text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
Noise in Cascaded Amplifiers
F1,G1
F2,G2
1
2
≡
3
1
F1+2,G
1+2
3
F
∆+
2 1
NS
1
1
NS
3
3
where
S
3
=
S G G
121
N
2
=
kT
G F
1 o
1
⎛
⎜
Recall
⎜
⎝
F
1
=
N
3
=
G
kT
2
G F
+
1
1 o
(
FG
22
⎞
⎟
⎟
⎠
NS
1
1
NS
2
2
) o
1
kT
−
amplified from “2”
excess from “2”
∴
NS
3
3
=
S G G
1 ... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
A
B
vs.
B
A
The better choice is not obvious.
F1,2
also depends on interstage impedance mismatches
and gain of the first amplifier, not just on F1
L2
Noise in superheterodyne receivers
0
f
fo
Gc,Fc
S1,N1 S2,N2
×
0
fi.f.
f
L.O.
“lower (rf) sideband”
Gi.f.,Fi.f.
∆
“i.f.”
“upper sideband”
“interme... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
. f.
i
+
amp.
2
−≅
(
3 ~ 8
−
9
)
dB
L4
Basic receiver types-Amplification
1) Simple detector
B
f
( )2
h(t)
vo(t)
2) RF Amplifier
3) Superheterodyne
OR
OR
×
N1
1)
2)
3)
4)
Basic receiver types-Combinors
vo(t) Total power radiometer
vo(t)
Dicke radiometer
×
vo(t) Multiplication (correlation) ... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
equation: b =
aS
Note:
1) The scattering matrix
S
is defined only when the n-port
is imbedded in a network
2) The phases of a
i and bi can be defined as some linear
combinations of those for voltage and current
.g.e
a
i
V
= +
2Y
o
=
(
V
+
)
I Z
o
, Z 8
o
b
i
=
(
V
−
)
I Z
o
Z 8
o
P1
Gain definition ... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
T1
kT2
1
?
2
3
symmetry
Losslessness
We want
S
=
k(T1 +T2)
∗
t
a
=•⇒
bba
∗
t
00
⎡
⎢
00
⎢
⎢
⎣
α
βαα
α
⎤
⎥
⎥
⎥
⎦
∗
t
b
b
=•
∗
(
aS
t
)
(
aS
•
)
•
=
∗
t
∗
t
Sa
aS
Therefore
∗
t
ISS
==
001
⎤
⎡
⎥
⎢
010
⎥
⎢
100
⎥
⎢
⎦
⎣
; if the combiner is lossless
P4
Example of constrained N-port netwo... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
ded in network)
For 3 matched ports:
S
=
0
⎡
⎢
α
⎢
β
⎢
⎣
βα
0
⎤
⎥
γ
⎥
0
⎥
⎦
γ
Does
∗
t
SS
I
?
=
(lossless passive constraint)
P6
Matched 3-port example
For 3 matched ports:
S
=
0
⎡
⎢
α
⎢
β
⎢
⎣
βα
0
⎤
⎥
γ
⎥
0
⎥
⎦
γ
Does
∗
t
SS
I
?
=
(lossless passive constraint)
∗
t
SS
=... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
R
+
S
L
)
>>
i(t)
i
diode:
sin
+ω
o
vt
s
sin
ω
s
t
i ≅
v
2
d
for
)t(v
d
v
≅ (cid:65)
+
small
high
-
order
terms
load line i =
diode
V
−
v
d
Th
R R
+
L
S
v (t
o
)
=
iR
L
(
∝ 2
v R
Th
L
v
k
+
1 Th
k
+
)
2
v
2
Th
0
diode i,vTh(t)
VTh
vd
k
+
3
v
3 Th
+
...
=
k
o
dominan... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
-port Model
T
R
=
T
SSB
=
T)1F(
−
o
=
T
o
(
t L
rc
−
)1
signal
for single sideband (SSB) operation
image
output
1
2
4
3
internal
noise
F
SBB
=
NS
1
1
NS
4
4
=
⎡
S S
1
⎢
⎢
S1
−
⎣
41
44
2
2
⎤
⎥
⎥
⎦
kTS
o
1
2
kT
2
S1
−
+
41
S
42
44
S
⎡
kT
o
⎢
⎢
⎣
kT
3
S
43
2
+
2
[No... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
2
44
It follows that
T
DSB
=
T
3
S
43
2
[
S
2
41
+
S
2
42
]
Often suggested:
SSBF
≅
F2
DSB
R4
Double-sideband Receiver
Therefore
T
SSB
=
T
2
S
2
42
2
S
41
+
T
3
2
signal
S
43
2
S
41
image
Often suggested:
SSBF
≅
F2
DSB
output
1
2
4
3
internal
noise
Let
S
2
2
=
S
42... | https://ocw.mit.edu/courses/6-661-receivers-antennas-and-signals-spring-2003/34219e23d858375d1c802669f2969dfc_lecture06.pdf |
6.825 Techniques in Artificial Intelligence
Resolution Theorem Proving:
Propositional Logic
• Propositional resolution
• Propositional theorem proving
• Unification
Lecture 7 • 1
Today we’re going to talk about resolution, which is a proof strategy. First, we’ll
look at it in the propositional case, then in the... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
¬
α v γ
• Resolution refutation:
Lecture 7 • 4
It turns out that that one rule is all you need to prove things. At least, to prove that a
set of sentences is not satisfiable. So, let's see how this is going to work. There's
a proof strategy called Resolution Refutation, with three steps. And it goes like
this.
4... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Lecture 7 • 7
And then we apply the Resolution Rule until either you can derive "false" -- which
means that the conclusion did, in fact, follow from the things that you had
assumed, right? If you assert that the negation of the thing that you're interested
in is true, and then you prove for a while and you manage t... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
. Furthermore, the process
is sound: the answer is always correct.
8
Propositional Resolution Example
Step
Formula
Derivation
Prove R
1
P v Q
2
R
P
→
Q
3
R
→
Lecture 7 • 9
So let's just do a proof. Let's say I'm given “P or Q”, “P implies R” and “Q implies
R”. I would like to conclude R from these three ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
2
3
4
P v Q
Given
P v R
Q v R
R
¬
¬
¬
Given
Given
Negated
conclusion
Lecture 7 • 13
Now we want to add one more thing to our list of given statements. What's it going
to be?
Not R. Right? We're going to assert the negation of the thing we're trying to prove.
We'd like to prove that R follows from these th... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
1
P v Q
2
R
P
→
Q
3
R
→
Step
Formula
Derivation
1
2
3
4
5
6
P v Q
P v R
Q v R
R
¬
¬
¬
Given
Given
Given
Negated
conclusion
Q v R
P
¬
1,2
2,4
And we can take lines 2 and 4, resolve away R, and get “not P.”
Lecture 7 • 16
16
Propositional Resolution Example
Prove R
1
P v Q
2
R
P
→
Q
3
R
→
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
P
→
Q
3
R
→
1
2
3
4
5
6
7
8
9
P v Q
P v R
Q v R
R
¬
¬
¬
Q v R
P
¬
Q
¬
R
•
Given
Given
Given
Negated
conclusion
1,2
2,4
3,4
5,7
4,8
Lecture 7 • 19
And finally, resolving away R in lines 4 and 8, we get the empty clause, which is
false. We’ll often draw this little black box to indicate that we’ve r... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
R
→
false v R
R v false
¬
false v false
1
2
3
4
5
7
8
9
P v Q
P v R
Q v R
R
¬
¬
¬
Q v R
P
Q
¬
R
•
Given
Given
Given
Negated
conclusion
1,2
3,4
5,7
4,8
Lecture 7 • 21
One of these steps is unnecessary. Which one? Line 6. It’s a perfectly good proof
step, but it doesn’t contribute to the final conclu... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
4
P
Z
P
¬
¬
•
Given
Given
Negated
conclusion
1,2
Note that (P Æ
P)
→
¬
Z is valid
Lecture 7 • 26
Because we can prove Z from “P and not P” using a sound proof procedure, then “P
and not P” entails Z. And, by the theorem relating entailment and validity, we
have that the sentence “P and not P implies Z” is ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
•
So, the first formula turns into “P or Q”.
(
¬
¬
P Æ
(
v
P
(
P v Q) v Q
Q) v Q
¬
Q) Æ (
¬
P
(
v
Q)
Q v Q)
Lecture 7 • 29
29
Example Problem
Convert to CNF
Prove R
1
2
3
(P
(P
→
→
Q)
→
Q
P)
→
R
(R
S)
(S
→
Q)
→
¬
→
•
•
•
•
•
•
•
(
¬
¬
P Æ
(
v
P
(
P v Q) v Q
Q) v Q
¬
Q)... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
(
¬
¬
(
P Æ
(
P
v
P v P) v R
P) v R
¬
R) Æ (
P v R)
¬
•
•
•
•
(
¬
¬
(
R Æ
(
R
v
(
v
R
R v S) v
(
¬
S) v (S Æ
¬
S v Q)
Q)
¬
S) Æ (
S) Æ (R v
¬
S v S) Æ(R v
Q) Æ(
¬
S v
Q) Æ(
Q)
¬
¬
¬
¬
S v
Q)
¬
¬
Finally, the last formula requires us to do a big expansion, but one of the terms is ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
3
→
S)
(S
→
¬
→
Q)
2
3
4
5
6
7
8
9
10
11
12
P v R
P v R
¬
R v S
R v
Q
¬
S v
Q
¬
R
¬
¬
S
Q
¬
P
R
•
Neg
4,7
6,8
1,9
3,10
7,11
Lecture 7 • 33
Here’s a sample proof. It’s one of a whole lot of possible proofs.
33
Proof Strategies
In choosing among all the possible proof steps that you can do ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
The set-of-support rule says you should involve the thing that you're trying to prove.
