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# Comparator

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In electronics, a comparator is a device that compares two voltages or currents and outputs a digital signal indicating which is larger. It has two analog input terminals ${\displaystyle V_{+}\,}$ and ${\displaystyle V_{-}\,}$ and one binary digital output ${\displaystyle V_{\rm {o}}\,}$. The output is ideally

## Contents

${\displaystyle V_{\rm {o}}={\begin{cases}1,&{\mbox{if }}V_{+}>V_{-}\\0,&{\mbox{if }}V_{+}

A comparator consists of a specialized high-gain differential amplifier. They are commonly used in devices that measure and digitize analog signals, such as analog-to-digital converters [ clarification needed ] (ADCs), as well as relaxation oscillators.

## Differential voltage

The differential voltages must stay within the limits specified by the manufacturer. Early integrated comparators, like the LM111 family, and certain high-speed comparators like the LM119 family, require differential voltage ranges substantially lower than the power supply voltages (±15 V vs. 36 V). [1] Rail-to-rail comparators allow any differential voltages within the power supply range. When powered from a bipolar (dual rail) supply,

${\displaystyle V_{S-}\leq V_{+},V_{-}\leq V_{S+}}$

or, when powered from a unipolar TTL/CMOS power supply:

${\displaystyle 0\leq V_{+},V_{-}\leq V_{\rm {cc}}}$

Specific rail-to-rail comparators with p-n-p input transistors, like the LM139 family, allow the input potential to drop 0.3 volts below the negative supply rail, but do not allow it to rise above the positive rail. [2] Specific ultra-fast comparators, like the LMH7322, allow input signal to swing below the negative rail and above the positive rail, although by a narrow margin of only 0.2 V. [3] Differential input voltage (the voltage between two inputs) of a modern rail-to-rail comparator is usually limited only by the full swing of power supply.

## Op-amp voltage comparator

An operational amplifier (op-amp) has a well balanced difference input and a very high gain. This parallels the characteristics of comparators and can be substituted in applications with low-performance requirements. [4]

A comparator circuit compares two voltages and outputs either a 1 (the voltage at the plus side; VDD in the illustration) or a 0 (the voltage at the negative side) to indicate which is larger. Comparators are often used, for example, to check whether an input has reached some predetermined value. In most cases a comparator is implemented using a dedicated comparator IC, but op-amps may be used as an alternative. Comparator diagrams and op-amp diagrams use the same symbols.

Figure 1 above shows a comparator circuit. Note first that the circuit does not use feedback. The circuit amplifies the voltage difference between Vin and VREF, and it outputs the result at Vout. If Vin is greater than VREF, then voltage at Vout will rise to its positive saturation level; that is, to the voltage at the positive side. If Vin is lower than VREF, then Vout will fall to its negative saturation level, equal to the voltage at the negative side.

In practice, this circuit can be improved by incorporating a hysteresis voltage range to reduce its sensitivity to noise. The circuit shown in Figure 1, for example, will provide stable operation even when the Vin signal is somewhat noisy.

In practice, using an operational amplifier as a comparator presents several disadvantages as compared to using a dedicated comparator: [5]

1. Op-amps are designed to operate in the linear mode with negative feedback. Hence, an op-amp typically has a lengthy recovery time from saturation. Almost all op-amps have an internal compensation capacitor which imposes slew rate limitations for high frequency signals. Consequently, an op-amp makes a sloppy comparator with propagation delays that can be as long as tens of microseconds.
2. Since op-amps do not have any internal hysteresis, an external hysteresis network is always necessary for slow moving input signals.
3. The quiescent current specification of an op-amp is valid only when the feedback is active. Some op-amps show an increased quiescent current when the inputs are not equal.
4. A comparator is designed to produce well limited output voltages that easily interface with digital logic. Compatibility with digital logic must be verified while using an op-amp as a comparator.
5. Some multiple-section op-amps may exhibit extreme channel-channel interaction when used as comparators.
6. Many op-amps have back to back diodes between their inputs. Op-amp inputs usually follow each other so this is fine. But comparator inputs are not usually the same. The diodes can cause unexpected current through inputs.

## Working

A dedicated voltage comparator will generally be faster than a general-purpose operational amplifier used as a comparator, and may also contain additional features such as an accurate, internal reference voltage, adjustable hysteresis, and a clock gated input.

A dedicated voltage comparator chip such as LM339 is designed to interface with a digital logic interface (to a TTL or a CMOS). The output is a binary state often used to interface real world signals to digital circuitry (see analog to digital converter). If there is a fixed voltage source from, for example, a DC adjustable device in the signal path, a comparator is just the equivalent of a cascade of amplifiers. When the voltages are nearly equal, the output voltage will not fall into one of the logic levels, thus analog signals will enter the digital domain with unpredictable results. To make this range as small as possible, the amplifier cascade is high gain. The circuit consists of mainly Bipolar transistors. For very high frequencies, the input impedance of the stages is low. This reduces the saturation of the slow, large P-N junction bipolar transistors that would otherwise lead to long recovery times. Fast small Schottky diodes, like those found in binary logic designs, improve the performance significantly though the performance still lags that of circuits with amplifiers using analog signals. Slew rate has no meaning for these devices. For applications in flash ADCs the distributed signal across eight ports matches the voltage and current gain after each amplifier, and resistors then behave as level-shifters.

The LM339 accomplishes this with an open collector output. When the inverting input is at a higher voltage than the non inverting input, the output of the comparator connects to the negative power supply. When the non inverting input is higher than the inverting input, the output is 'floating' (has a very high impedance to ground). The gain of op amp as comparator is given by this equation V(out)=V(in)

## Key specifications

While it is easy to understand the basic task of a comparator, that is, comparing two voltages or currents, several parameters must be considered while selecting a suitable comparator:

### Speed and power

While in general comparators are "fast," their circuits are not immune to the classic speed-power tradeoff. High speed comparators use transistors with larger aspect ratios and hence also consume more power. [6] Depending on the application, select either a comparator with high speed or one that saves power. For example, nano-powered comparators in space-saving chip-scale packages (UCSP), DFN or SC70 packages such as MAX9027, LTC1540, LPV7215, MAX9060 and MCP6541 are ideal for ultra-low-power, portable applications. Likewise if a comparator is needed to implement a relaxation oscillator circuit to create a high speed clock signal then comparators having few nano seconds of propagation delay may be suitable. ADCMP572 (CML output), LMH7220 (LVDS Output), MAX999 (CMOS output / TTL output), LT1719 (CMOS output / TTL output), MAX9010 (TTL output), and MAX9601 (PECL output) are examples of some good high speed comparators.

### Hysteresis

A comparator normally changes its output state when the voltage between its inputs crosses through approximately zero volts. Small voltage fluctuations due to noise, always present on the inputs, can cause undesirable rapid changes between the two output states when the input voltage difference is near zero volts. To prevent this output oscillation, a small hysteresis of a few millivolts is integrated into many modern comparators. [7] For example, the LTC6702, MAX9021 and MAX9031 have internal hysteresis desensitizing them from input noise. In place of one switching point, hysteresis introduces two: one for rising voltages, and one for falling voltages. The difference between the higher-level trip value (VTRIP+) and the lower-level trip value (VTRIP-) equals the hysteresis voltage (VHYST).

If the comparator does not have internal hysteresis or if the input noise is greater than the internal hysteresis then an external hysteresis network can be built using positive feedback from the output to the non-inverting input of the comparator. The resulting Schmitt trigger circuit gives additional noise immunity and a cleaner output signal. Some comparators such as LMP7300, LTC1540, MAX931, MAX971 and ADCMP341 also provide the hysteresis control through a separate hysteresis pin. These comparators make it possible to add a programmable hysteresis without feedback or complicated equations. Using a dedicated hysteresis pin is also convenient if the source impedance is high since the inputs are isolated from the hysteresis network. [8] When hysteresis is added then a comparator cannot resolve signals within the hysteresis band.

### Output type

Because comparators have only two output states, their outputs are either near zero or near the supply voltage. Bipolar rail-to-rail comparators have a common-emitter output that produces a small voltage drop between the output and each rail. That drop is equal to the collector-to-emitter voltage of a saturated transistor. When output currents are light, output voltages of CMOS rail-to-rail comparators, which rely on a saturated MOSFET, range closer to the rail voltages than their bipolar counterparts. [9]

On the basis of outputs, comparators can also be classified as open drain or push–pull. Comparators with an open drain output stage use an external pull up resistor to a positive supply that defines the logic high level. Open drain comparators are more suitable for mixed-voltage system design. Since the output is high impedance for logic level high, open drain comparators can also be used to connect multiple comparators to a single bus. Push-pull output does not need a pull up resistor and can also source current, unlike an open drain output.

