
of G. For instance, if G is a forest then its blocks are its edges. Two different blocks
of G intersect in at most one vertex. Figure 5 shows a graph with eight blocks, five
of which consist of | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
of G intersect in at most one vertex. Figure 5 shows a graph with eight blocks, five
of which consist of a single edge. The following proposition is straightforward to
prove (Exercise 16).
56
R. STANLEY, HYPERPLANE ARRANGEMENTS
Figure 5. A graph with eight blocks
Proposition 4.11. Let G be a graph with blocks G1, . . . , Gk . Then
LG ∪= LG1 × · · · ×
LGk
.
It is also easy to see that if L1 and L2 are geometric lattices, then L1 and
L2 are supersolvable if and only if L1
L2 is supersolvable (Exercise 18). Hence
in characterizing supersolvable graphs G (i.e., graphs whose bond lattice LG is
supersolvable) we may assume that G is doubly connected. Note that for any
connected (and hence a fortiori doubly connected) graph G, any coatom β of LG
has exactly two blocks.
×
Proposition 4.12. Let G be a doubly connected graph, and let β =
be a
coatom of the bond lattice LG, where #A #B. Then β is a modular element of
→
LG if and only if #A = 1, say A =
, and the neighborhood N (v) (the set of
}
vertices adjacent to v) forms a clique (i.e., any two distinct vertices of N (v) are
adjacent).
A, B
v
{
}
{
Proof. The proof parallels that of Proposition 4.9, which is | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
are
adjacent).
A, B
v
{
}
{
Proof. The proof parallels that of Proposition 4.9, which is a special case. Suppose
B such
that #A > 1. Since G is doubly connected, there exist u, v
≤
that u = v, u� = v , uu
E(G), and vv�
�
u .
v)
∅
≤
}
−
If G has n vertices then rk(β) = rk(π) = n
π) = n
4.
−
Hence β is not modular.
E(G). Set π =
{
2, rk(β π) = n
⇒
−
≤
u�)
(A
1, and rk(β
A and u�, v
v, (B
−
−
∅
∈
≤
�
�
�
Assume then that A =
need to show that β is not modular. Let π =
A
v
{
. Suppose that av, bv
}
E(G) but ab
a, b, v
⇔≤
. Then
≤
a, b
E(G). We
β = ˆ
1,
π
⇒
π
∈
β =
A
{
{
,
}
{
, a, b, v
− {
a, b
− {
}
}
}}
rk(π) = rk(β) = n
2, rk(π
β) = n
1, rk(π
β) = n
4.
−
∈
−
⇒
−
Hence β is not modular.
Conversely, let β =
E(G).
It is then straightforward to show (Exercise 8) that β is modular, completing the
�
proof.
As an immediate consequence of Propositions 4.10(b) and 4.12 we obtain a
. Assume that if av, bv
}
E(G) then ab
A, v
≤
≤
{
characterization | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
that if av, bv
}
E(G) then ab
A, v
≤
≤
{
characterization of supersolvable graphs.
Corollary 4.10. A graph G is supersolvable if and only if there exists an ordering
E(G),
v1, v2, . . . , vn of its vertices such that if i < k, j < k, vivk
E(G) and vj vk
≤
≤
⇔
⇔
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
57
then vivj
the neighborhood of vi is a clique.
≤
E(G). Equivalently, in the restriction of G to the vertices v1, v2, . . . , vi,
Note. Supersolvable graphs G had appeared earlier in the literature under the
names chordal, rigid circuit, or triangulated graphs. One of their many characteri
zations is that any circuit of length at least four contains a chord. Equivalently, no
induced subgraph of G is a k-cycle for k
4.
⊂ | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1: Integral Form of Maxwell’s Equations
I. Maxwell’s Equations in Integral Form in Free Space
1. Faraday’s Law
d
∫(cid:118) E ds = -
dt ∫ µ0 H i da
i
C
S
Circulation
of E
Magnetic Flux
0
µ = 4π ×10-7 henries/meter
[magnetic permeability of
free space]
(Kirchoff’s Voltage Law, conservative electric
EQS form: E ds
= 0
i(cid:118)
∫
C
field)
MQS circuit form: v = L
(Inductor)
di
dt
2.
Ampère’s Law (with displacement current)
(cid:118)
∫
H ds
i
=
C
i
∫
J da
+
S
d
dt
ε
0
∫
S
i
E
da
Circulation Conduction Displacement
of H
Current
Current
MQS form: H ds = ∫ J da
i
(cid:118)∫
i
C
EQS circuit form: i = C
(capacitor)
S
dv
dt
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1
Page 1 of 6
3. Gauss’ Law for Electric Field
(cid:118)∫ ε 0E da = ρ dV
∫
i
S
V
≈ 8.854 ×10-12 farads/meter
ε 0 ≈
10-9
36π
1
ε µ 0 0
free space)
c =
3≈
× 108 meters/second (Speed of electromagnetic waves in
4. Gauss’ Law for Magnetic Field
(cid: | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
3≈
× 108 meters/second (Speed of electromagnetic waves in
4. Gauss’ Law for Magnetic Field
(cid:118)∫ µ 0H da = 0
i
S
In free space:
B = µ H0
magnetic
flux
density
(Teslas)
magnetic
field
intensity
(amperes/meter)
5. Conservation of Charge
Take Ampère’s Law with displacement current and let contour C → 0
lim (cid:118)∫ H i ds = 0 = (cid:118)∫ J i da +
C 0→
C
S
d
dt S
(cid:118)∫ ε 0E i da
dVρ ∫
V
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1
Page 2 of 6
J i da +
(cid:118)∫
S
d
dt ∫
V
ρ dV = 0
Total current
leaving volume inside volume
through surface
Total charge
6. Lorentz Force Law
f = q E + v × µ H)
(
0
II. Electric Field from Point Charge
(cid:118)∫
S
ε0E i da = ε0E 4 π r = q
2
r
E =
r
q
4π ε0r
2
T sin θ = f =
c
2
q
4π ε0r
2
T cos θ = Mg
tan θ =
2
q
4π ε0r
2
Mg
=
r
2l
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1
Page 3 of 6
3
Mg
⎡2π ε0r
q = ⎢
l
⎣
2
1
⎤
⎥
⎦
III. Faraday Cage
J da = i = -
i
∫(cid:118)
S
d
dt
∫ ρ dV = -
d | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
da = i = -
i
∫(cid:118)
S
d
dt
∫ ρ dV = -
d
dt
(
) =
-q
dq
dt
∫
idt = q
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1
Page 4 of 6
IV. Boundary Conditions
1. Gauss’ Continuity Condition
ε0E i da = ∫σsdS ⇒ ε0 (E2n
(cid:118)∫
- E1n
) dS = σ dS
s
S
S
ε0 (E - E ) = σ ⇒ n i ⎡ε0 (E - E1 )⎤ = σ
1n
2n
2
s
s
⎣
⎦
2. Continuity of Tangential E
E i ds = (E - E2t ) dl = 0 ⇒ E - E2t = 0
1t
1t
(cid:118)∫
C
n× E1 - E2 ) = 0
(
Equivalent to = Φ2 along boundary
Φ1
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1
Page 5 of 6
3. Normal H
∇ µ H = 0 ⇒
i 0
(cid:118)∫ µ0
S
H i da = 0
4. Tangential H
µ (H - Hbn
an
0
) A = 0
H = Hbn
an
n i ⎡
⎣
H a - H b
⎤ = 0
⎦
(cid:118)
∇ × H = J ⇒ ∫ H i ds = J i da
∫
C
S
H ds - Hatds = Kds
bt
H - Hat = K
bt
n × ⎡
⎣
H a - H b
⎤ = K
⎦
∂ρ | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
n × ⎡
⎣
H a - H b
⎤ = K
⎦
∂ρ
∇ i J + = 0
∂t
5. Conservation of Charge Boundary Condition
i(cid:118)∫
d
∫
J da + ρdV = 0
dt V
S
n i ⎣
⎡ J a - Jb ⎦
⎤ +
∂
t
∂
σ = 0
s
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1
Page 6 of 6 | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
18.433 Combinatorial Optimization
NP-completeness
November 18
Lecturer: Santosh Vempala
Up to now, we have found many efficient algorithms for problems in Matchings, Flows,
Linear Programs, and Convex Programming. All of these are polynomial-time algorithms.
But there are also problems for which we have found no polynomial-time algorithms. The
theory of NP-completeness unifies these failures. Roughly speaking, an NP-complete prob-
lem is one that is as hard as any problem in a large class of problems. For example, the
Traveling Salesman Problem (TSP), Integer Programming (IP), the Longest Cycle, and
Satisfiability (SAT) are all hard problems. NP-completeness tells us that they are all, in a
precise sense, equally hard. Let’s look at each problem in a little more detail.
1. The Traveling Salesman Problem
Let’s say that there exist a salesman that has to visit n cities and there exists a
distance wi,j between cities i and j. He wants to make sure to minimize his traveling
time by visiting every city exactly once. In other words, there is a complete graph
G = (V, E) with lengths wi,j between nodes i and j. The question we must ask is:
What is the shortest cycle that visits every node exactly once?
2. Integer Linear Programming
Suppose that you have a linear program such as the following:
min cT x
Ax ≤ b for xi ≥ 0
This is your typical linear program. Now, if you decide to add an integrality constraint
on xi such that it is forced to be a positive integer, then you have an Integer Linear
Program (ILP).
3. Boolean Satisfiability
The satisfiability problem (SAT) uses boolean expressions such as the following
f = (x1 ∨ x2 ∨ x4) ∧ (x3 ∨ ¯x4)....
1
with xi = {True,False} and using well known boolean identities. Does F have a
satisfying assignment? Can we find values of xi such that every clause in F is equal
1?
4. Longest Cycles
Given a graph G = (V, E), find the longest cycle.
5. Cliques
A clique is a complete subgraph. Given a graph G | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
graph G = (V, E), find the longest cycle.
5. Cliques
A clique is a complete subgraph. Given a graph G = (V, E), find a clique of maximum
cardinality (vertices).
As different as these examples might seem, they have two main properties in common:
A) None of them is known to have a polytime algorithm.
B) If any one of them has a polytime algorithm, then they all do.
1 Optimization vs Decision
While property A seems trivial to us all by the inspection of each problem, property B is
not as easy to see. To understand this property, we first formulate the decision versions of
these optimization problems.
Find the optimum among a set of feasible solutions F with cost function c
vs
Is there a feasible solution of cost ≤ L?
If the Optimal (OPT) is solved, then the Decision (DEC) is also solved. Namely, DEC
reduces to OPT. Now, is OPT reducible to DEC? Well, using the TSP as an example, we
ask: Is there a tour ≤ L? Then, we proceed to do a binary search in order to find the
length of the shortest tour, say S. But, we still don’t know what the tour is. One way to
figure this out is to use the following algorithm:
Take out an edge e.
Ask if the same graph still has a tour ≤ S.
If it does, then we don’t need that edge and can delete it.
2
If it doesn’t, then we keep that edge because it will be part of our tour.
Repeat this algorithm for all the edges.
In the ILP example, we can ask: Is there an x of cost ≤ L? Well, one way to do this would
be to set xi = 0 and if optimum stays the same, then we can fix that particular xi to 0.
For the maximum cliques problem, the OPT problem would be: Find the largest clique,
while the DEC problem would be: Is there a clique of size ≤ k? To find the optimal size
k∗, again we do a binary search. We then consider the graph with vi and all its neighbors.
If the optimum in this graph remains the same, then save that vertex, then we can delete
all other vertices. Else, delete vi because it is | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
.
If the optimum in this graph remains the same, then save that vertex, then we can delete
all other vertices. Else, delete vi because it is not in the max clique.
2 P and NP
2.1 Definitions
P : Class of decision problems that can be solved in polytime.
N P : Decision problems that have a short proof (certificate) for YES answers. The proof
has length bounded by a polynomial in the size of the input, and its correctness can be
verified in polytime.
Note that problems in P have short proofs for both YES and NO answers. This means that
P ⊆ N P . Let’s look at a problem in P:
Linear Programming: Is the minimum less than some c?
YES: Give a feasible solution ≤ c
NO: Use the Dual of the problem to give a lower bound.