It might be that you can derive conclusions all day long about the solutions to
chess games and stuff from the axioms, but once you're trying to prove
something about what way to run, it doesn't matter. So, to direct your “thought... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
39
Let’s try to get some intuition through an example. Imagine you knew “for all X, P
of X implies Q of X.” And let's say you also knew P of A. What would you be
able to conclude?
Q of A, right? You ought to be able to conclude Q of A.
39
First-Order Resolution
Q(x)
∀
x. P(x)
→
P(A)
Q(A)
Syllogism:
All me... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
al
resolution
Lecture 7 • 42
Next, we could substitute the constant A in for the variable x in the universally
quantified sentence. By the semantics of universal quantification, that’s
allowed. And now, we can apply the propositional resolution rule.
42
First-Order Resolution
Q(x)
∀
x. P(x)
→
P(A)
Q(A)
Sy... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
• 46
A substitution is a set of variable-term pairs, written this way. It says that whenever
you see variable I, you should substitute in term I. There should not be more
than one entry for a single variable.
46
Substitutions
P(x, F(y), B) : an atomic sentence
Substitution
instances
P(z, F(w), B)
Substitution... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
(x, F(y), B) : an atomic sentence
Substitution
instances
P(z, F(w), B)
Substitution
{v1 /t1,…, vn /tn}
{x/z, y/w}
P(x, F(A), B)
{y/A}
P(G(z), F(A), B)
{x/G(z), y/A}
Comment
Alphabetic
variant
P(C, F(A), B)
{x/C, y/A}
Ground instance
Lecture 7 • 50
Here’s one more -- P(C, F(A), B). It’s sort of interesting, ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
{v1 /t1,…, vn /tn}
{x/z, y/w}
P(x, F(A), B)
{y/A}
P(G(z), F(A), B)
{x/G(z), y/A}
Comment
Alphabetic
variant
P(C, F(A), B)
{x/C, y/A}
Ground instance
Applying a substitution:
subst({A/y},P(x, F(y), B)) = P(x, F(A), B)
P(x, F(y), B) {A/y} = P(x,F(A),B)
Lecture 7 • 52
We’ll use the notation of an expression foll... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
the following are unifiers
s
ω1 s
ω2 s
So, let’s look at some unifiers of the expressions x and y. Since x and y are both
variables, there are lots of things you can do to make them match.
Lecture 7 • 55
55
Unification
• Expressions ω1 and ω2 are unifiable iff there exists a
substitution s such that ω1 s = ω2 s... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Unification
• Expressions ω1 and ω2 are unifiable iff there exists a
substitution s such that ω1 s = ω2 s
• Let ω1 = x and ω2 = y, the following are unifiers
s
{y/x}
{x/y}
ω1 s
ω2 s
x
y
x
y
{x/F(F(A)), y/F(F(A))}
F(F(A))
F(F(A))
{x/A, y/A}
A
A
Or, you could substitute some constant, like A, in for both x and y... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
two expressions equal.
60
Most General Unifier
g is a most general unifier of ω1 and ω2 iff for all
unifiers s, there exists s0
and ω2 s = (ω2 g) s0
such that ω1 s = (ω1 g) s0
ω1
P(x)
ω2
P(A)
MGU
{x/A}
So, let's do a few more examples together, and then you can do some as recitation
problems. So, what’s a m... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
G(y))
P(F(x), z, G(x))
{y/x, z/x}
Lecture 7 • 63
Okay. What about this one? It’s a bit tricky. You can kind of see that, ultimately,
all of the variables are going to have to be the same. Matching the arguments to
G forces y and x to be the same, And since z and y have to be the same as well
(to make the middle ar... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
))
P(F(x), x, G(x))
{y/x} or {x/y}
P(F(x), y, G(y))
P(F(x), z, G(x))
{y/x, z/x}
P(x, B, B)
P(A, y, z)
{x/A, y/B, z/B}
P(G(F(v)), G(u))
P(x, x)
{x/G(F(v)), u/F(v)}
Here’s a tricky one. It looks like x is going to have to simultaneously be G(F(v))
and G(u). How can we make that work? By substituting F(v) in for u.
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
, there exists s0
and ω2 s = (ω2 g) s0
such that ω1 s = (ω1 g) s0
ω1
P(x)
ω2
P(A)
MGU
{x/A}
P(F(x), y, G(x))
P(F(x), x, G(x))
{y/x} or {x/y}
P(F(x), y, G(y))
P(F(x), z, G(x))
{y/x, z/x}
P(x, B, B)
P(A, y, z)
{x/A, y/B, z/B}
P(G(F(v)), G(u))
P(x, x)
{x/G(F(v)), u/F(v)}
P(x, F(x))
P(x, x)
No MGU!
Lecture 7 • 67
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
a failure, we return s.
Lecture 7 • 69
69
Unification Algorithm
unify(Expr x, Expr y, Subst s){
if s = fail, return fail
else if x = y, return s
If x is equal to y, then we don’t have to do any work and we return fail.
