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Networks
Special topic: Optical Networks, Wireless networks
Final Exam during final exam week. Date and time to be announced.
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Eytan Modiano
Slide 5
Network Applications
• Resource sharing
– Computing
– Mainframe computer (old days)
Today, computers cheaper than comm (except LANS)
Printers, peripherals
–
Information
DB access and updates
E.g., Financial, Airline reservations, etc.
• Services
– Email, FTP, Telnet, Web access
– Video conferencing
– DB access
– Client/server applications
Eytan Modiano
Slide 6
Network coverage areas
• Wide Area Networks (WANS)
– Span large areas (countries, continents, world)
– Use leased phone lines (expensive!)
1980’s: 10 Kbps, 2000’s: 2.5 Gbps
User access rates: 56Kbps – 155 Mbps typical
– Shared comm links: switches and routers
E.g, IBM SNA, X.25 networks, Internet
• Local Area Networks (LANS)
– Span office or building
– Single hop (shared channel) (cheap!)
– User rates: 10 Mbps – 1 Gbps
E.g., Ethernet, Token rings, Apple-talk
• Metro Area networks (MANS)
• Storage area networks
Eytan Modiano
Slide 7
Network services
• Synchronous
– Session appears as a continuous stream of traffic (e.g, voice)
– Usually requires fixed and limited delays
• Asynchronous
– Session appears as a sequence of messages
– Typically bursty
– E.g., Interactive sessions, file transfers, email
• Connection oriented services
– Long sustained session
– Orderly and timely delivery of packets
– E.g., Telnet, FTP
• Connectionless services
– One time transaction (e.g., email)
• QoS
Eytan Modiano
Slide 8
Switching Techniques
• Circuit Switching
– Dedicated resources
• Packet Switching
– Shared resources
– | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf |
8
Switching Techniques
• Circuit Switching
– Dedicated resources
• Packet Switching
– Shared resources
– Virtual Circuits
– Datagrams
Eytan Modiano
Slide 9
Circuit Switching
• Each session is allocated a fixed fraction of the capacity on each
link along its path
– Dedicated resources
– Fixed path
–
If capacity is used, calls are blocked
E.g., telephone network
• Advantages of circuit switching
– Fixed delays
– Guaranteed continuous delivery
• Disadvantages
– Circuits are not used when session is idle
–
– Circuit switching usually done using a fixed rate stream (e.g., 64
Inefficient for bursty traffic
Kbps)
Difficult to support variable data rates
Eytan Modiano
Slide 10
Problems with circuit switching
• Many data sessions are low duty factor (bursty),
(message transmission time)/(message interarrival time) << 1
Same as: (message arrival rate) * (message transmission time) << 1
• The rate allocated to the session must be large enough to meet the
delay requirement. This allocated capacity is idle when the session
has nothing to send
If communication is expensive, then circuit switching is
uneconomic to meet the delay requirements of bursty traffic
•
• Also, circuit switching requires a call set-up during which
resources are not utilized. If messages are much shorter than the
call set-up time then circuit switching is not economical (or even
practical)
– More of a problem in high-speed networks
Eytan Modiano
Slide 11
Circuit Switching Example
L = message lengths
λ = arrival rate of messages
R = channel rate in bits per second
X = message transmission delay = L/R
– R must be large enough to keep X small
– Bursty traffic => λx << 1 => low utilization
• Example
λ = 1 message per second
– L = 1000 bytes (8000 bits)
–
– X < 0.1 seconds (delay requirement)
– => R > 8000/0.1 = 80,000 bps
Utilization = 8000/80000 = 10%
• With packet | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf |
8000/0.1 = 80,000 bps
Utilization = 8000/80000 = 10%
• With packet switching channel can be shared among many
sessions to achieve higher utilization
Eytan Modiano
Slide 12
Packet Switched Networks
Messages broken into
Packets that are routed
To their destination
PS
PS
PS
Packet Network
PS
PS
PS
PS
Buffer
Packet
Switch
Eytan Modiano
Slide 13
Packet Switching
• Datagram packet switching
– Route chosen on packet-by-packet basis
– Different packets may follow different routes
– Packets may arrive out of order at the destination
– E.g., IP (The Internet Protocol)
• Virtual Circuit packet switching
– All packets associated with a session follow the same path
– Route is chosen at start of session
– Packets are labeled with a VC# designating the route
– The VC number must be unique on a given link but can change from
link to link
Imagine having to set up connections between 1000 nodes in a mesh
Unique VC numbers imply 1 Million VC numbers that must be represented
and stored at each node
– E.g., ATM (Asynchronous transfer mode)
Eytan Modiano
Slide 14
Virtual Circuits Packet Switching
•
•
For datagrams, addressing information must uniquely distinguish
each network node and session
– Need unique source and destination addresses
For virtual circuits, only the virtual circuits on a link need be
distinguished by addressing
– Global address needed to set-up virtual circuit
– Once established, local virtual circuit numbers can then be used to
represent the virtual circuits on a given link: VC number changes from
link to link
• Merits of virtual circuits
– Save on route computation
Need only be done once
at start of session
– Save on header size
– Facilitate QoS provisioning
– More complex
– Less flexible
3
VC7
VC13
VC3
6
5
VC4
VC3
VC7
8
9
2
Node 5 table
(3,5) VC13 -> (5,8) VC3
(3,5) VC7 -> (5,8) VC4
(6,5) VC3 -> (5,8) VC7
Eytan | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf |
-> (5,8) VC4
(6,5) VC3 -> (5,8) VC7
Eytan Modiano
Slide 15
Circuit vs packet switching
• Advantages of packet switching
– Efficient for bursty data
– Easy to provide bandwidth on demand with variable rates
• Disadvantages of packet switching
– Variable delays
– Difficult to provide QoS assurances (Best-effort service)
– Packets can arrive out-of-order
Switching Technique
Network service
Circuit switching
Packet switching
Virtual circuits
Datagram
=>
=>
=>
=>
Synchronous (e.g., voice)
Asynchronous (e.g., Data)
Connection oriented
Connectionless
Eytan Modiano
Slide 16
Circuit vs Packet Switching
• Can circuit switched network be used to support data traffic?
• Can packet switched network be used for connection oriented
traffic (e.g., voice)?
• Need for Quality of service (QoS) mechanisms in packet networks
– Guaranteed bandwidth
– Guaranteed delays
– Guaranteed delay variations
– Packet loss rate
– Etc...
Eytan Modiano
Slide 17
7 Layer OSI Reference Model
Application
Presentation
Session
Transport
Virtual network service
Virtual session
Virtual link for end to end messages
Virtual link for end to end packets
Application
Presentation
Session
Transport
Network
Network
Network
Network
Virtual link for
reliable packets
Virtual bit pipe
DLC
DLC
DLC
DLC
phys. int. phys. int.
phys. int. phys. int.
Data link
Control
physical
interface
Physical link
subnet
node
subnet
node
External
site
Data link
Control
physical
interface
External
Site
Eytan Modiano
Slide 18
Layers
• Presentation layer
– Provides character code conversion, data encryption, data compression,
etc.
• Session layer
– Obtains virtual end to end message service from transport layer
– Provides directory assistance, access rights, billing functions, etc.
• Standardization has not proceeded well here, since transport to
application are all in the operating system and don't really need
standard interfaces
• Focus: Transport layer and lower
Eytan Modiano
Slide 19
Transport Layer
• The network layer provides a virtual end to end packet pipe to | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf |
Transport layer and lower
Eytan Modiano
Slide 19
Transport Layer
• The network layer provides a virtual end to end packet pipe to the
transport layer.
• The transport layer provides a virtual end to end message service
to the higher layers.
• The functions of the transport layer are:
1) Break messages into packets and reassemble
packets of size suitable to network layer
2) Multiplex sessions with same source/destination nodes
3) Resequence packets at destination
4) recover from residual errors and failures
5) Provide end-to-end flow control
Eytan Modiano
Slide 20
Network layer
• The network layer module accepts incoming packets from the
transport layer and transit packets from the DLC layer
•
It routes each packet to the proper outgoing DLC or (at the
destination) to the transport layer
• Typically, the network layer adds its own header to the packets
received from the transport layer. This header provides the
information needed for routing (e.g., destination address)
Each node contains one network
Layer module plus one
Link layer module per link
Transport
layer
Network
layer
DLC layer
link 1
DLC layer
link 2
DLC layer
link 3
Eytan Modiano
Slide 21
Link Layer
• Responsible for error-free transmission of packets across a
single link
– Framing
Determine the start and end of packets
– Error detection
Determine which packets contain transmission errors
– Error correction
Retransmission schemes (Automatic Repeat Request (ARQ))
Eytan Modiano
Slide 22
Physical Layer
• Responsible for transmission of bits over a link
• Propagation delays
– Time it takes the signal to travel from the source to the destination
Signal travel approximately at the speed of light, C = 3x108 meters/second
– E.g.,
LEO satellite: d = 1000 km => 3.3 ms prop. delay
GEO satellite: d = 40,000 km => 1/8 sec prop. delay
Ethernet cable: d = 1 km => 3 µs prop. delay
• Transmission errors
– Signals experience power loss due to attenuation
– Transmission is impaired by noise
– Simple channel model: Binary Symmetric Channel
P = bit error probability
Independent from bit to bit
–
In reality | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf |
is impaired by noise
– Simple channel model: Binary Symmetric Channel
P = bit error probability
Independent from bit to bit
–
In reality channel errors are often bursty
0
1
1-P
P
P
1-P
0
1
Eytan Modiano
Slide 23
Internet Sub-layer
• A sublayer between the transport and network layers is required
when various incompatible networks are joined together
• This sublayer is used at gateways between the different networks
It looks like a transport layer to the networks being joined
It is responsible for routing and flow control between networks, so
looks like a network layer to the end-to-end transport layer
In the internet this function is accomplished using the
Internet Protocol (IP)
– Often IP is also used as the network layer protocol, hence only one
protocol is needed
•
•
•
Eytan Modiano
Slide 24
Internetworking with TCP/IP
FTP
client
TCP
IP
IP Protocol
FTP Protocol
TCP Protocol
ROUTER
IP
FTP
server
TCP
IP Protocol
IP
Ethernet
driver
Ethernet
Protocol
Ethernet
driver
token ring
driver
token ring
Protocol
token ring
driver
Ethernet
token ring
Eytan Modiano
Slide 25
Encapsulation
user data
Appl
header
user data
TCP
header
application data
TCP segment
IP
header
TCP
header
IP datagram
application data
Ethernet
header
14
IP
header
20
TCP
header
20
application data
Ethernet
trailer
4
Ethernet frame
46 to 1500 bytes
Eytan Modiano
Slide 26
Application
TCP
IP
Ethernet
driver
Ethernet | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf |
18.409 An Algorithmist’s Toolkit
September 22, 2009
Lecturer: Jonathan Kelner
Scribe: Dan Iancu (2009)
Lecture 4
1 Random walks
Let G = (V, E) be an undirected graph. Consider the random process that starts from some vertex v ∈ V (G),
and repeatedly moves to a neighbor of the current vertex chosen uniformly at random.
For t ≥ 0, and for u ∈ V (G), let pt(u) denote the probability that you are at vertex u at time t. We can
think of pt as a vector in Rn. Clearly,
(cid:88)
u∈V (G)
pt(u) = 1
Observe that you are at a vertex u at time t, then at time t + 1 you are at each neighbor v of u with
probability 1/d(u), where d(u) denotes the degree of u. So,
pt+1(v) =
(cid:88)
Pr[at u at time t] · Pr[’go to v at time t + 1’ given ’at u at time t’]
=
(u,v)∈E(G)
(cid:88)
(u,v)∈E(G)
pt(u) ·
1
d(u)
We can write this using matrix notation as follows. Define the matrix W = WG:
[WG]i,j =
(cid:26) 1
d
0
(j)
(i, ) ∈ (G)
E
if
j
otherwise
Note that [WG]i,j is the probability of going from j to i. We have
WG = A · D−1,
where A is the adjacency matrix of G, and D is the diagonal matrix with [D]i,i the degree of the i-th vertex
of G.
2 Stationary distribution
We define a probability vector π which corresponds to the stationary distribution of the random walk. Let
π(u) =
d(u)
v∈V (G) d(v)
(cid:80)
.
Claim
1 π is a probability distribution.
Proof We have
(cid:88)
π(i) =
(cid:88)
u∈
V (G)
u∈
V (G)
d(u
)
(cid:80)
v∈V (G) d(v) (cid:80)
= u∈V (G) d(u)
v� | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
(cid:80)
v∈V (G) d(v) (cid:80)
= u∈V (G) d(u)
v∈V (G) d(v)
(cid:80)
= 1.
We next show that, if the random walk follows the distribution π at time t, then it has the same
distribution at time t + 1. This is expressed using matrix notation in the following claim.
4-1
Claim 2 W · π = π.
Proof Let k ∈ V (G). We have
n �
[W · π]k = Wk,iπi = �
i=1
1
�
v∈V (G)
d(v)
(i,k)∈E(G)
1
d(k)
· d(k) = �
1
v∈V (G)
· d(k) = π(k).
d(v)
This statement is equivalent to the matrix W having eigenvalue 1, with corresponding eigenvector π
(note that, since π is a multiple of the vector of node degrees, D · 1, we could also take the latter as the
eigenvector).
The natural next step at this point would be to claim that the random walk of a graph G always converges
to the stationary distribution π. This however turns out to be false. It is easy to see that for a bipartite
graph G. Consider for example the case G = C6, the cycle on 6 vertices, and let the vertex set of G be
V (G) = {1, 2, . . . , 6}. Assume without loss of generality that the random walk starts at time t0 = 1 at vertex
6. Then, at time t, the current vertex is odd if and only if t is odd. Therefore, the walk does not converge
to any distribution. | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
if and only if t is odd. Therefore, the walk does not converge
to any distribution.
3
Lazy Random Walks
There is an easy way to fix the above periodicity problem. We introduce a modified version of the original
walk, which we call lazy random walk. In a lazy random walk at time t:
• we take a step of the original random walk with probability 1/2,
• we stay at the current vertex with probability 1/2.
We can show that the above modification breaks the periodicity of the random walk. The transition proba
bilities are encoded in the following matrix:
W (cid:3) = (W + I)/2 = (I + A · D−1)/2,
where I denotes the identity matrix.
The fact that W and W (cid:3) are not symmetric matrices makes their analysis complicated. We will thus
define new matrices. The normalized walk matrix is defined as
N = D−1/2 · W · D1/2 = D−1/2 · A · D−1/2 .
The normalized lazy walk matrix is defined as
N (cid:3) = D−1/2 · W (cid:3) · D1/2 = (I + D−1/2 · A · D−1/2)/2.
Claim 3 The matrices N and W have the same eigenvalues and related eigenvectors.
Proof Suppose that v is an eigenvector of N , with eigenvalue λ. Let q = D1/2 · v. Then,
N · v = λ · v = D−1/2 · W · D1/2 · v = D−1/2 · W · q.
Multiplying by D1/2 on the left we obtain
Therefore, q is an eigenvector of W with eigenvalue λ. | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
Multiplying by D1/2 on the left we obtain
Therefore, q is an eigenvector of W with eigenvalue λ.
W · q = λ · D1/2 · v = λ · q.
Observe that, by Claim 2, W has eigenvector D · 1, with eigenvalue 1. Therefore, by Claim 3, the
normalized walk matrix N has eigenvector D1/2 · 1, with eigenvalue 1.
4-2
4 Connections to Laplacians
We’ve used the Laplacian L. The normalized Laplacian L is defined as
L = D−1/2 · L · D−1/2 .
Claim 4 N = I − L.
Therefore, the eigenvalues of N are given by 1 − (eigenvalues of L). So, it makes sense to order them in the
opposite way
1 = μ1 ≥ μ2 ≥ . . . ≥ μn
We can now translate our theorems about the eigenvalues of Laplacians to theorems about μis. We have
• For each i, μi ∈ [−1, 1].
• If G is connected, then μ2 < 1.
• The −1 eigenvalues occur only for bipartite graphs.
Let μ(cid:3)
i be the eigenvalues of N (cid:3) . Then
• For each i, μ(cid:3)
i
∈ [0, 1].
• If G is connected, then μ(cid:3)
2
< 1.
5
�2 Convergence
Define the spectral gap to be
λ := 1 − μ(cid:3)
2.
For probability distributions p, q, we define their (cid:4)2 distance to be
(cid:4)p − q(cid:4)2 =
� �
(p(i) − q(i))2 .
i
The following theorem gives a bound on the rate of convergence of the lazy random walk to the stationary
distribution π.
Theorem 5 Let p | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
following theorem gives a bound on the rate of convergence of the lazy random walk to the stationary
distribution π.
Theorem 5 Let p0 be an arbitrary initial distribution, and pt be the distribution after t steps of the lazy
random walk. Then,
(cid:4)pt − π(cid:4)2 ≤ (1 − λ)t ·
�
maxx d(x)
miny d(y)
.
[Proof for regular graphs] Observe that for a matrix M = Q−1 · Λ · Q, we have M k = Q−1 · Λk · Q.
Proof
Thus, for an eigenvector v of M , M k · v = λk · v.