### Internal reference

The most frequent application for comparators is the comparison between a voltage and a stable reference. Most comparator manufacturers also offer comparators in which a reference voltage is integrated on to the chip. Combining the reference and comparator in one chip not only saves space, but also draws less supply current than a comparator with an external reference. [9] ICs with wide range of references are available such as MAX9062 (200 mV reference), LT6700 (400 mV reference), ADCMP350 (600 mV reference), MAX9025 (1.236 V reference), MAX9040 (2.048 V reference), TLV3012 (1.24 V reference) and TSM109 (2.5 V reference).

### Continuous versus clocked

A continuous comparator will output either a "1" or a "0" any time a high or low signal is applied to its input and will change quickly when the inputs are updated. However, many applications only require comparator outputs at certain instances, such as in A/D converters and memory. By only strobing a comparator at certain intervals, higher accuracy and lower power can be achieved with a clocked (or dynamic) comparator structure, also called a latched comparator. Often latched comparators employ strong positive feedback for a "regeneration phase" when a clock is high, and have a "reset phase" when the clock is low. [10] This is in contrast to a continuous comparator, which can only employ weak positive feedback since there is no reset period.

## Applications

### Null detectors

A null detector identifies when a given value is zero. Comparators are ideal for null detection comparison measurements, since they are equivalent to a very high gain amplifier with well-balanced inputs and controlled output limits. The null detector circuit compares two input voltages: an unknown voltage and a reference voltage, usually referred to as vu and vr. The reference voltage is usually on the non-inverting input (+), while the unknown voltage is usually on the inverting input (). (A circuit diagram would display the inputs according to their sign with respect to the output when a particular input is greater than the other.) Unless the inputs are nearly equal (see below), the output is either positive or negative, for example ±12 V. In the case of a null detector the aim is to detect when the input voltages are nearly equal, which gives the value of the unknown voltage since the reference voltage is known.

When using a comparator as a null detector, accuracy is limited; an output of zero is given whenever the magnitude of the voltage difference multiplied by the gain of the amplifier is within the voltage limits. For example, if the gain is 106, and the voltage limits are ±6 V, then an output of zero will be given if the voltage difference is less than 6 μV. One could refer to this as a fundamental uncertainty in the measurement. [11]

### Zero-crossing detectors

For this type of detector, a comparator detects each time an ac pulse changes polarity. The output of the comparator changes state each time the pulse changes its polarity, that is the output is HI (high) for a positive pulse and LO (low) for a negative pulse squares the input signal. [12]

### Relaxation oscillator

A comparator can be used to build a relaxation oscillator. It uses both positive and negative feedback. The positive feedback is a Schmitt trigger configuration. Alone, the trigger is a bistable multivibrator. However, the slow negative feedback added to the trigger by the RC circuit causes the circuit to oscillate automatically. That is, the addition of the RC circuit turns the hysteretic bistable multivibrator into an astable multivibrator. [13]

### Level shifter

This circuit requires only a single comparator with an open-drain output as in the LM393, TLV3011 or MAX9028. The circuit provides great flexibility in choosing the voltages to be translated by using a suitable pull up voltage. It also allows the translation of bipolar ±5 V logic to unipolar 3 V logic by using a comparator like the MAX972. [9]

### Analog-to-digital converters

When a comparator performs the function of telling if an input voltage is above or below a given threshold, it is essentially performing a 1-bit quantization. This function is used in nearly all analog to digital converters (such as flash, pipeline, successive approximation, delta-sigma modulation, folding, interpolating, dual-slope and others) in combination with other devices to achieve a multi-bit quantization. [14]

### Window detectors

Comparators can also be used as window detectors. In a window detector, a comparator is used to compare two voltages and determine whether a given input voltage is under voltage or over voltage.

### Absolute value detectors

Comparators can be used to create absolute value detectors. In an absolute value detector, two comparators and a digital logic gate are used to compare the absolute values of two voltages. [15]

## See also

This article incorporates  public domain material from the General Services Administration document "Federal Standard 1037C" .

## Related Research Articles

An amplifier, electronic amplifier or (informally) amp is an electronic device that can increase the power of a signal. It is a two-port electronic circuit that uses electric power from a power supply to increase the amplitude of a signal applied to its input terminals, producing a proportionally greater amplitude signal at its output. The amount of amplification provided by an amplifier is measured by its gain: the ratio of output voltage, current, or power to input. An amplifier is a circuit that has a power gain greater than one.

An operational amplifier is a DC-coupled high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output. In this configuration, an op-amp produces an output potential that is typically hundreds of thousands of times larger than the potential difference between its input terminals. Operational amplifiers had their origins in analog computers, where they were used to perform mathematical operations in many linear, non-linear, and frequency-dependent circuits.

In electronics a relaxation oscillator is a nonlinear electronic oscillator circuit that produces a nonsinusoidal repetitive output signal, such as a triangle wave or square wave. The circuit consists of a feedback loop containing a switching device such as a transistor, comparator, relay, op amp, or a negative resistance device like a tunnel diode, that repetitively charges a capacitor or inductor through a resistance until it reaches a threshold level, then discharges it again. The period of the oscillator depends on the time constant of the capacitor or inductor circuit. The active device switches abruptly between charging and discharging modes, and thus produces a discontinuously changing repetitive waveform. This contrasts with the other type of electronic oscillator, the harmonic or linear oscillator, which uses an amplifier with feedback to excite resonant oscillations in a resonator, producing a sine wave. Relaxation oscillators are used to produce low frequency signals for applications such as blinking lights and electronic beepers and in voltage controlled oscillators (VCOs), inverters and switching power supplies, dual-slope analog to digital converters, and function generators.

A differential amplifier is a type of electronic amplifier that amplifies the difference between two input voltages but suppresses any voltage common to the two inputs. It is an analog circuit with two inputs and and one output in which the output is ideally proportional to the difference between the two voltages

A preamplifier is an electronic amplifier that converts a weak electrical signal into an output signal strong enough to be noise-tolerant and strong enough for further processing, or for sending to a power amplifier and a loudspeaker. Without this, the final signal would be noisy or distorted. They are typically used to amplify signals from analog sensors such as microphones and pickups. Because of this, the preamplifier is often placed close to the sensor to reduce the effects of noise and interference.

In electronics, a Schmitt trigger is a comparator circuit with hysteresis implemented by applying positive feedback to the noninverting input of a comparator or differential amplifier. It is an active circuit which converts an analog input signal to a digital output signal. The circuit is named a "trigger" because the output retains its value until the input changes sufficiently to trigger a change. In the non-inverting configuration, when the input is higher than a chosen threshold, the output is high. When the input is below a different (lower) chosen threshold the output is low, and when the input is between the two levels the output retains its value. This dual threshold action is called hysteresis and implies that the Schmitt trigger possesses memory and can act as a bistable multivibrator. There is a close relation between the two kinds of circuits: a Schmitt trigger can be converted into a latch and a latch can be converted into a Schmitt trigger.

A buffer amplifier is one that provides electrical impedance transformation from one circuit to another, with the aim of preventing the signal source from being affected by whatever currents that the load may be produced with. The signal is 'buffered from' load currents. Two main types of buffer exist: the voltage buffer and the current buffer.

Almost all integrated circuits (ICs) have at least two pins that connect to the power rails of the circuit in which they are installed. These are known as the power-supply pins. However, the labeling of the pins varies by IC family and manufacturer.

Delta-sigma modulation is a method for encoding analog signals into digital signals as found in an analog-to-digital converter (ADC). It is also used to convert high bit-count, low-frequency digital signals into lower bit-count, higher-frequency digital signals as part of the process to convert digital signals into analog as part of a digital-to-analog converter (DAC).

This article illustrates some typical operational amplifier applications. A non-ideal operational amplifier's equivalent circuit has a finite input impedance, a non-zero output impedance, and a finite gain. A real op-amp has a number of non-ideal features as shown in the diagram, but here a simplified schematic notation is used, many details such as device selection and power supply connections are not shown. Operational amplifiers are optimised for use with negative feedback, and this article discusses only negative-feedback applications. When positive feedback is required, a comparator is usually more appropriate. See Comparator applications for further information.

Crossover distortion is a type of distortion which is caused by switching between devices driving a load. It is most commonly seen in complementary, or "push-pull", Class-B amplifier stages, although it is occasionally seen in other types of circuits as well.

A constant fraction discriminator (CFD) is an electronic signal processing device, designed to mimic the mathematical operation of finding a maximum of a pulse by finding the zero of its slope. Some signals do not have a sharp maximum, but short rise times .

A flash ADC is a type of analog-to-digital converter that uses a linear voltage ladder with a comparator at each "rung" of the ladder to compare the input voltage to successive reference voltages. Often these reference ladders are constructed of many resistors; however, modern implementations show that capacitive voltage division is also possible. The output of these comparators is generally fed into a digital encoder, which converts the inputs into a binary value.

The operational transconductance amplifier (OTA) is an amplifier whose differential input voltage produces an output current. Thus, it is a voltage controlled current source (VCCS). There is usually an additional input for a current to control the amplifier's transconductance. The OTA is similar to a standard operational amplifier in that it has a high impedance differential input stage and that it may be used with negative feedback.