Now, let’s look at the following examples of NP problems:
1. TSP, Is there a tour ≤ L?
YES: Give a tour
NO: ?
2. SAT, Does there exist a satisfying assignment?
YES: Give a satifying assignment
3
NO: ?
3. Min ILP, Is the minimum ≤ c?
YES: Give a feasible solution that is ≤ c
NO: ?
This leads to the question: Is P = N P ?
2.2 Reductions
A reduction from a problem A to a problem B is a function f : A → B such that for all
instances x
x ∈ A ⇔ f (x) ∈ B.
If the function f can be computed in polynomial time, then it is called a polynomial-time
reduction. An implication of this is the following:
If there exists a polytime algorithm for B, then there exists one for A.
A problem B is NP-hard if every problem in NP has a polytime reduction to B.
addition, B is in NP, then it is NP-complete.
If, in
Thus if A is NP-complete, and it has a reduction to another problem B in NP, then B is
also NP-complete.
2.3 Examples of Reduction
SAT is NP-complete (we will not prove this in class).
1. ILP is NP-complete Let’s take the following SAT problem and see if it can be solved
by an ILP.
F = (x1 ∨ x2 ∨ ... ∨ ¯xi) ∧ (x4 ∨ ¯x | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
by an ILP.
F = (x1 ∨ x2 ∨ ... ∨ ¯xi) ∧ (x4 ∨ ¯x5) ∧ ... ∧ (xa ∨ xb ∨ ... ∨ xc)
This SAT problem can also be written in the following way
x1 + x2 + ... + ¯xi ≥ 1
x4 + ¯x5 ≥ 1
xa + xb + ... + ¯xc ≥ 1
4
X2 + X3 + X4 + X5
X2 + X4
X1 + X3 + X5
X2, 1
X3, 1
X4, 1
X5, 1
X2, 2
X1, 3
X4, 2
X3, 3
X5, 3
Figure 1: clique is NP-complete
(cid:1)
xi =
...
1,
0,
then true
then false
Since SAT can be reduced to an ILP, ILP is NP-complete.
2. Clique is NP-complete
SAT can be reduced to clique by the following construction. Suppose we have a
formula F with m clauses.
1) Vertices are going to be of the form < xa, i > where xa is a literal that occurs in
clause Ci
2) Edges are going to be of the form {< xa, i >, < xb, j >} for all xa (cid:9)= ¯xb and i (cid:9)= j.
By defining the vertices and edges this way, we ensure that all the connected vertices
are compatible, since their truth values won’t overlap. If we find a clique of size m in
this graph, F is satisfiable. Refer to Fig. 1.
5 | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
MIT OpenCourseWare
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18.727 Topics in Algebraic Geometry: Algebraic Surfaces
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ALGEBRAIC SURFACES, LECTURE 5
LECTURES: ABHINAV KUMAR
1. Examples
→
(1) If S ⊂ Pn, p ∈ S, then projection from p gives a rational map S ��� Pn−1
defined away from p extending to BlpS = S˜
Pn−1 . For instance, if
Q is a smooth quadric in P2 , we get a birational map Q ��� P2 with
|tildeQ → P2 a morphism. It contracts the two lines passing through p,
so Q = P2(2 − 1).
(2) A birational map of P2 to itself is called a plane Cremona transformation
e.g. quadratic transformation. One example is φ : P ��� P2 given by
(x : y : z) �→ ( 1 : 1 : 1 ) It is clearly birational and its own inverse.
Let p = (1 : 0 : 0), q = (0 : 1 : 0), r = (0 : 0 : 1). These are the
3 base points of φ, and φ blows up these points and then blows down
the three lines joining them. Similarly, we could take a linear system of
3 independent conics passing through three point p, q, r (non-collinear).
Generally, 2 conics passing through p, q, r would have a unique 4th point
of intersection, gives | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
non-collinear).
Generally, 2 conics passing through p, q, r would have a unique 4th point
of intersection, gives the birational map.
z
x
y
(3) Linear systems of cubics: let p1, . . . , pr be r distinct points in the plane
(r ≤ 6) in general position, i.e. no 3 of them are collinear and no six lie on
a conic. Let πr : Pr → P2 be the blowup of p1, . . . , pr. Let d = q − r. The
linear system of cubics passing through p1, . . . , pr defines an embedding
Pd, and Sd = j(Pr) is a surface of degree d in Pd, called a del
j : Pr
Pezzo surface of degree d. e.g. S1 is a
→
Note. Contracting other curves and singularities: let f : Y
X be a resolution
of a normal surface singularity p ∈ X (i.e. X is normal at p). Then p ⊂ X is
called a rational singularity— if R1f∗OY = O and Y → X is an isomorphism
away from Y � {f −1(p)} → X � {p}, e.g. can include nonsingular p as a rational
singularity.
→
Example. The duVal singularities are examples of rational singularities.
An x2 + y2 + zn+1 = 0
Dn x2 + y2z + zn−1 = 0.
E6 x2 + y3 + z4 = 0.
1
2
LECTURES: ABHINAV KUMAR
E6 x2 + y3 + yz3 = 0.
E8 x2 + y3 + z5 = 0.
If | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
2 + y3 + yz3 = 0.
E8 x2 + y3 + z5 = 0.
If you resolve these, you get the corresponding Dynkin diagrams for the dual
graph of the exceptional curves.
Theorem 1 (Artin Contraction). A connected set of curves {Ci} on a surface
Y is the exceptional locus of a rational singularity p ∈ X iff (a) the intersection
matrix (Ci, Cj ) is negative definite, and (b) pa(D) ≤ 0 for every D supported on
Ci. Note that pa(D) = 1 − χ(OD) by definition, 2pa − 2 = D (D + K).
·
2. Ruled Surfaces
Definition 1. A surface X is ruled if it birational to P1 × B for a singular
projective curve B.
Let X be a surface, B a nonsingular projective curve.
Definition 2. A pencil of curves with base B on X is a dominant rational map
π : X ��� B s.t. k(B) is alg. closed in k(X).
Note that this map π is defined on the complement of a finite number of points
x1, . . . , xn. If π is not regular at these points, they are called base points of the
pencil, and the fibers {π−1(y)|y ∈ B} is the family of curves of the pencil π. For
η the generic point of B, π−1(η) is called the generic curve of the pencil π.
→
Definition 3. A smooth morphism X
over B if the fibers are all isomorphic to P1 .
Theorem 2 (Noether-Tsen). Let π : X ��� B be a pencil of curves s.t. the
generic | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
Let π : X ��� B be a pencil of curves s.t. the
generic curve has arithmetic genus zero. Then X is birational to P1 × B (and
the generic fiber of π is ∼
B is called a geometrically ruled surface
k(B)). In particular, X is a ruled surface.
= P1
Definition 4. Let K be a field. K has property Cr (r ≥ 0) if for every homo
geneous polynomial of degree d ≥ 1 in n ≥ 2 variables, it has a nonzero solution
in K n whenever dr < n.
Remark. Note that K has property C0 iff K is alg. closed, and finite fields have
property C1. Moreover, if K has property Cr, then K has property Cs for s ≥ r.
Lemma 1. If K has property C1, so does every alg. extension of L of K.
Proof. We can assume that L/K is finite. Let F (x) be a homogeneous polynomial
of degree d in n variables (d < n) coefficients in L. Let f (x) = NormL/K F (x).
By choosing a basis e1, . . . , em (m = [L : K]) of L/K, and setting x = x1e1 +
+ xmem we see that f can be expressed as a homogeneous polynomial of
· · ·
degree md in mn variables. Since d < n, md < m, we have a nontrivial solution
�
NL/K (F (x)) = 0 = F (x) = 0.
⇒
ALGEBRAIC SURFACES, LECTURE 5
3
Proposition | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
= 0.
⇒
ALGEBRAIC SURFACES, LECTURE 5
3
Proposition 1. Let k be algebraically closed. Then k(T ) (purely transcendental
extension in one variable) has property C1.
Proof. Let f (X1, . . . , Xn) be a homogeneous polynomial of degree d < n in
X1, . . . , Xn with coefficients in k(T ). We may as well assume that the coeffi
cients are in k[T ]. We’ll show ∃ a nontrivial solution in k[T ]. Let f (x1, . . . , Xk) =
�
in for ci1···in ∈ k[T ]. Let µ = max deg ci1···in over all coefficients
ci1···in X1
of f . Write
i1 · · · Xn
(1)
f = f0(X1, . . . , Xn) + Y f1(X1, . . . , Xn) + + T µfµ(X1, . . . , Xn)
· · ·
where fi ∈ k[X1, . . . , Xn]. For new variables Y10, . . . , Yns (s to be chosen later),
write
(2)
and let
(3)
φ(Y10,
· · ·
Xi = Yi0 + Yi1T +
· · ·
+ YisT s
, Yns) = f (
s
� �
Y1j T j ,
Y2j T j , . . . ,
Ynj T j )
�
This has degree sd + µ in T . Write it as
j=0
(4) φ = φ0(Y10,
· · ·
, Yns) + T φ1(Y10,
· · ·
, Yns) + + T sd+µφsd+µ(Y10,
· · ·
· · ·
, Yns)
, | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
, Yns) + + T sd+µφsd+µ(Y10,
· · ·
· · ·
, Yns)
, Yns. Since
i.e. have ds+µ+1 homogeneous polynomials φj of degree d in Y10,
n > d, for large enough s, n(s + 1) > ds + µ + 1 and there are more variables
�
than equations. Because k is alg. closed, we have a solution in k.
· · ·
Proposition 2. Let k be a field, k its alg. closure. Let X be an algebraic curve,
proper over k.
Proof. Riemann-Roch on KX , straightforward.
�
Lemma 2. If, in addition to the hypothesis of proposition, X also has a k-
rational point, then X is k-isomorphic to P1
k.
Corollary 1. Let X have property C1, and let X be geometrically integral, proper
curve over k of arithmetic genus 0. Then X ∼
=k P1
.
of Noether-Tsen. Let η be the generic point of B. By the above, the field k(η) =
k(B) has property C1. By assumption, Xη = π−1(η) has arithmetic genus 0.
Blowing up X enough times, we get φ : X �
B
completing π φ. Note that this does not change the generic fiber. By assumption,
k(B) is algebraically closed in k(X). We see Xη = (η φ)(η) is geometrically
◦
(Xη) ∼ k(η)(t) for t an
integral, and therefore is k(η)-isomorphic to P1
k(η). So k
=
�
independent variable over k(η), | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
(η)-isomorphic to P1
k(η). So k
=
�
independent variable over k(η), and X is birational to P1 × B.
X and a morphism X �
→
→
◦
4
LECTURES: ABHINAV KUMAR
B be a surjective morphism from a surface X to a
Theorem 3. Let π : X
nonsingular, projective curve B s.t. for some closed point b ∈ b, π−1(b) ∼ P1 =
.
Then ∃ a section σ : B X, an open subset W ⊂ B, b ∈ W , and an isomorphism
f : π−1(W ) → P1 × W
→
s.t. the following diagram commutes
→
(5)
f
π−1(W )���������
π
� �
P
1 × W
��
� �
pr2
� �
W
Proof. B is a nonsingular curve, and π∗(OX ) is a torsion-free coherent OB
module, locally free of finite rank (π is flat and H 1(π−1(b), Oπ−1(b)) = 0). By
the base change theorem, we see that H 1(π−1(b�), Oπ−1(b�)) = 0 for b� in a neigh
borhood V of B, and π∗OX ⊗ k(b) → H 0(π−1(b�), Oπ−1(b�)) is an isomorphism for
b� ∈ V .
(6)
π−1(b) ∼
= P1
= ⇒ dim H 0(π−1(b�), Oπ−1(b�)) = 1
so π∗OX is locally free of rank 1, i.e. is OB. Thus, k(B) is alg. closed in k(X), | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
OX is locally free of rank 1, i.e. is OB. Thus, k(B) is alg. closed in k(X),
and ∃U ⊂ V containing b s.t. Fb� = π−1(b�) is geometrically integral for b� ∈ U .
Fb = P1, and the arithmetic genus of Fb� does not depend on b�, so the generic
fiber has arithmetic genus 0 and the closed fibers are P1 .
Thus, Fη
∼
= P1
∼
k(η).