Lecture 7 • 70
70
Unification Algorithm
unify(Expr x, Expr y, Subst s){
if s = fail, retu... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Expr x, Expr y, Subst s){
if s = fail, return fail
else if x = y, return s
else if x is a variable, return unify-var(x, y, s)
else if y is a variable, return unify-var(y, x, s)
else if x is a predicate or function application,
if y has the same operator,
return unify(args(x), args(y), s)
return fail
else
If n... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
substitution that unifies var and x in the context of s. What makes this tricky is
that we have to first keep applying the existing substitutions in s to var, and to x,
if it is a variable, before we’re down to a new concrete problem to solve.
76
Unify-var subroutine
Substitute in for var and x as long as possible... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Substitute in for var and x as long as possible, then add new
binding
unify-var(Variable var, Expr x, Subst s){
if var is bound to val in s,
return unify(val, x, s)
else if x is bound to val in s,
return unify-var(var, val, s)
else if var occurs anywhere in x, return fail
else return add({var/x}, s)
}
Finally... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
F(F(y, A), F(z, F(B,z)))
Lecture 7 • 82
Please do at least half of these problems before you go on to the next lecture, and all
of them before the next recitation.
82
Inference using Unification
x.
¬
∀
P(x) v Q(x)
P(A)
Q(A)
For universally quantified variables, find MGU {x/A} and
proceed as in propositiona... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/3439d481b8e3d5abecc1403caf1ca89d_Lecture7FinalPart1.pdf |
Massachusetts Institute of Technology
Department of Materials Science and Engineering
77 Massachusetts Avenue, Cambridge MA 02139-4307
3.21 Kinetics of Materials—Spring 2006
February 21, 2006
Lecture 4: Interdiffusion. Effects of electical potential, capillarity, and stress on diffusion potential.
References
1. ... | https://ocw.mit.edu/courses/3-21-kinetic-processes-in-materials-spring-2006/3453d577b53c280708513740833c4932_ls4.pdf |
different rates. In a substitutional alloy this requires vacancy creation and destruction. These processes
can take place quite efficiently provided the material contains sufficient numbers of vacancy sources
and sinks. However, it is not unusual to find Kirkendall porosity on one side of the interdiffusion zone.
• The... | https://ocw.mit.edu/courses/3-21-kinetic-processes-in-materials-spring-2006/3453d577b53c280708513740833c4932_ls4.pdf |
()%
£tAV *
eCxo
i
- t
.
-.. 1..
-2.
-"~~i~;e~a~i·-
30
ESL-, MSL4
. I
-
.. �
.e
A
�F,
El- bm-r-0
tj\ -J~(ti, Z o
wrr~s 14%
-,L
kt,1-6
(C§(2
) .
same
'S%;SA
Sq
Not
~or
sl,
c
-----------
�I
�---�--�------
----
·-
M~
·---
;~~
-.
^
AS-
4~~~~~~~~~~~~F
Yo -. ..------ -.
.P.~~~~~~~N~A
''-
~
... | https://ocw.mit.edu/courses/8-322-quantum-theory-ii-spring-2003/3469fc3f8ad9f1562fbc172aaab9efcd_83228HigherMultiple.pdf |
LECTURE 19
Brauer Groups
In this lecture, we present an overview of Brauer groups. Our presentation
will be short on proofs, but we will give precise constructions and formulations of
claims. For complete proofs, see [Mil13, Ser79, Boy07]. Our motivating question
is: “what was all that stuff about Hamiltonian algebras?”... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
L2/L1/K, with Gal(Li/K) = Gi for i = 1, 2. Then we have a short exact sequence
0 → Gal(L2/L1) → G2 → G1 → 0,
and maps
Brcoh(L1/K) = H 2(G1, L×
1 ) → H 2(G2, L×
1 ) → H 2(G2, L×
2 ) = Brcoh(L2/K)
since invariance with respect to G2 implies impvariance with respect to G1, and via
the embedding L×
2 . This motivates the f... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
:122)
H 2(Gal(L),K×)
,
which we’ll justify next time.
We now turn to the algebraic perspective, which provides the classical definition
of the Brauer group.
Proposition 19.7. Let K be a field, and let A be a finite-dimensional K-
algebra with center K. Then the following are equivalent:
(1) A is simple, that is, it has no... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
same
dimension).
Remark 19.11. The isomorphism class of A depends only on the isomorphism
class of D.
19. BRAUER GROUPS
83
Proposition 19.12. If A and B are csas over K, then A ⊗K B is also a csa
(note that tensor products multiply dimension, so A ⊗K B also has square dimen-
sion). Up to equivalence, this only depends... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
9.14.
(1) If K is algebraically closed, then Brcsa(K) = {0} is
clearly trivial.
(2) Brcsa(R) = {R, H} = Z/2Z, since the only other division algebra over R
is C which is not central. Note that the Hamiltonians H split over C and
have dimension 4 = 22.