In the field of electronics, a bootstrap circuit is one where part of the output of an amplifier stage is applied to the input, so as to alter the input impedance of the amplifier. When applied deliberately, the intention is usually to increase rather than decrease the impedance. Generally, any technique where part of the output of a system is used at startup is described as bootstrapping.

A fully differential amplifier (FDA) is a DC-coupled high-gain electronic voltage amplifier with differential inputs and differential outputs. In its ordinary usage, the output of the FDA is controlled by two feedback paths which, because of the amplifier's high gain, almost completely determine the output voltage for any given input.

A log amplifier is an amplifier for which the output voltage Vout is K times the natural log of the input voltage Vin. This can be expressed as,

The input offset voltage is a parameter defining the differential DC voltage required between the inputs of an amplifier, especially an operational amplifier (op-amp), to make the output zero.

The Millers theorem refers to the process of creating equivalent circuits. It asserts that a floating impedance element, supplied by two voltage sources connected in series, may be split into two grounded elements with corresponding impedances. There is also a dual Miller theorem with regards to impedance supplied by two current sources connected in parallel. The two versions are based on the two Kirchhoff's circuit laws.

A comparator is an electronic component that compares two input voltages. Comparators are closely related to operational amplifiers, but a comparator is designed to operate with positive feedback and with its output saturated at one power rail or the other. An op-amp can be pressed into service as a poorly performing comparator if necessary, but its slew rate will be impaired.

## References

1. LM111/LM211/LM311 datasheet. Texas Instruments. August 2003. Retrieved 2014-07-02.
2. LM139/LM239/LM339/LM2901/LM3302 datasheet. Texas Instruments. August 2012. Retrieved 2014-07-02.
3. LMH7322 datasheet. Texas Instruments. March 2013. Retrieved 2014-07-02.
4. Malmstadt, Enke and Crouch, Electronics and Instrumentation for Scientists, The Benjamin/Cummings Publishing Company, Inc., 1981, ISBN   0-8053-6917-1, Chapter 5.
5. Ron Mancini, "Designing with comparators," EDN, March 29, 2001.
6. Rogenmoser, R.; Kaeslin, H, "The impact of transistor sizing on power efficiency in submicron CMOS circuits," Solid-State Circuits, IEEE Journal of Volume 32, Issue 7, Jul 1997 Page(s):1142–1145.
7. Ron Mancini, "Adding Hysteresis to comparators Archived 2005-02-21 at the Wayback Machine ," EDN, May 3, 2001.
8. AN3616, Maxim Integrated Products, Adding Extra Hysteresis to Comparators.
9. AN886, Maxim Integrated Products, Selecting the Right Comparator.
10. Pedro M. Figueiredo, João C. Vital (2009). Offset Reduction Techniques in High-Speed Analog-to-Digital Converters: Analysis, Design and Tradeoffs. Springer. pp. 54–62. ISBN   978-1-4020-9715-7.
11. Malmstadt, Howard V.; Enke, Christie G.; Crouch, Stanley R. (January 1981), , The Benjamin/Cummings Publishing Co, pp.  108–110, ISBN   978-0-8053-6917-5
12. Electronics and Instrumentation for Scientists. Malmstadt, Enke, and Crouch, The Benjamin/Cummings Publishing Co., In., 1981, p.230.
13. Paul Horowitz and Winfield Hill: The Art of Electronics, Cambridge University Press, Second edition, Cambridge 1989, pp.284–285.
14. Phillip Allen and Douglas Holberg: CMOS Analog Circuit Design, Oxford University Press, Second edition, Oxford 2002.
15. "Iranmanesh, S., Rodriguez-Villegas, E. (2016). CMOS implementation of a low power absolute value comparator circuit. IEEE NEWCAS, June 2016". doi:10.1109/NEWCAS.2016.7604807.Cite journal requires |journal= (help)


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# Euler and Lefschetz numbers

(Redirected from Euler-Poincare formula)

## 1 The Euler characteristic

Recall the Euler formula for a convex polyhedron: $$\# {\rm vertices} - \# {\rm edges} + \# {\rm faces} = 2.$$

A remarkable result, especially when you realize that it has nothing to do with convexity! Indeed, we can simply turn protrusions into indentations without changing this number:

We already know, and will prove below, that the meaning of the formula is topological.

Definition. The Euler characteristic $\chi (K)$ of an $n$-dimensional cell complex $K$ is the alternating sum of the number of cells in $K$ for each dimension: $$\chi (K) = \# 0 \text{-cells} - \# 1\text{-cells} + \# 2\text{-cells} - ... \pm \# n\text{-cells}.$$

Exercise. Compute the Euler characteristic of the circle, the cylinder, the Mobius band, the torus, and the Klein bottle.

To reveal the topological nature of this number, we will need to link it to homology. This will require a more formal, algebraic approach which will prove to be very powerful when applied to the Euler characteristic and related concepts. In particular, counting cells in the definition is replaced with the following linear algebra.

Proposition. If $C_k(K)$ is the $k$-chain group of cell complex $K$, then $$\# k \text{-cells} = \dim C_k(K).$$ Consequently, $$\chi (K) = \sum_{k} (-1)^k \dim C_k(K).$$

We assume that all chains and homology below are over the reals.

We will need this well-known result.

Theorem (Inclusion-Exclusion Formula). For sets $A,B \subset X$, we have: $$\#(A \cup B) = \# A + \# B - \#(A \cap B).$$

It has an analogue for the Euler characteristic. But we need to apply this formula to bases of the vector spaces (rather than the cells).

Proposition. For two subspaces $U,V$ of a finitely dimensional vector spaces, we have $$\dim < U \cup V > = \dim U + \dim V - \dim U \cap V.$$

Exercise. Prove the proposition. Hint: $\dim W=\#$ basis of $W$.

Theorem (Inclusion-Exclusion Formula for Euler Characteristic). Suppose $K, L,$ and $K \cap L$ are subcomplexes of finite cell complex $K \cup L$, then $$\chi (K \cup L) = \chi (K) + \chi (L) - \chi (K \cap L).$$

Proof. By the above proposition, we have: $$\begin{array}{ccc} \dim < C_0(K) \cup C_0(L) > = \dim C_0(K) + \dim C_0(L) - \dim C_0(K) \cap C_0(L), \\ \dim < C_1(K) \cup C_1(L) > = \dim C_1(K) + \dim C_1(L) - \dim C_1(K) \cap C_1(L), \\ \vdots \\ \dim < C_n(K) \cup C_n(L) > = \dim C_n(K) + \dim C_n(L) - \dim C_n(K) \cap C_n(L). \\ \end{array}$$ Since for each $i=0,1,...,n$, we have $$\begin{array}{cccc} < C_i(K) \cup C_i(L) > & = C_i(K \cup L), \\ C_i(K) \cap C_i(L) & = C_i(K \cap L), \end{array}$$ the alternating sum of the above equations gives us the formula. $\blacksquare$

The Euler characteristic behaves well under products.

Theorem (Product Formula for Euler Characteristic). The Euler characteristic is multiplicative. For two finite cell complexes $K$ and $L$, we have: $$\chi (K \times L) = \chi (K) \cdot \chi(L).$$

Exercise. Prove the theorem. Hint: $\dim (U\oplus V)=\dim U + \dim V$.

Exercise. Compute the Euler characteristic of the $n$-dimensional torus ${\bf T}^n$.

## 2 The Euler-Poincaré Formula

The Euler characteristic of a cell complex is computed from: $$c_k(K) := \dim C_k(K).$$ These numbers resemble the Betti numbers of the cell complex: $$\beta_k(K) := \dim H_k(K).$$ What if we compute an alternating sum of the latter numbers rather than the former?

Implicitly, we have already done this, in the Euler formula. On the one hand, we have: $$\chi ({\bf S}^2) = 2,$$ which holds regardless of the triangulation. On the other, the alternating sum of the Betti numbers of the sphere $$\beta_0({\bf S}^2) = 1,\ \beta_1({\bf S}^2) = 0,\ \beta_2({\bf S}^2) = 1,\ \beta_3({\bf S}^2) = 0,...$$ also produces $2$!

Is this a coincidence? Let's check out other complexes.

Example (point). First, the point $P$: $$\begin{array}{llll} \chi(P) = 1 - 0 = 1,\\ \beta_0(P) - \beta_1(P) = 1-0=0. \end{array}$$ Once again, the Betti numbers add up to the Euler number.

Next, the circle $C$: $$\begin{array}{llll} \chi(C) = 1 - 1 + 0 = 0,\\ \beta_0(C) - \beta_1(C) +\beta_2(C) = 1-1+0= 0! \end{array}$$ $\square$

Example (torus). Consider the torus ${\bf T}^2$. We have:

• $0$-cells: $1$,
• $1$-cells: $2$,
• $2$-cells: $1$.