Recall that N (cid:3) = (I + D−1/2 · A · D−1/2)/2. Since G is regular, D = d · I, for some integer d >0. Thus,
and the stationary distribution is simply the uniform distribution on V (G)
1
N (cid:3) = I + A d
π =
1
n
· 1.
4-3
Let ci = vi
T p0, where vi denotes the eigenvector corresponding to the i-th eigenvalue. We have
N (cid:3)k · p0 =
�
n
ci · μk
i
i=1
Since c1 = v1
T p0 = 1/n, it follows that
· vi = c1 · v1 +
ci · μk
i
· vi
�
n
i=2
(cid:4)pk − π(cid:4)2 = (cid:4)
ci · μk
i
n
�
i=2
�
n
�
�
�
· vi(cid:4)2 = �
n
�
2 · μ2
ci
i
�
�
�
�
k ≤ μk
2
n
�
2
ci
i=2
i=2
≤ μk
2
(vi
T p0)2 | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
�
2
ci
i=2
i=2
≤ μk
2
(vi
T p0)2 ≤ μk
2 = (1 − λ)k .
i=1
Using a similar argument, we can also show an analogous bound for (cid:4)∞ convergence.
Theorem 6 For any vertex v ∈ V (G),
|pt(v) − π(v)| ≤ (1 − λ)t ·
�
d(v)
miny d(y)
6 Conductance
Cheeger’s inequality carries over too, by replacing the isoperimetric number by a new parameter, which we
call conductance Φ.
Definition 7 (Conductance) For S ⊆ V (G), let
Φ(S) =
��
min
v∈S
e(S)
d(v),
�
� .
v∈S¯ d(v)
We define the conductance to be
Φ(G) = min Φ(S).
S⊆V
Using the above definition, Cheeger’s inequality now becomes:
Θ(1) · Φ2(G) ≤ 1 − μ(cid:3)
2
≤ Θ(1) · Φ(G).
The parameter Φ(G) is related to the rate of convergence to the stationary distribution. In particular,
bounds on Φ(G) let us prove that a walk mixes quickly.
The intuitive interpretation of the connection between conductance and the rate of convergence is as
follows. If a graph has high conductance, it is well-connected. Therefore, a large amount of probability mass
can very quickly move from one part of the graph to another.
7
Introduction to Monte Carlo methods
Assume that we want to estimate π = 3.1415 . . . by throwing darts in the following dartboard:
4-4
Assume that the square corresponds to [−1, 1] × [−1, 1]. If you pick | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
that the square corresponds to [−1, 1] × [−1, 1]. If you pick a point in the square uniformly at
random, the probability that you pick one inside the circle is equal to π/4. Suppose that you pick n points
in [−1, 1] × [−1, 1], uniformly at random. Then,
E[number of points inside circle] = n · π/4
So, you can return the estimate
A natural question is how close this estimate would be to the right answer.
πˆ = (number of points inside circle) · 4/n.
In order to answer the above question, we will introduce the Chernoff bound. Suppose we have a random
variable r ∈ {0, 1}, such that Pr[r = 1] = p, and Pr[r = 0] = 1 − p. Assume that we draw n independent
samples r1, . . . , rn, and let R =
ri. By the linearity of expectation, we have
�
i
E[R] = E[
� �
ri] =
E[ri] = n · p
We will say that R (cid:5)-approximates E[R] if
i
i
(1 − (cid:5))E[R] ≤ R ≤ (1 + (cid:5))E[R]
This is a multiplicative error measure.
Theorem 8 (One version of the Chernoff bound) The probability that R fails to (cid:5)-approximate E[R]
is
Pr [|R − E[R]| ≥ (cid:5)E[R]] ≤ 2e −np(cid:2)2/12 = 2e −E[R](cid:2)2/12 .
Some notes on the above bound:
• The bound is near tight.
• It is necessary for the trials to be independent, in order for the bound to hold.
• It provides a multiplicative, but not an additive error guarantee.
• For fixed (cid:5), it | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
• It provides a multiplicative, but not an additive error guarantee.
• For fixed (cid:5), it falls off exponentially in n. So, if we have failure probability 1/2, we can improve it to
1/2k by performing m = n · k trails.
• Therefore, smaller n requires more trials.
• If we want (cid:5)-approximation with probability 1 − δ, then we need
N ≥ Θ
�
log(1/δ)
p(cid:5)2
�
.
That is, we need enough trials to get Θ(log(1/δ)/(cid:5)2) successes.
4-5
Back to the dartboard example, if we want to estimate π within, say, 5%, with probability at least 0.99,
then we have (cid:5) = 0.05, δ = 1/100. Therefore, we need
�
�
N ≥ Θ
log(100)
(π/4)(0.05)2
Observe that it is easy to make δ smaller, but it is harder to make (cid:5) smaller.
If we are bad darts, then we run into trouble. This happens if we have a big dartboard, and a small
circle.
In particular, if p is exponentially small, then we need exponentially many trials to expect a constant number
of successes.
We can also run into trouble if it is hard to throw darts at all. That is, if it is hard to draw samples
uniformly at random from the ambient space. We will develop some techniques for fixing the above problems
in certain scenarios.
4-6
MIT OpenCourseWare
http://ocw.mit.edu
18.409 Topics in Theoretical Computer Science: An Algorithmist's Toolkit
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/100377025e8520aab9f61d8585e71cc5_MIT18_409F09_scribe4.pdf |
DIVISOR CLASSES ON THE MODULI SPACE OF CURVES
1. The cohomology of the moduli space of pointed genus zero curves
In this section we discuss the Chow rings of the moduli spaces of n-pointed
genus zero curve M0,n. Recall that we are working over the complex numbers C.
The cohomology and Chow groups of M0,n turn out to be isomorphic. The main
statement is that the Chow/cohomology ring of M0,n is generated by the classes of
boundary divisors. The main reference for this section is [Kee].
The basic strategy for determining the Chow/cohomology ring of M0,n is to
exhibit M0,n as a sequence of blow-ups of the product P1 × · · · × P1 of n − 3 copies
of P1 along smooth centers. One then inductively calculates the Chow ring at each
stage of the blow-up process using the following basic theorem.
Theorem 1.1 (The Chow ring of blow-ups). Let X be a codimension d smooth
subvariety of a smooth variety Y with normal bundle NX/Y . Let i : X � Y denote
the inclusion of X in Y . Let Y˜ be the blow-up of Y along X. Assume that
i� : A(Y ) � A(X)
is surjective. Then
A�(Y˜ ) �
=
A�(Y )[λ]
< λ ker(i�), λ d + λ d−1c1(NX/Y ) + · · · + λcd−1(NX/Y ) + cd(NX/Y ) >
where −λ is the class of the exceptional divisor.
Now we introduce the generators of the Chow ring. Let S be a subset of
{1, . | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
introduce the generators of the Chow ring. Let S be a subset of
{1, . . . , n} with the property that both S and its complement have at least two
elements. We will denote the number of elements of S by #S. Given such a set
we can define the class ζS on M0,n as the class of the divisor �S of stable curves
C that have a separating node that divides C into C1 ∞ C2 where the labelings of
the points on C1 are precisely the elements of S and the labelings of the points on
C2 are precisely the elements of S c . The divisor �S is a normal crossings divisor
isomorphic to
�
�S = M 0,S≤{r} × M 0,Sc ≤{s}
obtained by the map that glues the marked points r and s.
The main theorem about the Chow ring of M0,n is the following:
Theorem 1.2 (Keel). The Chow/cohomology ring of M0,n is generated by the
classes ζS where #S ∼ 2 and #Sc ∼ 2 subject to the following relations:
(1) ζS = ζSc .
(2) For any four distinct elements i, j, k, l ⊕ {1, . . . , n}
�S =
ζS =
ζS .
i,j≥S,k,l /
⎪
≥S
i,k≥S,j,l /
⎪
≥S
i,l≥S,j,k /
⎪
≥S
1
(3) For two subsets S and T
ζS ζT = 0
unless S ⊗ T, T ⊗ S, | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
subsets S and T
ζS ζT = 0
unless S ⊗ T, T ⊗ S, S ⊗ T c or T c ⊗ S.
�
Example 1.3. Since M0,4 = P1, the classes of the three boundary divisors �{1,2},
�{1,3} and �{1,4} are linearly equivalent. If we specialize the statement of the
theorem to n = 4 we recover the cohomology of P1 .
Remark 1.4. It is easy to see that the claimed relations are satisfied. The divisor
classes ζS and ζSc are equal since the divisors they represent are equal.
To prove the relation
consider the map
ζS =
ζS
i,j≥S,k,l /≥S
⎪
i,k≥S,j,l /≥S
⎪
ψi,j,k,l : M0,n � M0,4
given by forgetting all the points, but the points labeled by i, j, k, l and stabilizing
the resulting curve. The pull-back of the divisor class ζ{i,j} on M0,4 is given by
ζS .
i,j≥S,k,l /≥S
⎪
The pull-back of the divisor class ζ{i,k} on M0,4 is given by
Since these divisors have to be linearly equivalent, the relation follows.
i,k≥S,j,l /≥S
⎪
ζS .
Finally to see that ζS ζT = 0 unless S ⊗ T, T ⊗ S, S ⊗ T c or T c ⊗ S note
that two divisors �S and �T contain the point represeting a curve C in their
intersection if and only if there are two nodes on C that divide C into C1, C2 and
∅ , C ∅ where | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
and only if there are two nodes on C that divide C into C1, C2 and
∅ , C ∅ where the labeling on C1 is S and the labeling on C ∅ is T . Observe that
C1
unless the conditions S ⊗ T, T ⊗ S, S ⊗ T c or T c ⊗ S are satisfied �S and �T are
disjoint, hence the product of their classes is zero in the Chow/cohomology ring.
1
2
Example 1.5. We can view M0,5 as the blow-up of P1 × P1 at the three points
(0, 0), (1, 1) and (→, →). Hence M0,5 is isomorphic to the Del Pezzo surface D5.
The 10 boundary divisors on M0,5 correspond to the 10 exceptional curves on D5.
We can recover the cohomology ring of D5 from Keel’s relations. Note that Keel’s
second set of relations in this case give us that for any distinct 4-tuple i, j, k, l:
ζi,j + ζk,l = ζi,k + ζj,l = ζi,l + ζj,k .
Multiplying these relations by ζi,j and using the third set of relations easily gives
that ζ2 = ζ2
k,l = −ζr,sζt,u for any i, j, k, l and distinct r, s, t, u. Finally all triple
i,j
products vanish. Note that one can give a very simple presentation of the coho
mology ring of D5 realizing it as the blow-up of P2 in four points. Sending the
divisors ζi,5 to the classes | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
-up of P2 in four points. Sending the
divisors ζi,5 to the classes of the four exceptional divisors E1, . . . , E4 and ζi,j to
H − Ek − El (where {k, l} is disjoint from {i, j, 5}) for the remaining i, j gives
a ring isomorphism. Here H denotes the hyperplane class on P2 . Hence, Keel’s
presentation is not necessarily the simplest presentation.
2
Exercise 1.6. Verify the claims made in the discussion of the previous example.
Exercise 1.7. Using the description of the cohomology ring of M0,n determine its
Betti numbers. Find the Euler characteristic of M0,n.
Now we can describe the main technical tool that allows one to compute the
Chow ring of M0,n. Consider the map
given by forgetting the last marked point. This morphism factors through
ψn+1 : M0,n+1 � M0,n
M0,n+1
M0,n × M0,4
�
�n+1
M0,n
id
pr1
M0,n
where pr2 is the projection onto the second factor and χ is induced by (ψn+1, ψ4,...,n)
where ψ4,...,n is the morphism that forgets all but the points marked 1, 2, 3, n + 1.
The calculation is based on the obervation that the morphism χ is in fact a sequence
of n − 3 blow-ups along explicit smooth centers.
Set X1 = M0,n × M0,4. If S is a subset of {1, . . . , n}, we can embed the divisors
�S into X1 by first mapping �S by the universal section corresponding to the i-th
point to M0,n+1, then following it with the map | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
S by the universal section corresponding to the i-th
point to M0,n+1, then following it with the map to X1. Let X2 be the blow-up of
X1 along �S where #Sc = 2 and S contains at most one of 1, 2, 3. Note that these
are disjoint in X1. Let X3 be the blow-up of X2 along the proper transform of the
�S with #Sc = 3 and S contains at most one of 1, 2, 3. Continue in this way where
Xk is the blow-up of Xk−1 along the proper transform of �S with #Sc = k such
that S contains at most one of 1, 2, 3. Then M0,n+1 = Xn−2 and the map
�
is the blowing-up just described.
χ : M0,n+1 � M0,n × M0,4
To finish the proof of Theorem 1.2 we simply have to inductively apply the
theorem describing the Chow ring of the blow-up repeatedly. This is messy but
straightforward (see [Kee]).
Remark 1.8. Note that the construction of M0,n as a blow-up of P1 × · · · × P1
implies that the Chow ring and the cohomology ring are isomorphic. In particular,
M0,n does not have any odd cohomology.
Remark 1.9. Observe that M0,n is an affine variety. Fixing three of the points
at 0, 1 and → we can view this space as the complement of hyperplanes in Cn−3 .
Hence, M0,n is affine of dimension n − 3. Recall that the hom | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
ence, M0,n is affine of dimension n − 3. Recall that the homology of an affine
manifold vanishes above half its real dimension.
Theorem 1.10. Let X be a smooth, complex affine variety of complex dimension
n, then Hk (X, Q) = 0 for k > n.
Milnor’s proof of this theorem using Morse theory is one of the most beautiful
proofs in mathematics (see [Mi]). We conclude that the cohomology Hk (M0,n, Q
vanishes for k > n − 3.
3
�
�
�
�
Note that Theorem 1.2 in particular determines second homology/the Picard
group of M0,n.
Corollary 1.11. The Picard group of M0,n is generated by the classes of boundary
divisors �S subject to the relations ζS = ζSc and for any four distinct elements
i, j, k, l ⊕ {1, . . . , n}
ζS =
ζS =
ζS .
i,j≥S,k,l /
⎪
≥S
i,k≥S,j,l /
⎪
≥S
i,l≥S,j,k /
⎪
≥S
Exercise 1.12. Determine the class of the canonical divisor of M0,n.
In fact, we can do better than the previous corollary.
Proposition 1.13. Let n ∼ 4. Fix three distinct indeces i, j, k. The second coho
mology group of M0,n has basis ζ{j,k}, ζS where i ⊕ S and #S � n − 3.
Proof. We can give an elementary proof of this | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
�S where i ⊕ S and #S � n − 3.
Proof. We can give an elementary proof of this result that does not depend on the
complicated combinatorics of Keel’s theorem. We already saw that the boundary
divisors generate the second cohomology (e.g. the complement of the boundary is
Cn−3 with some hyperplanes removed) and that the relations in Keel’s theorem are
satisfied. We need to show that we can express all boundary divisors in terms of
these and that these are independent. The only boundary divisors not on the list
are those of the form ζu,v where neither of u, v is i and the pair is not j, k. Writing
the boundary relation for i, w, u, v we see that
ζ{i,w} +
ζS + ζ{u,v} = ζ{i,v} +
ζS + ζ{u,w}.
i,w≥S,u,v /
≥S,3�#S�n−3
⎪
i,v≥S,u,w /
≥S,3�#S�n−3
⎪
Hence ζ{u,v} = ζ{u,w}. Taking v = j and then applying the relation again to replace
u by k, we see that the given boundary divisors generate.
We prove the fact that they are independent by induction. Suppose there was a
relation among them. Look at the morphism forgetting a point other than i, j, k.
It immediately follows that all the coefficients of the relation have to be zero. �
Remark 1.14. Note that the following proposition implies that the rank of the
second cohomology group is
2n−1 −
n2 − n + 2
.
2
2. The second homology group of the moduli space of curves
Originally Harer determined the second homology group | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
The second homology group of the moduli space of curves
Originally Harer determined the second homology group of the moduli space of
curves by computing the second homology group of the mapping class group. Some
good references for Harer’s work on this computation is Harer’s original paper [Har1]
and Harer’s C.I.M.E. notes [Har2]. Here we will outline Arbarello and Cornalba’s
algebraic approach to the computation of the second homology group [AC2].
We begin by introducing some divisor classes on Mg,n. Let
ψn+1 : Mg,n+1 � Mg,n
4
denote the morphism that forgets the last marked point. Let ��
be the relattive
dualizing sheaf. ψn+1 has n sections given by the marked points p1, . . . , pn. Denote
the images of these sections πi by �i. The class � in this notation is defined by
n+1
� = ψn+1 �(c1(��n+1 (
�i))2).
n
i=1
⎪
The classes of the n cotangent lines ξi for 1 � i � n are defined by
The sum
ξi = πi
n
i=1 ξi is often denoted by ξ.
�(��n+1 ).
⎩
Finally there are the classes of the boundary divisors. Let ζirr be the class of
the divisor of curves �irr that contain a non-separating node. Let 0 � h � g be
an integer and let S be a subset of {1, . . . , n}. Let ζh,S be the class of the divisor
�h,S of curves that contain a node which separates the curve into two components
of genus h with marked points pi for i ⊕ S and genus g − h and marked points pi
for i ⊕ Sc | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
genus h with marked points pi for i ⊕ S and genus g − h and marked points pi
for i ⊕ Sc . If h (respectively, g − h) is zero, we require that #S ∼ 2 (#S c ∼ 2).