√
b), which are the usual commutative
(3) Ha,b splits over K(
subalge... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
⊗K L → EndL(D) = Mn(L)
with L acting on D via right multiplication and D acting on itself via left mul-
tiplication; the two actions commute. This map is injective (if dxl = x for some
d ∈ D, l ∈ L, and all x ∈ D, then dl = 1, hence dx = xl−1 = xd for all x ∈ D and
therefore d ∈ K, so d ⊗ l = 1 ⊗ 1), and therefore surj... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
gh]
for each x ∈ L and g, h ∈ G. We now check that the 2-cocyle identity is equivalent
to associativity:
[g1]([g2][g3]) = [g1]ϕ(g2, g3)[g2g3]
= g1ϕ(g2, g3)[g1][g2g3]
= g1ϕ(g2, g3)ϕ(g1, g2g3)[g1g2g3]
= ϕ(g1g2, g3)ϕ(g1, g2)[g1g2g3]
= ϕ(g1, g2)[g1g2][g3]
= ([g1][g2])[g3],
for all g1, g2, g3 ∈ G. We claim, but do not prove... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
, we’ve seen that both horizontal maps should have non-trivial
kernel; when L = K, Br(L) is trivial. Then a central division algebra of degree
n2 has order n in Br(K), i.e., is n-torsion, since it splits over its degree-n maxi-
mal commutative subalgebra. Alternatively, H 2(G, −) is #G-torsion, as proven in
Problem 2(c... | https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/346c269899e7aef6cd2da8960a965a61_MIT18_786S16_lec19.pdf |
Asymptotic Expansions of Integrals
Lecture 12
The Laplace Method
We begin with integrals of the form
IΩRæ : X
b
a
e?RvΩxæhΩxædx,
(8.8)
where vΩxæ and hΩxæ are independent of the parameter R. The variable of integration x is real. We
shall show how to find the asymptotic form of IΩRæ as R î K.
We note that the ... | https://ocw.mit.edu/courses/18-305-advanced-analytic-methods-in-science-and-engineering-fall-2004/3477ac43d18b5f901fe9bae124e85ce6_twelve.pdf |
R. Then we have
IΩRæ u
1 X
R
K
?K
e?[2 d[ :
Z
R
.
(8.10)
In the above, we have made use of the fact that the Gaussian integral XK
given in Appendix A of this chapter.
?K
e?[2 d[ is equal to Z , as is
If one makes a small mistake in an approximation, the answer one gets can be far off the mark.
Thus it is ... | https://ocw.mit.edu/courses/18-305-advanced-analytic-methods-in-science-and-engineering-fall-2004/3477ac43d18b5f901fe9bae124e85ce6_twelve.pdf |
2 (The integrals are given on p. 218);
Problem 3;
Find also the entire asymptotic series of each of the integrals in Problem 3.
— 2 — | https://ocw.mit.edu/courses/18-305-advanced-analytic-methods-in-science-and-engineering-fall-2004/3477ac43d18b5f901fe9bae124e85ce6_twelve.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.014 Calculus with Theory
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-014-calculus-with-theory-fall-2010/34a75af33a916b46c2ad007e5c110502_MIT18_014F10_ChInotes.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.102 Introduction to Functional Analysis
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
8
LECTURE NOTES FOR 18.102, SPRING 2009
Lecture 3. Tuesday, 10 Feb.
Recalled the proof from last time that the bounded o... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
of the
MIT calendar (but up later today). This problem may seem rather heavy sledding
but if you can work through it all you will understand, before we get to it, the
main sorts of arguments needed to prove most of the integrability results we will
encounter later.
Let V be a normed space with norm � · �V . A comp... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
sequences,
and multiply by constants, by doing the operations on each component:
(3.3)
t1{uk} + t2{u�
k} = {t1uk + t2u�
k}.
This always gives an absolutely summable series by the triangle inequality:
(3.4)
�
�t1uk + t2u�
k� ≤ |t1|
�uk� + |t2|
k
k
k
�
�
�u�
k�.
LECTURE NOTES FOR 18.102, SPRING 2009
9
Within... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
is a Banach space with the norm (3.7).
So, first that (3.7) is a norm. The limit on the right does exist since the limit
of the norm of a Cauchy sequence always exists – namely the sequence of norms
is itself Cauchy but now in R. Moreover, adding an element of S to {uk} does not
change the norm of the sequence of pa... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
�
uk� + �
� )� ∀ n ≥ N = ⇒
uk
k=1
k=1
A − � ≤ �b�B + �b��B ∀ � > 0 =⇒
�b + b��B ≤ �b�B + �b��B .
Now the norm of the element I(v) = v, 0, 0, · · · , is the limit of the norms of the
sequence of partial sums and hence is �v�V so �I(v)�B = �v�V and I(v) = 0
therefore implies v = 0 and hence I is also injective. ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
>
�
lim
N→∞
n
N
�
�
k=1
u(n)
k �V .
n
�
n
So, we want to show that
bn = b converges, and to do so we need to find the
limit b. It is supposed to be given by an absolutely summable series. The ‘problem’
is that this series should look like � �
n) in some sense – because it is supposed
u(
k
to represent th... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
�B | ≤ �,
u(
k
n)�V ≤ �.
�u(
k
k≤N
k≥N
�
�
Then in fact we can choose successive Nj < Nj−1 (remember that little n is fixed
here) so that
(3.13)
|�
N )�V − �bn�B | ≤ 2−j �,
u(
k
n)�V ≤ 2−j �.