So, $$\chi({\bf T}^2) = 1 - 2 + 1 = 0.$$ Because this cell complex representation is so efficient, each of the cells also represents an element of a basis element of the corresponding homology group: $$\begin{array}{lll} H_0({\bf T}^2) = < A > & \Rightarrow \beta_0({\bf T}^2) = 1, \\ H_1({\bf T}^2) = < a,b > & \Rightarrow \beta_1({\bf T}^2) = 2, \\ H_2({\bf T}^2) = < \tau > & \Rightarrow \beta_2({\bf T}^2) = 1, \\ H_3({\bf T}^2) = 0 & \Rightarrow \beta_3({\bf T}^2) = 0. \end{array}$$ The formula holds: $$\chi({\bf T}^2) = \beta_0(C) - \beta_1(C) + \beta_2(C) - ...$$ $\square$

Since we are just counting the cells, it's no surprise that the formula holds. Of course, it will always hold when there is one cell per homology generator. We need to show that this formula holds for all cell complexes.

Theorem (Euler-Poincaré Formula). If $K$ is a finite cell complex, then its Euler characteristic is equal to the alternating sum of the Betti numbers of each dimension: $$\chi(K) = \displaystyle\sum_k(-1)^k \beta_k(K).$$

Proof. The proof will require a couple of facts from linear algebra.

Proposition.

• (1) If $M, L$ are vector spaces and $A: M \to L$ is a linear operator, then

$$M /\ker A \simeq \operatorname{Im} A.$$

• (2) If $Y$ is a subspace of vector space $X$, then

$$\dim X / Y = \dim X - \dim Y.$$

The proposition implies the following:

Corollary. If $M, L$ are vector spaces and $A: M \to L$ is a linear operator, then $$\dim M - \dim \ker A = \dim \operatorname{Im} A.$$

Now, there are four vector spaces involved in the computation of the Betti numbers of complex $K$, for each $k=0,1,2,...$: $$\begin{array}{lllll} \text{chain group: } & C_k , & c_k := \dim C_k;\\ \text{cycle group: } & Z_k = \ker \partial_k, & z_k := \dim Z_k;\\ \text{boundary group: } & B_k =\operatorname{Im} \partial_{k+1}, & b_k := \dim B_k;\\ \text{homology group: } & H_k = Z_k / B_k, & \beta_k := \dim H_k, \end{array}$$ where $$\partial_k : C_k \to C_{k-1}$$ is the boundary operator.

Applying the corollary to the boundary operator, we obtain: $$\dim C_k - \dim \ker \partial_k = \dim \operatorname{Im} \partial_k.$$ In other words, we have the following.

Lemma A. $c_k - z_k = b_{k-1}.$

Next, from part (2) of the proposition, we have the following.

Lemma B. $\beta_k = z_k - b_k.$

Finally, suppose $n$ is the highest dimension of a cell in $K$. Then the right-hand side of the formula is: $$\begin{array}{llllll} = &\beta_0 &- \beta_1 &+ \beta_2 - ... &+ (-1)^k \beta_n\\ &&&\text{ substitute from Lemma B} \\ = &(z_0 - b_0) &- (z_1 - b_1) &+(z_2 - b_2) -... &+ (-1)^n(z_n - b_n) \end{array}$$ $$\begin{array}{llllll} = &z_0 &- b_0 - z_1 &+b_1 +z_2 - ... &+ (-1)^n z_n - (-1)^n b_n\\ &&&\text{ substitute from Lemma A}\\ = &z_0 &- (c_1 - z_1) - z_1 &+ (c_2 - z_2) +z_2 + ... &+ (-1)^n z_n - (-1)^n(c_{n+1} - z_{n+1})\\ &&&\text{ cancel the z's} \\ = &z_0 &- c_1 &+ c_2 - ... & + (-1)^{n+1}c_{n+1} - (-1)^n z_{n+1} \\ = &c_0 &- c_1 &+ c_2 - ... &+ 0 &- 0 \\ = &\chi(K). \end{array}$$

$\blacksquare$

Exercise. Prove the propositions.

The invariance of homology then gives us the following.

Corollary. The Euler characteristic is a topological invariant of polyhedra.

Because so much information is lost, the Euler characteristic is poor man's homology; however, ...

Exercise. Under what circumstances can the homology of $X$ be derived from its Euler characteristic for (a) $X$ a graph, (b) $X$ a surface?

## 3 Fixed points

If you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position:

And, it doesn't matter if the stretching isn't uniform or if there is some shrinking, or even folding! Such a point is called a fixed point of the transformation.

This problem is described mathematically as follows.

Fixed Point Problem. If $X$ is a topological space and $f:X \to X$ is a self-map, does $f$ have a fixed point: $x\in X$ such that $f(x)=x$?

The result above is, therefore, about a fixed point of a map $f:[a,b] \to [a,b]$.

Exercise. Prove this result by using the Intermediate Value Theorem.

Exercise. Demonstrate that both continuity of the map and the compactness of the interval are crucial.

Exercise. Under what circumstances is the set of fixed points closed?

A similar problem about stretching a disk or a ball requires more advanced techniques (to be proven later).

Theorem (Brouwer Fixed Point Theorem). Any map on a closed ball has a fixed point. In other words, if $f: {\bf B}^n\to {\bf B}^n$ is a map then there is $x\in {\bf B}^n$ such that $f(x)=x$.

An informal interpretation of the $2$-dimensional version of this theorem is: a crumpled copy of a page placed on it will have two identical words located one exactly over the other.

Exercise. Prove the theorem. Hint:

Of course, the theorem holds if the ball is replaced with anything homeomorphic to it, such a simplex.

Though technically not a fixed point theorem, the following result is closely related to the last theorem.

Theorem (Borsuk-Ulam Theorem). Every map $f:{\bf S}^n \to {\bf R}^n$ from an $n$-sphere to $n$-dimensional Euclidean space maps some pair of antipodal points to the same point: $$f(x)=f(-x).$$

A common interpretation of the case $n=1$ is that at any moment there is always a pair of antipodal points on the Earth's surface with exactly equal temperatures, or with equal barometric pressures.

Furthermore, an interpretation for $n=2$ is that there is always a pair of antipodal points with exactly equal temperatures and equal barometric pressures, simultaneously.

Exercise. Explain how the theorem implies that at any moment there is always a pair of antipodal points on the Earth's surface with identical wind.

Exercise. Derive the case $n=1$ of the theorem from the Intermediate Value Theorem.

Exercise. Prove the theorem. Hint: consider this map: $$g(x)=\frac{f(x)-f(-x)}{||f(x)-f(-x)||}.$$

## 4 An equilibrium of a market economy

Suppose we have a map $F:S \to S$. Applied consecutively, it creates a discrete dynamical system: $$X_{k+1}=F(X_k),$$ where $X_k \in S$ is the current state, at time $k$, of the system if it started at time $k=0$ at the state $X_0\in S$. Then the meaning of the equilibrium of the system is simple: $F(a)=a$. It's a fixed point!

A typical interpretation of such a system is a particle moving through a stream. We will consider instead the dynamics of prices in a simple market economy. Is it possible for them to stay constant? In other words, does every price dynamics allow an equilibrium?

Recall that we have $n$ commodities freely traded with possible prices that form a price vector $p=(p_1,...,p_n),$ at each moment of time. Then all positive, or even non-negative, prices are considered possible, i.e., the set of prices is $$P:={\bf R}_+^{n}.$$ The dynamics of prices is captured by a continuous function $F:P\to P$. However, we know that this isn't enough to conclude that there is a fixed point since $P$ isn't compact!

Let's make some additional assumptions in order to be able to guarantee an equilibrium.

How can we make the set of prices compact? One approach is to normalize the prices: $$p\mapsto \frac{p}{||p||},$$ under some norm so that the new prices form the $n$-simplex $\sigma$.

Exercise. Show that this is an equivalence relation. What assumption would guarantee that the price function is well-defined?

A more natural (if not more realistic) assumption is that the money supply is fixed. Suppose there are $n$ commodities. We assume that this is a pure exchange economy, i.e., there is no production of goods (other than making up for possible consumption). Then, there is only a fixed amount of each of the commodities: $$\bar{c}:=\sum_i c^i \in {\bf R}_+^n.$$ Then, the total cost of all the commodities, i.e., the total wealth $W>0$, is fixed too, for any combination of prices $p$: $$\langle p,c \rangle=W.$$ Then the space of prices can be set to be $$P:=\{p\in {\bf R}_+^{n}:\langle p, c \rangle =W\}.$$

Proposition. The set $P$ of prices with a fixed money supply is a convex polyhedron.

Exercise. Prove the proposition. What kind of polyhedron is it?

Therefore, the price map $F:P\to P$ has a fixed point by the Brouwer theorem.

Proposition. With a fixed money and commodity supply, a continuous price dynamics of an exchange economy has an equilibrium.