There is one exception to this definition. When we define the class ζ1,∗ = ζg−1,n, we
need to be careful because a general member of this divisor has an automorphism
of order 2. When we define this class, we take it to be half the class of the locus
of the class of the boundary divisor. In terms of this notation the main theorem of
this section is the following:
Theorem 2.1. Let g and n be non-negative integers such that 2g − 2 + n > 0 and
g > 0. The second cohomology group H 2(Mg,n, Q) is generated by the classes �, ξi
for 1 � i � n and the classes ζirr and ζh,S such that 0 � h � g and 2h − 2 + #S > 0
and 2(g − h) − 2 + #Sc > 0.
(1) If g > 2, the relations among these classes are generated by
(2) If g = 2, there is the additional relation
ζh,S = ζg−h,Sc .
(3) If g = 1, there are the following two additional relations
5� = 5ξ + ζirr − 5ζ0 + 7ζ1.
� = ξ − ζ0, 12ξp = ζirr + 12
ζ0,S .
p≥S,#S�2
⎪
Since Theorem 1.2 already determines the genus zero case we will omit it from | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
p≥S,#S�2
⎪
Since Theorem 1.2 already determines the genus zero case we will omit it from our
discussion.
The strategy of the proof is to do induction on the genus and the number of
marked points. We now explain the mechanism that allows us to do this induction.
Recall that since the coarse moduli scheme Mg,n is an orbifold, Poincar´e duality
holds for it provided that we work with rational coefficients.
We need to know the vanishing of the k-th homology groups of Mg,n for large k.
Recall Harer’s theorem which states that the moduli space Mg,n has the homotopy
type of a finite cell complex of dimension 4g − 4 + n for n > 0. Since the homology
groups of a finite cell complex vanish in dimension bigger than the dimension of
the cell complex, we can deduce that
5
Furthermore, a spectral sequence argument implies that
Hk (Mg,n) = 0, k > 4g − 4 + n, n > 0
Hk (Mg,0, Q) = 0, k > 4g − 5.
Combining this vanishing with Poincar´e duality and the long exact sequence of
cohomology
Hc
k (Mg,n, Q) � H k (Mg,n, Q) � H k(ζMg,n, Q) � Hc
k+1(Mg,n, Q)
we conclude the following proposition.
Proposition 2.2. The map H k(Mg,n, Q) � H k (ζMg,n, Q) is an isomorphism when
k < d(g, n) and injective when k = d(g, n), where d(g, n) is defined by
n − 4
d(g, n) = 2g − 2
2g − | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
is defined by
n − 4
d(g, n) = 2g − 2
2g − 3 + n
if g = 0
if n = 0
if g, n > 0.
�
⎧
�
⎧
�
This proposition gives us hope to do induction on the genus and the number of
marked points. Recall that
�irr = Mg−1,P ≤{r,s}
�
where the isomorphism is obtained by attaching the marked points r and s to obtain
a curve of arithmetic genus g. Similarly
�h,S = Mh,S≤{r} × Mg−h,Sc ≤{s}
�
where the isomorphism is obtained by attaching the two curves along the last
marked points. The problem is while we can inductively understand each irre
ducible component of boundary of the moduli space, these boundary components
intersect. However, the next proposition guarantees that this does not effect the
small cohomology groups.
Proposition 2.3. Let Xi, i ⊕ I, denote all the irreducible components of the
boundary of Mg,n. The map
H k (Mg,n, Q) � ≥i≥I H k(Xi, Q)
is injective if k � d(g, n).
Sketch. This proposition follows from the fact that the map
H k (Mg,n, Q) � H k(ζMg,n, Q)
is a morphism of Hodge structures. Since the map is an injection in the claimed
range and H k(Mg,n, Q) is pure of weight k, the cohomology injects to the weight
k part of the cohomology. The proposition follows from a result of Deligne which
asserts that if f | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
part of the cohomology. The proposition follows from a result of Deligne which
asserts that if f : X � Y is a proper, surjective morphism from a smooth variety to
a proper variety, then the weight k quotient of H k (Y, Q) is the image of H k (Y, Q)
in H k (X, Q). Taking X to be the disjoint union of the irreducible components of
the boundary and Y to be the boundary, Deligne’s result (at least its modification
�
for orbifolds) implies the proposition.
6
Proposition 2.4. Let � : Mg−1,n+2 � Mg,n be the morphism that glues the last
two marked points. Then the induced map
�� : H 2(Mg−1,n+2, Q) � H 2(Mg,n, Q)
is injective if g ∼ 2.
Exercise 2.5. Prove the Proposition 2.4 by induction on the number of marked
points and the genus. Use K¨unneth decomposition, Proposition 2.3 and the fact
that H 1(Mg,n, Q) = 0 for every g and n. There are many ways of proving the last
statement. It follows, for example, from the fact that Mg,n is simply connected.
We will see an elementary proof in the next section.
2.1. The relations among tautological classes. In this subsection we indicate
how tautological divisor classes pull-back under special morphisms. Let
be the morphism that forgets the n + 1st marked point.
ψn+1 : Mg,n+1 � Mg,n
Exercise 2.6. Prove the following formulae:
(1) ψ�
(2) ψ�
(3) ψ�
(4) � | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
e:
(1) ψ�
(2) ψ�
(3) ψ�
(4) ψ�
n+1(�) = � − ξn+1.
n+1(ξi) = ξi − ζ0,{i,n+1} for i � n.
n+1(ζirr ) = ζirr .
n+1(ζh,S ) = ζh,S + ζh,S≤{n+1}.
Let
be the morphism that glues the two points x, y.
� : Mg−1,n≤{x,y} � Mg,n
Exercise 2.7. Show that � pulls back the tautological classes as follows:
(1) ��(�) = �.
(2) ��(χi) = χi for i � n.
(3) ��(ζirr ) = ζirr − ξx − ξy +
x≥S,y / ζg,S
≥S
(4) ��(ζh,S ) =
⎩
ζh,S
ζh,S + ζh−1,S≤{x,y} otherwise
if g = 2h, n = 0
⎨
Finally, we need to know the pull-backs of tautological classes by the morphism
ath,S : Mg−h,n−S≤{x} � Mg,n
obtained by attaching a fixed curve of genus h and marking S ∞ {y} to curves in
Mg−h,n−S≤{x} by identifying x and y.
Exercise 2.8. Show that the following relations hold:
(1) at�
h,S (�) = �.
(2) at�
h,S (χi) =
χi
0
⎨
if i ⊕ S
otherwise
(3) at�
h,S (ζirr ) = ζirr .
7
( | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
S
otherwise
(3) at�
h,S (ζirr ) = ζirr .
7
(4) If S = {1, . . . , n}, then
�
ath,S (ζk,T ) =
ζ2h−g,S≤{x} − ξx
ζk,T + ζk+h−g,T ≤{x} otherwise
⎨
if k = h, #T = n, or k = g − h, #T = 0
(5) If S ∈= {1, . . . , n}, then
�
ath,S (ζk,T ) =
−ξx
ζk,T
ζk+h−g,(T \Sc )≤{x}
0
�
⎧
⎧
⎧
�
if (k, T ) = (h, S) or (k, T ) = (g − h, S c)
if T ⊗ S and (k, T ) ∈= (h, S)
if Sc ⊗ T and (k, T ) ∈
otherwise
= (g − h, Sc)
Using the previous three exercises we can obtained the claimed relations in The
orem 2.1. Recall that the Hodge class ω is the first chern class of the Hodge bundle.
⎧
⎧
⎧
�
Lemma 2.9 (Mumford’s relation). On any Mg,n there is the following relation
� = 12ω − ζ + ξ.
Proof. It suffices to prove the formula when n = 0. The general case follows
by pulling-back via the relations given by the forgetful morphisms. We use the
Grothendieck - Riemann - Roch (GRR) formula to see the case n = 0. Set
� = �1
.
Mg,1 /Mg
Recall the GRR formula reads
ch(ψ1! | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
. Set
� = �1
.
Mg,1 /Mg
Recall the GRR formula reads
ch(ψ1!F ) = ψ1�(ch(F ) · T odd(�)).
Set F = �Mg,1 /Mg
the degree one term of the GRR formula we obtain
. Since R1ψ1� of the relative dualizing sheaf is trivial, solving for
ω = c1(ψ1�F ) = ψ1�(
c1(�)2 + c2(�)
12
−
c1(F )c1(�)
2
+
c1(F )
2
)
A local calculation shows that
c1(�1
) = c1(�1
Mg,1 /Mg
where Sing denotes the singular locus. This follows from the exact sequence
Mg,1 /Mg
Mg,1 /Mg
), c2(�1
) = [Sing]
0 � �1
� �1
Mg,1 /Mg
Mumford’s formula immediately follows.
Mg,1 /Mg
� �1
Mg,1 /Mg
≤ OSing � 0.
�
Now all the relations follow when we observe that on M2 we have the relation
10ω = ζ0 + 2ζ1.
To prove this relation, for instance, consider the following test families.
(1) To a fixed genus 1 curve attach a fixed point of a genus 1 curve at a variable
point.
(2) On a genus 1 curve identify a variable point with a fixed point.
(3) Identify a fixed point of a fixed genus 1 curve with a pencil of plane cubics.
Exercise 2.10. By calculating the intersections of these families with ζ0, ζ1 and ω
prove the claimed equality.
8
Exercise 2.11. Deduce the relations in Theorem 2.1 from the relations in this
section.
Remark 2.12. By intersecting with test families it | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
the relations in this
section.
Remark 2.12. By intersecting with test families it is not hard to show that the
relations in Theorem 2.1 are the only relations among tautological divisors.
2.2. Sketch of the proof of Theorem 2.1. In this subsection we will sketch
the proof of Theorem 2.1. We would like to show that H 2(Mg,n, Q) is tautologi
cal. Assume that the tautological classes generate the second cohomology of Mh,m
whenever h < g or h = g and m < n. Suppose that the genus is at least 3 for now.
Let d ⊕ H 2(Mg,n, Q) be any class. Consider
��d ⊕ H 2(Mg−1,n≤{x,y}, Q)
where � : Mg−1,n≤{x,y} � Mg,n is the morphism that identifies the two points
x, y. Since by induction H 2(Mg−1,n≤{x,y}, Q) is tautological ��d may be expressed
as a linear combination of tautological classes. Moreover, since the morphism is
symmetric under exchanging x and y, the expressions of divisors involving x and y
need to be symmetric. Hence, ��d is a linear combination of �, ξi, i � n, ξx + ξy ,
ζirr, ζh,S , ζh,S≤{x,y} and ζh,S≤{x} + ζh,S≤{y}.
We can find a tautological class dt in H 2(Mg,n, Q) such that ��(d − dt) can be
expressed only in terms of ξx + ξy , ζh,S≤{x,y} and ζh,S� | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
expressed only in terms of ξx + ξy , ζh,S≤{x,y} and ζh,S≤{x} + ζh,S≤{y}. To conclude
that all the coefficients vanish we further pull-back ��(d − dt) by the morphism
ellg−2 : Mg−2,n≤{x,y,z} � Mg−1,n≤{x,y}
obtained by attaching a fixed elliptic tail at the marked point
pull-back d − dt to Mg−2,n≤{x,y,z} in a different order, first by the map
z.
We could also
ellg−1 : Mg−1,n≤{z} � Mg,n
that attaches a fixed elliptic curve at the point z, then by the map
�g−2 : Mg−2,n≤{x,y,z} � Mg−1,n≤{z}
that identifies the points x and y. The classes of these two pull-backs have to
coincide. This gives a relation that shows that ��(d − dt) must be identically zero.
Since by Proposition 2.4, the map �� is injective, we conclude that d is tautological.
To conclude the proof then one needs to analyze the cases of genus 1 and 2 in
greater detail. This is straightforward but tedious. We leave you to read the details
in [AC2].
9
3. The first, third and fifth cohomology groups of moduli space
The purpose of this section is to sketch an elementary proof of the vanishing
of the first, third and fifth cohomology groups of Mg,n following Arbarello and
Cornalba [AC2].
Theorem 3.1. H k | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
of Mg,n following Arbarello and
Cornalba [AC2].
Theorem 3.1. H k (M g,n, Q) = 0 for k = 1, 3, 5.
The proof proceeds by reducing the general case to checking the vanishing for
finitely many M g,n with g and n small and carrying out these verifications explicitly.
As in the previous section set
n − 4
d(g, n) = 2g − 2
�
⎧
�
2g − 3 + n
if g = 0
if n = 0
if g, n > 0.
Recall that H k(M g,n, Q) injects into ≥iH k(Xi, Q) where the Xi denote all the
irreducible components of the boundary. Like in the previous section we have the
following Reduction Lemma.
⎧
�
Lemma 3.2 (Reduction Lemma). Let k be an odd integer. Suppose that
H q (M g,n, Q) = 0
for all odd q � k, and for all g and n such that q > d(g, n), then
for all odd q � k and all g and n.
H q (M g,n, Q) = 0
In other words, as long as all the odd cohomology for j < k vanishes, to conclude
vanishing of the k-th cohomology it suffices to verify it for finitely many special
values, namely those values for which q > d(g, n).
Proof. The proof is by induction on k. Suppose H q (M g,n, Q) vanishes for all odd
q � k. We can assume d(g, n) ∼ k. By the previous lemma we conclude that
H k(M g | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
assume d(g, n) ∼ k. By the previous lemma we conclude that
H k(M g,n, Q) injects into H k (Xi, Q). Each Xi is of the form M g−1,n+2 or a product
of M a,A and M b,B where either a < g or a = g and |A| < n. (Similarly for b and
B). Using the K¨unneth formula, we conclude that H k(M g,n, Q) injects into a direct
sum of H k (M g−1,n+2, Q) and H l(M a,A, Q) ≤ H m(M b,B , Q) with l + m = k. Since
either l or m must be odd, all these spaces vanish by the induction hypothesis
except possibly for k = m or k = l. In this case either the genus is smaller than
g or if the genus is equal to g the number of marked points is smaller than n. A
�
double induction concludes the proof.
Proof of vanishing of the first cohomology. By the Reduction Lemma to prove that
the first cohomology groups of M g,n vanish we need to check the cases
M 0,3, M 0,4, M 1,0.
M 0,3 consists of a single point. M 0,4 and M 1,1 are isomorphic to the projective
line. The first cohomology of all these spaces vanish. This concludes the proof that
�
H 1(M g,n, Q) = 0 for all g and n.
10
Remark 3.3. H 1(M g,n, Q) = 0 also follows from the fact that Mg,n is simply
connected. However, note that Mg,n is not simply | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
the fact that Mg,n is simply
connected. However, note that Mg,n is not simply connected. This is one reason
why computing the cohomology of the compactified moduli space is simpler. For
example, we can identify M0,4 with P1 with three points removed. Fix the three
marked points at 0, 1, →. The fourth fixed point is free to vary on the sphere
except it cannot be one of the other three marked points. The fundamental group
of P1 − {0, 1, →} is the free group on two letters. In particular, the first cohomology
group of P1−{0, 1, →} has rank 2. In contrast we saw above that all odd cohomology
groups of M 0,n vanish.
To emphasize the point, observe that the Euler characteristic of M0,n is given
by the formula
α(M0,n) = (−1)(n−3)(n − 3)!.
To prove this formula consider the map M0,n � M0,n−1 given by forgetting one
of the marked points. This is a fibration with each fiber given by a sphere with
n − 1 points removed. We conclude that the Euler characteristic of M0,n is (3 −
n)α(M0,n−1). The result follows by induction. The Euler characteristic of M0,n
is negative for even n. At least for those n, the odd cohomology groups cannot
vanish.
Proof of the vanishing of third cohomology. To conclude that H 3 vanishes for all
M g,n we need to check | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
third cohomology. To conclude that H 3 vanishes for all
M g,n we need to check the cases
(1) g = 0 and 3 � n � 6,
(2) g = 1 and 1 � n � 3, and
(3) g = 2 and n = 0 or 1.
We already observed that the odd cohomology of M0,n vanishes. In this range,
this is easy to check directly.) M0,3 is a point so H 3 clearly vanishes. Both M0,4
and M1,1 are isomorphic to P1, hence their third cohomology clearly vanishes.
The moduli spaces M0,5 and M1,2 both have complex dimension 2 or real dimen
sion 4. By Poincar´e duality we conclude that the dimension of H 3 is equal to the
dimension of H 1 . Since H 1 vanishes we conclude that H 3 vanishes.
To show the vanishing of the third cohomology groups of M 2,0 and M 2,1, we
observe that they admit surjective morphisms from M 0,6 and M 0,7, respectively.
This suffices to show the vanishing of the third cohomology. Recall that genus 2
curves are all hyperelliptic. They are a double cover of P1 ramified at six points.
Given a Riemann sphere with six marked points take the hyperelliptic curve ramified
over these six points. Similarly given a Riemann sphere with seven marked points
take the hyperelliptic curve of genus 2 ramified at the first six with one of the points
above the seventh point as marked. (Note that since the | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
the first six with one of the points
above the seventh point as marked. (Note that since the hyperelliptic involution
takes one sheet of the covering to the other, the choice is immaterial.) We conclude
that the third cohomology groups of these two spaces vanish.