�u(
k
k≤Nj
k≥Nj
Now, ‘resum the series’ defining instead v1
N1
(n) = �
k=1
u(
k
n), vj
(n) =
Nj
�
k=Nj−1
n) an... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
to do it first. Given
an element b ∈ B we need to find elements in V, vk such that �I(vk) − b�B → 0 as
k → ∞. Take an absolutely summable series uk representing b and take vj =
where the Nj ’s are constructed as above and check that I(vj )
→
b by computing
Nj
�
k=1
uk
(3.16)
�I(vj ) − b�B = lim
p→∞
�
�
up�V ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
‘central interval [1/3, 2/3). This leave C1 = [0, 1/3) ∪ [2/3, 1). Then
remove the central interval from each of the remaining two intervals to get C2 =
[0, 1/9) ∪ [2/9, 1/3) ∪ [2/3, 7/9) ∪ [8/9, 1). Carry on in this way to define successive
sets Ck ⊂ Ck−1, each consisting of a finite union of semi-open intervals. Now... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
The area of a rectangle is (b1 − a1) × (b2 − a2).
(1) We may subdivide a rectangle by subdividing either of the intervals – re
placing [a1, b1) by [a1, c1) ∪ [c1, b1). Show that the sum of the areas of
rectangles made by any repeated subdivision is always the same as that of
the original.
(2) Suppose that a finite ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
interval into 2n equal pieces and define the step functions to take infimim
of the continuous function on the corresponding interval. Then use uniform
convergence.
(2) By using the ‘telescoping trick’ show that any continuous function on [0, 1)
can be written as the sum
�
fj (x) ∀ x ∈ [0, 1)
(3.17)
where the fj a... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
means writing out the proof that this is a linear space and that the three
conditions required of a norm hold.
Solution:- We know that the functions from any set with values in a linear space
form a linear space – under addition of values (don’t feel bad if you wrote this out,
it is a good thing to do once). So, to... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
this is a direct consequence of
the usual triangle inequality:
�
�
�a + b�1 =
|ai + bi| ≤
(|ai| + |bi|) = �a�1 + �b�1.
(3.20)
i
i
For 1 < p < ∞ it is known as Minkowski’s inequality. This in turn is deduced
from H¨older’s inequality – which follows from Young’s inequality! The latter says
if 1/p + 1/q = 1, s... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
≤
|ai||bi|/�a�p�b�q ≤
�
�
i|
a
|
p
�
a
�
p
�
i
q
i|
b
|
qq
�
b
�
q
+
p
= 1
(3.23)
i
i
�
= ⇒ |
aibi| ≤ �a�p�b�q.
i
Of course, if either �a�p = 0 or �b�q = 0 this inequality holds anyway.
Now, from this Minkowski’s inequality follows. Namely from the ordinary trian
gle inequality and then Minkowski’s ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
p
is a norm on CN
j=1
would that help?
Solution:- Yes indeed it helps. If we know that for each N
(3.26)
⎛
⎝
N
�
j=1
⎞ 1 ⎛
|aj + bj | ⎠ ≤ ⎝
p
p
N
�
|aj |p
p
⎞ 1 ⎛
⎠ + ⎝
N
�
p
⎞ 1
|bj | ⎠
p
j=1
j=1
then for elements of lp the norms always bounds the right side from above, meaning
⎞ 1
p
(3.27)
|a... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
respect to the norm in
Problem 1.2 on CN . Show that this is the same as being Cauchy in CN in the usual
sense (if you are doing p = 2 it is already the usual sense) and hence, this cut-off
sequence converges. Use this to find a putative limit of the Cauchy sequence and
then check that it works.
Solution:- So, suppo... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
i=1.
(3.30)
N
�
(n)|p ≤ Ap
|ai
and we can pass to the limit here as n → ∞ since there are only finitely many
terms. Thus
i=1
(3.31)
N
�
|ai|p ≤ Ap ∀ N =⇒ �a�p ≤ A.
i=1
Thus, a ∈ lp as we hoped. Similarly, we can pass to the limit as m → ∞ in the
finite inequality which follows from the Cauchy conditions
N
�... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
is closed.
(2) Recall the sequential (so not the open covering definition) characterization
of compactness of a set in a metric space (e.g. by checking in Rudin).
(3) Show that S is not compact by considering the sequence in lp with kth
element the sequence which is all zeros except for a 1 in the kth slot. Note
th... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
3.37)
�u� ≤ �u − v� + �v�, �v� ≤ �u − v� + �u�
which implies that
(3.38)
|�u� − �v�| ≤ �u − v�.
This shows that � · � is continuous, indeed it is Lipschitz continuous.
� | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/34aa7af3453590196e4c834ad62b8129_MIT18_102s09_lec03.pdf |
18.311:
Principles
of
Applied
Mathematics
Lecture
5
Rodolfo
Ros
ales
Spring
2014
servation
laws
and
pde:
igher
order
(TRANSPORT)
effects
beyond
quasi
Time
it
takes
to
reach
R
=
5
cm
[salt
in
water].