We have proven the existence of an equilibrium: there is such a combination of prices that, once acquired, will remain unchanged and all trading will happen in perfect harmony, forever... In reality, the outside events will be changing the function $D$ itself.

Exercise. What difference would price controls have?

Of course, this simple topological analysis won't tell us what this special combination of prices is, or how many of these are possible, or whether one of them will be reached (or even approached) from the current state.

Next, we consider the trading dynamics in this model. There are $m$ “agents” in the market and $i$th agent owns $c_j^i$ of $j$th commodity. These amounts form the $i$th commodity vector $c^i \in {\bf R}_+^n.$ The dynamics of distribution of commodities is captured by a continuous function $G:C\to C,$ where $$C:=\{(c^1,...,c^m):\ c^i\in {\bf R}_+^n\} = \left( {\bf R}_+^n \right)^m.$$ Once again, we know that this isn't enough to conclude that there is a fixed point since $C$ isn't compact!

We have to assume again that there is no production or consumption. Then, we can set instead: $$C:=\{(c^1,...,c^m):\ c^i\in {\bf R}_+^n,\ \sum_i c^i=\bar{c} \}.$$

Proposition. The set $C$ of commodities with a fixed commodity supply is a convex polyhedron.

Exercise. Prove the proposition. What kind of polyhedron is it?

Exercise. What if one is allowed to borrow?

Once again, there is a fixed point by the Brouwer theorem.

Proposition. With a fixed commodity supply, a continuous commodity dynamics of an exchange economy has an equilibrium.

In other words, there might be no trading at all!

We can combine these variables by defining the state of the system to be a combination of the prices and a distribution of the commodities. Then the state space is $$S:=P\times C,$$ and the state dynamics is given by a continuous function $D:S\to S$. Therefore, $S=P\times C$ is also a convex polyhedron and the map $D:S\to S$ has a fixed point by the Brouwer theorem.

How can we incorporate other quantities into this model? Any new quantity can be added to this model by simply extending the state space $S$ via the product construction. And there still be an equilibrium as long as:

• the next value of this quantity is determined by the circumstances not time,
• this dependence is continuous, and
• the range of its values is a priori bounded.

Exercise. What happens if we add interest rates to this model?

## 5 Chain maps of self-maps

We are to study fixed points of a map $f:|K|\to |K|$ by considering its homology map $f_*:H(K)\to H(K)$. However, the latter is built from the former via a simplicial approximation $g:K'\to K$, where $K'$ is an appropriate subdivision of $K$. However, $g$ cannot possibly be a self-map!

The invariance of homology can be restated as follows.

Proposition. Suppose $K'$ is a subdivision of a simplicial complex $K$, then $H(K') \cong H(K)$.

Then, two cycles are homologous in $K'$ if and only if their “counterparts” in $K$ are also homologous. What is the correspondence between the finer cycles in $K'$ and the coarser cycles in $K$?

It is easy to find the counterpart in $K^m$ of a chain in $K$. We define the subdivision map $$s:K\to K'$$ on each $k$-simplex $x$ in $K$ by $$s(x):=\sum_ix_i,$$ with $x$ subdivided into the collection of $k$-simplices $\{x_i\}\subset K'$ oriented compatibly with each other and $x$.

Proposition. The subdivision map extended linearly to chains, $$s:C(K)\to C(K'),$$ is a chain map.

Exercise. Prove the proposition.

A simplicial approximation of the identity can be constructed as follows. Suppose, initially, that $K$ is an $n$-simplex and $K'$ is its barycentric subdivision. We define $I:K'\to K$ by first assigning to a vertex in $K'$ a vertex in $K$ following the rule: if vertex $A$ in $K'$ is the barycenter of simplex $A_0...A_m$ in $K$, we let $I(A)$ to be one of these vertices.

Proposition. There is exactly one $n$-simplex of $K'$ that is cloned (to the whole $K$) by $I$.

Proof. Suppose $C\in K'$ is the barycenter of the $n$-simplex $K$ and $I(C)=B\in K$. Then there is a face $\tau = B_0...B_{n-1}$ of $K$ the vertices of which are taken by $I$ to other vertices: $I(B_i)\ne B$. Next, let $C$ be the barycenter of the face $\tau$ and repeat the process. $\blacksquare$

Exercise. (a) Prove the uniqueness. (b) Show how the construction and the proof applies to the general case of a simplicial complex $K$ and its subdivision $K'=K^p$.

Theorem. The subdivision map generates the identity map on homology, $$s_*=\operatorname{Id}:H(K)\to H(K').$$

Proof. From the last proposition it follows that $I_{\Delta}s=\operatorname{Id}:C(K)\to C(K)$. $\blacksquare$

Thus, every self-map $f:|K|\to |K|$ does indeed have a matching chain self-map $sf:C(K')\to C(K')$ in the sense that $f_*=s_*f_*:H(K)\to H(K)$.

## 6 The Lefschetz number

The analogue of a fixed point free map is a cell map that is fixed cell free. It is a self-map $g:K \to K$ of a cell complex $K$ with $$g(s)\ne s, \ \forall s\in K.$$ What does it mean algebraically? Let's consider the chain maps of $g$, $$g_k:C_k(K)\to C_k(K), \ k=0,1,2,....$$ Then the above condition simply means that the diagonal elements of the matrix $A$ of $g_k$ are all zero! It follows, in particular, that the trace of this matrix, i.e., the sum of its diagonal elements: $$\operatorname{Tr}(A):=\sum_{i}A_{ii},$$ is also zero. We develop this idea further and convert it fully from the chain level to the homology level by means of a generalization of the Euler-Poincare formula.

Definition. The Lefschetz number of a self-map $f:S \to S$ of polyhedron $S$ is the alternating sum of the traces of the homology maps $[f_k]:H_{k}(S) \to H_{k}(S)$ of $f$: $$\Lambda (f):=\sum_{k}(-1)^{k}\operatorname{Tr}([f_k]).$$

With the alternating sum over the dimensions of the space, the formula resembles the Euler characteristic. This is not a coincidence: the latter is equal to the Lefschetz number of the identity function of $S$: $$\chi (S) = \Lambda (\operatorname{Id}_S).$$ This fact follows from the simple observation: $$\operatorname{Tr} (\operatorname{Id}:{\bf R}^n \to {\bf R}^n) = n.$$

Exercise. (a) Compute the Lefschetz numbers of $\delta P$ and $P \delta$, where $P:{\bf S}^2\to {\bf S}^2$ is the projection and $\delta:{\bf S}^2 \to {\bf S}^2 \times {\bf S}^2$ is the diagonal map. (b) Replace ${\bf S}^2$ with $M$, a path-connected compact surface.

The main result is below.

Theorem (Lefschetz Fixed Point Theorem). If a map $f:S \to S$ of a polyhedron $S$ has non-zero Lefschetz number, $$\Lambda (f) \ne 0,$$ then $f$ has a fixed point, $f(a)=a$, and so does any map homotopic to $f$.

If the polyhedron $S$ happens to be acyclic, we have $$[f_0]=\operatorname{Id}, \ [f_1]=0,\ [f_2]=0,\ ...,$$ and, therefore, $\Lambda (f) =1\ne 0$. The Brouwer Fixed Point Theorem follows.

Corollary. Suppose $S$ is a polyhedron with $\chi (S)\ne 0$ (such as a sphere) and a map $f:S \to S$ is homotopic to the identity, $f\simeq \operatorname{Id}_S$. Then $f$ has a fixed point.

The Lefschetz number only tells us that the set of fixed points is non-empty. The converse of course isn't true.

Example. Let $f$ be the flip of the figure eight, ${\bf S}^1 \vee {\bf S}^1$, one loop onto the other. Then $$[f_0]=1,\ [f_1]=\left[ \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right],\ [f_2]=0,...$$ Therefore, $$\Lambda (f) =1-0=1.$$ Not only there is a fixed point, as we can see, but also any map homotopic to $f$ has a fixed point.

$\square$

Example. We consider the flip of the “kissing spheres” space, ${\bf S}^2 \vee {\bf S}^2$, one sphere onto the other:

Then $$[f_0]=1,\ [f_1]=0,\ [f_2]=\left[ \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right],\ [f_3]=0,....$$ Therefore, $$\Lambda (f) =1-0+0=1.$$

$\square$

Exercise. Consider the flip of the “double banana”.

Exercise. Consider the other flip of the figure eight, the “kissing spheres”, and the “double banana”.

Exercise. Suggest a map with $\Lambda (f) \ne 0$ but $\Lambda (f)=0$ modulo $2$.

The beginning of the proof of the Lefschetz theorem is similar to that of the Euler-Poincare formula, except, we will manipulate the traces of linear operators rather than the dimensions of subspaces (after all $\dim V=\operatorname{Tr}(\operatorname{Id}_V)$). We rely on the following fact from linear algebra.

Exercise. Prove the lemma.

Exercise. Show that the lemma implies the proposition in subsection 1.