We are left to consider the case g = 1 and n = 3. One way to check the
vanishing of cohomology groups is to use Euler characteristic considerations. If Y
is a quasi-projective variety which has a filtration by closed subvarieties Y i
Y = Y d ⊗ Y d−1 ⊗ · · · ⊗ Y 1 ⊗ Y 0
11
so that Yi = Y i\Y i−1 is empty or of pure dimension i for every i, then by the exact
sequence of cohomology with compact supports the Euler characteristic of
Y with
cohomology with compact supports is the sum of those of Yd and Y d−1. Repeating
the process and using Poincar´e duality we conclude that the Euler characteristic of
M g,n is the sum of the Euler characteristics of open strata where we stratify M g,n
according to graph type.
M 1,3 has complex dimension 3 or real dimension 6. We already know that its first
and by Poincar´e duality its fifth cohomology groups vanish. The second cohomology
group is generated by
�, ξ1, ξ2, ξ3, ζirr, ζ0,{1,2}, ζ0,{1,3}, ζ0,{2,3}, ζ0,{1,2,3}.
There are 4 independent linear relations among these. Hence the rank of the second
(and by Poincar´e duality fourth) cohomology groups are 5. If we can show that the
Euler characteristic of M 1,3 is twelve, it follows that the third cohomology group
has to vanish.
0,4
Let us compute | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
twelve, it follows that the third cohomology group
has to vanish.
0,4
Let us compute that the Euler characteristic of M 1,3 is 12. This is done by
splitting M 1,3 to its strata according to topological type. In this computation we
∅
,5, where M ∅
∅
∅
and M0
,4, M0
need the Euler characteristics of M1,2, M1,3, M0
,5
denote the space obtained by taking the quotients of M0,4 and M0,5 under the
operation of interchanging the labeling of two marked points. To calculate the
Euler characteristics of the latter two we note that we have morphisms from M0,4
and M0,5 to these spaces. Both morphisms have degree 2 since the fiber over a point
has two points corresponding to the two different ways of ordering the identified
∅
marked points. The morphism from M0,4 to M0
,4 is ramified at one point. If there
is only one point over (0, →, 1 x), then there must be an automorphism of the
sphere permuting 1 and x and keeping 0 and → fixed. This can only happen if
x = −1 and the automorphism is multiplication by −1. By the Riemann-Hurwitz
∅
∅
formula we conclude that α(M0
,5 is unramified
,4) = 0. The map in the case of M0
∅
and therefore α(M0
,5) = 1.
The Euler characteristics of M1,n can be computed inductively. | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
∅
and therefore α(M0
,5) = 1.
The Euler characteristics of M1,n can be computed inductively. First, M1,1 is the
affine line, so its Euler characteristic is 1. It is a fundamental theorem in the theory
of elliptic curves that the group of automorphisms fixing a point is a group of order
2 except in two cases. In one case the elliptic curve can be realized as ramified
over the points 0, 1, −1, → of the sphere and it has the extra automorphism coming
from rotating the sphere by ψ along the 0 − → axis (multiplication by −1). In the
other case the elliptic curve can be realized as ramified over the cube roots of unity
and →. Its automorphism group has order 6 and it can be generated by the usual
involution and by multiplication by a cube root of unity (rotation of the underlying
sphere around the 0 − → axis by an angle of 2ψ/3).
Consider the morphism from M1,2 to M1,1 given by forgetting the second marked
point. The fiber over each point of M1,1 is an affine line. Hence, the Euler char
acteristic of M1,2 is 1. Next, consider the morphism from M1,3 to M1,2. Here we
need to break M1,2 up to pieces over which the fibers have nice descriptions. First,
consider the case where p2, the second marked point, is a 2-torsion point with re
∅
,4 and the fiber of the map over such
spect to p1. Observe that this space is M0
a point is the sphere with two points removed. Next, there is the case when C is
the special curve | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
sphere with two points removed. Next, there is the case when C is
the special curve whose automorphism group has order 6 and p2 lies above 0. In
12
this case the fiber is also a sphere with two points removed. Finally, there is the
case when p2 is not a 2-torsion point and not the special point considered in the
previous case. In this case the fiber is an elliptic curve with two points removed.
Adding up the various Euler characteristics we conclude that α(M1,3) = 0. This
information together with an enumeration of the strata of M 1,3 suffices to calculate
that the Euler characteristic is 12. Since the Euler characteristic is 12, the third
cohomology group must vanish. By the reduction lemma this completes the proof
�
that all the third cohomology groups of M g,n vanish.
The technique for showing that the fifth cohomology groups of M g,n vanish is
similar. The cases that need to be checked in this case are
(1) g = 0 and n � 8
(2) g = 1 and n � 5
(3) g = 2 and n � 3
(4) g = 3 and n � 1.
We already know the case g = 0. The case g = 1 and n � 4 are easy. The remaining
cases are more challenging.
Remark 3.4. Arbarello and Cornalba’s approach outlined here cannot be applied
directly to the odd cohomology groups for k ∼ 11 since these groups do not always
vanish. For example, H 11(M 1,11, Q) does not vanish. Their | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
not always
vanish. For example, H 11(M 1,11, Q) does not vanish. Their inductive argument
breaks down.
Problem 3.5. Determine H 7(Mg,n, Q) and H 9(Mg,n, Q).
4. The Picard group of the moduli functor
In this section we will determine the Picard group of the moduli functor following
[AC1]. A very good introduction to Picard groups of moduli functors is contained
in [Mum].
Let Mg,n denote the moduli functor of genus g stable curves with n marked
points. Let (C � S, π1, . . . , πn) denote a family of stable curves of genus g and
n marked points parameterized by S. A line bundle on the moduli functor Mg,n
is an assignment of a line bundle LC to the base of the family S for every family
C � S and isomorphisms between LD = κ�(LC) for every fiber diagram
�
D
C
T
�
S
satisfying the cocycle condition.
Similarly let Mg,n denote the moduli functor of genus g smooth curves with n
marked points. The Picard group of the functor Mg,n is defined the same way.
The Hodge class ω and the classes of the boundary divisors ζirr, ζ1, . . . , ζ∪g/2∈
are elemements of P ic(Mg). Recall that the Hodge class ω is defined as the class
of the determinant of the Hodge bundle which is the push-forward of the relative
dualizing sheaf on any family. The class ζirr is the class of | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
relative
dualizing sheaf on any family. The class ζirr is the class of the divisor of curves
13
�
�
�
�
with a non-separating node. The class ζi is the class of the divisor of curves that
contain a node that separates the curve to a subcurve of genus i and genus g − i.
Similarly ω, ξ1, . . . , ξn, ζirr, ζh,S are elements of P ic(Mg,n). Recall that ω is the
Hodge class. The class ξi is the class of the cotangent line at the i-th marked
point and is formally defined by the pull-back of the relative dualizing sheaf by the
section giving the i-th marked point. The classes ζh,S are the classes of boundary
divisors of curves containing a node that separates the curve to a subcurve of genus
1 � h � ∪g/2⊂ with the marked points pi for i ⊕ S ⊗ {1, . . . , n} and a residual
curve of genus g − h with the remaining marked points. Of course, for the curve to
be stable #S ∼ 2 if h = 0.
Theorem 4.1. Let g ∼ 3. The Picard group P ic(Mg) is freely generated by the
classes ω, ζirr, ζ1, . . . , ζ∪g/2∈. The Picard group P ic(Mg) is freely generated by ω.
In the rest of the course we will only use Theorem 4.1. However, similar tech
niques also prove the following more general theorem.
Theorem 4.2. Let g ∼ 3. The Picard group P ic(Mg,n) | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
Theorem 4.2. Let g ∼ 3. The Picard group P ic(Mg,n) is freely generated by
the classes ω, ξ1, . . . , ξn and the classes of boundary divisors. The Picard group
P ic(Mg,n) is freely generated by ω and ξ1, . . . , ξn.
Sketch of the proof of Theorem 4.1. We first remark that P ic(Mg) is torsion free
and contains P ic(Mg) as a finite index subgroup. To see that P ic(Mg) is torsion
free one uses Teichm¨uller theory. Suppose P ic(Mg) had a torsion element L of
prime order p. Since the p-th power of L is trivial, we can take the p-th root of
a nowhere vanishing section to get an unramified Z/pZ covering of any family. In
particular, we get an unramified covering of Teichm¨uller space which must split
completely. It follows that L has a section over the automorphism free locus. This
extends to a holomorphic, nowehere vanishing section of L since the p-th power
does. Hence L is trivial. Any class in P ic(Mg) whose restriction to Mg is trivial is
an integral linear combination of the boundary classes. The boundary classes are
independent, hence P ic(Mg) is torsion free.
By the calculation of the second homology group of P ic(Mg), we can express
any divisor class as a linear combination of
ω, ζirr, ζ1, . . . , ζ∪g/2 | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
any divisor class as a linear combination of
ω, ζirr, ζ1, . . . , ζ∪g/2∈ .
The point is to show that it may be expressed as an integral linear combination.
The strategy is to construct two different sets of one-parameter families of curves
F1, . . . , F∪g/2∈+2 and G1, . . . , G∪g/2∈+2 such that their intersection matrices with
respect to
ω, ζirr, ζ1, . . . , ζ∪g/2∈
are non-singular and have relatively prime determinant. Since the determinant of
these matrices times the coefficients of the expressions of any divisor class in terms
of
ω, ζirr, ζ1, . . . , ζ∪g/2∈
have to be integral, the theorem follows.
The required families are obtained as follows:
14
Let Kh be the family consisting of a pencil of hyperplane sections of a K3 surface
of degree 2h − 2 to which a fixed curve of genus g − h is attached at a base point
of the pencil. It is easy to see that
Kh · ζirr = 18 + 6h, Kh · ζh = −1, Kh · ζi = 0 if i ∈= h.
The degree of ω on Kh is h + 1.
Let Fh be the family consisting of three curves C1, C2, E of genus h, g − h − 1
and 1, respectively. Attach C2 to E at a fixed point, then attach C1 to E at a
fixed point of E1, but a variable point of E. The degree of ω on this familiy is zero.
All the intersections with the boundary divisors | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
of E. The degree of ω on this familiy is zero.
All the intersections with the boundary divisors vanish unless i = 1, h or h + 1.
The degree of ζ1 on Fh is 1 if h > 1, 0 if g − h − 1 > h = 1 and −1 if g = 3
and g − h − 1 = h = 1. The degree of ζh on Fh is −1 if g − h − 1 > h = 1 or if
g − h − 1 = h = 1, 0 if g − h − 1 > h = 1 and −2 if g − h − 1 = h > 1.
Let C be the family obtained by attaching a fixed genus g − 3 curve at fixed 4
points to the base points of a pencil of conics. The degree of ω and ζi on this family
is zero. The degree of ζirr is −1.
Finally let CE be the family obtained by attaching a genus g − 3 curve at three
of the base points of a pencil of conics and a genus one curve at the fourth base
point. All the degrees except for the degree of ζ1 vanish on this family. The latter
degree is −1.
The theorem follows from these computations. If the genus is 2m + 1, the inter
section matrix for the families
Kh, C, F1, . . . , Fm
has determinant (−1)m+1(h + 1) if m ∼ h ∼ 2. Taking h = 2 and h = 3 gives two
relatively prime determinants. If the genus is 2m + 2, the intersection matrix for
the families
Kh, C, CE, F1, . . . , Fm | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
m + 2, the intersection matrix for
the families
Kh, C, CE, F1, . . . , Fm
has determinant (−1)m+1(h + 1) if m ∼ h ∼ 2. Again taking h = 2 and h = 3 gives
�
two relatively prime determinants.
5. The Tautological ring of Mg
In this course we will not have time to discuss the tautological ring. In this section
I will give a few references to where you may learn more about it. Many people have
worked on it, including Faber, Looijenga, Pandharipande, Graber, Vakil, Getzler,
Ionel to mame very few (see, for example, [Fab], [Lo], [FaP1], [FaP2], [GP], [GV1],
[V], [GV2]). .
Usually when a moduli space is defined with respect to a universal property, it
contains certain tautologically defined Chow classes. The prime example of such
Chow classes are the chern classes of the universal tautological and quotient bundles
on Grassmannians. The Chow ring of the Grassmannian is generated by these
tautological classes.
For the moduli space of curves Mg, it is also possible to define tautological
classes. Consider the universal curve
ψ1 : Mg,1 � Mg.
15
The first chern class of the relative dualizing sheaf leads to a sequence of classes on
. These are classes
Mg . More precisely, let K = c1(�Mg,1 /Mg ). Define �l
in Al (Mg ). Also on Mg there is a rank g locally free sheaf called the Hodge bundle
E. The Hodge bundle is defined by E = ψ1��Mg,1 /Mg . The chern classes ωl = cl (E)
also | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
by E = ψ1��Mg,1 /Mg . The chern classes ωl = cl (E)
also define classes in Al (Mg ). Ths subring of the Chow ring generated by these
classes is called the tautological ring.
= ψ1�K l+1
One of the first things to observe is that the cohomology of Mg is not in general
tautological. There are many ways to see this. The simplest is to observe that
tautological classes are even cohomology classes. Since we have computed the Euler
characteristic of the moduli spaces, we can see that the moduli space of curves has
odd cohomology classes. There are also explicit constructions of non-tautological
classes.
Faber has very detailed conjectures about the structure of the tautological ring.
Roughly these conjectures say that the tautological ring of Mg exhibits properties
that one would expect the algebraic cohomology ring of a smooth projective variety
of dimension g − 2 to exhibit. For instance that it is Gorenstein with socle in degree
g − 2, satisfies Hard Lefschetz and Hodge Positivity with respect to the class �1.
Furthermore, Faber conjectures that
generate the ring and gives some explicit relations among these generators. I refer
you to the papers cited above for detailed statements and what is known.
�1, . . . �∪g/3∈
References
[AC1] E. Arbarello and M. Cornalba. The Picard groups of the moduli spaces of curves. Topology
26(1987), 153–171.
[AC2] E. Arbarello and M. Cornalba. Calculating cohomology groups of moduli spaces of curves
via algebraic geometry. Inst. Hautes Etudes Sci. Publ. Math. (1998), 97–127 (1999).
´
[Fab] C. Faber. A | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
l. Math. (1998), 97–127 (1999).
´
[Fab] C. Faber. A conjectural description of the tautological ring of the moduli space of curves.
In Moduli of curves and abelian varieties, Aspects Math., E33, pages 109–129. Vieweg,
Braunschweig, 1999.
[FaP1] C. Faber and R. Pandharipande. Logarithmic series and Hodge integrals in the tautological
ring. Michigan Math. J. 48(2000), 215–252. With an appendix by Don Zagier, Dedicated
to William Fulton on the occasion of his 60th birthday.
[FaP2] C. Faber and R. Pandharipande. Hodge integrals, partition matrices, and the �g conjec
ture. Ann. of Math. (2) 157(2003), 97–124.
[GP] T. Graber and R. Pandharipande. Constructions of nontautological classes on moduli
spaces of curves. Michigan Math. J. 51(2003), 93–109.
[GV1] T. Graber and R. Vakil. On the tautological ring of M g,n. Turkish J. Math. 25(2001),
237–243.
[GV2] T. Graber and R. Vakil. Relative virtual localization and vanishing of tautological classes
on moduli spaces of curves. Duke Math. J. 130(2005), 1–37.
[Har1] J. Harer. The second homology group of the mapping class group of an orientable surface.
Invent. Math. 72(1983), 221–239.
[Har2] J. L | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
surface.
Invent. Math. 72(1983), 221–239.
[Har2] J. L. Harer. The cohomology of the moduli space of curves. In Theory of moduli (Mon
tecatini Terme, 1985), volume 1337 of Lecture Notes in Math., pages 138–221. Springer,
Berlin, 1988.
[Kee] S. Keel. Intersection theory of moduli space of stable n-pointed curves of genus zero. Trans.
[Lo]
[Mi]
Amer. Math. Soc. 330(1992), 545–574.
Eduard Looijenga. On the tautological ring of Mg . Invent. Math. 121(1995), 411–419.
J. Milnor. Morse theory. Based on lecture notes by M. Spivak and R. Wells. Annals of
Mathematics Studies, No. 51. Princeton University Press, Princeton, N.J., 1963.
16
[Mum] D. Mumford. Picard groups of moduli problems. In Arithmetical Algebraic Geometry
(Proc. Conf. Purdue Univ., 1963), pages 33–81. Harper & Row, New York, 1965.
R. Vakil. The moduli space of curves and its tautological ring. Notices Amer. Math. Soc.
50(2003), 647–658.
[V]
17 | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/10078cf3d80a1cdf4ff766a80eee8b32_picard.pdf |
18.409 An Algorithmist’s Toolkit
10/8/2009
Lecturer: Jonathan Kelner
Lecture 9
At the end of the previous lecture, we began to motivate a technique called Sparsification. In this lecture,
we describe sparsifiers and their use, and give an overview of Combinatorial and Spectral Sparsifiers. We
also define Spectral Sparsifiers, and create tools and language with which to construct and analyze them.
1 Sparsification
Suppose we are given a graph G = (V, E). We would like to solve some cut problem (i.e. min-cut, s-t min
cut, sparsest cut) and so on. The running time of algorithms for these problems typically depends on the
number of edges in the graph, which might be as high as O(n2). Is there any way to approximate our graph
with a sparse graph G(cid:2) in which all cuts are approximately of the same size?