→
Fη and therefore Spec OB,η
This implies that Fη has a rational point over k(η) = k(B), and ∃ a morphism
X a B-morphism, giving us a ra
Spec k(B)
tional section σ : B ��� X. B is a nonsingular curve and X is projective, so σ
X is a section (π σ = idB ). Let D = σ(B).
extends to a morphism. σ : B
Then D Fb� = 1 for b� ∈ B. Let X � = π−1(U ). Since the fibers of π� are P1, and
∼
= OFb� (1), we have dim k(b�)H 0(OX (D) ⊗ k(b�)) = 2 for b� ∈ U .
OX � (D) ⊗ OFb�
Again applying the base change theorem, we have E = π∗(OX � (D)) a locally free
O+U -module of rank 2 and the canonical homomorphism
→
→
·
◦
(7)
π∗OX � (b) ⊗ k(b�) → H 0(OX (D) ⊗ OFb� )
is an isomorphism for b� ∈ U . Thus π∗π∗OX � (D) = | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
an isomorphism for b� ∈ U . Thus π∗π∗OX � (D) = π∗(E) → OX � (D) is sur
jective. By the universal property of P(E), we have a unique U -morphism
u∗(OP(E)(D)) ∼
u : → P(E) s.t.
= OX � (D). It is clear that u is an iso-
∼ P1(k(b�))) and
→
morphism since it is an isomorphism fiber by fiber (ub : Fb�
�
take b ∈ W ⊂ U small enough to trivialize P(E).
X � | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
MIT OpenCourseWare
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18.306 Advanced Partial Differential Equations with Applications
Fall 2009
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Stability of Numerical Schemes for PDE’s.
Rodolfo R. Rosales
.
(cid:3)
MIT, Friday February 12, 1999.
Abstract
The purpose of these notes is to give some examples illustrating how naive numerical approx-
imations to PDE’s may not work at all as expected. In addition, the following two important
notions are introduced: (I) von Neumann stability analysis | helps identify when (and
if ) numerical schemes behave properly. (II) Arti(cid:12)cial viscosity | a tool in stabilizing nu-
merical schemes. These notes should be read in conjunction with the use of the MatLab
scripts (in the Athena 18311-Toolkit at MIT) whose names end with the acronym GBNS (for
Good-Bad-Numerical-Schemes).
Contents
1 Naive Scheme for the Wave Equation.
2
2 von Neumann stability analysis for PDE’s.
7
3 Numerical Viscosity and Stabilized Scheme.
12
4 Reference.
12
List of Figures
1.1 Naive scheme, cosine initial data with 40 points. . . . . . . . . . . . . . . . . . . . .
3
1.2 Naive scheme, cosine initial data with 57 points. . . . . . . . . . . . . . . . . . . . .
4
1.3 Naive scheme, cosine initial data with 80 points. . . . . . . . . . . . . . . . . . . . .
5
1.4 Naive scheme, periodic Gaussian initial data. Small corner. . . . . . . . . . . . . . .
6
1.5 Naive scheme, periodic Gaussian initial data. Sharper corner. . . . . . . . . . . . . .
7
3.1 Corrected scheme, cosine initial data with 55 points. . . . . . . . . . . . . . . . . . .
13
3.2 Corrected scheme, cosine initial data with 190 points.
. . . . . . . . . . . . . . . | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
13
3.2 Corrected scheme, cosine initial data with 190 points.
. . . . . . . . . . . . . . . . .
14
(cid:3)
MIT, Department of Mathematics, room 2-337, Cambridge, MA 02139.
1
Stability of Numerical Schemes for PDE’s.
2
MIT, Friday February 12, 1999 | Rosales.
1 Naive Scheme for the Wave Equation.
We will illustrate the points we want to make with the wave equation (in one space dimension)
2
2
@
u
@
u
(cid:0)
= 0 :
(1.1)
2
2
@ t
@x
Since this equation is second order in time, it needs two initial conditions. For example:
u(x; 0) = u
(x)
and
(x; 0) = v
(x) :
(1.2)
0
0
@ t
@u
We will assume here that both u
and v
are periodic, with some period T > 0. Then the solution
0
0
of (1.1) is periodic in x with the same period: u(x + T ; t) = u(x; t).
Remark 1.1 We note that, in fact, we can write the solution of this problem explicitly
1
x
t
+
u =
u
(x (cid:0) t) + u
(x + t) +
v
(s)ds
:
0
0
0
2
(cid:0)
x
t
(cid:18)
(cid:19)
Z
However, this is not the point here (see below).
Operate now as if (1.1) were complicated enough that we needed to solve the equation numerically.
For this purpose introduce a numerical grid fx
; t
g | where n and j are integers, as follows
n
j
x
= x
+ n(cid:1)x and t
= j(cid:1)t :
(1.3)
n
j
0
Here (cid:1)x and (cid:1)t are some \small" positive constants and x
is arbitrary. Next replace the function
0
u = u(x; t) of the continuum variables x and t by a discrete double sequence fu
g, where
n
j
j
u
= u(x | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
t) of the continuum variables x and t by a discrete double sequence fu
g, where
n
j
j
u
= u(x
; t
) :
(1.4)
n
n
j
Finally, introduce the new variable v =
to re-write equation (1.1) as a (cid:12)rst order in time system
@u
@ t
@u
@ v
@
u
2
= v
and
=
:
(1.5)
@ t
@ t
@x
2
j
j
In view of (1.4) it is now clear that u
(and the similarly de(cid:12)ned v
) should satisfy
n
n
j
j
+1
j
j
j
+1
j
j
u
(cid:0) u
v
(cid:0) v
u
(cid:0) 2u
+ u
n
n
n
n
+1
n
1
j
n
n
2
(cid:0)
= v
+ O((cid:1)t) and
=
+ O((cid:1)t; ((cid:1)x)
) ;
(1.6)
n
2
(cid:1)t
(cid:1)t
((cid:1)x)
which can be checked by expanding u
, u
, . . . in Taylor series centered at (x
; t
) | using (1.4)
n
j
n
+1
n
j
+1
j
| and substituting the expansions in (1.6). This suggests the following numerical scheme, allowing
simple calculation of the solution at time t = t
(once it is known at time t = t
)
j
+1
j
j
j
j
j
j
j
+1
+1
(cid:1)t
j
j
u
= u
+ (cid:1)t v
and v
= v
+
u
(cid:0) 2u
+ u
;
(1.7)
n
n
n
n
n
+1
n
1
n
n
(cid:0)
2
((cid:1)x)
2
(cid:16)
(cid:17)
where the errors should be of size O((cid:1)t | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
)
2
((cid:1)x)
2
(cid:16)
(cid:17)
where the errors should be of size O((cid:1)t; ((cid:1)x)
), that is: small.
Upon implementation one quickly discovers that this algorithm is disastrously bad. The MatLab
scripts: InitGBNS, lectureGBNS, demoGBNS, movieGBNS and the help (cid:12)le readmeGBNS in the Athena
18311-Toolkit all deal with this scheme and another one to be introduced later in these notes. In
particular, lectureGBNS goes through and explains a series of calculations showing the details of
how the scheme fails. We illustrate here the problem with a couple of examples.
Stability of Numerical Schemes for PDE’s.
3
MIT, Friday February 12, 1999 | Rosales.
Example 1.1 Consider the fol lowing initial data (with period T = 2) for equation (1.5):
u(x; 0) = u
(x) =
(1 + cos((cid:25) x))
and v (x; 0) = v
(x) (cid:17) 0 :
(1.8)
0
0
2
1
The exact solution: u =
(2 + cos((cid:25) (x (cid:0) t)) + cos((cid:25) (x + t))) =
(1 + cos((cid:25) x) cos((cid:25) t)) | see
1
1
4
2
remark 1.1 | is clearly also periodic in time of period 2 (a standing wave). For the numerical
solution we take (cid:1)x = 2 (cid:1)t = 2=N (for some \large" N ) and x
= (cid:0)1 in (1.3). Then we im-
0
plement (1.7) for 1 (cid:20) n (cid:20) N (the periodicity of the solution means that the indexes n + N and n
are equivalent) and solve the equations over one time period: 0 (cid:20) t (cid:20) 2.
Numerical solution u with N = 40 points
.
)
t
,
x
(
u
=
u
n
o
i
t
u
o
S
l | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
40 points
.
)
t
,
x
(
u
=
u
n
o
i
t
u
o
S
l
1.5
1
0.5
0
-0.5
2
1.5
1
0.5
1
0.5
0
-0.5
Time t --- dt=1/N.
0
-1
Space x --- dx=2/N.
Figure 1.1: Solution of (1.5) with initial data (1.8) using (1.7) with 40 points in the
space grid. To avoid an over-dense graph not all the points in the numerical grid are
plotted. However, enough points to show all the relevant details are kept.
Figure 1.1 shows the result of this calculation using N = 40. Note that the periodicity in time fails
to hold. In fact, after one time period the numerical method appears to have ampli(cid:12)ed the initial
Stability of Numerical Schemes for PDE’s.
4
MIT, Friday February 12, 1999 | Rosales.
data by about 30%! However, maybe this is not so bad (or is it?); after al l the value of N being
used is not that large and the numerical solution looks otherwise quite reasonable.
Let us now check what happens as we increase the resolution (larger N). Any reasonable numerical
scheme ought to give a better approximation when we do this. Figure 1.2 shows the result of in-
creasing N to N = 57 (a rather smal l increase). The new approximation is not only not better; it
is a disaster. By time t (cid:25) 2, O(1) grid scale (i.e. wavelength = 2 (cid:1)x) oscil lations appear in the
numerical solution, making it useless. As we wil l soon see, the scheme is amplifying the errors; the
30% ampli(cid:12)cation of the initial cosine wave seen when using N = 40 was just a forewarning of what
happens for larger N . As N is made even larger, the oscil lations generated become huge (in fact,
Numerical solution u with N = 57 points
.
)
t
,
x
(
u
=
u
n
o | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
(in fact,
Numerical solution u with N = 57 points
.
)
t
,
x
(
u
=
u
n
o
i
t
u
o
S
l
1.5
1
0.5
0
-0.5
2
1.5
1
0.5
1
0.5
0
-0.5
Time t --- dt=1/N.
0
-1
Space x --- dx=2/N.
Figure 1.2: Solution of (1.5) with initial data (1.8) using (1.7) with 57 points in the
space grid. To avoid an over-dense graph not all the points in the numerical grid are
plotted. However, enough points to show all the relevant details are kept.
their size increases exponential ly with N , as we wil l soon show). This is il lustrated by (cid:12)gure 1.3,
which corresponds to N = 80. Here (instead of a 3D graph) we plot the numerical solution at time
t = 2. Grid scale (wavelength = 2(cid:1)x) oscil lations is al l that can be seen in this graph | notice the
(very large) vertical scale on this (cid:12)gure!
Stability of Numerical Schemes for PDE’s.
5
MIT, Friday February 12, 1999 | Rosales.
Final ly, we point out that if (instead of increasing N ) we compute for longer times, the same e(cid:11)ect
of large amplitude grid scale oscil lations arising (which grow exponential ly in time) is observed.
.
)
t
,
x
(
u
=
u
n
o
i
t
u
o
S
l
7 Numerical solution u with N = 80 points
x 10
1.5
1
0.5
0
-0.5
-1
-1.5
-1
-0.8
-0.6
0
Space x --- dx=2/N. Solution for time t = 2
-0.4
-0.2
0.4
0.6
0.2
0.8
1
Figure 1.3: Solution of (1.5 | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
2
0.4
0.6
0.2
0.8
1
Figure 1.3: Solution of (1.5) with initial data (1.8) using (1.7) with 80 points
in the space grid. Notice the large amplitude grid scale oscillations generated by
the scheme. There is nothing but numerical noise in this picture!
Example 1.2 In a second example we take the fol lowing Gaussian initial data for equation (1.5)
u(x; 0) = u
(x) = exp((cid:0)a ln(10) x
)
and v (x; 0) = v
(x) (cid:17) 0 ;
(1.9)
0
0
2
for (cid:0)1 (cid:20) x (cid:20) 1, where a > 0 is a constant. We extend this to periodic initial data (of period T = 2)
by repeating the above pro(cid:12)les over each interval (2n (cid:0) 1) (cid:20) x (cid:20) (2n + 1), with n integer. These
initial values are not smooth | as were the ones in the prior example. There is a smal l corner in
u
(x), whenever x is an odd integer (in particular for x = (cid:6)1). This is because at these points there
0
is a cut-o(cid:11) from a Gaussian centered at x (cid:0) 1 to one centered at x + 1. Notice that the size of the
miss-match in the derivatives of u
goes down very rapid ly as a increases.