Next, we apply this lemma, for each $k=0,1,2,...$, to

• the chain map $f_k:C_k \to C_k$,
• its restriction $f^Z_k$ to the subspace of cycles $Z_k$,
• its restriction $f^B_k$ to the subspace of boundaries $B_k$, and
• the homology map $[f_k]$ induced by $f$.

Exercise. Apply the lemma to $$U=Z_k \hookrightarrow V=C_k \xrightarrow{\quad \partial_k \quad} W=B_{k-1}.$$

Exercise. Apply the lemma to $$U=B_k \hookrightarrow V=Z_k \xrightarrow{\quad q \quad} W=H_k,$$ where $q$ is the quotient map.

The conclusions are: $$\operatorname{Tr} (f_k) = \operatorname{Tr} (f^Z_k) + \operatorname{Tr} (f^B_{k-1}),$$ $$\operatorname{Tr} (f_k^Z) = \operatorname{Tr} ([f_k]) + \operatorname{Tr} (f^B_{k}).$$ Next, we use these two identities to simplify the following alternating sum: $$\sum_k (-1)^k \left( \operatorname{Tr} (f_k) - \operatorname{Tr} (f_{k-1}^B) \right)=\sum_k (-1)^k \left( \operatorname{Tr} ([f_k]) + \operatorname{Tr} (f^B_{k}) \right).$$ The terms with $f^B$ in the left-hand side and the right-hand side cancel. The result is this generalization of the Euler-Poincare formula.

Theorem (Hopf Trace Formula). $$\sum_k (-1)^k \operatorname{Tr} (f_k) =\sum_k (-1)^k \operatorname{Tr} ([f_k]).$$

The result is purely algebraic. In fact, similar to the Euler-Poincare formula, the algebra of the formula is identical on the left and right but applied to two different sequences of linear operators. This observation justifies the definition of the Lefschetz number of an arbitrary sequence of linear (self-)operators $h=\{h_k:\ k=0,1,2...\}$, as follows: $$L(h):=\sum_k (-1)^k \operatorname{Tr} (h_k).$$ Then the trace formula becomes: $$L(h)=L([h]).$$ At this point, the latter is recognized as $\Lambda (f)$ if we choose $h=f_{\Delta}$.

Exercise. Show that $L$ is a linear operator.

Now we finish the proof of the fixed point theorem.

Suppose map $f:S\to S$ has no fixed points. Since $S$ is compact, there is an $\epsilon >0$ such that $$d(x,f(x))>\epsilon$$ for all $x\in S$. Suppose $S=|K|$ for some simplicial complex $K$. We subdivide $K$ enough times, $p$, so that $\operatorname{mesh}^*(K^p)<\epsilon$. Then $$d(\sigma,f(\sigma))>\operatorname{mesh}^*(K^p)$$ for all $\sigma \in K^p$.

We rename $K^p$ as $K$ and proceed to the next step. According to the Simplicial Approximation Theorem, we can acquire a simplicial approximation $g:K^m\to K$ of $f$: $$f(N_A)\subset N_{g(A)}$$ for any vertex $A$. We do so for finer and finer subdivisions of $K$ -- until we have $$\sigma \cap g(\sigma)=\emptyset$$ for all $\sigma \in K^m$. It follows that $$sg_{\Delta}(\sigma)\ne\sigma,$$ where $s$ is the subdivision chain map from the last subsection. Then, $$L(sg_{\Delta}):=\sum_k (-1)^k \operatorname{Tr} (s_kg_k) = \sum_k (-1)^k 0 = 0.$$ By the Hopf Trace Formula, we have $$\Lambda (f) :=\Lambda (g_*) =L (sg_{\Delta}) =0.$$

This finishes the proof of the Lefschetz theorem.

Exercise. Prove that $\Lambda (hfh^{-1}) = \Lambda (f)$ for any homeomorphism $h$.

## 7 The degree of a map

The degree is initially introduced for maps of the circle: $$f:{\bf S}^1 \to {\bf S}^1,$$ as the number of times the first circle wraps around the second.

The degree is also defined for maps $$f:{\bf S}^1 \to {\bf R}^2 \setminus \{0\},$$ as the “winding number”. It is computed by using a polar coordinate parametrization, then measuring the change in the angle, and setting $$\deg f:=\frac{\theta (1) -\theta (0)}{2\pi}.$$

We, instead, approach the idea homologically.

Let's consider the homology map induced by our map: $$[f_1]: {H}_1({\bf S}^1) \to {H}_1({\bf S}^1).$$ Since both of the groups are isomorphic to ${\bf Z}$, group theory tells us that such a homomorphism must be the multiplication by an integer. This integer is the degree of $f$.

The idea and the construction are now applied to a more general case.

Definition. Given a map between any two $n$-dimensional compact, path-connected manifolds $$f:M^n \to N^n,$$ the degree $\deg f$ of $f$ is the integer that satisfies: $$[f_n](x)=\deg f \cdot x,$$ where $$[f_n]: H_n(M^n) \cong {\bf Z} \to H_n(N^n)\cong {\bf Z}.$$ is the homology map of $f$.

Exercise. Prove the following.

• 1. The degree of the constant map is $0$.
• 2. The degree of the identity map is $1$.
• 3. The degree antipodal map of the $n$-sphere ${\bf S}^n \subset {\bf R}^{n+1}$ is $(-1)^{n+1}$.

We will accept the following without proof (Bredon, p. 187).

Theorem. The degree of a reflection of the $n$-sphere ${\bf S}^n \subset {\bf R}^{n+1}$ through an $(n+1)$-dimensional hyperplane containing $0$, such as $$f(x_1,x_2,...,x_n,x_{n+1})=(-x_1,x_2,...,x_n,x_{n+1}),$$ is $-1$.

Exercise. Any map $f: {\bf S}^n \to {\bf S}^n$ can be used as an attaching map of a cell complex. Suppose two such maps $f,g$ of degrees $k,m$ are used to attach two copies of ${\bf B}^{n+1}$ to ${\bf S}^n$. Compute the homology of the resulting cell complex for all $k,m$. Hint: start with $n=1,k=1,m=2$.

The following theorems are derived from the properties of homology maps.

Theorem. The degree is multiplicative under compositions, i.e., $$\deg(fg) = \deg(f)\cdot\deg(g).$$

Theorem. The degree is homotopy invariant, i.e., $$f\simeq g \Longrightarrow \deg(f) = \deg(g).$$

Theorem. The degree of a homeomorphism, or even a homotopy equivalence, is $\pm 1$.

Exercise. Prove the theorems.

The following is an easy computation.

Theorem. For a map $f: {\bf S}^n \to {\bf S}^n$ of spheres, we have $$\Lambda (f) = 1 + (-1)^{n+1}\deg f.$$

Corollary. A map $f: {\bf S}^n \to {\bf S}^n$ with $\deg f \ne (-1)^n$ has a fixed point.

It follows that a map of spheres not homotopic to a homeomorphism will have a fixed point.

The theorem below provides a partial solution of the problem of surjectivity.

Theorem. A map $f: {\bf S}^n \to {\bf S}^n$ of nonzero degree is onto.

Proof. If $a\in {\bf S}^n \setminus \operatorname{Im} f$, the image is contractible as a subset of a contractible set ${\bf S}^n \setminus \{a\}$, the latter fact proven via the stereographic projection. Therefore, $f$ is homotopic to the constant map and $\deg f =0$, a contradiction. $\blacksquare$

Notice that ${\bf T}^2 \setminus \{a\}$ isn't contractible and that's why the proof fails if applied to an arbitrary pair of manifolds. A more general result will be provided later.

Exercise. Provide details of the proof and then make both the proof and the statement of the theorem as general as possible.

Exercise. What is the converse of the theorem? Hint: don't try to prove it.

Let's recall also the argument we made about the solvability of the Extension Problem for maps of spheres.

Theorem. A map $f: {\bf S}^n =\partial {\bf B}^{n+1} \to {\bf S}^n$ of $n$-spheres of non-zero degree cannot be extended to a map on the whole ball, $F: {\bf B}^{n+1} \to {\bf S}^n$.

Exercise. Provide details of the proof and then make both the proof and the statement of the theorem as general as possible.

## 8 Zeros of vector fields

First, we will see what the Brouwer Fixed Point Theorem tells us about zeros of vector fields.

Suppose a vector field $V$ is defined on a subset $R\subset {\bf R}^n$ which is homeomorphic to ${\bf B}^m, \ m<n$. We know that $V$ determines a map $f$ but is it a self-map? If it is, the map will have a fixed point.

We know that we can just follow the direction of $V(x)$ to define $f(x)$ but how far? We will say that $V$ points inside $R$ if for every $x\in \partial R$ there is $\epsilon >0$ such that $\epsilon V(x) \in R$.

Exercise. Prove that, in that case, there is $\epsilon >0$ such that $\epsilon V(x) \in R$ for all $x\in R$.