We will describe two ways of “sparsifying” our graph. The first is the method of Benczur-Karger, and
relies on random sampling of edges. The second technique is Spectral Sparsification, and uses spectral
techniques to improve upon Benczur-Karger’s algorithm.
1.1 First Try
Our first attempt at sparsifying will use random sampling. Let’s start by sampling each edge with probability
p. Then, if a cut has c edges crossing it in G, the expected value of edges crossing it in the new graph G(cid:2) is
pc. Our algorithm will solve the cut problem in G(cid:2). Say the answer is a cut with value S(cid:2); then our algorithm
will output the estimate S = S(cid:2)/p for the original graph G.
Denoting the number of edges between S and | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
our algorithm
will output the estimate S = S(cid:2)/p for the original graph G.
Denoting the number of edges between S and S ¯ by e(S) = pc, we have the following concentration result
due to Chernoff’s inequality:
So our result will be close to the correct answer provided pc is large. In particular, picking
P (|eG� (S) − pc| ≥ (cid:2)pc) ≥ e −(cid:2)2 pc/2 .
(1)
p = Ω(
d log n
(cid:2)2c
),
will make the right side of Eq. (1) at most n−d . Summarizing, we can choose p to get an (cid:2) multiplicative
approximation with probability at least 1 − n−d .
Is it possible to choose p to get this multiplicative approximation for all cuts, rather than just one as
above? The answer is yes; the main ingredient is a result of Karger that the number of small cuts in a graph
is not too large:
Theorem 1 (Karger) If G has a min-cut of size c, then the number of cuts of value αc or less is at most
2αn .
1.2 Second try
The problem with this proposal is that it breaks for small cuts. Say c is small, but an edge e is only involved
in cuts of size ≥ k. What we want to do is to sample these edges with a small probability of failure.
The idea that we use is to sample edges, but with a “weight” of 1/p. This method is called importance
sampling. To do this, we need a slightly modified version of the Chernoff bound:
9-1
Theorem 2 (Chernoff Bound) Let X1, . . . , Xn be random variables so that Xi ∈ [0, 1], and let X = Xi. | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
1, . . . , Xn be random variables so that Xi ∈ [0, 1], and let X = Xi.
Then,
(cid:2)
P r[|X − E[X]| ≥ (cid:2)X] ≤ 2e −Θ(1)(cid:2)2E[X]
Proof The only difference here is that the random variables Xi are no longer discrete variables, but lie in
the interval [0, 1]. The proof is carried out the same as with the regular Chernoff bound.
What this allows us to do is to scale our random variables without changing the error bounds. Returning
to our case, we assign to every edge e a random variable Ye and a weight we. If e is in a cut of size c, we
require that we ≤ c. We will set Ye = 1 with probability p/we; and Ye = 0 with probability 1 − p/we. Instead
of counting how many edges cross a cut (S, S¯), we will compute a weighted sum:
(cid:3)
YS =
weYe
The expectation is still correct; if there are c edges across the cut (S, S¯) in G, then
e∈∂(S,S¯)
(cid:3)
E[YS ] =
e∈∂(S,S¯)
we
p
we
= pc.
This scheme gives us an advantage: if an edge is present in only cuts of large size, we can keep it with low
probability, which corresponds to setting we to be large. On the other hand, if an edge is present in cuts of
small size, we will keep it with high probability, which corresponds to setting we to be small. In this way, we
can approximate cut problems while throwing away more edges which are present in only cuts of high size.
Thus, a natural choice for we would be the size of the smallest cut containing e. Unfortunately, we
do not know we; however, | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
size of the smallest cut containing e. Unfortunately, we
do not know we; however, it is possible to approximate it quickly. The final result is an (cid:2) multiplicative
approximation based on this scheme. We refer the reader to [1] for details.
2
Spectral Sparsifiers
The construction shown above is known as a Combinatorial Sparsifier. In the upcoming section and following
lecture, we will see how to improve upon it with the spectral methods that we have been learning.
Let G = (V, E) be our original graph. Recall that the laplacian has the property that
(cid:3)
x T LGx =
(xi − xj )2 ,
(i,j)∈E
for some x ∈ Rn, and the sum is being taken over all edges in G. If x takes value 1 on the set S and −1 on
the S¯, this equation becomes
x T LGx = 4e(S).
Let G(cid:2) be a combinatorial sparsifier of the graph G. The condition that all cuts in G are approximated
with a multiplicative error of at most (cid:2) by cuts in G(cid:2) can be restated as
(1 − (cid:2))x T LG� x ≤ x T LGx ≤ (1 + (cid:2))x T LG� x,
(2)
for all x that take on only the values 1 and −1. This is true for all such discrete values of x.
On the other hand, consider if Eq. (2) is true for all x ∈ Rn. Note that in this case we can limit ourselves
to the instances x ∈ [−1, 1]n by normalization. We now have a good definition for a spectral version of
sparsification:
Definition 3 A Spectral Sparsifier G(cid:2) of a graph G is one for which the | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
ral Sparsifier G(cid:2) of a graph G is one for which the relation
(1 − (cid:2))x T LG� ≤ x T LGx ≤ (1 + (cid:2))x T LG� x
for all x ∈ [0, 1]n
9-2
It is clear from this definition that spectral sparsifiers are combinatorial sparsifiers. A natural question
is then to ask if all combinatorial sparsifiers also spectral sparsifiers.
The answer is no, and we provide a proof by counterexample. Consider the graph G(cid:2) with vertex set
{1, 2, . . . , n} and an edge between i, j when i − j mod n ≤ k. G is G(cid:2) with the edge (1, n/2) added. The
graph looks something like the figure below.
Then, for an appropriate (cid:2), G(cid:2) is a combinatorial sparsifier of G. Indeed, the min cut in G cuts Θ(k)
edges; the min cut in G(cid:2) cuts one less. With (cid:2) = Θ(1/k), we have that G(cid:2) is a combinatorial sparsifier of G.
On the other hand, G(cid:2) is not a spectral sparsifier of G. Let
(cid:4)
x =
0
1
. . . n/2 − 1 n/2 − 1
. . . 1
0
(cid:5)
.
Then, we have that
since each vertex contributes Θ(
x T LG� x = Θ(nk3)
(cid:2)
k
=1 k2) to the sum. On the other hand,
i | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
= Θ(nk3)
(cid:2)
k
=1 k2) to the sum. On the other hand,
i
x T LGx = Θ(nk3) + (
n − 1)2
2
If k is constant, we get that we need (cid:2) = Θ(1/n) for G(cid:2) to be a spectral sparsifier of G.
2.1 Order Relations on Laplacians
In order to define spectral approximations, we first need to define the appropriate vocabulary. Earlier, we
made error approximations based on cut size. In the spectral case, we will be using the laplacian of the
graph instead - so a nice way to compare laplacians would be idea. That is to say, we want a good relation
(cid:5) on symmetric matrices that is an ordering on them, and also is somewhat consistent with the notions of
cuts.
How will we define this ordering? An immediate idea is the following:
M (cid:5) N ⇔ mi,j ≥ ni,j ∀i, j
Upon second thought, we realize that this is no good for our purposes. For one, spectral graph theory is
all about eigenvalues, and this relation tells us nothing about the eigenvalues of the matrix! Furthemore, the
values of individual entries are highly dependent on choice of basis, which would be bad. If such a definition
were used, a process like diagonalizing the Laplacians could possibly affect the graph orders.
We try again with another definition:
M (cid:5) N if the ith eigenvalue of M is ≥ the ith eigenvalue of N for all indices i
This is better in that it is basis independent - but it is too basis independent. Under this definition, we
have both
9-3
as well as
(cid:6)
1
0
0 −1
(cid:6)
(cid:7)
1
(cid: | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
as well as
(cid:6)
1
0
0 −1
(cid:6)
(cid:7)
1
(cid:5) √
2
1
1
1 −1
(cid:6)
1
√
2
1
1
1 −1
(cid:7) (cid:6)
(cid:5)
1
0
0 −1
(cid:7)
(cid:7)
After this experimentation, we claim that the following is the “right” definition of order.
Definition 4 We write that M (cid:5) N if
Note that this definition of order has the following properties:
x T M x ≥ X T N x ∀x ∈ Rn
1. If M (cid:5) N and N (cid:5) M , then M = n
2. M (cid:5) 0 if M is a positive semidefinite matrix.
3. M (cid:5) N if M − N is positive semidefinite
4. If M1 (cid:5) N1 and M2 (cid:5) N2, then
M1 + M2 (cid:5) N1 + N2
These properties suffice for our purposes, and with this, we can define an associated order on graphs as
well.
Definition 5 Given graphs G and H, say that G (cid:5) H if LG (cid:5) LH .
Claim 6 Let G = (V, EG, wG) and H = (V, EH , wH ) be weighted graphs on the same vertex set such that
wG(i, j) ≥ wH (i, j) for all edges (i, j) ∈ E. Then, G (cid:5) H
2.2 Towards Spectral Sparsification
With this order relation on graphs, we can now | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
2 Towards Spectral Sparsification
With this order relation on graphs, we can now restate the goal of spectral sparsification: Given a dense
graph G, we want to create a sparse graph H where
Lh (cid:9) LG (cid:9) (1 + (cid:2))LH
By “sparse,” we mean that H has polylog(n) edges, where n is the number of nodes. We will show in this
and the next lecture how to construct spectral sparsifiers with O(nlogn) edges in Polynomial time. This
can actually be improved to a linear time construction, but will use geometric techniques that we will learn.
Moreover, it is possible to construct O(n) edge sparsifiers in polynomial time. The benefits of this are that
the problem is more geometrically flavored. It is also a nice example of how generalizing can make things
easier sometimes.
The algorithm that we propose is very simple. It is similar in structure to the B-K algorithm, but we use
different probabilities for sampling the edges.
• Compute probability pe for each edge e.
• Sample each edge uniformly with probability pe, and if an edge is selected, include it with weight 1/pe.
These probabilities are based on a linear algebra sense of importance, and have a nice interpretation in terms
of effective resistance of circuits. To proceed with our analysis, however, we need to develop the ideas of
pseudoinverses, calculating effective resistances, and a matrix version of the Chernoff Bound.
9-4
2.3 Pseudoinverses
In our analysis, we will come across the need to “invert” a singular matrix. Since this is obviously not
possible | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
come across the need to “invert” a singular matrix. Since this is obviously not
possible, we redefine our question to one that makes more sense. Let M be a n × n symmetric matrix. We
can diagonalize M :
M =
T
λivivi
(cid:3)
n
i=1
If all the eigenvalues are nonzero, then it obviously invertible, and M −1 =
The case we worry about is when there is a zero eigenvalue. But this is okay too: when M is degenerate,
we define the pseudoinverse by throwing away the zero eigenvalues and eigenvectors. In that case, we have
(cid:3) 1
λi
M + =
T
vivi
t
vivi
i|λi(cid:4)=0
(cid:2)
n
1
=1 λ
i
i
The pseudoinverse has many nice properties. Of these, we use:
• ker(L) = ker(L+
• M M + =
(cid:2)
i|λi(cid:4)
T = the projection onto the nonzero eigenvectors.
=0 viv
i
It is easy to see that M M + = I when restricted to the image of M .
2.4 Effective Resistance
We mentioned earlier that Spectral Sparsification also samples edges with different probability. It turns
out that the correct way to do this is to sample each edge with probability proportional to its “effective
resistance.”
The basic idea is to treat each edge as a resistor with resistance 1. If the edge had a capacity of c, we give
it a resistance of 1/c. After calculating these values, we sample the edge (u, v) with probability proportional
to the effective resistance between nodes u and v.
Students may recall learning methods to solve circuits from their previous classes. For example, students
may use a combination of Ohm’s law and Kircho | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
Students may recall learning methods to solve circuits from their previous classes. For example, students
may use a combination of Ohm’s law and Kirchoff’s law, as well as the rules for calculating effective resistances
of resistors in series and parallel. To those who are comfortable with solving circuits, this may be a good
way to think about the problem. However, the students who don’t like solving circuits are in luck too: now
that we have the tools of Spectral Graph Theory, we can solve circuits with only linear algebra! In fact, we
will combine our frequent use of the graph Laplacian with the pseudoinverse defined above.
Let U be the edge-vertex adjacency matrix, C be the diagonal matrix with the various capacitances, and
re = 1/ce.
That is, we define U as in:
U (e, v) = −1
⎧
⎨ 1
⎩
0
if v is the head of e
if v is the tail of e
otherwise
Then, we have that L = U T CU . From ohm’s law, we have i = CU v for i ∈ RE , and v ∈ Rv . From the
conservation of current, we have iext = U T i, for iext ∈ RV . Finally, we have iext = Lv, and v = L+iext
We define U (e, v) to be the adjacency matrix with ±1 values. Let ue be the eth row, and v = L+iext.
We have
and as a result,
T
Ref f (e) = ueL+ ue
Ref f (e) = (U L+U T )e,e
Thus, calculating the effective resistance of an edge is as simple as calculating the pseudoinverse of the
Laplacian. Simple!
9-5
2.5 Error | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
calculating the pseudoinverse of the
Laplacian. Simple!
9-5
2.5 Error Bounds
The last tool that we need to build is a way to define error bounds for matrices. In particular, we will use
the following theorem.
Theorem 7 For distributions on vectors y where (cid:10) y (cid:10)≤ t and (cid:10) Eyyt (cid:10)2≤ 1 (where we are using the l2
norm) then:
E (cid:10) EyyT −
1
q
q
(cid:3)
i=1
yiyi
T (cid:10)2≤ kt
(cid:11)
log q
q
This is a “concentration of measure theorem, and we claim that it is similar to the Chernoff bound.
Now, onto approximation. For our sparisifier H to approximate the original dense graph G, we want that
for all vectors x. Rather, it is sufficient to show that
1 − (cid:2) ≤
xT LH x
xT LGx
≤ 1 + (cid:2)
zT M T LH M z
zT M T LGM z
for all vectors z, provided that x ⊥ (LG) ⇒ x ∈ range(M ). Choose M so that M T LGM is a projection.
Then, it suffices to show that
≤ 1 + (cid:2)
1 − (cid:2) ≤
(cid:10) M T LH M − M T LGM (cid:10)2≤ (cid:2)
From before, we have that LG = U T CU . Choose M = L+U T C 1/2 . Then, we have
G
Π = M T LGM = C 1/2U L+U T C 1/2 | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
Then, we have
G
Π = M T LGM = C 1/2U L+U T C 1/2 = ΠΠ
G
Now, recall that LG = U T CU . If we let de be the weight of e in the sparsifier H, set Se,e = d
e . Then,
ce
we can write
yielding
LH = U T CSU = U T C 1/2SC 1/2U
M T LH M = ΠSΠ
We need to choose a diagonal S such that the number of nonzero elements of S is O(nlogn/(cid:2)2) With this
choice, we have
(cid:10) ΠSΠ − Π (cid:10)2≤ (cid:2)
Define πe as the eth column of Π: that is, πe = Π(·, e). Then, ΠSΠ =
(cid:10) πe (cid:10)2= Πe,e = ceRef f (e)
(cid:2)
Se,eπeπe
T , so
(this is because Π = Π2 = C 1/2(U L+U T )C 1/2)
(cid:12)
G
√
We then set τe =
Recall that
n−1
ceRef f (e) πe with (cid:10) τe (cid:10)= n − 1. Choose edges with probability pe =
c2Ref f (e)
n−1
.
(cid:3)
ceRef f (e) =
(cid:3)
Πe,e = n − 1
e
e
Then, we find that
E[τeτe
T ] =
(cid:3)
peτeτe
T =
(cid:3)
πeπe
T = Π
Sample q times with replacement, and set S(e, e) = | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
πeπe
T = Π
Sample q times with replacement, and set S(e, e) =
Then, from the theorem above, we have
e
e
1
qceRef f (e)
× the number of times that e is chosen.
9-6
E[(cid:10) Π − ΠSΠ (cid:10)2] ≤ k n − 1
(cid:11)
√
logq
q
≤ (cid:2)/2
for q = O(n log n/(cid:2)2). Thus, we see that our construction yields a spectral sparsifier as desired.
From the algorithmics of the construction, it is easy to see that this is a poly-time procedure. The whole
procedure is constructive, and uses the standard linear algebra operations. The bottleneck in this procedure
comes from computing effective resistances, and in particular, the matrix inversions and multiplications. We
claim that the procedure can be improved to nearly linear time. Doing so would involve two components:
• Close to linear algorithms for solving linear equations of the form Lx = b for a laplacian L.
• A way to compute all the effective resistances by solving logarithmically many linear systems. This
uses the Johnson-Lindenstrauss Lemma.
References
[1] “Randomized Approximation Schemes for Cuts and Flows in Capacitated Graphs, ” A. Benczur, D. Karger,
manuscript.
9-7
MIT OpenCourseWare
http://ocw.mit.edu
18.409 Topics in Theoretical Computer Science: An Algorithmist's Toolkit
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/105bb5d8da4f17bc9ed3099824e0e59e_MIT18_409F09_scribe9.pdf |
6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture 2-1
Lecture 2 - Carrier Statistics in Equilibrium
February 8, 2007
Contents:
1. Conduction and valence bands, bandgap, holes
2. Intrinsic semiconductor
3. Extrinsic semiconductor
4. Conduction and valence band density of states
Reading assignment:
del Alamo, Ch. 2, §§2.1-2.4 (2.4.1)
Announcement:
Go to http://ilab.mit.edu and register. Select member
ship in the 6.720 group. You will need this to access the
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Lecture 2-2
Key questions
• What are these ”energy band diagrams”?