Con
H
Not
fl
ux
appe
•
•
e
be... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
in
traffic
fl
ids.
flow
in
a
silo.
ow.
xample
of
a
higher
order
transport
effect
is
viscosity:
Forces
proportional
Another
e
to
the
flow
velocity
gradient.
REASON:
the
flow
velocity
is
a
macroscopic
"average"
variabl... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
vation
osity.
D
rivation
of
e
-‐law
d
erive
TO
PIC:
Solution
of
1st
order
(scalar)
quasilinear
equations
by
characteristics.
Examples:
traffic
flow
and
river
waves.
Tra
ffic
Flow
equations.
Traffic
density,
flow
and
car
velocity.
r ... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
=
coefficient
0.
s.
solve
ut+c0*ux
=
0.
But
first,
note
that:
for
traffi
Density
waves
reach
cars
from
ahead.
River
waves
move
faster
than
the
flow.
c
flow
u
>
c
and
for
river
flows
u
<
c.
Interpret
... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
into
statements
about
the
directional
derivatives
of
the
solutions.
Such
equations
are
called
hyperbolic.
1
18.311:
Principles
of
Applied
M... | https://ocw.mit.edu/courses/18-311-principles-of-applied-mathematics-spring-2014/34ccec51ce5d47f7bd3e6fd81ae192f7_MIT18_311S14_Lecture5.pdf |
18.782 Introduction to Arithmetic Geometry
Lecture #17
Fall 2013
11/05/2013
Throughout this lecture k denotes an algebraically closed field.
17.1 Tangent spaces and hypersurfaces
For any polynomial f ∈ k[x1, . . . , xn] and point P = (a1, . . . , an) ∈ An we define the affine
linear form
fP (x1, . . . , xn) :=
(P )(xi
− ai... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
of I(V ). Applying the product rule and the fact that
fi,P (P ) = 0 yields
gP =
(cid:88)(cid:0)hi(P )fi,P + hi,P fi(P )
(cid:1) =
(cid:88)
hi(P )fi,P ,
i
i
(1)
which is an element of the ideal (f1,P , . . . , fm,P ). Thus I(Tp(V )) = (f1,P , . . . , fm,P ).
When considering the tangent space of a variety at a particula... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
. This follows immediately from Lemma 17.2 and its proof; the set of linear forms in
I(TP (V )) is precisely the set of linear forms that vanish at every point in Tp(V ), which, by
∨. Moreover, we see from (1) that every linear
definition, is the orthogonal complement TP
form in I(TP (V )) is a k-linear combination of f... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
rank JP and dim TP (V ) = n − rank JP .
In particular, the rank of JP does not depend on the choice of generators for I(V ) and P
is a smooth point of V if and only if dim TP = dim V .
Remark 17.5. For projective varieties V we defined smooth points P as points that are
smooth in all (equivalently, any) affine part contai... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
�, so degxi f > 0 for some xi,
say x1. If dim V < n − 1 then the transcendence degree of k(V ) is less than n − 1, therefore
φ(x2), . . . , φ(xn) must be algebraically dependent as elements of k(V ). Thus there exists
g ∈ k[x2, . . . , xn] such that g(φ(x2), . . . , φ(xn)) = 0. But then φ(g) = 0, so g ∈ ker φ = (f ).
⊆... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
over k(γ1, . . . , γn), and we claim that in fact m = n + 1 and we are also done.
Suppose m > n + 1. Let f ∈ k[x1, . . . , xn+1] be irreducible with f (γ1, . . . , γn+1) = 0;
such an f exists since γ1, . . . , γn+1 are algebraically dependent. We must have ∂f /∂xi = 0
for some xi;
if not than we must have char(k) = p >... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
if all their nonempty
affine parts are, and the projective closure of a hypersurface is a hypersurface, so it suffices
to consider affine varieties. Recall from Lecture 15 that varieties are birationally equivalent
if and only if their function fields are isomorphic, and it follows from Theorem 17.8 that
every function field a... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
/∂xj), the determinant of each of its r × r minors is a polynomial in k[x1, . . . , xn], and
Sing(V) is the intersection of V with the zero locus of all these polynomials. Thus Sing(V)
is an algebraic set, hence closed.
Recall the one-to-one correspondence between points P = (a1, . . . , an) in An and max-
imal ideals ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
if and only if ∂f /∂xi(0) = 0 for i = 1, . . . , n, and this occurs precisely
when every term in f has degree at least 2, equivalently, f ∈ M 2
P . It follows that
M /M 2 (cid:39) (kn
P
P
)∨.