0
Stability of Numerical Schemes for PDE’s.
6
MIT, Friday February 12, 1999 | Rosales.
For the numerical solution we take x
= (cid:0)1, (cid:1)x = 0:02 and (cid:1)t = 0:01 in (1.3) | this corresponds
0
to N = 100 in the notation of example 1.1 | and use (1.7) to solve the equations for 0 (cid:20) t (cid:20) 0:5.
This is very similar to what we did in the prior example, except that here we | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
cid:20) t (cid:20) 0:5.
This is very similar to what we did in the prior example, except that here we vary the initial
conditions (by changing the parameter a) instead of changing the resolution with variations in N .
In the (cid:12)rst calculation, we take a relatively large a, namely a = 10. Figure 1.4 shows the result of
this calculation, which appears quite reasonable.
Numerical solution u with a = 10.
1
.
)
t
,
x
(
u
=
u
n
o
i
t
u
o
S
l
0.8
0.6
0.4
0.2
0
0.5
0.4
0.3
0.2
Time t --- dt=0.01.
0.1
0
-1
1
0.5
0
-0.5
Space x --- dx=0.02.
Figure 1.4: Solution of (1.5) with initial data (1.9) using (1.7) with 100 points in
the space grid and a = 10. To avoid an over-dense graph not all the points in the
numerical grid are plotted (enough points to show all the relevant details are kept).
In the second calculation, we take a smal ler value a = 6. This makes the corners more substantial
(though stil l pretty weak). Figure 1.5 shows the result of this last calculation, which is now not
reasonable at al l.
It is quite clear that, just as in the prior example, the smal l errors that are
triggered by the corners are ampli(cid:12)ed by the scheme (so we observe grid scale oscil lations near
x = (cid:6)1 towards the end of the run).
Final ly, we point out that, if the calculations are run for times longer than 0 (cid:20) t (cid:20) 0:5, even the one
with a = 10 eventual ly shows grid scale oscil lations. These grow exponential ly in time and pretty
soon dominate the whole solution (not just the neighborhood of x = (cid:6)1) with huge amplitudes.
Stability of Numerical Schemes for PDE’s.
7
MIT, Friday February | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
:6)1) with huge amplitudes.
Stability of Numerical Schemes for PDE’s.
7
MIT, Friday February 12, 1999 | Rosales.
Numerical solution u with a = 6.
.
)
t
,
x
(
u
=
u
n
o
i
t
l
u
o
S
1
0.5
0
-0.5
-1
0.5
0.4
0.3
0.2
Time t --- dt=0.01.
0.1
0
-1
1
0.5
0
-0.5
Space x --- dx=0.02.
Figure 1.5: Solution of (1.5) with initial data (1.9) using (1.7) with 100 points in
the space grid and a = 6. To avoid an over-dense graph not all the points in the
numerical grid are plotted (enough points to show all the relevant details are kept).
The next section gives a detailed explanation of why this is happening.
2 von Neumann stability analysis for PDE’s.
In this section we introduce the von Neumann stability analysis technique, that can be used to
analyze numerical schemes and predict when the behavior observed in the prior section will occur.
There are two basic concepts useful in understanding numerical schemes. These are the notions of
consistency and stability. For a numerical scheme to be useful it must be both consistent and
stable. It is very important to realize that these two notions are independent.
Consistency simply means that, as (cid:1)x and (cid:1)t vanish, the solutions of the equation must satisfy
the numerical scheme with errors that vanish. This is in fact what equation (1.6) tells us about
the scheme in (1.7). Consistency guarantees that the scheme truly approximates the equation we
intend to solve with it (and not something else).
Stability of Numerical Schemes for PDE’s.
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MIT, Friday February 12, 1999 | Rosales.
Stability simply means that the scheme does not amplify errors. Obviously this is very important,
since errors are impossible to avoid in any numerical calculation. In fact, even in the ideal case
of in(cid:12)nite precision, we still have to deal with discretization errors | i.e. | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
. In fact, even in the ideal case
of in(cid:12)nite precision, we still have to deal with discretization errors | i.e. the O terms in (1.6).
Clearly, if errors are ampli(cid:12)ed, pretty soon they will dominate any computation (making it useless).
As it turns out, for linear constant coeÆcient schemes such as (1.7), a complete stability
analysis is possible, because the numerical algorithm equations can be solved exactly by separa-
tion of variables. This means then that any solution of the scheme can be written as a superposition
of Fourier modes. These Fourier modes are solutions of the form
j
j
ikn
j
j
ikn
u
= U G
e
and v
= V G
e
;
(2.1)
n
n
where U , V , G and k are constants (with k real). Generally double sequences like this will be solu-
tions provided G, U and V are restricted by some functional relations of the form G = G(k ; (cid:1)x; (cid:1)t),
U = U (k ; (cid:1)x; (cid:1)t) and V = V (k ; (cid:1)x; (cid:1)t) | below we carry through the calculations for the speci(cid:12)c
example of (1.7).
G is called the Growth Factor. It is clear that:
for stability kGk (cid:20) 1 is needed for all k.
(2.2)
Else some modes will be ampli(cid:12)ed by a factor G in each time step, eventually dominating the
solution. A scheme is called stable if the stability condition kGk (cid:20) 1 can be satis(cid:12)ed with (perhaps)
a restriction on the time step of the form 0 < (cid:1)t (cid:20) (cid:28) ((cid:1)x), where (cid:28) is a positive function of its
argument. Notice that restrictions of this latter form allow arbitrarily small time and space steps,
which are needed to be able to compute the solution with any required degree of accuracy (how
small is determined by how well consistency is satis(cid:12)ed, which determines the size of the errors for
any given (cid:1)t and ( | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
determined by how well consistency is satis(cid:12)ed, which determines the size of the errors for
any given (cid:1)t and (cid:1)x).
Remark 2.1 The parameter k is the wavenumber of the mode, related to the wavelength (cid:21) in
1
space
by (cid:21) = (2(cid:25)(cid:1)x)=k. For the particular case of periodic problems (such as the ones consid-
ered in examples 1.1 and 1.2), the Fourier modes (2.1) must also satisfy the periodicity condition.
That is, one must have (cid:21) = T =‘, where ‘ is an integer and T is the period in space. Since in this
case one would normal ly take (cid:1)x = T =N , where N is a large natural number, the acceptable values
for k end up restricted to the set
2 (cid:25) (cid:1)x
2(cid:25)
T
k = k
=
‘ =
‘
and (cid:21) = (cid:21)
=
; with
0 (cid:20) ‘ < N :
(2.3)
‘
‘
T
N
‘
Here the upper bound N on ‘ fol lows from the fact that k
and k
give the same Fourier mode in
‘
‘
N
+
(2.1); thus there is no reason to keep both.
We note that (due to the fact that the numerical scheme only samples the solution at a discrete set
fx
g of points in space) there is a certain trickiness in the interpretation of the wavelengths
n
(cid:21)
above. Clearly, ‘ = 0 corresponds to a solution independent of x and ‘ = 1 corresponds to the
‘
fundamental mode with wavelength T in x. As ‘ continues to increase harmonics of this fundamental
mode appear, with wavelengths T =2, T =3 . . . However, this process cannot continue forever, since
1
k
Write the argument
in the exponentials in (2.1) as
=
(
), using (1.3).
n
0
kn
kn
x
x
(cid:0)
x
(cid:1)
Stability of Numerical Schemes for PDE’s.
9
MIT, Friday February 12, 1999 | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
)
x
(cid:1)
Stability of Numerical Schemes for PDE’s.
9
MIT, Friday February 12, 1999 | Rosales.
the numerical grid cannot resolve arbitrarily smal l wavelengths. In fact, the shortest wavelength that
can be resolved corresponds to ‘ = N=2 with (cid:21)
= 2 (cid:1)x (grid size oscillations, with period 2 in n: the
‘
solution alternates between two values on the grid). To see this recal l that k
and k
give the same
‘
‘
N
+
Fourier mode in (2.1). Thus the mode (N (cid:0) ‘) has the same wavelength as the mode (cid:0)‘, i.e. T =‘.
This means that, after ‘ = N=2 the wavelengths start increasing, to reach back the fundamental
mode at ‘ = N (cid:0) 1. Each wavelength then actual ly appears twice in the range 1 < ‘ < N .
We should not be too surprised by the fact that each wavelength appears twice in the range 1 < ‘ < N .
Notice that the modes in (2.1) are complex valued (except when k is a multiple of 2(cid:25)). Thus, to be
real valued any solution should include both the modes and their complex conjugates. However, the
mode conjugate to the one with k = k
above in (2.3) is the mode with k = k
, which is precisely
‘
‘
(cid:0)
the same as the mode with k = k
.
N
‘
(cid:0)
In any numerical calculation it is the modes with wavelengths of the order of the grid size (cid:1)x
(i.e. ‘ close to N=2) that are worrisome in terms of instabilities. These modes cannot be expected
to represent accurately any true feature of the real solution one is trying to compute
and should
2
not have any signi(cid:12)cant presence in the numerical solution. Thus, it is very important that they
not be ampli(cid:12)ed by the scheme. In fact, general ly it is desirable to have them damped, since they
mostly represent numerical "noise" generated by al l the approximations implicit in any numerical
calculation.
On the other hand, the modes with wavelengths much bigger | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
mostly represent numerical "noise" generated by al l the approximations implicit in any numerical
calculation.
On the other hand, the modes with wavelengths much bigger than (cid:1)x (that is, ‘ (cid:25) 0 or ‘ (cid:25) N in
(2.3)) should be treated "accurately" by the scheme. By this we mean that their time evolution
(given by the factors G
in (2.1)) should be as close as possible to the one provided by the PDE the
j
scheme approximates. This is what consistency is al l about.
Consider now the special case of the algorithm (1.7). To see under which conditions (2.1)
is a solution, substitute this form into (1.7). Dividing by the common factor G
e
it follows that
j
ikn
G U = U + (cid:1)t V
and G V = V +
(e
(cid:0) 2 + e
) U :
2
((cid:1)x)
(cid:1)t
(cid:0)
ik
ik
Clearly an eigenvalue equation A Y = G Y, with eigenvalue G, eigenvector Y = (U; V )
and matrix
T
of coeÆcients
1
(cid:1)t
A =
:
(cid:1)
t
2
k
(cid:0)4
sin
(
)
1
2
((cid:1)
)
2
x
!
From the characteristic equation det(A (cid:0) G) = 0, then
(cid:1)t
1
G = 1 (cid:6) 2 i
sin(
k) :
(2.4)
(cid:1)x
2
It is clear that, for (1.7) there is no stability, since (2.4) yields
2
(cid:1)t
1
2
kGk
= 1 +
2
sin(
k)
;
(2.5)
(cid:1)x
2
(cid:18)
(cid:19)
which is always bigger than one.
2
Recall (1.4), which makes sense in terms of approximating the solution only if (cid:1)
is much smaller than any
x
distance over which the solution changes signi(cid:12)cantly.
Stability of Numerical Schemes for | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
is much smaller than any
x
distance over which the solution changes signi(cid:12)cantly.
Stability of Numerical Schemes for PDE’s.
10
MIT, Friday February 12, 1999 | Rosales.
Notice that the maximum ampli(cid:12)cation for the scheme (1.7) occurs | as follows from (2.5)
| for k = (cid:25) . This corresponds to ‘ = N=2 in (2.3), i.e.: grid size oscillations with (cid:21) = 2 (cid:1)x.
In this case
where (cid:17) = ((cid:1)t=(cid:1)x)
: For (1.7), the amplitude of the grid size oscillations grows like G
. Thus
M
2
j
q
kGk = G
=
1 + 4 (cid:17) ;
(2.6)
M
we can write for the ampli(cid:12)cation factor A
= A
(t) (for the period 2(cid:1)x mode)
2
2
ln(G
)
M
A
= exp(t
) ;
(2.7)
2
(cid:1)t
where we have used j = t=(cid:1)t. In particular (in examples 1.1 and 1.2 earlier) we took (cid:1)x = 2 (cid:1)t
and (cid:1)t = 1=N , so that
ln 2
N t
2
A
= exp(
N t) = 2
:
(2.8)
2
2
We will now use these results to explain the behavior observed earlier in (cid:12)gures 1.1 through 1.5.