•
What are these ”holes”?
• In a perfectly pure semiconductor, how many electrons and holes
are there?
• Can one engineer the electron and hole concentrations in a semi
conductor?
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Lecture 2-3
1. Conduction and valence bands, bandgap, holes
y
g
r
e
n
e
Ec
Ev
conduction band
bandgap
↓
valence band
Ec
Ev
Eg
↓
space coordinate
Conduction and valence bands:
• bonding electrons occupy states in valence band | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/10707b9dced1bed9ea148baae9689b0b_lecture2.pdf |
space coordinate
Conduction and valence bands:
• bonding electrons occupy states in valence band
• ”free” electrons occupy states in conduction band
• holes: empty states in valence band
•
CB electrons and VB holes can move around: ”carriers”
+
--
+
--
a)a)
b)
b)
c)
c)
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Lecture 2-4
∧
y
y
g
g
r
r
e
e
n
n
e
e
n
n
o
o
r
r
t
t
c
c
e
e
e
e
l
l
y
y
g
g
r
r
e
e
n
n
e
e
l
l
e
e
o
o
h
h
↓↓
--
EEcc
EEvv
++
--
--
--
++
++
++
Elements of energy band diagrams:
• at edges of bands, kinetic energy of carriers is zero
• electron energies increase upwards
• hole energies increase downwards
--
EEcc
++
EEvv
υ = Eg
hhυ = Eg
y
y
g
g
r
r
e
e
n
n
e
e
n
n
o
o
r
r
t
t
c
c
e
e
e
e
l
l
--
++
a)
υ > Eg
hhυ > Eg
--
++
hhυ >υ > EgEg
b)
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6.720J Integrated Microelectronic Devices, Spring 2007.
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Lecture 2-5
2. Intrinsic semiconductor
Define intrinsic semiconductor, or ”ideal” semiconductor:
• perfectly crystalline (no perturbations to periodic lattice)
• perfectly pure (no foreign atoms)
•
no surface effects
Question: How many electrons and holes are there in an intrinsic
semiconductor in thermal equilibrium at a given temperature?
Answer requires fairly elaborate model [lecture 3], but key dependen
cies can be readily identified.
Define:
no ≡ equilibrium (free) electron concentration in conduction band [cm−3]
po ≡ equilibrium hole concentration in valence band [cm−3]
Certainly in intrinsic semiconductor:
no = po = ni
ni ≡ intrinsic carrier concentration [cm−3]
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Lecture 2-6
Key dependencies of ni:
• Temperature:
• Bandgap:
T
↑⇒
ni
Eg ↑⇒
ni
What is detailed form of dependencies?
Use analogy of chemical reactions.
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Integrated Microelectronic Devices, Spring 2007.
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6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture 2-7
Electron-hole formation can be thought of as chemical reaction:
bond �� e− + h+
similar to water decomposition reaction:
H2O �� H + + OH −
Law-of-mass action relates concentration of reactants and reaction
products. For water:
K =
[H +][OH −]
[H2O]
∼ exp(−
E
kT
)
E is energy consumed or released in reaction.
This is a ”thermally activated” process
by need to overcome energy barrier E (activation energy).
rate of reaction limited
⇒
In analogy, for electron-hole formation:
K =
nopo
[bonds]
∼ exp(−
Eg
kT
)
[bonds] is concentration of unbroken bonds.
Note: activation energy is Eg.
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Lecture 2-8
In general, relatively few bonds are broken. Hence:
and
Then:
Two important results:
• First,
[bonds] � no, po
[bonds] � constant
nopo ∼ exp(−
Eg
kT
)
ni ∼ exp(−
Eg
2kT
)
As expected:
T
↑⇒
ni ↑
Eg ↑⇒ ni ↓
To get prefactor, need detailed model | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/10707b9dced1bed9ea148baae9689b0b_lecture2.pdf |
��
ni ↑
Eg ↑⇒ ni ↓
To get prefactor, need detailed model [lecture 3].
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6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture 2-9
Arrhenius plot for Si [experiments of Misiakos and Tsamakis, 1993]:
300 K
)
3
-
m
c
(
i
n
1E+15
1E+10
1E+05
1E+00
1E-05
1E-10
1E-15
1E-20
30
50
0.612 eV
70
90
1/kT (eV-1)
110
130
150
In Si at 300 K, ni � 1.1 × 1010 cm−3 .
• Second important result:
2
nopo = ni
Equilibrium np product in a semiconductor at a certain temper
ature is a constant specific to the semiconductor.
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Lecture 2-10
3. Extrinsic semiconductor
Can electron and hole concentrations be engineered?
Insert dopants in substitutional positions in the lattice:
•
Donors: introduce electrons to conduction band without intro
ducing holes to valence band
• Acceptors: introduce holes to valence band without | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/10707b9dced1bed9ea148baae9689b0b_lecture2.pdf |
ducing holes to valence band
• Acceptors: introduce holes to valence band without introduc
ing electrons to conduction band
If any one carrier type overwhelms ni
⇒
extrinsic semiconductor
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Lecture 2-11
Donor in Si, atom from column V (As, P):
--
+
--
a) neutral donor
b) ionized donor
Acceptor in Si, atom from column III (B):
++
--
a) neutral acceptor
b) ionized acceptor
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Lecture 2-12
Representation of donor and acceptor states in energy band diagram:
Ed
Ec
Ev
Ea
ED
EA
Ed, Ea ∼ 40 − 60 meV , for common dopants
� Near room temperature, all dopants are ionized:
ND
+ � ND
− � NA
NA
Typical doping levels:
NA, ND ∼ 1015 − 1020 cm−3
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Lecture 2-13
� n-type semiconductor
no � ND
po �
2ni
ND
These equations are valid at intermediate temperatures.
log no
intrinsic regime
full ionization
ND
-Eg/2
extrinsic regime
-Ed/2
carrier�
freeze-out
1/kT
� p-type semiconductor
po � NA
no �
2ni
NA
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Lecture 2-14
4. Conduction and valence band density of states
Image removed due to copyright restrictions.
Figure 1b) on p. 468 in Laux, S. E., M. V. Fischetti, and D. J. Frank.
"Monte Carlo Analysis of Semiconductor Devices: The DAMOCLES Program."
IBM Journal of Research and Development 34, no. 4 (Jul. 1990): 466-494.
Can also be found in Fischetti, M. V., and S. E. Laux. “Monte Carlo Simulation of
Submicron Si MOSFETs.” In Simulation of Semiconductor Devices and Processes.
Vol. 3.: Proceedings of the Third International Conference on Simulation of Semiconductor
Devices and Processes, held at the University of Bologna, Bologna, Italy, on September
26th-28th, 1988. Edited by G. Baccarani and M. Rudan. Bologna, Italy: Technoprint, 1988,
pp. 349-368.(cid:13)
Close to edges:
√
gc(E) ∝
E − Ec
E ≥ Ec
√
gv(E) ∝
Ev − E
E ≤ Ev
g(E)
∧
gv(E)
gc(E)
0
Ev
Ec
> | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/10707b9dced1bed9ea148baae9689b0b_lecture2.pdf |
Ev − E
E ≤ Ev
g(E)
∧
gv(E)
gc(E)
0
Ev
Ec
>
E
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(cid:10)
6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture 2-15
g(E) ∧
gv(E)
gc(E)
0
Ev
Ec
>
E
Common expressions for DOS:
gc(E) = 4π
⎛
⎝
2m∗
de
h2
⎞
⎠
3/2
√
E − Ec
E ≥ Ec
gv(E) = 4π
⎛
⎝
2m∗
dh
h2
⎞
⎠
3/2
√
Ev − E
E ≤ Ev
m∗
m∗
de ≡ density of states electron effective mass
dh ≡ density of states hole effective mass
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Lecture 2-16
Key conclusions
• Concept of (free) electron: electron in conduction band.
• Concept of hole: empty state in valence band.
• Intrinsic semiconductor: ideally pure semiconductor.
no = po = ni ∼ exp(−
Eg
2kT
)
• To first order, for a given semiconductor nopo is a constant that
only | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/10707b9dced1bed9ea148baae9689b0b_lecture2.pdf |
)
• To first order, for a given semiconductor nopo is a constant that
only depends on T :
2
nopo = ni
• Equilibrium carrier concentrations can be engineered through
shallow dopants
⇒
extrinsic semiconductor.
– n-type semiconductor:
no � ND,
po �
– p-type semiconductor:
po � NA,
no �
2ni
ND
2ni
NA
• Around edges, conduction and valence bands in semiconductors
feature DOS ∼
E.
√
• Order of magnitude of key parameters for Si at 300 K:
– intrinsic carrier concentration: ni ∼ 1010 cm−3
– typical doping level range: ND, NA ∼ 1015 − 1020 cm−3
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Lecture 2-17
Self study
• Charge neutrality
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Lecture 2
8.821/8.871 Holographic duality
Fall 2014
8.821/8.871 Holographic duality
MIT OpenCourseWare Lecture Notes
Hong Liu, Fall 2014
Lecture 2
1.2: BLACK HOLE THERMODYNAMICS
1.2.1: IMPORTANT SCALES
Planck scale
We can construct physical units using fundamental constants l (reduced Planck constant), GN (gravitational
constant), c (speed of light):
�
lc
GN
lGN
c3
�
mp =
lp =
≈ 1.2 × 1014GeV /c2 = 2.2 × 10−5 g;
≈ 1.6 × 10−33 cm;
tp = ≈ 5.4 × 10−44 s
lp
c
This quantity is called “Planck scale”, and represents the energy scale at which the quantum effects of gravity
become strong.
Strength of gravity
Let us first compare the strength of gravity and strength of electro-magnetic (EM) interaction. In the EM case,
to be the smallest distance
interaction takes the form VEM =
between particles, because this distance can be thought as the fundamental limitation on measuring the positions
of a particle, taking quantum mechanics and special relativity into account. Using the unit of particle static mass,
the EM interaction has the effective strength:
. We take the reduced Compton wavelength rc =
:
mc
2 e
r
On the other hand, we can also get the effective strength of gravity:
λEM =
VEM (rC )
mc2
=
2
e
lc
= α =
1
137
λG =
VG(rc) GN m 1
=
2
2
m
l/mc mc2 m2
p
=
=
l2
p
r2
c
mc2
Then λG « 1, for m « mp. | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/107c04c3759a2e244b8778e9bd616a72_MIT8_821S15_Lec2.pdf |
p
=
=
l2
p
r2
c
mc2
Then λG « 1, for m « mp. For example, in the case of electron, me = 5 × 10−4GeV /c2, we have
λG ∼ 10−43
λEM
The gravity effect is quite weak in this case. But if the mass is at Planck mass scale mp, then λG ∼ O(1), which
means quantum gravity effects become significant (the corresponding length scale will be lp).
Schwarzschild radius
Now taking a step back from the quantum gravity effects, we can even ask a simpler question: for an object of mass
m, at what distance rs from it, the classical gravity becomes strong? To answer this question, we can consider a
probe mass m', then the classical gravity becomes strong means that
GN mm'/rs
'c2m
∼ 1 ⇒
rs ∼
GN m
c2
1
Lecture 2
8.821/8.871 Holographic duality
Fall 2014
So now for an object of mass m, we have two important scales:
rc =
rs =
l
mc
2GN m
:
c2
: Reduced Compton wavelength
Schwarzschild radius
The pre-factor 2 of rs comes from a GR computation of a Schwarzschild black hole.
From ∼
rs
rc
2
m
mp
2 , we can conclude
1. m » mp, rs » rc: classical gravity (quantum effects not important);
2. m « mp, rs « rc: rs is not relevant, gravity effect is weak and not important;
3. m ∼ mp, rc ∼ rs, quantum gravity effects are important.
If this were the whole story, life would be much simpler, but much less interesting. However, black holes can make
quant | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/107c04c3759a2e244b8778e9bd616a72_MIT8_821S15_Lec2.pdf |
s are important.
If this were the whole story, life would be much simpler, but much less interesting. However, black holes can make
quantum gravity effects manifest at macroscopic level, at length scales of O(rs), we will discuss this later.
p
Remark: lp can be thought as the minimal localization strength. In non-gravitational physics, the probing length
scale l ∼ : , in principle, can be as small as one wants if one is powerful enough to get sufficiently large p. But
with gravity, when E ∼ p » mp, then rs
give larger length scales, lp is the minimal scale one can probe. Alternatively, consider uncertainty principle
δp ∼
takes over as the minimal scale. Since rs ∝ p, so larger energies
, then δx > GN δp
, so we obtain δx >
∼ GN p
c3
p
= lp.
∼ GN : 1
δx
c
3
3
c
:GN
3
c
:
δx
Now let us summarize various regimes of gravity for fixed energy scales we are interested in:
• Classical gravity: l → 0, GN finite;
• QFT in a fixed spacetime (including curved): l finite, GN → 0;
• Quantum gravity: GN , l finite; and in the semi-classical regime for quantum gravity, we take l finite and
expand the theory in GN .
1.2.2 Classical black hole geometry
Black hole geometry is the solution of Einstein equation with zero cosmology constant. The spacetime is due to
an object of mass M. If we consider the object to be spherically symmetric, non-rotational, neutral, we have the
Schwarzschild metric solution:
ds2 = −f dt2 + dr2 + r 2(dθ2 + sin θ2dφ2),
1
f
f = 1 −
2GN M
r
= | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/107c04c3759a2e244b8778e9bd616a72_MIT8_821S15_Lec2.pdf |
2(dθ2 + sin θ2dφ2),
1
f
f = 1 −
2GN M
r
= 1 −
rs
r
(1)
Note that from now on, we have adapted the convention to take c = 1.
The event horizon is defined at
r = rs = 2GN M
where gtt = 0, grr = ∞. When r goes across the event horizon, f changes sign, r and t switch role.
Here are some simple facts on this metric:
1. It is time-reversal invariant, i.e. invariant under t → −t.
It does not describe a black hole formed from gravitational collapse which is clearly not time-reversal
symmetric, but it is a mathematical idealization of such a black hole.
2. The spacetime is non-singluar at the horizon, as one can check this by computing curvature invariants. It is
only a coordinate singularity (not an intrinsic singularity), where t (Schwarzschild time) and r coordinates
become singular at the horizon.
3. At r = rs, the surface is a null hypersurface.
2
Lecture 2
8.821/8.871 Holographic duality
Fall 2014
4. The horizon is a surface of infinite redshift.
Consider an observer Oh at the hypersurface r = rh ≈ rs and another observer O∞ at the hypersurface
r = ∞. At r = ∞: ds2 → −dt2 + dr2 + r2dΩ2
ds2 = −f (rh)dt2 + · · · = −dτ 2 + · · · . We have dτh = f 1/2(rh)dt, with τh to be the proper time for Oh. Then
2, t is the proper time for O∞. On the other hand, at r = rh:
h
dτh
dt
= (1 −
1
2
rs
)
rh
As rh → rs,
Consider some event of energy Eh happening at r = rh, to O∞ this event has energy
→ | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/107c04c3759a2e244b8778e9bd616a72_MIT8_821S15_Lec2.pdf |
As rh → rs,
Consider some event of energy Eh happening at r = rh, to O∞ this event has energy
→ 0, i.e. compared to the time at r = ∞, the time at r = rh becomes infinitely slow.
dτh
dt
E∞ = Ehf
2 (rh)
1
i.e. for fixed local proper energy Eh, E∞ → 0 as rh → rs, we call it infinitely redshifted.
5. It takes a free-fall traveler a finite proper time to reach the horizon, but infinite Schwarzschild time.
6. Once inside the horizon, a traveler cannot send signals to outside, nor can she/he escape.
7. Two intrinsic geometric quantities of the horizon:
• Area of a spatial section
• Surface gravity
A = 4πr2 = 16πG2
s
N M 2
The acceleration of a stationary observer at the horizon as measured by an observer at infinity is given
by
More details can be found in Ref. [1]
K =
1
f ' (rs) =
2
1
4GN M
References
[1] Robert M. Wald, General Relativity, University Of Chicago Press (1984).
3
MIT OpenCourseWare
http://ocw.mit.edu
8.821 / 8.871 String Theory and Holographic Duality
Fall 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/107c04c3759a2e244b8778e9bd616a72_MIT8_821S15_Lec2.pdf |
6.252 NONLINEAR PROGRAMMING
LECTURE 3: GRADIENT METHODS
LECTURE OUTLINE
• Quadratic Unconstrained Problems
• Existence of Optimal Solutions
• Iterative Computational Methods
• Gradient Methods - Motivation
• Principal Gradient Methods
• Gradient Methods - Choices of Direction
QUADRATIC UNCONSTRAINED PROBLEMS
min f (x) = 1
x∈(cid:2)n(cid:160)
2 x(cid:3)Qx − b(cid:3)x,
where Q is n × n symmetric, and b ∈ (cid:2)n.
• Necessary conditions:
∇f (x ∗) = Qx∗ − b = 0,
∇2f (x ∗) = Q ≥ 0 : positive semidefinite.