Then $f(x):=\epsilon _x V(x)$ is well defined as a map $f:R\to R$. Therefore $f$ has a fixed point, $V$ has a zero, and, if $V$ defines an ODE, there is a stationary point.

We can do more with the Lefschetz Fixed Point Theorem, without requiring that $V$ points inside $R$.

We know that, if the forward propagation map $Q_t:R \to S$ of $V$ is well-defined for some $S\subset {\bf R}^n$, and some $t>0$, and $R$ is a deformation retract of $S$, then we have: $$rQ_t \simeq \operatorname{Id}_R,$$ where $r:S\to R$ is the retraction. We also know that $$\Lambda (rQ_t) = \Lambda (\operatorname{Id}_R) = \chi (R).$$ Therefore, when $\chi (R) \ne 0$, map $rQ_t$ has a fixed point by the corollary to the Lefschetz Fixed Point Theorem. Whether this means that $V$ has a zero depends on $r$.

It is easy to design a non-zero vector field for a circle. In fact, we can have a unit vector field:

We run into problems if try this idea for a sphere. If we are to assign vectors of equal length to points of equal latitude, we realize that their lengths would have to diminish as we approach the pole to assure continuity:

Theorem (Hairy Ball Theorem). A continuous vector field tangent to the even-dimensional sphere has a zero.

For dimension $2$, it reads: “you can't comb the hair on a coconut”.

Exercise. Prove the theorem.

Example (cyclone). Another commonly known interpretation of the theorem is the following. The atmosphere forms a layer around the Earth and the air moves within this layer.

Assuming that this layer is “thin”, what we have is a sphere. Therefore, the vector field of the wind has a stationary point, which is a point where the wind speed is $0$. That point might be a cyclone.

$\square$

What if next we want to be able to distinguish the following behaviors of vector fields:

We can consider only the behavior of the vector field $V$ on the boundary of the neighborhood. In the $2$-dimensional case, we can simply evaluate the rotation of $V$ along this curve. It is $0$ for the first one and $2\pi$ for the rest of them. This approach to detecting zeros applies to every vector field with no zeros on the boundary. Problem solved!

What about higher dimensions? The boundary of an $n$-ball is an $(n-1)$-sphere, not a curve. What is the analog of the rotation then?

A convenient way to compute the rotation of a vector field $V:{\bf B}^2 \to {\bf R}^2$ with $V(u) \ne 0, \ \forall u\in \partial {\bf B}^2$, is to normalize it first: $$\phi(u):=\frac{V(u)}{||V(u)||}.$$ The result is a new function, $\phi :{\bf S}^{1} \to {\bf S}^{1}$:

Then the rotation of $V$ along the boundary of the disk is simply $2\pi\cdot\deg \phi$!

Suppose $D$ is a closed ball neighborhood in ${\bf R}^n$ and vector field $V$ has no zeros on its boundary $\partial D$.

Definition. We define the Gauss map of $V$ on $D$, $$\phi_D:\partial D \to {\bf S}^{n-1},$$ by $$\phi_D(z):=\frac{V(z)}{||V(z)||}.$$

Now, if $V$ has no zeros on the whole $D$, the Gauss map can be extended to the whole $D$ with the same formula. But by the last theorem from the last subsection, such an extension is only possible if $\deg \phi _D =0$.

Definition. We define the index of a vector field $V$ on $D$, $\operatorname{Index}_D (V)$, to be the degree of the Gauss map: $$\operatorname{Index}_D (V):=\deg \phi _D.$$

Exercise. Prove that the index is independent from the choice of $D$.

We have the following.

Theorem. If $\operatorname{Index}_D (V)\ne 0$ then vector field $V$ has a zero in $D$.

The following is a generalization (it follows from the Lefschetz Fixed Point Theorem) we will state without proof (see Bredon).

Theorem (Poincaré–Hopf Index Theorem). Suppose a continuous vector field $V$ has only isolated zeroes in domain $Q\subset {\bf R}^n$ and suppose that $V$ points in the outward normal direction along the boundary of $Q$. Then we have the formula: $$\sum_i \operatorname{Index}_{D_i}(V) = \chi(Q),$$ where the sum is over neighborhoods of all the zeroes of $V$.

Therefore, if $\chi (S)\ne 0$, there is a zero just as in the previous theorem.

## 9 Social choice: gaming the system

Let's recall the two hikers that are about to go into the woods and they need to choose (or we may suggest to them) a procedure on how to decide where to set up their camp, as a fair compromise.

This procedure should be equally applicable to the future in case they decide to take a trip again. The answer is simple: take the middle point between the two locations, if possible. It is always possible when the forest is convex.

Now, what if one of the participants decides to manipulate the situation and, while his (true but possibly secret) preference point remains the same, declares his choice to be a point farther away from the other person's choice?

Then the mid-point, compromise point moves closer to the first person's preferred location!

Seeing this, the other person would move his declared choice in the opposite direction as well. They respond to to each other in this fashion and this turns into a hostile game!

Example. To be able to fully analyze such a game, let's consider its simpler, $1$-dimensional version: the choices are limited to a straight segment of a road in the forest. Suppose the players counteract each other's moves by moving backwards continuously. In that case, this game will go on until one of them reaches the end of the road (i.e., the edge of the forest). Then the other person continues to gain ground until the “compromise” location coincides with his or otherwise until he also reaches the end of the road.

The three possible outcomes share an interesting property: neither player can improve his position by a unilateral action! Such a state is called a Nash equilibrium.

Let's confirm this analysis with a quick computation.

Suppose the road is the interval $[0,1]$ and the preferred locations of the two players are $A_1$ and $A_2$ respectively. We assume that $A_1<A_2$ and they remain unchanged. However, the players choose points $x_1$ and $x_2$. Either makes his selection as he tries to decrease the distance from the preferred location to the compromise, midpoint location. In other words, $i$th goal is to

• minimize the harm $h_i:=|A_i-\frac{1}{2}(x_1+x_2)|$,

or to maximize some related utility. We have: $$0 \le x_1 \le A_1 < A_2 \le x_2 \le 0.$$ If $x_1$ is chosen and fixed, the response of the second player would be to try to get the prefect hit and choose $x_2$ so that $$\frac{1}{2}(x_1+x_2)=A_2,$$ if possible. Then $$x_2=r_2(x_1):=\min\{1,\ 2A_2-x_1\}.$$ Similarly, if $x_2$ is chosen and fixed, the response of the first player is $$x_1=r_1(x_2):=\max\{0,\ 2A_1-x_2\}.$$ These maps are continuous!

To have an equilibrium, we need $x_1=r_1(x_2)$ and $x_2=r_2(x_1)$. However, we can't have both $x_1$ and $x_2$ within $(0,1)$; otherwise, $$x_1=2A_1-x_2, \ x_2=2A_2-x_1,$$ which leads to a contradiction: $A_1=A_2$. Let's assume that $x_1=0$. Then $x_2=2A_2$ or $1$, just as depicted.

$\square$

We summarize this analysis with the following.

Definition. For a pair of maps $$r_1:X_2\to X_1,\ r_2:X_1\to X_2,$$ its Nash equilibrium is a fixed point of $R:X_1\times X_2 \to X_1\times X_2$, where $$R(x_1,x_2):=(r_1(x_2),r_2(x_1))$$ is a function that may be called the combined response function.

This function can also be defined to be: $$R:=\tau (r_2\times r_1),$$ where $\tau :X_1\times X_2 \to X_2\times X_1$ is the transposition: $\tau(x_1,x_2)=(x_2,x_1)$.

This function can be illustrated as a vector field. In particular, for the example above with $A_1=.15,A_2=.45$, the vector field looks like this:

The path taken by the pair of players is also shown.

The Brouwer Fixed Point Theorem implies the following.

Theorem. Suppose $X$ is an acyclic polyhedron, then any pair of self-maps of $X$ allows a Nash equilibrium.

The key to this theorem is not only the acyclicity of the space of choices but also the continuity of the response map. The continuity assumption is natural but the uniqueness of a possible response is not. For example, when the above game is played not on a segment of a road but in a forest, the response isn't unique (one can more deviate from the straight line) and the simple theorem above won't apply.

Exercise. Prove that the response is unique in the case of a convex forest.

Let's follow the example and assume that either player wants to maximize some utility of his own: $$u_1:X\times X\to {\bf R},\ u_2:X\times X\to {\bf R}.$$ Then, in general, the maxima of these functions are attained for multiple values of $x$.

One way to avoid having to deal with set-valued maps is thinking of this game as a dynamical system, as follows. Suppose $X$ is a closed domain in ${\bf R}^n$ and the utilities $u_1,u_2$ are defined on some open neighborhood of $X$. Then to find a maximum, all we need to do is to follow the gradients. The gradients of the utilities define a vector field $V$ (just as illustrated above): $$V_x=\operatorname{grad}_x u_1(x,y),\ V_y=\operatorname{grad}_y u_2(x,y).$$ Then the Poincaré–Hopf Index Theorem applied to $V$ implies the following.