Remark 2.2 Consider (cid:12)rst example 1.1, with the initial data for scheme (1.7) given by
0
0
1
2n(cid:25)
u
=
1 (cid:0) cos(
)
and v
= 0 :
n
n
2
N
(cid:18)
(cid:19)
These data correspond to a superposition of just three modes in (2.1), with k = k
, k = k
and
0
1
k = k
(cid:24) k
in (2.3 | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
), with k = k
, k = k
and
0
1
k = k
(cid:24) k
in (2.3). Thus, the exact solution for the scheme equations is rather simple
1
1
N
(cid:0)
(cid:0)
and has the form
j
j
1
g
+ (cid:22)g
2n(cid:25)
g
(cid:0) (cid:22)g
2n(cid:25)
(cid:25)
j
j
j
j
u
=
1 (cid:0)
cos(
)
and v
=
^v cos(
) ;
for
g = 1 + i sin(
) ; (2.9)
n
n
2
2
N
2i
N
N
!
where ^v is a constant and (cid:22)g denotes the complex conjugate of g. Of course, g and (cid:22)g are the values
G in (2.4) takes for k = k
= 2(cid:25)=N .
1
Notice that the exact solution (2.9) does not exhibit any catastrophic growth of grid size oscil lations,
as was observed in example 1.1. However, the results displayed in (cid:12)gures 1.1 through 1.3 do not
correspond to the exact solution above but to actual computations using the scheme in (1.7) | which
were done using double precision (cid:13)oating point arithmetic (MatLab’s default). The round o(cid:11) errors
introduced by the (cid:12)nite precision of the calculations introduces (very smal l) perturbations into the
exact solution above, which the scheme then evolves in time just as if they were part of the solution.
To understand what the scheme does with the perturbations introduced by the (cid:12)nite precision, de-
compose them into a sum over the modes in (2.1). This sum wil l general ly include all the modes,
in particular the highly ampli(cid:12)ed ones with grid size wavelengths. Consider then what would happen
with the solution of the scheme if we add to the initial data above
a smal l amount of the component
3
corresponding to the maximum ampli(cid:12)cation rate above in (2.6). | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
a smal l amount of the component
3
corresponding to the maximum ampli(cid:12)cation rate above in (2.6). Let the amplitude of this compo-
nent be (cid:15), where (cid:15) has (roughly) the size of the expected errors. Actual ly, (cid:15) should be a little smal ler
than the round o(cid:11) errors that occur, since not al l the errors get projected into the fastest growing
modes. Thus take (cid:15) = O(10
) as a good ballpark (cid:12)gure for the calculations in section 1
(cid:0)
17
and use (2.8) above to explain the behavior observed in (cid:12)gures 1.1 through 1.3, as fol lows:
3
Which has only components corresponding to
= 0,
= 1 and
=
1 in (2.3).
‘
‘
‘
N
(cid:0)
Stability of Numerical Schemes for PDE’s.
11
MIT, Friday February 12, 1999 | Rosales.
1. First, for N = 40, (2.8) gives A
(cid:25) 1:1 (cid:2) 10
for the (cid:12)nal time t = 2. This is not enough to
2
12
compensate for the smal lness of (cid:15) and the numerical solution is wel l described by (2.9).
Notice that (2.9) is not periodic in time; since the wave amplitude in u behaves like Re(g
),
j
which grows as j grows. In fact, 2 N = 80 steps are needed to reach the (cid:12)nal time t = 2 and
it is easy to check that
80
80
(cid:25)
Re(g
) = Re
1 + i sin(
)
(cid:25) 1:28 :
(
)
40
(cid:18)
(cid:19)
This agrees quite wel l with the (cid:25) 30% growth in the wave amplitude observed in (cid:12)gure 1.1.
2. Second, for N = 57, (2.8) gives A
(cid:25) 1:4 (cid:2) | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
.
2. Second, for N = 57, (2.8) gives A
(cid:25) 1:4 (cid:2) 10
for the (cid:12)nal time t = 2. This is about the
2
17
(cid:0)
1
same as (cid:15)
and agrees with the fact that grid oscil lations of O(1) amplitude are observed in
(cid:12)gure 1.2.
3. Third, for N = 80, (2.8) gives A
(cid:25) 1:2 (cid:2) 10
for the (cid:12)nal time t = 2. This is about 10
2
24
7
(cid:0)
1
times bigger than (cid:15)
, which (again) agrees pretty wel l with the observed amplitude of the grid
size oscil lations in (cid:12)gure 1.3.
4. Final ly, it is not just the mode with ‘ = N=2 in (2.3) that gets a large ampli(cid:12)cation factor by
the scheme. Al l the ones with ‘ (cid:25) N=2 do and should thus be present in the solution. It is
wel l known that when sinusoidals with close wavenumbers are added, "beats" with wavenumbers
equal to the di(cid:11)erence in wavenumbers occur. Thus, in this case we should observe "beats" with
wavenumbers low multiples of k
= 2(cid:25)=N | which, indeed, are quite obvious in (cid:12)gure 1.3.
1
Remark 2.3 Now consider example 1.2, where N = 100 and 0 (cid:20) t (cid:20) 0:5. Then, for the time
t = 0:5, equation (2.8) gives A
(cid:25) 3:4 (cid:2) 10
:
2
7
In this case the initial data has components in al l the modes 0 (cid:20) ‘ < N in (2.3).
In fact, be-
cause of the corners at x = (cid:6)1, the amplitude present in the higher modes | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
in (2.3).
In fact, be-
cause of the corners at x = (cid:6)1, the amplitude present in the higher modes is relatively large. The
strength of these corners can be measured by the jump in the derivative of the initial data there:
J (a) = 4 a ln(10) 10
. For moderate
size a, J (a) pretty much determines how much amplitude
(cid:0)
a
4
there is in the higher modes. Now J (10) (cid:25) 9:2 (cid:2) 10
and J (6) (cid:25) 5:5 (cid:2) 10
. Thus, from the value
(cid:0)
(cid:0)
9
5
of A
above, it should be clear why in (cid:12)gure 1.4 (corresponding to a = 10) the solution exhibits no
2
detectable oscil lations, while in (cid:12)gure 1.5 (corresponding to a = 6) they show up.
Notice that in this case it is also true that it is not just the mode with ‘ = N=2 in (2.3) that gets a
large ampli(cid:12)cation factor by the scheme. Al l the neighboring ones are also present. However, now
their amplitudes and phases are al l correlated because they (mostly) are generated by the corner in
the initial data. Thus they interfere with each other in ways subtler than the mere beating observed in
the prior example; i.e.: the pattern of grid size oscil lations has a clear maximum near the positions
of the corners in (cid:12)gure 1.5.
In the next section we will discuss a simple strategy to stabilize numerical schemes, to get rid
of numerical oscillations and other undesirable e(cid:11)ects. The strategy is based on the introduction
of arti(cid:12)cial (numerical) dissipation to (selectively) damp the higher modes, without signi(cid:12)cantly
a(cid:11)ecting the lower modes (where a consistent scheme should behave properly | see remark 2.4).
4
a
When
is large, the corner is very weak and the dominant contribution to the mode amplitudes comes from the
smooth | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
.4).
4
a
When
is large, the corner is very weak and the dominant contribution to the mode amplitudes comes from the
smooth part of the initial data (which yields very little amplitude in the high modes).
Stability of Numerical Schemes for PDE’s.
12
MIT, Friday February 12, 1999 | Rosales.
Remark 2.4 Final ly, going back now to the last paragraph in remark 2.1, consider the behavior of
G in (2.4) for k smal l. Namely
G = 1 (cid:6) i
k + O
k
:
(2.10)
(cid:1)t
(cid:1)t
3
(cid:1)x
(cid:1)x
(cid:18)
(cid:19)
This should be compared with the behavior of the exact solution for the wave equation (1.1) | see
remark 1.1 | which evolves Fourier modes according to the rule
u / exp
i
(x
(cid:6) t
)
/ exp
i
kn (cid:6)
kj
:
n
j
k
(cid:1)t
(
)
(cid:1)x
(cid:1)x
(cid:26)
(cid:18)
(cid:19)(cid:27)
Thus the exact evolution corresponds to a factor G given by
G
= exp
(cid:6)i
k
= 1 (cid:6) i
k + O
(
k)
:
(2.11)
exact
(cid:18)
(cid:19)
(cid:18)
(cid:19)
(cid:1)x
(cid:1)x
(cid:1)x
(cid:1)t
(cid:1)t
(cid:1)t
2
This should be compared with (2.10) above. It is clear then that (for k smal l) G is correct up to
smal l terms in k, which is an alternative way of verifying that the scheme (1.7) is consistent.
3 Numerical Viscosity and Stabilized Scheme.
FILL IN HERE THE GOOD SCHEME EQUATIONS.
(3.1)
Notation used for Good Scheme in MatLab: (cid:17) = ((cid:1)t=(cid:1)x)
and (cid:23) = (cid:1)t=(cid:1)x
.
2
2
Next the | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
)t=(cid:1)x)
and (cid:23) = (cid:1)t=(cid:1)x
.
2
2
Next the (cid:12)gures that go with the good scheme.
4 Reference.
For more information regarding stability of numerical schemes (and many other useful numerical
topics) a good all-around practical reference is Numerical Recipes, The Art of Scienti(cid:12)c Computing
by W. H. Press, S. A. Teukolsky, W. T. Vetterling and B. P. Flannery. Cambridge U. Press, New
York, 1992.
Stability of Numerical Schemes for PDE’s.
13
MIT, Friday February 12, 1999 | Rosales.
Numerical solution u with N = 55 points
1
.
)
t
,
x
(
u
=
u
n
o
i
t
u
o
S
l
0.8
0.6
0.4
0.2
0
2
1.5
1
0.5
1
0.5
0
-0.5
Time t --- dt=1/N.
0
-1
Space x --- dx=2/N.
Figure 3.1: Solution of (1.5) with initial data (1.8) using the corrected scheme (3.1)
with 55 points in the space grid. To avoid an over-dense graph not all the points
in the numerical grid are plotted. However, enough points to show all the relevant
details are kept.
Stability of Numerical Schemes for PDE’s.
14
MIT, Friday February 12, 1999 | Rosales.
Numerical solution u with N = 190 points
1
.
)
t
,
x
(
u
=
u
n
o
i
t
u
o
S
l
0.8
0.6
0.4
0.2
0
2
1.5
1
0.5
1
0.5
0
-0.5
Time t --- dt=1/N.
0
-1
Space x --- dx=2/N.
Figure 3.2: Solution of (1.5) with initial data (1.8) using the corrected scheme (3.1)
with 190 points in | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
.2: Solution of (1.5) with initial data (1.8) using the corrected scheme (3.1)
with 190 points in the space grid. To avoid an over-dense graph not all the points
in the numerical grid are plotted. However, enough points to show all the relevant
details are kept. | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
Lecture 10: Solving the Time-Independent Schr¨odinger Equation
Contents
1 Stationary States
2 Solving for Energy Eigenstates
3 Free particle on a circle.
1 Stationary States
B. Zwiebach
March 14, 2016
1
3
6
Consider the Schr¨odinger equation for the wavefunction Ψ(x, t) with the assumption that the potential
energy V is time independent:
i(cid:126)
∂Ψ
∂t
ˆ= HΨ(x, t) =
(cid:18)
(cid:0)
−
(cid:126)2 ∂2
2m ∂x2
(cid:19)
+ V (x) Ψ(x, t) ,
(1.1)
where we displayed the form of the Hamiltonian operator H with the time independent potential
V (x). Stationary states are a very useful class of solutions of this differential equation. The signature
property of a stationary state is that the position and the time dependence of the wavefunction
factorize. Namely,
ˆ
Ψ(x , t) = g(t) ψ(x) ,
(1.2)
for some functions g and ψ. For such a separable solution to exist we need the potential to be time
independent, as we will see below. The solution Ψ(x, t) is time-dependent but it is called stationary
because of a property of observables. The expectation value of observables with no explicit time
dependence in arbitrary states has time dependence. On a stationary state they do not have time
dependence, as we will demonstrate.
