text stringlengths 11 2.41k | source stringlengths 82 201 |
|---|---|
P + (
−→
P Q +
−→
QR) = (P +
−→
QR to P ,
−→
QR
−→
P Q +
−→
P Q) +
−→
QR
= Q +
= R.
On the other hand, there is at most one vector �v such that when we
�
add it P we get R, namely the vector
−→
QR =
−→
P Q +
−→
P R.
−→
P R.
So
Note that (1.6) expresses the geometrically obvious statement that
if one goes... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
to rep
resent forces; every force has both a magnitude and a direction. The
combined effect of two forces is represented by the vector sum. Sim
ilarly we can use vectors to measure both velocity and acceleration.
The equation
F� = m�a,
is the vector form of Newton’s famous equation.
Note that R3 comes with three ... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
parametrise lines in R3 . Suppose we are given
two different points P and Q of R3 . Then there is a unique line l
containing P and Q. Suppose that R = (x, y, z) is a general point of
2
ˆ
−→
P R is parallel to the vector
−→
P Q, so that
the line. Note that the vector
−→
−→
P R is a scalar multiple of
P Q.
−→
P R... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
q3 − p3) = t(v1, v2, v3),
where (v1, v2, v3) = (q1 − p1, q2 − p2, q3 − p3). We can always rewrite
this as,
(x, y, z) = (p1, p2, p3) + t(v1, v2, v3) = (p1 + tv1, p2 + tv2, p3 + tv3).
Writing these equations out in coordinates, we get
x = p1 + tv1
y = p2 + tv2
and
z = p3 + tv3.
Example 1.7. If P = (1, −2, 3) and... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
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 ... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
= −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 w... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
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 thr... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
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) = (a... | 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 (Th... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
{
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 b... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
= 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 ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
2(�) = 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, known as Philip Hall’s theorem, for the M¨obius function
value µ(ˆ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 o... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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
1, and let P � = P
0 and ˆ
ζ)k (ˆ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
·
� (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 ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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.... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 alge... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 permutati... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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}
{
∗
−
· · ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
ne 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 clai... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
fied. 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... | 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 � x... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 interpreta... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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
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 ) ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
)).
(
−
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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
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 nothin... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
= 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−... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
→
⇔→
⇒
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)
· · ·
the proof of Claim 1 follows by induction.
· · ·
⊂
⊂
Claim 2. The broken circuit comple... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 circ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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,... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
}
∅ {
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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. Mod... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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) =... | 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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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).
Fig... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
spaces. 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
⊕
∈
⇒
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 latti... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 i... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
ive 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 , 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,... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
n, 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 inte... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 CIRCUITS AND MODULAR ELEMENTS
49
r... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
�
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
|
= 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.
⇒
�
∈
y = ˆ
0 and x
In a finite lattice L, a complement of x
y = ˆ
L such that
1. For i... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 ar... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
(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 defini... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 Exampl... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
). 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
modu... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
)(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 :=
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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. Clea... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
(v
∈
⇒
y)) = rk(z) + rk(v
= rk(z) + rk(v
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
)
= 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) 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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
��
⎡
�
⎡
�
⎡
⎡
⎢
�
�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 con... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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
Le... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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, . . . ,... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 4... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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
ma... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
−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 con... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
�
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
q . 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
)
· · ·
−... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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
ran... | 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)!. Compar... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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.)
No... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
erao [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) =
(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) =... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
1, . . . , 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(... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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,
+ qnQn, where qi
Q = q1Q1 +
· · ·
then the basis
can be chosen to be homogeneous, i.e., all coordinates
Q1, . . . , Qn}
{
of each Qi are ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
; 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 = ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 i... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 Propo... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
}
−
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... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
− {
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 ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
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 h... | 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 ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
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 Field... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
/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: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’... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
. 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 =... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
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
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 Conditi... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
⎡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 ⇒ ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
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
∇ 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, Elec... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
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 polynomi... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
,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 Linea... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
atisfying 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 = (V, E), find a clique of maximum
cardinality (vertices).
As different as these examples might seem, they ha... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
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 ... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
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 t... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
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 mi... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
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 S... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
5
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) ... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
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
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 5
LECTURES: ABHINAV KUMAR
1. Examples
→
(1) If S ⊂ Pn, p ∈ S, then projection ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
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 ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
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 the birational map.
z
x
y
(3) Linear systems of cubics: let p1, . . . , pr be r distinct points in the plane
(r ≤ 6) in... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
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... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
2 + 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 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 surfac... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
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