The restriction map (kn)∨ → (TP )∨ that restricts the domain of a linear form on kn to
TP (V ) is surjective, and composing this... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
= dim k[V ]. The key
4 If R is any
point is that we now have a completely algebraic notion of smooth points.
affine algebra, the maximal ideals m of R correspond to smooth points on a variety with
coordinate ring R, and we can characterize the “smooth” maximal ideals as those for which
dimk m/m2 = dim R, where k = Rm/m i... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/34da4eafb6fbe4d016d08aa7914a4be7_MIT18_782F13_lec17.pdf |
6 Young diagrams and q-binomial coefficients.
0 satisfying �1
0 is a sequence � = (�1, �2, . . .) of integers
A partition � of an integer n
i�1 �i = n. Thus all but finitely
and
�i ←
many �i are equal to 0. Each �i > 0 is called a part of �. We sometimes
suppress 0’s from the notation for �, e.g., (5, 2, 2, 1), (... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
Ø, 1, 2, 3, 11, 21, 31, 22, 32, 33
L(2, 3) =
}
{
the unique partition (0, 0, . . .) with no parts.) If � = (�1, �2, . . .) and µ =
(µ1, µ2, . . .) are partitions, then define �
µi for all i. This makes
the set of all partitions into a very interesting poset, denoted Y and called
Young’s lattice (named after the Br... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
21��
����
�
�
�
� �
���1
22 ��
��
11��
� �
�� 31
����
�
�
�2
��
� �
�� 3
��
� Ø
There is a nice geometric way of viewing partitions and the poset L(m, n).
The Young diagram (somtimes just called the diagram) of a partition � is a
left-justified array of squares, with �i squares in the ith row. For instance,
... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
6.2 Proposition. L(m, n) is graded of rank mn and rank-symmetric.
The rank of a partition � is just � (the sum of the parts of � or the number
|
of squares in its Young diagram).
|
Proof. As in the proof of Proposition 5.6, we leave to the reader every
thing except rank-symmetry. To show rank-symmetry, consider the... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
same northwest corner, as explained
fit inside an m
previously). Though not really necessary for this goal, it is nonetheless in
teresting to obtain some information on these numbers pi(m, n). First let us
consider the total number L(m, n) of elements in L(m, n).
×
|
6.3 Proposition. We have L(m, n) = m+n .
|
|
|
m... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
reader) that the above correspondence gives
a bijection between Young diagrams D fitting in an m
n rectangle R, and
sequences of m N ’s and n E’s. Hence the number of diagrams is equal to
m+n , the number of sequences. �
m
×
�
�
We now consider how many elements of L(m, n) have rank i. To this end,
38
... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
is called a q-binomial coefficient (or Gaussian coefficient ).
One sometimes says that
k
j
�
�
k
j q=1
=
k
j
� �
.
� ⎟
is a “q-analogue of the binomial coefficient k .”j
� �
6.4 Example. We have
[why?]. Moreover,
k
j =
�
�
=
�
k
0
� ⎟
k
k−j
�
= 1
k
k
� ⎟
k
�
=
1
⎟
−
[4][3][2][1]
[2][1][2][1]
k
k
1
�... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
the “initial conditions” 0 = 1,
j > k (the same intial conditions satisfied by the binomial coefficients
k
j = 0 if j < 0 or
k
�
j ).
� �
Proof. This is a straightforward computation. Specifically, we have
�
�
0
[k
[j]![k
1]!
1
j]!
+ q k−j
k
1
−
j
+ q k−j k
−
j
�
−
1
1
⎟
⎟
�
=
=
=
=
=
−
−
[k
−
1]![k ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
triangle recurrence.
�
�
�
�
We can regard equation (26) as a recurrence relation for the q-binomial
coefficients. Given the initial conditions of Lemma 6.5, we can use (26) induc
tively to compute
for any k and j. From this it is obvious by induction
k
j
�
�
is a polynomial in q with nonnegative inte
that the q-b... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
also satisfies (29). Moreover, the initial conditions of Lemma 6.5 show that
�
m+n also satisfies (28). Hence (28) and (29) imply that P (m, n) = m+n ,
m
so to complete the proof we need only establish (28) and (29).
−
m
�
�
Equation (28) is clear, since L(0, n) consists of a single point (the empty
i�0 pi(0, n)xi = ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
−n(m
1, n) such �, and the proof follows [why?]. �
n which fits in an (m
1)
−
×
−
×
−
−
−
Note that if we set q = 1 in (27), then the left-hand side becomes L(m, n)
and the right-hand side m+n , agreeing with Proposition 6.3.
m
|
|
�
�
Note: There is another well-known interpretation of
k
j , this time not
of i... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
an m
n rectangle of
squares. For instance, R35 is given by the 15 squares of the diagram
×
We now define the group G = Gmn as follows. It is a subgroup of the group
S
R of all permutations of the squares of R. A permutation λ in G is allowed
to permute the elements in each row of R in any way, and then to permute ... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
just defined a group Gmn of permutations of the set Rmn of
squares of an m
n rectangle. Hence Gmn acts on the boolean algebra BR of
all subsets of the set R. The next lemma describes the orbits of this action.
×
Every orbit
6.8 Lemma.
of the action of Gmn on BR contains
R such that D
exactly one Young diagram D... | https://ocw.mit.edu/courses/18-318-topics-in-algebraic-combinatorics-spring-2006/351d525546a333048bca0b2f436ea256_young.pdf |
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