Let us use the ansatz (1.2) for Ψ in the Schr¨odinger equation. We then find
(cid:19)
(cid:18) dg(t)
dt
i(cid:126)
ψ(x) = g(t) Hψ(x) ,
ˆ
(1.3)
because g(t) can be moved across H. We can then divide this equation by Ψ(x, t) = g(t)ψ(x), giving
ˆ
i(cid:126)
1 dg(t)
g(t) dt
=
1
ψ(x)
ˆ
Hψ(x) .
(1.4)
The left side is a function of only t, while the right side is a function of only | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
�
Hψ(x) .
(1.4)
The left side is a function of only t, while the right side is a function of only x (a time dependent
potential would have spoiled this). The only way the two sides can equal each other for all values of t
and x is for both sides to be equal to a constant E with units of energy because H has units of energy.
We therefore get two separate equations. The first reads
ˆ
i(cid:126) = Eg .
dg
dt
1
(1.5)
This is solved by
g(t) = e(cid:0)iEt/(cid:126) ,
−
(1.6)
and the most general solution is simply a constant times the above right-hand side. From the x-
dependent side of the equality we get
ˆHψ(x) = Eψ(x) .
(1.7)
This equation is an eigenvalue equation for the Hermitian operator H. We showed that the eigenvalues
of Hermitian operators must be real, thus the constant E must be real. The equation above is called
the time-independent Schr¨odinger equation. More explicitly it reads
ˆ
(cid:18) (cid:126)2 d2
2m dx2
−
(cid:0)
(cid:19)
+ V (x)
ψ(x) = Eψ(x) ,
(1.8)
Note that this equation does not determine the overall normalization of ψ. Therefore we can write
the full solution without loss of generality using the g(t) given above:
Stationary state: Ψ(x, t) = e(cid:0)iEt/ ψ(x) , with E 2 R and Hψ = Eψ .
∈
−
ˆ
(cid:126)
(1.9)
Note that not only is ψ(x) an eigenstate of the Hamiltonian operator H, the full stationary state is
also an H eigenstate
ˆ
ˆ
ˆHΨ(x, t) = EΨ(x, t) ,
(1.10)
since the time dependent function in Ψ cancels out.
We have noted that the energy E must be real. If it was not we would also have trouble normalizing
the stationary state consisten
tly. The normalization condition for Ψ, if E is not real | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
it was not we would also have trouble normalizing
the stationary state consisten
tly. The normalization condition for Ψ, if E is not real, would give
(cid:90)
∗
1 = dx Ψ(cid:3)
= ei(E∗(cid:0)E)t/(cid:126) (cid:90)
−
(x, t)Ψ(x, t) =
(cid:90)
dx eiE t/ e(cid:0)iEt/ ψ(cid:3)(x)ψ(x)
−
∗
∗ (cid:126)
(cid:126)
dx ψ(cid:3)
∗
(cid:126)
(x)ψ(x) = e2 Im(E)t/
(cid:90)
dx ψ(cid:3)(x)ψ(x).
∗
(1.11)
The final expression has a time dependence due to the exponential.
On the other hand the normal-
ization condition states that this expression must be equal to one. It follows that the exponent must
be zero, i.e., E is real. Given this, we also see that the normalization condition yields
(cid:90) 1
∞
(cid:0)1
−∞
dx ψ(cid:3)(x)ψ(x) = 1 .
∗
(1.12)
How do we interpret the eigenvalue E? Using (1.10) we see that the expectation value of H on the
ˆ
state Ψ is indeed the energy
(cid:90)
h ˆ i
(cid:3)
∗
H Ψ = dx Ψ (x, t)HΨ(x, t) =
(cid:105)
(cid:104)
ˆ
(cid:90)
dxΨ(cid:3)(x, t)EΨ(x, t) = E
∗
(cid:90)
dxΨ(cid:3)(x, t)Ψ(x, t) = E,
∗
(1.13)
Since the stationary state is an eigenstate of H, the uncertainty ∆H of the Hamiltonian in a stationary
state is zero.
ˆ
ˆ
There are two important observations on stationary states:
2
(1) The expectation value of any time-independent operator Q on a stationary state Ψ is time-
ˆ
independent:
(cid:90) | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
(1) The expectation value of any time-independent operator Q on a stationary state Ψ is time-
ˆ
independent:
(cid:90)
(cid:105)
hQi
(cid:104)
Ψ(x,t)
=
ˆ
dx Ψ(cid:3)(x, t)QΨ(x, t) =
∗
(cid:90)
(cid:90)
= dx e
e
iEt/(cid:126) (cid:0)iEt/(cid:126) (cid:3)
∗
−
∗
(cid:3)
ψ (x)Qψ(x) = dx ψ (x)Qψ(x) = hQiψ(x) ,
(cid:104)
(cid:105)
ˆ
ˆ
dx eiEt/(cid:126)
(cid:90)
∗
ψ(cid:3)
(x)Qe(cid:0)iEt/(cid:126)ψ(x)
ˆ
−
(1.14)
since the last expectation value is manifestly time independent.
(2) The superposition of stationary states with different energies not stationary. This is clear because
a stationary state requires a factorized solution of the Schr¨odinger equation:
if we add two
factorized solutions with different energies they will have different time dependence and the
total state cannot be factorized. We now show that that a time-independent observable Q may
have a time-dependent expectation values in such a state. Consider a superposition
ˆ
(cid:126)
(cid:126)
Ψ(x, t) = c e(cid:0)iE1t/ ψ1(x) + c2e(cid:0)iE2t/
−
−
1
ψ2(x),
(1.15)
where ψ1 and ψ2 are H eigenstates with energies E1 and E2, respectively. Consider a Hermitian
operator Q. With the system in state (1.15), its expectation value is
ˆ
ˆ
(cid:90) 1
(cid:90) ∞
(cid:0)1
−∞
1
∞
(cid:90)
(cid:0)1
−∞
1
∞
(cid:90)
hQiΨ =
(cid:105)
(cid:104) | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
(cid:0)1
−∞
1
∞
(cid:90)
hQiΨ =
(cid:105)
(cid:104)
=
=
(cid:0)1
−∞
dx Ψ(cid:3)
∗
ˆ
(x, t)QΨ(x, t)
(cid:0) (cid:3) iE1t/(cid:126) (cid:3)
∗
∗
dx c1e
(cid:3) iE2t/(cid:126) (cid:3)
∗
∗
ψ2(x)
ψ1(x) + c2e
(cid:1)(cid:0)
(cid:0)iE1t/(cid:126) ˆ
−
c1e
Qψ1(x) + c
−
(cid:1)
(cid:126) ˆ
2e(cid:0) 2t/ Qψ2(x)
iE
(cid:16)
dx
c1 ψ1Qψ1 + c2 ψ2Qψ2 + c1c2ei(E1(cid:0)E2)t/ ψ(cid:3)Qψ + c(cid:3)c e(cid:0)i(E1
j
|
∗
2 1
j2 (cid:3) ˆ
∗
|
j2 (cid:3) ˆ
∗
|
∗
1
ˆ
−
−
(cid:3)
∗
2
j
|
(cid:126)
(cid:0)E2)t/
−
(cid:126)
(cid:17)
ˆ
(cid:3)Qψ1
∗
ψ2
(1.16)
We now see the possible time dependence arising from the cross terms. The first two terms are
simple time-independent expectation values. Using the hermitically of Q in the last term we
then get
ˆ
hQiΨ = jc1j2hQi
(cid:105)
|
(cid:104)
| (cid:104)
(cid:105)
2
ψ1 + jc2j hQiψ2
(cid:105)
| (cid:104)
|
(cid:0)E2)t/(cid:126) (cid:90)
1
∞ | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
:105)
| (cid:104)
|
(cid:0)E2)t/(cid:126) (cid:90)
1
∞
−
1c2ei(E1
+ c(cid:3)
∗
(cid:3)Qψ2 + c1c(cid:3)
∗
dx ψ1
ˆ
∗ −
2e(cid:0)i(E1(cid:0)E2)t/(cid:126)
−
(cid:90)
1
∞
(cid:0)1
−∞
ˆ
dx ψ1(Qψ2)(cid:3)
∗
(1.17)
(cid:0)1
−∞
The last two terms are complex conjugates of each other and therefore
hQiΨ = jc1j2hQiψ1 + jc2j2hQiψ2 + 2 Re
(cid:104)
| (cid:104)
| (cid:104)
(cid:105)
(cid:105)
(cid:105)
|
|
(cid:20)
c(cid:3)
1c ei(E1(cid:0)E2)t/
−
∗
2
(cid:90)
1
∞
(cid:126)
(cid:0)1
−∞
ˆ
(cid:3)Qψ2
∗
dx ψ1
(cid:21)
.
(1.18)
We see that this expectation value is time-dependent if E1 = E2 and (ψ1, Qψ2) is nonzero. The
full expectation value hQiΨ is real, as it must be for any Hermitian operator.
(cid:54)=
(cid:104)
(cid:105)
2 Solving for Energy Eigenstates
We will now study solutions to the time-independent Schr¨odinger equation
ˆHψ(x) = E ψ(x).
3
(2.19)
6
ˆ
For a given Hamiltonian H we are interested in finding the eigenstates ψ and the eigenvalues E,
which happen to be the corresponding energies. Perhaps the most interesting feature of the above
equation is that generally the value of E cannot be arbitrary. Just like finite size matrices have a set
of eigenvalues, the above, time-independent Schr¨odinger equation may have a discrete set | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
. Just like finite size matrices have a set
of eigenvalues, the above, time-independent Schr¨odinger equation may have a discrete set of possible
energies. A continuous set of possible energies is also allowed and sometimes important. There are
indeed many solutions for any given potential. Assuming for convenience that the eigenstates and
their energies can be counted we write
ψ1(x) ,
ψ2(x) ,
.
.
.
E1
E2
.
.
.
(2.20)
Our earlier discussion of Hermitian operators applies here. The energy eigenstates can be organized
to form a complete set of orthonormal functions:
(cid:90)
(cid:3)(x)ψj(x) = δij .
∗
ψi
Consider the time-independent Schr¨odinger equation written as
d2ψ
dx2
= (cid:0)
−
2m
(cid:126)2
(E V
(cid:0) (x)) ψ .
−
(2.21)
(2.22)
The solutions ψ(x) depend on the properties of the potential V (x). It is hard to make general state-
ments about the wavefunction unless we restrict the types of potentials. We will certainly consider
continuous potentials. We also consider potentials that are not continuous but are piece-wise contin-
uous, that is, they have a number of discontinuities. Our potentials can easily fail to be bounded.
We allow delta functions in one-dimensional potentials but do not consider powers or derivatives of
delta functions. We allow for potentials that become plus infinity beyond certain points. These points
represent hard walls.
We want to understand general properties of ψ and the behavior of ψ at points where the potential
V (x) may have discontinuities or other singularities. We claim: we must have a continuous
wavefunction. If ψ is discontinuous then ψ0 contains delta-functions and ψ00 in the above left-hand
side contains derivatives of delta functions. This would require the right-hand side to have derivatives
of delta functions, and those would have to appear in the potential. Since we have declared that our
potentials contain no derivatives of delta functions we must indeed have a continuous ψ.
(cid:48)(cid:48)
(cid:48)
Consider now four possibilities concerning the potential:
(1) V (x) is continuous. In this | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
(cid:48)(cid:48)
(cid:48)
Consider now four possibilities concerning the potential:
(1) V (x) is continuous. In this case the continuity of ψ(x) and (2.22) imply ψ00 is also continuous.
(cid:48)(cid:48)
This requires ψ0 continuous.
(cid:48)
(2) V (x) has finite discontinuities. In this case ψ00 has finite discontinuities: it includes the product of
(cid:48)(cid:48)
a continuous ψ against a discontinuous V . But then ψ0 must be continuous, with non-continuous
derivative.
(cid:48)
(3) V (x) contains delta functions. In this case ψ00 also contains delta functions: it is proportional
(cid:48)(cid:48)
to the product of a continuous ψ and a delta function in V . Thus ψ0 has finite discontinuities.