• Q ≥ 0 ⇒ f : convex, nec. conditions are also
sufficient, and local minima are also global
• Conclusions:
− Q : not ≥ 0 ⇒ f has no local minima
− If Q > 0 (and hence invertible), x ∗ = Q−1b
is the unique global minimum.
− If Q ≥ 0 but not invertible, either no solution
or ∞ number of solutions
y
y
α > 0, β > 0
(1/α, 0) is the unique
global minimum
α = 0
There is no global minimum
0
1/α
x
0
x
y
α > 0, β = 0
{(1/α, ξ) | ξ: real} is the set of
global minima
y
α > 0, β < 0
There is no global minimum
0
1/α
x
0
1/α
x
Illustration of the isocost surfaces of the quadratic cost
function f(cid:160): (cid:1)2 (cid:2)→ (cid:1) given by
f (x, y) = 1 αx2 + βy2 − x
(cid:1)
2
(cid:2 | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/10cddea946dfe20a44ed08128ef2077d_6252_slides03.pdf |
f (x, y) = 1 αx2 + βy2 − x
(cid:1)
2
(cid:2)
for various values of α(cid:160)and β.
EXISTENCE OF OPTIMAL SOLUTIONS•
Consider
min f (x)
x∈X
Two possibilities:
(cid:3)
• The set f (x) | x ∈ X
and there is no optimal solution
• The set f (x) | x ∈ X is bounded below
(cid:3)
(cid:4)
(cid:4)
is unbounded below,
− A global minimum exists if f is continuous
and X is compact (Weierstrass theorem)
− A global minimum exists if X is closed, and
f is coercive, that is, f (x) → ∞ when (cid:8)x(cid:8) →
∞
GRADIENT METHODS - MOTIVATION•
∇f(x)
x
f(x) = 1
c
c
f(x) = 2 < c1
c
f(x) = 3 < c2
xα = x - α∇f(x)
If ∇f (x) (cid:5)= 0, there is an
interval (0, δ) of stepsizes
such that
(cid:1)
f x(cid:160)− α∇f (x) < f (cid:160)(x)
(cid:2)
x - δ∇f(x)
for all α(cid:160)∈ (0, δ).
f(x) = 1
c
f(x) = 2 < c1
c
f(x) = 3 < c2
c
∇f(x)
x
xα = x + αd
If d(cid:160) makes an angle with
∇f (x) that is greater than
90 degrees,
∇f(cid:160)(x)
�
d < (cid:160)0,(cid:160)
x + δd
d
there is an interval (0, δ)
of stepsizes such that f(cid:160)(x+
α | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/10cddea946dfe20a44ed08128ef2077d_6252_slides03.pdf |
d
d
there is an interval (0, δ)
of stepsizes such that f(cid:160)(x+
αd) < f (cid:160)(x) for all α(cid:160) ∈
(0, δ).
PRINCIPAL GRADIENT METHODS•
xk+1 = xk + αkdk ,
k = 0, 1, . . .
where, if ∇f (xk) (cid:9)= 0, the direction dk satisfies
∇f (xk)(cid:3)dk < 0,
and αk is a positive stepsize. Principal example:
xk+1 = xk − αkDk∇f (xk),
where Dk is a positive definite symmetric matrix
• Simplest method: Steepest descent
xk+1 = xk − αk∇f (xk),
k = 0, 1, . . .
• Most sophisticated method: Newton’s method
xk+1 = xk−αk
∇2f (xk)
(cid:1)
(cid:2)−1 ∇f (xk),
k = 0, 1, . . .
STEEPEST DESCENT AND NEWTON’S METHOD•
x0
Slow convergence of steep
est descent
f(x) = 1
c
Quadratic Approximation of f at x0
c
f(x) = 2 < c1
f(x) = 3 < c2
c
.
x0
x1
.
.
x2
Quadratic Approximation of f at x1
Fast convergence of New-
ton’s method w/ αk(cid:160) = 1.
Given xk(cid:160), the method ob
tains xk+1 as the minimum
of a quadratic approxima
tion of f(cid:160) based on a sec
ond order Taylor expansion
around xk(cid:160).
OTHER CHOICES OF DIRECTION
• Diagonally Scaled Steepest Descent
Dk = Diagonal approximation to
(cid:1)
∇2f (xk)
(cid | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/10cddea946dfe20a44ed08128ef2077d_6252_slides03.pdf |
caled Steepest Descent
Dk = Diagonal approximation to
(cid:1)
∇2f (xk)
(cid:2)−1
• Modified Newton’s Method
Dk = (∇2f (x0))
−1 ,
k = 0, 1, . . . ,
• Discretized Newton’s Method
(cid:1)
H(xk)
(cid:2)−1
,
Dk =
k = 0, 1, . . . ,
where H(xk) is a finite-difference based approxi-
mation of ∇2f (xk),
• Gauss-Newton method for least squares prob-
lems minx∈(cid:2)n(cid:160)
1
2
(cid:8)g(x)(cid:8)2. Here
(cid:2)−1
,
(cid:1)
∇g(xk)∇g(xk)(cid:3)
Dk =
k = 0, 1, . . . | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/10cddea946dfe20a44ed08128ef2077d_6252_slides03.pdf |
More Symple Types
Progress And Preservation
Armando Solar-Lezama
Computer Science and Artificial Intelligence Laboratory
M.I.T.
September 28, 2015
September 28, 2015
L06-1
Formalizing a Type System
Recap
September 28, 2015
Static Semantics
• Typing rules
– Typing rules tell us how to derive typing judgments
– Very similar to derivation rules in Big Step OS
• Ex. Language of Expressions
L06-3
Ex. Language of Expressions
• Show that the following Judgment is
valid
L06-4
Simply Typed 𝜆 Calculus (F1)
• Basic Typing Rules
• Extensions
L06-5
Example
• Is this a valid typing judgment?
• How about this one?
L06-6
Example
• What’s the type of this function?
(𝜆 f. 𝜆 x. if x = 1 then x else (f f (x-1) ) * x)
– Hint: This IS a trick question
L06-7
Simply Typed 𝜆 Calculus (F1)
• We have defined a really strong type
system on 𝜆-calculus
– It’s so strong, it won’t even let us write non-
terminating computation
– We can actually prove this!
L06-8
Progress and Preservation
September 28, 2015
What makes a type system “correct”
• “Well typed programs never go wrong”
• Inductive argument
– Preservation: If a program is well typed it will stay well
typed in the next step of evaluation
– Progress: If a program is well typed now, it won’t go
wrong in the next step of evaluation
• What do we mean by “step of evaluation”?
September 28, 2015
L06-10
Preservation
• Using Big-Step semantics we can argue
global preservation
Γ ⊢ 𝑒1: 𝜏 ∧ 𝑒1 → 𝑒2 ⇒ Γ ⊢ 𝑒2: 𝜏
• Prove by induction on the structure of
derivation of � | https://ocw.mit.edu/courses/6-820-fundamentals-of-program-analysis-fall-2015/10f472f16fcff7bceab5d9b40aa8bfa3_MIT6_820F15_L06.pdf |
��2 ⇒ Γ ⊢ 𝑒2: 𝜏
• Prove by induction on the structure of
derivation of 𝑒1 → 𝑒2
September 28, 2015
L06-11
Proof by induction on Structure
of Evaluation
• Base cases: trivial
• Inductive case is a little trickier
𝑒1 → 𝜆𝑥. 𝑒1
′ 𝑒2 𝑥 → 𝑒3
′ 𝑒1
𝑒1 𝑒2 → 𝑒3
September 28, 2015
L06-12
Induction on the Structure of the
Derivation
• Inductive case
𝑒1 → 𝜆𝑥. 𝑒1
′ 𝑒2 𝑥 → 𝑒3
′ 𝑒1
𝑒1 𝑒2 → 𝑒3
– Given we want to show that
– By our typing rule, we have
– And by the IH, we have that
– Which again by the typing rule
– Now, we need to show that
Γ, 𝑥: 𝜏′ ⊢ 𝑒1
′: 𝜏𝑒12 ∧ Γ ⊢ 𝑒2: 𝜏′ ⇒ Γ ⊢ 𝑒1
′ 𝑒2 𝑥 : 𝜏𝑒12
– And from our IH
Γ ⊢ 𝑒1
′ 𝑒2 𝑥 : 𝜏𝑒12 ⇒ Γ ⊢ 𝑒3: 𝜏𝑒12
L06-13
Small Step Semantics
• Big step goes directly from initial
program to result
• Small Step evaluates one step at a time
September 28, 2015
L06-14
Small Step Example
• Contexts
H ::= o | H e1 | H + e | n + H |
if H then e1 else e2 |
H == e1 | n == H
• Local Reduction Rules
– n1 + n2 n (where n = plus n1 n2) | https://ocw.mit.edu/courses/6-820-fundamentals-of-program-analysis-fall-2015/10f472f16fcff7bceab5d9b40aa8bfa3_MIT6_820F15_L06.pdf |
n == H
• Local Reduction Rules
– n1 + n2 n (where n = plus n1 n2)
– n1 == n2 b (where b =(equals n1 n2))
– if true then e1 else e2 e1
– if false then e1 else e2 e2
– (𝜆x:𝜏.e1) v2 [v2/x] e1
• Global Reduction Rules
– H[r] H[e] iff r e
L06-15
The proof strategy
• Progress Theorem
If ├ e:𝜏 and e is not a value, then there is an e’ s.t.
e e’
• We can prove this through a
decomposition lemma
– If ├ e:𝜏 and e is not a value, then there are H and r
s.t. e = H[r]
– This guarantees one step of progress
L06-16
Proving the Progress Theorem
If ├ e:𝜏 and e is not a value, then there is an e’ s.t.
e e’
or equivalently, e = H[r]
• Proved by induction on the derivation of
├ e:𝜏
• Base case:
– Irreducible values
L06-17
Proving the Progress Theorem
• Inductive case
– by the IH, e can be irreducible,
• in which case it must be true or false and the
whole thing is a redex
– Or, it can be decomposed into H[r]
• in which case if H then e1 else e2 is a valid
context.
L06-18
MIT OpenCourseWare
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6.820 Fundamentals of Program Analysis
Fall 2015
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-820-fundamentals-of-program-analysis-fall-2015/10f472f16fcff7bceab5d9b40aa8bfa3_MIT6_820F15_L06.pdf |
3.15
Photoconductors, Photovoltaics and Photodetectors
C.A. Ross, Department of Materials Science and Engineering
Reference: Pierret, chapter 9.2 and 9.3
Photoconductors – conductivity a function of light
Photovoltaics – generate power from light
Photodetectors – use a pn junction to detect light
Photoconducting materials:
CdS, ZnS for camera lightmeters, amorphous As,Se,Te for photocopiers
Photodiode and Photovoltaic (PV):
Carriers created within Ln or Lp of junction contribute to reverse current:
I = Io + IG
Photodiode operates in reverse bias. A PIN diode has a wide depletion
region; operates much faster than a pn junction photodetector because it
doesn not rely on diffusion.
A PV operates in the fourth quadrant (positive V, negative I). When
connected to a load (e.g. a battery charger or a lightbulb) with resistance RL,
V = I (RPV + RL)
I = Io (exp(eV/kT) - 1) + IG
also
these two relations define the voltage and current that the PV produces.
Power = IV
Solar Cells: the PV must respond to the visible spectrum (400 – 700 nm,
or 2 – 3 eV; note that l (mm) = 1.24/E (eV))
Ideally we would use a bandgap of about 1.2 eV, but Si does not absorb light
well because it has an indirect band gap.
Direct and indirect gap
On an E-k plot: m* = h
Momentum of an electron typically p/a ~ 1010 m-1
Momentum of a photon = 2p/l ~ 107 m-1
(∂2 E /∂k 2)-1
2
†
If the band gap is indirect, a phonon plus a photon are eeded to make an e-h
pair, so light absorption (and emission) is less efficient.
Amorphous Si: uncertainly principle DxDp ≥ h
-the localization of carriers gives them an uncertain momentum, so direct
absorption of light can occur. Use PIN design because mobility is low.
Handout 5
†
1
Scanned article removed due to copyright restrictions | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/110ed969f7442b8ed4f4ff7a8e3d9026_lecture9.pdf |
of light can occur. Use PIN design because mobility is low.
Handout 5
†
1
Scanned article removed due to copyright restrictions.
Please See "This Month in Physics History, October 22, 1938: Invention of Xerography." APS News 12 (2003): 2.
Handout 5
2
ASE Americas, Inc. Solar Panel, Model BC-16-DG
Voltage per Cell (Volts)
0.0
0.2
0.4
0.6
Darkened l-V Behavior
0.04
0.02
0.00
-0.02
)
2
m
c
/
s
p
m
A
(
y
t
i
s
n
e
D
t
n
e
r
r
u
C
0
19.8
28.5
37.0
43.8
41.6
Power (Watts)
-0.04
Fully Illuminated
0
12.0
23.0
32.9
)
s
p
m
A
(
t
n
e
r
r
u
C
4
2
0
-2
-4
-5
0
5
10
15
20
25
Voltage -- 36 Cells in Series (Volts)
Figure by MIT OCW.
Handout 5
4 | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/110ed969f7442b8ed4f4ff7a8e3d9026_lecture9.pdf |
Lecture 7
8.821/8.871 Holographic duality
Fall 2014
8.821/8.871 Holographic duality
MIT OpenCourseWare Lecture Notes
Hong Liu, Fall 2014
Lecture 7
In fact, any orientable two dimensional surface is classified topologically by an integer h, called the genus. The
genus is equal to the number of “holes” that the surface has (Fig. 1).
Figure 1: sphere (genus-0), torus (genus-1) and double torus(genus-2).
An topological invariant of the manifold is the Euler character:
χ = 2 − 2h
Here we make some claims:
1. For any non-planar diagram, there exists an integer h, such that the diagram can be straightened out (i.e.
non-crossing) on a genus-h surface, but not on a surface with a smaller genus.
2. For any non-planar diagram, the power of N that comes from contracting propagators is given by the number
of faces on such a genus-h surface, i.e. the number of disconnected regions separated by the diagram.
Both claims are self-evident after a bit practices.
In general, a vacuum diagram has the following dependence on g2 and N:
A ∼ (g2)E(g2)−V N F
where E is the number of propagators, V is the number of vertices, F is the number of faces. This does not give a
sensible N →∞ limit or 1/N expansion, since there is no upper limit on F. However, ’t Hooft suggests that we can
take the limit N →∞ and g2→0 but keep λ = g2N fixed. Then
where L is the number of loops. The relation χ = F + V − E is guaranteed by the following theorem.
A ∼ (g2N )E−V N F +V −E = λL−1N χ = λL−1N 2−2h
Theorem: Given a surface composed of polygons with F faces, E edges and V vertices, the Euler character satisfy
χ = F + V − E = 2 − 2h
Since each Feynman diagram can be considered as a partition of the surface separating it into polygons, then the
above | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/1119bf97378aff76fb5c02ae928d930d_MIT8_821S15_Lec7.pdf |
= 2 − 2h
Since each Feynman diagram can be considered as a partition of the surface separating it into polygons, then the
above theorem also works for our counting in N.
Thus in this limit, to the leading order in N is the planar diagrams
N 2(c0 + c1λ + c2λ2 + · · · ) = N 2f0(λ)
Because log Z evaluates the sum of all vacuum diagrams, we can conclude, including higher order 1/N 2 corrections:
log Z =
∞
(cid:88)
h
=0
fh(λ) = N 2f0(λ) + f1(λ) +
1
N 2
f2(λ) +
· · ·
The first term comes from the planar diagrams, second term from the genus-1 diagrams, etc.
1
Lecture 7
8.821/8.871 Holographic duality
Fall 2014
There is a heuristic way to understand log Z = O(N 2) + · · · . Since Z =
Lagrangian as
L =
N
λ
Tr
(cid:20) 1
2
(∂Φ)2 +
1
4
Φ4
(cid:21)
´
DΦeiS[Φ] and we can rewrite the
The trace also gives a factor of N, thus L ∼ O(N 2), we have log Z ∼ O(N 2).
Clearly our discussion only depends on the matrix nature of the fields. So for any Lagrangian of matrix valued
fields of the form
L =
Tr (
· · · )
N
λ
we would have
log Z =
∞
(cid:88)
h
=0
N 2−2hfh(λ)
To summarize, in the ’t Hooft limit, 1/N expansion is the same as topological expansion in terms of topology of
Feynman diagrams.
General observables
Now we have introduced two theories:
(a)
L = −
1
g2 Tr
(cid:20)
1
g2
Y M
(cid:20) 1
2
1
4
−
(∂Φ)2 +
(cid:21)
Φ4
1
4
Tr FµνF µν
(cid | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/1119bf97378aff76fb5c02ae928d930d_MIT8_821S15_Lec7.pdf |
−
(∂Φ)2 +
(cid:21)
Φ4
1
4
Tr FµνF µν
(cid:21)
− iΨ(D/ − m)Ψ
(b)
L =
(a) is invariant under the global U (N ) transformation: Φ→U ΦU † with U constant U (N ) matrix, i.e. the theory
has a global U (N ) symmetry. (b) is invariant under local U (N ) transformation:
Aµ→U (x)AµU †(x) − i∂µU (x)U †(x) with U (x) any U (N ) matrix, the theory has a U (N ) gauge symmetry.