Theorem. Suppose the utilities are continuously differentiable on an open neighborhood of a domain $Q\subset {\bf R}^n$ and suppose that their gradients point in the outward normal direction along the boundary of $Q$. Then, provided $\chi(Q) \ne 0$, the utilities have critical points in $Q$.

If the utilities are also concave down, these points are the maxima. We have an analogue of a Nash equilibrium.

Exercise. Without the convexity condition, what can we say about how the game is played around these points?


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## Algebra 2 (1st Edition)

We simplify to find: $7! = (7 * 6) * (5 * 4) * (3 * 2) * 1$ or $42*20*6$ or $120*42$ or $5040$.


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Browse Questions

# Find the general solution $(e^x+e^{-x})\;dy-(e^x-e^{-x})\;dx=0$

$\begin{array}{1 1} y = \log(e^x + e^{-x}) + C \\ y = \log(e^x - e^{-x}) + C \\ y = \large\frac{x}{\log(e^x )} + C\\y = 1 - \log(e^x + e^{-x}) + C \end{array}$

Toolbox:
• If the numerator of a given function is the derivative of the denominator we can follow the substitution method.
Step 1:
Given:$(e^x + e^{-x})dy - (e^x - e^{-x}) = 0$
$(e^x + e^{-x})dy = (e^x - e^{-x})dx$
seperating the variables we get
$dy =\large\frac{ (e^x - e^{-x})dx}{(e^x + e^{-x})}$
Step 2:
Using the information from the tool box
Let us follow the substitution method
Let $t = e^x + e^{-x}$ then $e^x - e^{-x} dx= dt$
$dy = \large\frac{dt}{t}$
Step 3:
Integrating on both sides
$y = \log t + C$
Substituting for t
$y = \log(e^x + e^{-x}) + C$
This is the required general equation.


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## Pages

### Chessboard and pyramid

Have you heard of the legend of the chessboard? The story goes like this. There was a wise man who invented the chess game and introduced it to a king. The king loved the game so much that he told the wise man that he could choose anything for a reward. The wise man then pointed to the chessboard and asked for 1 grain of wheat for the first chess square, 2 grains of wheat for the second square, 4 grains of wheat for the third square, 8 grains of wheat for the fourth square, and repeat this doubling pattern. This sounds like a very little reward but at the end the king didn't have enough wheat to reward the wise man.

Today, we will calculate to see how much wheat the wise man actually requested. Is it a lot? Is it little? I'll give you a clue, it's HUGE!

To see how big this reward is, we will calculate the number of pyramids that we could build up using all this wheat.

Step 1. Find the total number of grains

A chessboard has $8 \times 8 = 64$ squares and the wise man asked for his reward the following:
• Square 1: number of grains $=1$
• Square 2: number of grains $=2$
• Square 3: number of grains $= 4=2^2$
• Square 4: number of grains $= 8=2^3$
• Square 5: number of grains $=16=2^4$
• Square $n$: number of grains $=2^{n-1}$
• Square 63: number of grains $=2^{62}$
• Square 64: number of grains $=2^{63}$

So the total number of grains is $$S = 1 + 2 + 4 + 8 + 2^4 + \dots + 2^{62} + 2^{63}$$

We will simplify the formula $S$ as follows. Multiply both sides of the formula $S$ by 2, we have $$2S = 2 + 4 + 8 + 16 + 2^5 + \dots + 2^{63} + 2^{64}$$
Subtract it to the original formula $$S = 1 + 2 + 4 + 8 + 2^4 + \dots + 2^{63}$$ we obtain $$S = 2^{64} - 1$$

Step 2. Estimate the volume of the wheat

We have calculated that the total number of grains is $S = 2^{64} - 1$. For simplicity, we will give the wise man one more grain of wheat to make $S = 2^{64}$. We will assume that a grain of wheat has a volume of 2 cubic millimetres so that the total volume of all the wheat is $V = 2^{65}$ cubic millimetres.

But how can we estimate the number $2^{65}$?

In computer science, we have
• 1 kilobyte = $2^{10}$ bytes $\approx$ 1000 bytes,
• 1 megabyte = $2^{10}$ kilobytes $\approx$ 1000 kilobytes,
• 1 gigabyte = $2^{10}$ megabytes $\approx$ 1000 megabytes,
• 1 terabyte = $2^{10}$ gigabytes $\approx$ 1000 gigabytes.

Thus, we have made the following approximation: $$2^{10} = 1024 >\approx 1000$$

Now we will use the same approximation $2^{10} \approx 1000$ to estimate the volume $V = 2^{65}$ (cubic millimetres): $$V = 2^{65} = 2^{5} \times 2^{60} = 32 \times 2^{60} >\approx 32 \times 1000^{6}$$
Therefore, $V \approx 32 \times 1000^{6}$ cubic millimetres $=32 \times 1000^{3}$ cubic metres.

The volume of all the wheat that the wise man asked for his reward is
$V\approx$ 32 billion cubic metres!

Step 3. Estimate the volume of a pyramid

The pyramids in the above picture are the Giza pyramids. They are located in the outskirts of Cairo - capital of Egypt. The pyramids are the burial tombs of Egypt pharaohs. The pyramid on the far side is the largest pyramid, 147 m tall, tomb of a pharaoh named Khufu. It was constructed first around 2550BC. The middle pyramid, 144 m tall, 2520 BC, was the tomb of a pharaoh named Khafre. The one in the front is the smallest, 65 m tall, tomb of a pharaoh named Menkaure, constructed around 2490BC.

We have a formula to calculate the volume of a pyramid. If a pyramid has a height of $h$ and its base is a square of side length $a$ then its volume is $$V = \frac{1}{3} a^2 h$$

Apply this formula, we have:
 height h base length a volume V Pyramid of Khufu 147 m 230 m 2,592,100 cubic metres Pyramid of Khafre 144 m 215 m 2,218,800 cubic metres Pyramid of Menkaure 65 m 105 m 238,875 cubic metres

The largest pyramid among these three, the pyramid of Khufu, has a volume of approximately 2.6 million cubic metres.

Step 4. Find the number of pyramids

Now we are ready to answer the question that we put forth in the beginning. We have estimated the volume of the wheat that the wise man requested: 32 billion cubic metres. We have calculated the volume of the largest Giza pyramid: 2.6 million cubic metres. If we could use all this wheat to form pyramids then the number of wheat pyramids would be $$\frac{32 \times 1000^3}{2.6 \times 1000^2} > \approx 12000$$
So the wise man had requested a gift of more than 12 thousand Giza pyramids of wheat. No wonder why the king did not have enough wheat to reward the wise man!!!

Let us stop here for now. The above pictures of Giza pyramids were taken from Google Maps. The picture of a chessboard at the top of our post is a screenshot from a movies by Cristóbal Vila - "Inspirations" - follow the link below to watch it on YouTube:

Homework.

1. Go to google.com and find out how large the moon surface is. Is it possible to cover the whole surface of the moon by the wise man's wheat?

2. Simplify the sum $$1 + \frac{1}{2} + \frac{1}{4} + + \frac{1}{8} + \dots + + \frac{1}{2^{64}}$$

3. Simplify the sum $$1 + 3 + 9 + 27 + \dots + 3^{64}$$


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# How would you calculate the average atomic mass of copper if 69.2% of copper has a mass of 62.93 amu and 30.8% has a mass of 64.93 amu?

Dec 5, 2015

$\text{63.546 u}$

#### Explanation:

The average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its naturally-occurring isotopes.

Simply put, each isotope will contribute to the average atomic mass of the element proportionally to its percent abundance.

$\textcolor{b l u e}{{\text{avg. atomic mass" = sum_i ("isotope"_i xx "abundance}}_{i}}$

As far as the actual calculations go, you will use decimal abundances, which are simply percent abundances divided by $100$.

So, you know that the atomic masses of these two copper isotopes are $\text{62.93 u}$ and $\text{64.93 u}$, respectively. Their decimal abundances will be $0.692$ and $0.308$, respectively.

The average atomic mass of copper will thus be

"avg. atomic mass" = overbrace("62.93 u" xx 0.692)^(color(red)(1^"st" "isotope")) + overbrace("64.93 u" xx 0.308)^(color(red)(2^"nd" "isotope"))

"avg. atomic mass " = color(green)(" 63.546 u")

I'll leave the answer is rounded to four sig figs, despite the fact that the values you have for the percent abundances justify only three sig figs.


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# Category:Complex Subtraction

This category contains results about Complex Subtraction.

Let $\struct {\C, +, \times}$ be the field of complex numbers.

The operation of subtraction is defined on $\C$ as:

$\forall a, b \in \C: a - b := a + \paren {-b}$

where $-b$ is the negative of $b$ in $\C$.

## Subcategories

This category has only the following subcategory.

## Pages in category "Complex Subtraction"

The following 3 pages are in this category, out of 3 total.