(cid:48)
4
(4) V (x) contains a hard wall. A potential that is finite immediately to the left of x = a and becomes
infinite for x > a is said to have a hard wall at x = a. In such a case, the wavefunction will
vanish for x (cid:21) a. The slope ψ0 will be finite as x ! a from the left, and will vanish for x > a.
Thus ψ0 is discontinuous at the wall.
→
≥
(cid:48)
(cid:48)
In the first two cases ψ0 is continuous, and in the second two it can have a finite discontinuity. In
(cid:48)
conclusion
Both ψ and ψ0 are continuous unless the potential has delta functions
or hard walls in which cases ψ0 may have finite discontinuities.
(cid:48)
(cid:48)
(2.23)
Let us give an slightly different argument for the continuity of ψ and dψ in the case of a potential
dx
with a finite discontinuity, such as the step shown in Fig. 1.
Figure 1: A potential V (x) with a finite discontinuity at x = a.
Integrate both sides of (2.22) a − (cid:15) to a + | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
with a finite discontinuity at x = a.
Integrate both sides of (2.22) a − (cid:15) to a + (cid:15), and then take (cid:15) ! 0. We find
d (cid:18) (cid:19)dψ
dx
dx
2m (cid:90) a+(cid:15)
(cid:126)2
(cid:90) a+(cid:15)
= −
dx (E − V (x))ψ(x) .
dx
→
a (cid:15)
−
a−(cid:15)
(2.24)
The left-hand side integrand is a total derivative so we have
dψ (cid:12)
(cid:12)
(cid:12)
dx a+(cid:15)
(cid:12)
−
(cid:12)dψ
(cid:12)
(cid:12)
dx
(cid:12)
a−(cid:15)
=
(cid:90)
2m a+(cid:15)
(cid:126)2
a−(cid:15)
dx (V (x) − E)ψ(x) .
(2.25)
By definition, the discontinu
side:
Back in (2.25) we then have
ity in the
derivative of ψ at x = a is the limit as (cid:15) ! 0 of the left-hand
→
∆a
(cid:18) dψ (cid:19)
dx
(cid:17) lim
≡
(cid:15)!0
→
(cid:16) dψ
(cid:12)
(cid:12)
(cid:12)
dx a+(cid:15)
(cid:12)
−
dψ
dx
(cid:12)
(cid:12)
(cid:12)
(cid:12)a−(cid:15)
(cid:17)
.
(2.26)
∆a
(cid:18) dψ (cid:19)
dx
2m (cid:90) a+(cid:15)
= lim
(cid:15)!0 (cid:126)2
→
a
−(cid:15)
dx (V (x)
− E)ψ(x) .
(2.27)
The potential V is discontinuous but not infinite around x = a, nor | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
)
− E)ψ(x) .
(2.27)
The potential V is discontinuous but not infinite around x = a, nor is ψ infinite around x = a and,
of course, E is assumed finite. As the integral range becomes vanishingly small about x = a the
integrand remains finite and the integral goes to zero. We thus have
∆a
(cid:19)
(cid:18) dψ
dx
= 0 .
5
(2.28)
There is no discontinuity in
. This gives us one of our boundary conditions.
dψ
dx
To learn about the continuity of ψ we reconsider the first integral of the differential equation. The
integration that led to (2.25) now applied to the range from x0 < a to x yields
dψ(x)
dx
=
dψ (cid:12)
(cid:12)
(cid:12)
dx x0
(cid:12)
2m (cid:90)
(cid:0) (cid:126)
−
x0
x
(E
(cid:0) V (x0))dx0.
(cid:48)
(cid:48)
−
(2.29)
Note that the integral on the right is a bounded
a + (cid:15). Since the first term on the right-hand side is a constant we find
function of x. We now integrate again from a (cid:0) (cid:15) to
−
ψ(a + (cid:15)) (cid:0) ψ(a (cid:0) (cid:15)) = 2(cid:15)
−
−
dψ
dx
(cid:12)
(cid:12)
(cid:12)
(cid:12)
x
0
(cid:0)
−
(cid:90)
2m a+(cid:15)
(cid:126)
a(cid:0)(cid:15)
−
dx
(cid:90)
x
x0
0
(cid:48)
dx
(E V (x0)).
(cid:0)
−
(cid:48)
(2.30)
Taking the (cid:15) ! 0
→
(cid:82)
to zero because
limit, the first term on the righ
x
(cid:48 | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
) ! 0
→
(cid:82)
to zero because
limit, the first term on the righ
x
(cid:48)
x0
−
(cid:48)
dx0 (E (cid:0) V (x0)) is a bounded function of x. As a result we have
t-hand side clearly vanishes and the second term goes
showing that the wavefunction is continuous at x = a. This is our second boundary condition.
∆aψ = 0 ,
(2.31)
3 Free particle on a circle.
Consider now the problem of a particle confined to a circle of circumference L. The coordinate along
the circle is called x and we can view the circle as the interval x 2 [0, L] with the endpoints identified.
It is perhaps clearer mathematically to think of the circle as the full real line x with the identification
∈
x (cid:24) x + L ,
∼
(3.1)
which means that two points whose coordinates are related in this way are to be considered the same
point. If follows that we have the periodicity condition
ψ(x + L) = ψ(x) .
(3.2)
From this it follows that not only ψ is periodic but all of its derivatives are also periodic.
The particle is assumed to be free and therefore V (x) = 0. The time-independent Schr¨odinger
equation is then
(cid:0)
−
(cid:126)2 d2ψ
2m dx2
= E ψ(x) .
(3.3)
Before we solve this, let us show that any solution must have E (cid:21) 0. For this multiply the above
equation by ψ(cid:3)(x) and integrate over the circle x 2 [0 , L). Since ψ is normalized we get
∈
≥
∗
(cid:0)
−
(cid:90) L
(cid:126)2
2m 0
ψ(cid:3)(x)
∗
2ψ
d
dx2
(cid:90)
dx = E
∗
ψ(cid:3)(x)ψ(x)dx = E .
The integrand on the left hand side can be rewritten as
(cid:0)
− | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
:3)(x)ψ(x)dx = E .
The integrand on the left hand side can be rewritten as
(cid:0)
−
(cid:126)2
(cid:90) L
2m
0
(cid:16)
ψ(cid:3)
∗
(cid:104) d
dx
(cid:17)
dψ
dx
(cid:0)
−
(cid:105)
dψ(cid:3) dψ
∗
dx dx
dx = E .
6
(3.4)
(3.5)
and the total derivative can be integrated
−
(cid:0)
(cid:126)2
(cid:104)(cid:16)
2m
∗
ψ(cid:3)
dψ
dx
(cid:17)(cid:12)
(cid:12)
(cid:12)
x=L
(cid:16)
− ∗
(cid:0) (cid:3)
ψ
dψ
dx
(cid:17)(cid:12)
(cid:12)
(cid:12)
x=0
(cid:105)
+
(cid:126)2
(cid:90)
2m
0
L dψ 2
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
dx
dx = E .
(3.6)
Since ψ(x) and its derivatives are p
are left with
eriodic, the con
tributions from
x =
L and x = 0 cancel out and we
E =
(cid:126)2
(cid:90)
2m
0
L dψ 2
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
dx
dx
≥
(cid:21)
0 ,
(3.7)
which establishes our claim. We also see that E = 0 requires ψ constant (and nonzero!).
Having shown that all solutions must have E (cid:21) 0 let us go back to the Schr¨odinger equation, which
≥
can be rewritten as
We can then define k via
d2ψ
−
dx2 (cid:0)
=
2mE
(cid:126)2
ψ .
k2
2mE
≡
(cid:17 | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
(cid:0)
=
2mE
(cid:126)2
ψ .
k2
2mE
≡
(cid:17) (cid:126) (cid:21) 0 .
≥
Since E (cid:21) 0, the constant k is real. Note that this definition is very natural, since it makes
≥
which means that, as usual, p = (cid:126)k. Using k2 the differential equation becomes the familiar
E =
(cid:126)2k2
2m
,
d2ψ
dx
2 (cid:0)k2ψ .
−
=
(3.8)
(3.9)
(3.10)
(3.11)
We could write the general solution in terms of sines and cosines of kx, but let’s use complex expo-
nentials:
ψ(x) (cid:24) eikx.
∼
(3.12)
This solves the differential equation and, moreover, it is a momentum eigenstate. The periodicity
condition (3.2) requires
eik(x+L) = eikx ! eikL = 1 ! kL = 2πn , n 2 Z .
→
→
∈
(3.13)
We see that momentum is quantized because the wavenumber is quantized! The wavenumber has
discrete possible values
≡
kn (cid:17)
2πn
L
,
n
2 Z.
∈
(3.14)
All integers positive and negative are allowed and are in fact necessary because they all correspond to
different values of the momentum pn = (cid:126)kn. The solutions to the Schr¨odinger equation can then be
indexed by the integer n:
ψn(x) = N eiknx ,
where N is a real normalization constant. Its value is determined from
1 =
(cid:90) L
0
∗
(cid:3)(x)ψn(x)dx =
ψn
(cid:90)
L
0
N 2dx = N 2L
! N = p ,
→
1
√
L
(3.15)
(3.16)
7
so we have | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
! N = p ,
→
1
√
L
(3.15)
(3.16)
7
so we have
1
ψn(x) = p eiknx
√
L
=
1
√
p
L
2πinx
e L
.
(3.17)
The associated energies are
(cid:126)24π2n2
2mL2
There are infinitely many energy eigenstates. We have degenerate states because En is just a
−
function of jnj and thus the same for n and (cid:0)n. Indeed ψn and ψ(cid:0)n both have energy En. The only
nondegenerate eigenstate is ψ0 = p1 , which is a constant wavefunction with zero energy.
2π2(cid:126)2n2
mL2
(cid:126)2k2
n
2m
(3.18)
En =
=
=
√
−
.
|
|
L
Whenever we find degenerate energy eigenstates we must wonder what makes those states differ-
ent, given that they have the same energy. To answer this one must find an observable that takes
different values on the states. Happily, in our case we know the answer. Our degenerate states can be
distinguished by their momentum: ψn has momentum 2πn and ψ
(cid:0)n has momentum ((cid:0)2πn ).L
−
(cid:126)
L
−
(cid:126)
Given two degenerate energy eigenstates, any linear combination of these states is an eigenstate
with the same energy. Indeed if
then
ˆ
Hψ1 = Eψ1 , Hψ2 = Eψ2 ,
ˆ
(3.19)
ˆ
H(aψ1 + bψ2) = aHψ1 + bHψ2 = aEψ1 + bEψ2 = E(aψ1 + bψ2) .
We can therefore form two linear combinations of the degenerate eigenstates ψn and ψ(cid:0)n to obtain
another description of the energy eigenstates:
(3.20)
ˆ
ˆ
−
− ∼
� | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
:0)n to obtain
another description of the energy eigenstates:
(3.20)
ˆ
ˆ
−
− ∼
ψn + ψ n (cid:24) cos(k x
n ) ,
(cid:0)
− − ∼
ψn (cid:0) ψ(cid:0)n (cid:24) sin(knx) .
While these are real energy eigenstates, they are not momentum eigenstates. Only our exponentials
are simultaneous eigenstates of both H and pˆ.
(3.21)
ˆ
The energy eigenstates ψn are automatically orthonormal since they are pˆ eigenstates with no
degeneracies (and as you recall eigenstates of a hermitian operator with different eigenvalues are
automatically orthogonal) :
(cid:90)
(cid:90)
L
L
∗
(cid:3)
ψn
(x)ψm(x)dx =
2πi(m− )x
L
n
e
dx = δmn.
1
L 0
(3.22)
0
e can then construct a
They are also complete: w
fact a Fourier series. For any Ψ(x, 0) that satisfies
general wavefunction as a superposition that is in
the periodicity condition, we can write
Ψ(x, 0) =
(cid:88)
n2Z
∈
an ψn(x),
where, as you should check, the coefficients an are determined by the integrals
an =
(cid:90) L
0
∗
(cid:3)(x) Ψ(x, 0) .
dx ψn
The initial state Ψ(x, 0) is then easily evolved in time:
Ψ(x, t) =
(cid:88)
n2Z
∈
an ψn(x)e
(cid:0) iEnt
−
(cid:126)
.