On the other hand, consider allowed operators in the two theories. In (a), operators like Φa
it is not invariant under global U (N ) symmetry. But in (b), allowed operators must be gauge invariant, so Φa
not allowed. So if we consider gauge theories: L = L(Aµ, Φ, · · · ), the allowed operators will be
b are allowed, although
b is
Single-trace operators : Tr (FµνF µν), Tr(Φn), · · ·
Multiple-trace operators : Tr (FµνF µν) Tr (Φ2), Tr (Φ2) Tr (Φn) Tr (Φn), · · ·
We denote single-trace operators as Ok, k = 1, · · · represents different operators. Then multiple-trace ones will be
like OmOn(x), Om1Om2Om3 (x), · · ·
So general observables will be correlation functions of gauge invariant operators, here we focus on local operators:
(cid:104)O1(x1)O2(x2) · · · On(xn)(cid:105)c
(1)
Note that it is enough to focus on single-trace operators since multiple-trace ones are products of them. Since we
are working in the t’ hooft limit, we want to know how correlation (Eq. 1) scales in the large N limit. There is a
trick, consider
ˆ
ˆ
Z [J1, · · · , Jn] =
DAµDΦ | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/1119bf97378aff76fb5c02ae928d930d_MIT8_821S15_Lec7.pdf |
ick, consider
ˆ
ˆ
Z [J1, · · · , Jn] =
DAµDΦ · · · exp (iSef f ) =
DAµDΦ · · · exp
iS0 + iN
Ji(x)Oi(x)
ˆ
(cid:88)
j
Then the correlation (Eq. 1) can be expressed as
(cid:104)O1(x1)O2(x2) · · · On(xn)(cid:105)c =
δn log Z
δJ1(x1) · · · δJn(xn)
|J1=···=Jn=0
1
(iN )n
(2)
2
Lecture 7
8.821/8.871 Holographic duality
Fall 2014
With Oi single-trace operators, Sef f has the form N Tr (· · · ). So we have
log Z [J1, · · · , J
n] =
∞
(cid:88)
h=0
N 2−2 fh(λ, · · · )
h
(cid:104)O1(x1)O2(x
2) · · · On(xn)(cid:105)c ∼ N −
2 n
(cid:20)
1 + O(
1
N 2
(cid:21)
) +
· · ·
Applying Eq. (2),
e.g.
(cid:104)1(cid:105) ∼ O(N 2) + O(N 0) + · · ·
(cid:104)O(cid:105) ∼ O(N ) + O(N −1) + · · ·
(cid:104)O1O2(cid:105)c ∼ O(N 0) + O(N −2) + · · ·
2O3(cid:105)c ∼ O(N − ) + O(N − ) + · · ·
1
3
(cid:104)O1O
All leading order contributions come from planar diagrams.
Physical implications:
1. In the large N limit, O(x)|0(cid:105) can be interpreted as creating a single-particle state (”glue ball”). Similarly
: O1 · · · On(x) : |0(cid:105) represents n | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/1119bf97378aff76fb5c02ae928d930d_MIT8_821S15_Lec7.pdf |
creating a single-particle state (”glue ball”). Similarly
: O1 · · · On(x) : |0(cid:105) represents n-particle state.
• since (cid:104)OiOj(cid:105) ∼ O(N 0), we can diagonalize them such that (cid:104)OiOj(cid:105) ∝ δi
• (cid:104)Oi(x)O2
j (y)(cid:105) ∼ O(N −1)→0 as N →∞, i.e. there is no mixing between single-trace and multiple-trace
j.
operators in the large N limit.
• (cid:104)O1O2(x)O1O2(y)(cid:105) = (cid:104)O1(x)O1(y)(cid:105)(cid:104)O2(x)O2(y)(cid:105) + (cid:104)O1O2(x)O1O2(y)(cid:105)c, the first term is the multiple
of independent propagators of O1 and O
2 states, the second term scales like O(N − ).
2
Note that it is not necessary there exists a stable on-shell particle associated with Oi(x)|0(cid:105).
2. The fluctuations of “glue balls” are suppressed:
√
suppose (cid:104)O(cid:105) =(cid:54)
(cid:104)O1O2(cid:105) = (cid:104)O1(cid:105)(cid:104)O2(cid:105) + (cid:104)O
the large N limit, it is more like a classical theory.
0 ∼ O(N ), the variance of (cid:104)O(cid:105) is (cid:104)O2(cid:105) − (cid:104)O(cid:105)2 = (cid:104)O2(cid:105)c ∼ O(1), i.e.
(cid:104)O2(cid:105)c N
(cid:104)O(cid:105) ∼ −1→0. Also
1O2(cid:105)c, the first term scales as O(N ) while the second term scales as O(1). Thus in
2
3
MIT OpenCourseWare
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) while the second term scales as O(1). Thus in
2
3
MIT OpenCourseWare
http://ocw.mit.edu
8.821 / 8.871 String Theory and Holographic Duality
Fall 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/1119bf97378aff76fb5c02ae928d930d_MIT8_821S15_Lec7.pdf |
Lecture Five: The Cacciopolli Inequality
1
The Cacciopolli Inequality
The Cacciopolli (or Reverse Poincare) Inequality bounds similar terms to the Poincare
inequalities studied last time, but the other way around. The statement is this.
Theorem 1.1 Let u : B2r → R satisfy u�u ≥ 0. Then
�
Br
2
|�u| ≤
4
2
r
�
B2r \Br
2
u .
(1)
First prove a Lemma.
Lemma 1.2 If u : B2r → R satisfies u�u ≥ 0, and φ : B2r R is nonnegative with
φ = 0 on ∂B2r, then
→
�
�
φ2|�u| ≤ 4
2
2
|u| |�φ|2 .
B2r
B2r
�
0 ≤
B2r
φ2 u�u.
(2)
(3)
�
B2r
�
φ2u�u + B2r �(φ2u) ·
φ2u�u · dS = 0, so apply Stokes’ theorem to get
Proof Consider
�
Clearly ∂B2r
�u = 0. From this
�
0 ≤ −
B2r
�(φ2 u)�u = −2
�
�
φu�φ · �u −
B2r
B2r
φ2|�u|
2
,
(4)
and so
�
B2r
φ2
�
2
|�u| ≤ −2
�
≤ 2
φu�φ�u
B2r
φ
|u||�φ||�u|
.
(5)
(6)
B2r
1
��
Recall the inequality f g ≤
of the CauchySchwarz inequality), and apply it above to get
f 2 �1 | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/11294d6e66f18a2a5dc4230e13a6d8cc_lecture5.pdf |
Recall the inequality f g ≤
of the CauchySchwarz inequality), and apply it above to get
f 2 �1/2 ��
�
g2 �1/2 for any functions f and g (this is one form
�
B2r
��
φ2|�u|2 ≤ 2
�1/2 ��
φ2|�u|2
�1/2
|u|2|�φ|2
.
B2r
B2r
Dividing and squaring then gives
�
B2r
�
φ2|�u|2 ≤ 4
|u|2|�φ|2 .
B2r
To complete the proof of theorem 1.1 pick
φ(x) =
�
1
2r−|x|
r
if |x| ≤ r;
if r < x ≤ 2r,
(7)
(8)
so |�φ|
the result, namely
= 0 on Br and |�φ = 1/r on B2r \ Br . Substitute this into the lemma to obtain
|
�
Br
2
|�u| ≤
4
2
r
�
B2r \Br
2
u .
(9)
2 Applications of the Cacciopolli Inequality
2.1 Bounding the growth of a harmonic function
One nice consequence of the Cacciopolli Inequality is the following inequality bounding the
rate at which a harmonic function can decay.
Proposition 2.1 There are strictly positive dimensional constants k(n) such that
�
B2r
2
u ≥ (1 + k(n))
�
Br
2
u
(10)
for all harmonic functions u : B2r R.→
Proof Let φ be a test function as before, and consider
�
B2r
|�(φu)|
2 =
=
�
B2r
�
B2r
2
|φ�u + u�φ|
φ2|�u|2 + u |�φ|2 + 2uφ�φ · | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/11294d6e66f18a2a5dc4230e13a6d8cc_lecture5.pdf |
|φ�u + u�φ|
φ2|�u|2 + u |�φ|2 + 2uφ�φ · �u.
2
Apply CauchySchwarz and lemma 1.2 to get
2
�
B2r
|�(φu)| ≤
2
�
φ2
�
|�u|2 +
��
u |�φ|2 + 2
2
�1/2 ��
2
|�u|
φ2
�1/2
|�φ|2
2
u
B2r
�
≤ 2
B2r
�
φ2|�u|2 + 2
2
|�φ|2
.
u
B2r
�
≤ 10
B2r
2
|�φ|2
.
u
B2r
B2r
B2r
Now make the same choice of φ as before to give
�
�
|�(φu)| ≤
2
B2r
10
2
r
2
u
B2r \Br
and apply DirichletPoincare to the left hand side to get
1
C(n)r2
�
B2r
φ2 u 2
≤
10
2
r
�
B2r \Br
u 2 .
Since (φu)2 is a positive function we can reduce the area of the integration, therefore
�
k(n)
�
φ2 2
u ≤
2
u .
Br
B2r \Br
for k(n) = 10C(n) . Finally note that φ = 1 on Br , so
1
�
k(n)
�
2
u ≤
2
u ,
Br
�
(1 + k(n))
B2r \Br
�
2
u ≤
2
u .
Br
B2r
and
This completes the proof.
(11)
(12)
(13)
(14)
(15)
2.2 Bounding the growth of the energy of a harmonic function
We will now prove a similar inequality for the Dirich | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/11294d6e66f18a2a5dc4230e13a6d8cc_lecture5.pdf |
Bounding the growth of the energy of a harmonic function
We will now prove a similar inequality for the Dirichlet energy of a harmonic function.
Proposition 2.2 There are dimensional constants c(n) such that
�
B2r
2
|�u|
�
≥ (1 + θ(n))
2
|�u|
.
Br
(16)
for all harmonic functions u : B2r R. .
→
3
Proof It suffices to show that
�
c(n)
�
2
|�u| ≤
Br
B2r \Br
|�u|
2 .
(17)
To do this we use two inequalities. Firstly we will state and use without proof the Neumann
Poincare inequality for an annulus, namely if A = volB2r \Br
�
B2r \Br
u then
1
�
�
(u − A)2 ≤ d(n)r
2
2
|�u|
.
(18)
B2r \Br
B2r \Br
Secondly we use Cacciopolli, noting that if �u = 0 then �(u+A) = 0, and �(u+A) = �u,
to give
�
r 2
Br
�
|�u|2 ≤ 4
(u − A)2 .
B2r \Br
1
4d(n)
�
Br
�
|�u|2 ≤
|�u|2
B2r \Br
Together (15) and (16) give
as required.
(19)
(20)
4 | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/11294d6e66f18a2a5dc4230e13a6d8cc_lecture5.pdf |
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
LECTURE 2: One-Dimensional ‘Traveling Waves’
Main Points
- Exponential and sine-wave solutions to the one-dimensional wave equation.
- The distributed compressibility and mass in acoustic plane waves are analogous with the
distributed capacitance and inductance in electrical transmission lines.
- Traveling waves vary in both space and time.
-
Interactions of waves with structures of different impedance that are of significant size compared
to a wavelength produce reflected waves.
The magnitude of the reflection depends on the relative impedance of the object and the media.
-
1. The One-Dimensional Wave Equation for Plane Waves:
S=z y
y
( 0 ,0 ,0 )
z
x
x+²x
Figure 2.1: A long duct of height y, width z and undetermined length. Our derivation of the wave
equation is based on a section of duct described by the interval x to x+∆x.
We saw in Lecture 1 that we can characterize the propagation of plane waves fairly simply, if we
make some generally reasonable assumptions:
a. the forces related to the viscosity of air are negligible, and
b. the rapid variations in pressure associated with sound don't allow heat transfer within the medium
or to the surround (the adiabatic condition),
c. the sound induced variations in the scalars p(x,t), ρ(x,t) and T(x,t) are small compared to their
static values.
d. the sound induced particle velocity vx(x,t) is small compared to the propagation velocity.
These assumptions together with considerations of Newton’s second law, conservation of mass
and consideration of the adiabatic compressibility of air lead to lossless acoustic equations (consistent
with a and b above) in which the distributed mass (the densityρ0) and distributed compliance (the bulk
modulus BA) of the air completely determine the relationship between vx (the magnitude of the x
component of the particle velocity) and p (the sound pressure) at any position (x) and time (t) in a one
dimensional system like Figure 2.1.
Newtons 2nd Law:
∂p(x, t)
∂x
= −ρ0
∂ | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
2.1.
Newtons 2nd Law:
∂p(x, t)
∂x
= −ρ0
∂vx (x,t)
∂t
Conservation of Mass-Compressibility Relationship:
∂v x (x,t)
∂x
= −
1
BA
∂p(x,t)
∂t
.
The Wave-Equation for Sound Pressure in a Plane Wave:
14-Sept-2004
(1.2.7)
(1.2.20)
page 1
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
∂2 p(x,t)
∂x 2
=
ρ0
BA
∂2 p(x,t)
∂t 2
, where
ρ0
BA
=
1
c 2
.
(1.2.21)
We can write a ‘matching’ equation to describe the variation in particle velocity as a function of
time by taking the partial of both sides of 1.2.7 with respect to time, and substituting 1.2.20 into the left-
hand side of the result:
∂2v(x,t)
∂x 2
=
ρ0
BA
∂2v(x,t)
∂t 2
.
(2.1)
2. Wave propagation.
The propagation of sound as described by the wave equation can be understood by a ‘distributed’
series of ‘lumped’ masses and compliance (spring-like) elements. The following figure is a
modification of Denes and Pinsons’s Figure 3, in which a perturbation in a string of springs and masses
causes a propagated wave of force and motion, (modified from Denes & Pinson “The Speech Chain”,
WH Freeman 1993).
A
B
B
B
B
B
B
B
B
B
C
C
C
C
C
C
C
C
C
A
A
A
A
A
A
A
A
D
D
D
D
D
D
D
E
D
D
E
E
E
E
E
E
E
E
14-Sept-2004
page | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
D
E
D
D
E
E
E
E
E
E
E
E
14-Sept-2004
page 2
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
In the top row the balls and springs are at rest
In the second row, ball A is displaced to the left, stretching the AB spring
In the third row, ball B has moved toward A, compressing AB and stretching BC
In the fourth row Ball B moves even closer to A compressing AB past its rest position
In the fifth row Ball B moves back to the left and settles into its new rest position, etc.
14-Sept-2004
page 3
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
3. Similarity to Transmission Line Equations
We can use the acoustic analog of what are known as transmission line equations to describe sound in a
one-dimensional plane wave. In such a system, the wave is described in terms of the interaction of a
distributed series of lumped electrical capacitors and inductances. In this analogy the voltage e(t,x) is
the analog of pressure, the current i(t,x) is analogous to the one-dimensional particle velocity, the
inductances are analogous to acoustic inertances per unit distance, and the capacitors are analogous to
acoustic compliance per unit distance.
Lx = is the electrical inductance per unit length (the electrical analog of density ρ0),
Cx = the electrical capacitance per unit length (the analog of the compressibility of air 1/BA),
I = a complex amplitude that describes the current (the analog of particle velocity), and
E = a complex amplitude that describes the voltage (the analog of sound pressure).
The inductance/length:
The compliance per length:
∂e(x,t)
∂x
= −Lx ∂i(x,t)
∂t
∂i(x,t)
∂x
= −C x ∂e(x,t)
∂t
.
(2.2)
(2.3)
The variations in voltage and current in time and space can be described by wave equations:
∂2e(x,t)
∂x 2 | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
The variations in voltage and current in time and space can be described by wave equations:
∂2e(x,t)
∂x 2
=
1
c 2
∂2e(x,t)
∂t 2
∂2i(x,t)
∂x 2
=
1
c 2
∂2i(x,t)
∂t 2
, and
c =
1
LxC x
, where:
.
(2.4,5 &6)
We can also define an electrical impedance in the transmission line where
Z = Lx
C x .
(2.7)
All of these equations are analogous to the equations we derived for plane-wave propagation of sound,
where:
e(t) → p(t)
i(t) → vx (t)
Lx → ρ0
C x → 1
BA
.
14-Sept-2004
page 4
Lecture 2
Acoustics of Speech and Hearing
6.551J / HST.714J
4. Solutions to the Wave Equation
A general solution for one-dimensional plane-wave propagation describes the pressure and
particle velocity at any time and as the sum of two traveling waves, one moving in the positive x
direction and the other negative going:
and
p(t,x) = f + t − x /c
(
)+ f − t + x /c
(
) ,
vx (t,x) =
1
z0
[
f + t − x /c
(
)− f − t + x /c
(
] ,
)
(2.8)
(2.9)
where the argument to the wave functions ( f+ and f–) is a time (t-x/c) determined by the absolute time t
and the time needed to travel to x, i.e. x/c.
You should note:
–vx and p in each wave are related by the characteristic impedance of the medium z0.
– The two scalar pressure terms add.
– Because of a difference in direction, the two velocity terms subtract.
– Because of the difference in the signs of the second terms, vx(t,x) and p(t,x) need not be
proportional.
An alternative form of this solution can be given in terms of absolute position and the distance
propagated in | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf |
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