(3.23)
(3.24)
(3.25)
Andrew Turner transcribed Zwiebach’s handwritten notes to create the first LaTeX version of this
document.
8
MIT OpenCourseWare
https://ocw.mit.edu
8.04 Quantum Physics I
Spring 2016
For information about citing these materials or our Terms of Use, visit: https | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
ocw.mit.edu
8.04 Quantum Physics I
Spring 2016
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
8. Geometric problems
Convex Optimization — Boyd & Vandenberghe
extremal volume ellipsoids
centering
classification
placement and facility location
•
•
•
•
8–1
Minimum volume ellipsoid around a set
L¨owner-John ellipsoid of a set C: minimum volume ellipsoid
s.t. C
parametrize
as
=
v
Av + b
E
≤
is proportional to det A−1; to compute minimum volume ellipsoid,
| �
∈
E
{
}
�2
E
; w.l.o.g. assume A
1
⊆ E
Sn
++
•
•
vol
E
minimize (over A, b)
subject to
log det A−1
supv∈C �
Av + b
1
�2
≤
convex, but evaluating the constraint can be hard (for general C)
finite set C =
x1, . . . , xm
:
}
{
minimize (over A, b)
subject to
log det A−1
Axi + b
�2
�
≤
1,
i = 1, . . . , m
also gives L¨owner-John ellipsoid for polyhedron conv
x1, . . . , xm
{
}
Geometric problems
8–2
Maximum volume inscribed ellipsoid
maximum volume ellipsoid
inside a convex set C
E
Bu + d
as
=
parametrize
vol
E
•
•
E
}
is proportional to det B; can compute
| �
≤
E
{
u
�2
by solving
E
1
; w.l.o.g. assume B
Sn
++
∈
Rn
⊆
maximize
subject to
log det B
supkuk2≤1 IC(Bu + d)
0
≤
(where IC(x | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
imize
subject to
log det B
supkuk2≤1 IC(Bu + d)
0
≤
(where IC(x) = 0 for x
C and IC(x) =
for x
C)
�∈
∞
∈
convex, but evaluating the constraint can be hard (for general C)
polyhedron
T x
ai
x
{
|
bi, i = 1, . . . , m
}
:
≤
maximize
subject to
log det B
Bai
T d
�2 + ai
�
(constraint follows from supkuk2≤1 aT (Bu + d) =
≤
i
bi,
i = 1, . . . , m
Bai
�2 + aT d)
i
�
Geometric problems
8–3
Efficiency of ellipsoidal approximations
Rn convex, bounded, with nonempty interior
⊆
L¨owner-John ellipsoid, shrunk by a factor n, lies inside C
C
•
maximum volume inscribed ellipsoid, expanded by a factor n, covers C
•
example (for two polyhedra in R2)
factor n can be improved to √n if C is symmetric
Geometric problems
8–4
Centering
some possible definitions of ‘center’ of a convex set C:
center of largest inscribed ball (’Chebyshev center’)
for polyhedron, can be computed via linear programming (page 4–19)
center of maximum volume inscribed ellipsoid (page 8–3)
•
•
xchebxcheb
xmve
MVE center is invariant under affine coordinate transformations
Geometric problems
8–5
Analytic center of a set of inequalities
the analytic center of set of convex inequalities and linear equations | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
of a set of inequalities
the analytic center of set of convex inequalities and linear equations
fi(x)
≤
0,
i = 1, . . . , m,
F x = g
is defined as the optimal point of
m
minimize
i=1
subject to F x = g
− �
log(
fi(x))
−
•
•
more easily computed than MVE or Chebyshev center (see later)
not just a property of the feasible set: two sets of inequalities can
describe the same set, but have different analytic centers
Geometric problems
8–6
analytic center of linear inequalities ai
T x
bi, i = 1, . . . , m
≤
xac is minimizer of
m
φ(x) =
−
�
i=1
log(bi
−
T x)
ai
xac
inner and outer ellipsoids from analytic center:
Einner
x
⊆ {
|
T x
ai
≤
bi, i = 1, . . . , m
} ⊆ Eouter
where
Einner =
Eouter =
(x
(x
x
x
{
{
|
|
−
−
xac)T
xac)T
∇
∇
2φ(xac)(x
2φ(xac)(x
xac)
xac)
≤
≤
−
−
1
}
m(m
1)
}
−
Geometric problems
8–7
Linear discrimination
separate two sets of points
{
a T xi + b > 0,
i = 1, . . . , N,
,
x1, . . . , xN
}
a T yi + b < 0,
y1, . . . , yM
}
{
by a hyperplane:
i = 1, . . . , M
homogeneous in a, b, hence equivalent to | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
hyperplane:
i = 1, . . . , M
homogeneous in a, b, hence equivalent to
a T xi + b
≥
1,
i = 1, . . . , N,
a T yi + b
1,
≤ −
i = 1, . . . , M
a set of linear inequalities in a, b
Geometric problems
8–8
Robust linear discrimination
(Euclidean) distance between hyperplanes
H1 =
H2 =
z
z
{
{
a T z + b = 1
}
a T z + b =
1
−
|
|
}
is dist(
H1,
H2) = 2/
�
a
�2
to separate two sets of points by maximum margin,
(1/2)
�
a
minimize
�2
subject to aT xi + b
≥
aT yi + b
≤ −
1,
1,
(after squaring objective) a QP in a, b
Geometric problems
i = 1, . . . , N
i = 1, . . . , M
(1)
8–9
Lagrange dual of maximum margin separation problem (1)
maximize
subject to 2
�
��
�
1T λ + 1T µ
N
i=1 λixi
1T λ = 1T µ,
M
i=1 µiyi
0, µ
�
−
λ
�
1
�
�
� 2 ≤
0
�
(2)
from duality, optimal value is inverse of maximum margin of separation
interpretation
change variables to θi = λi/1T λ, γi = µi/1T µ, t = 1/(1T λ + 1T µ)
invert objective to minimize 1/(1T λ + 1T µ) | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
1/(1T λ + 1T µ)
invert objective to minimize 1/(1T λ + 1T µ) = t
•
•
minimize
t
subject to
�
��
�
θ
�
N
i=1 θixi
0,
�
−
1T θ = 1,
M
i=1 γiyi
γ
�
�
� 2 ≤
0,
�
t
1T γ = 1
optimal value is distance between convex hulls
Geometric problems
8–10
Approximate linear separation of non-separable sets
1T u + 1T v
minimize
subject to aT xi + b
aT yi + b
0,
u
�
i = 1, . . . , N
i = 1, . . . , M
ui,
−
1 + vi,
0
1
≥
≤ −
v
�
an LP in a, b, u, v
at optimum, ui = max
0, 1
aT xi
b
, vi = max
0, 1 + aT yi + b
}
can be interpreted as a heuristic for minimizing #misclassified points
−
−
{
{
}
•
•
•
Geometric problems
8–11
Support vector classifier
minimize
�
subject to aT xi + b
aT yi + b
0,
u
�2 + γ(1T u + 1T v)
a
1
ui,
−
≥
1 + vi,
≤ −
0
v
�
�
i = 1, . . . , N
i = 1, . . . , M
produces point on trade-off curve between inverse of margin 2/
classification error, measured by total slack 1T u + 1T v
�2 and
a
� | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
cation error, measured by total slack 1T u + 1T v
�2 and
a
�
same example as previous page,
with γ = 0.1:
Geometric problems
8–12
Nonlinear discrimination
separate two sets of points by a nonlinear function:
f (xi) > 0,
i = 1, . . . , N,
f (yi) < 0,
i = 1, . . . , M
choose a linearly parametrized family of functions
f (z) = θT F (z)
F = (F1, . . . , Fk) : Rn Rk are basis functions
→
solve a set of linear inequalities in θ:
•
•
θT F (xi)
≥
1,
i = 1, . . . , N,
θT F (yi)
1,
≤ −
i = 1, . . . , M
Geometric problems
8–13
quadratic discrimination: f (z) = zT P z + qT z + r
x T P xi + q T xi + r
i
1,
≥
y T P yi + q T yi + r
i
1
≤ −
can add additional constraints (e.g., P
I to separate by an ellipsoid)
� −
polynomial discrimination: F (z) are all monomials up to a given degree
separation by ellipsoid
separation by 4th degree polynomial
Geometric problems
8–14
Placement and facility location
N points with coordinates xi
R2 (or R3)
∈
some positions xi are given; the other xi’s are variables
for each pair of points, a cost function fij(xi, xj)
•
•
•
placement problem
minimize
�
i6
=j | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
a cost function fij(xi, xj)
•
•
•
placement problem
minimize
�
i6
=j
fij(xi, xj)
variables are positions of free points
interpretations
points represent plants or warehouses; fij is transportation cost between
facilities i and j
points represent cells on an IC; fij represents wirelength
•
•
Geometric problems
8–15
example: minimize
�
(i,j)∈A h(
�
xi
−
xj
�2), with 6 free points, 27 links
optimal placement for h(z) = z, h(z) = z2 , h(z) = z4
1
0
−1
−1
0
1
1
0
−1
−1
0
1
histograms of connection lengths
xi
�
−
xj
�2
4
3
2
1
0
0
4
3
2
1
0
0
0.5
1
1.5
2
−1
0
1
1
0
−1
6
5
4
3
2
1
Geometric problems
8–16
0.5
1
1.5
0
0
0.5
1
1.5
MIT OpenCourseWare
http://ocw.mit.edu
6.079 / 6.975 Introduction to Convex Optimization
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.727 Topics in Algebraic Geometry: Algebraic Surfaces
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ALGEBRAIC SURFACES, LECTURE 9
LECTURES: ABHINAV KUMAR
1. Castelnuovo’s Criterion for Rationality
Theorem 1. Any surface with q = h1(X, OX ) = 0 and p2 = h0(X, ω⊗2) = 0 is
rational.
X
Note. Every rational surface satisfies these: they are birational invariants which
vanish for P2 .
Reduction 1: Let X be a minimal surface with q = p2 = 0. It is enough to
show there is a smooth rational curve C on X with C 2 ≥ 0.
2
·
Proof. First, observe that 2g(C) − 2 = −2 = C (C + K) and χ(OX (C)) =
χ(OX ) + 1 C(C − K). Since p2 = 0, p1 = h0(X, ω) = h2(X, OX ) = 0 and
1 C(C − K),
χ(OX ) = 1. Since h2(C) = h0(K − C) ≤ h0(K) = 0, h0(C) ≥ 1 + 2
so h0(C) ≥ 2 + C 2 ≥ 2. Choose a pencil inside this system containing C, i.e. a
subspace of dimension 2. The pencil has no fixed component (the only possibility
is C, but C moves in the pencil): after blowing up finitely many base points, we
P1 with a fiber isomorphic to C ∼ P1 . Therefore, by the | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01d543f81d0b743f06deed68d9eec77d_lect9.pdf |
we
P1 with a fiber isomorphic to C ∼ P1 . Therefore, by the
get a morphism X˜
˜
Noether-Enriques theorem, X is ruled over P1 and X is rational (as is X). �
→
=
˜
Reduction 2: Let X be a minimal surface with q = p2 = 0. It is enough to
show that ∃ an effective divisor D on X s.t. |K + D| = ∅ and K · D < 0.
Proof. This implies that some irreducible component C of D satisfies K C < 0.
Clearly, |K + C| ⊂ |K + D|. Using Riemann-Roch for K + C gives
·
0 = h0(U + C) + h0(−C) = h0(K + C) + h2(K + C)
(1)
≥ 1 + (K + C) C = g(C)
·
1
2
We thus obtain a smooth, rational curve C on X: −2 = 2g − 2 = C(C + K)
and C · K < 0 = ⇒ C 2 ≥ −1. Since X is minimal, C 2 =� −1, so C 2 ≥ 0 as
�
desired.
We now prove our second statement. There are three cases:
1
2
LECTURES: ABHINAV KUMAR
Case 1 (K 2 = 0): Riemann-Roch gives
h0(−K) = h0(−k) + h0(2K) = h0(−K) + h2(−K)
(2)
≥ 1 +
1
2
K · 2K = 1 + K 2 = 1
·
so |−K = ∅. Take a hyperplane section H | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01d543f81d0b743f06deed68d9eec77d_lect9.pdf |
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