Feb 16

Proposition​

A statement that could bear a truth value, i.e. can be true or false.

• Non-e.g. Linus
• Non-e.g. A cat
• E.g. Linus is a cat.

Atomic proposition​

The smallest unit of language that could take a truth value. Roughly, this means a complete sentence.

Conjunction​

In English, this roughly correspond to the natural language use of "and".

Conjunction compounds two props s.t. the conjunction is true if and only if both props are true.

Symbol: $\land$, $\&$.

Conjunction Truth Table​

Disjunction​

Disjunction compounds two props s.t. the disjunction is true if and only if at least one of the props is true.

Symbol: $\lor$

Roughly corresponds to the natural language use of "or" but is NOT the same as English "or"!

• English "or" is sometimes exclusive: either $A$, or $B$.

• Exclusive or logic representation: $[A\lor B]\land \lnot [A\land B]$, where $\lnot$ or $\sim$ is the negation operator.

• However, logical disjunction is inclusive: $A\lor B =$ true when either one is true, but also when both are true.

Inclusive and exclusive disjunction​

E.g. Basic vs. comparative forms of adjectives

a. Exclusive

Mike is angry at either Alex or at Tre

b. Inclusive

Mike is angrier than either Alex or Tre

Disjunction truth table​

(Inclusive)

Negation​

The negation of a proposition flips the truth of a proposition.

Roughly equivalent to no in English.

Material implication (if... then...)​

Compounds two propositions A and B and evaluates to true if and only if whenever A is true, we have B is true.

Symbol: $\to$ or $\implies$ or $\supset$ (has nothing to do with superset though)

Note that there are no causal/entailment relationship expressed in the propositional setting.

Material implication truth table:​

When A is not true, we are not making any statements about B. Therefore, when A is false, whether B is true or not, the statement that "whenever A is true, we have B is true" cannot be falsified. Therefore, it takes the truth value of true.

Biconditional connective (material equivalence)​

Two props are materially equivalent if they have the same truth value, either both true or both false.

Symbols: $\leftrightarrow$, $\iff$, $\equiv$

Feb 21

Quantifiers​

Universal​

Symbol: $\forall$

Roughly equivalent to 'for all', 'every' or 'for any' in natural language.

e.g. $\forall x (\text{pomona-student}(x)\implies \text{stressed}(x))$

Every Pomona student is stressed.

Existential​

Symbol: $\exists$

Roughly equivalent to 'there exists'.

e.g. $\exists x(\text{professor}(x)\land \text{enthrall}(x, \text{the students}))$

A professor enthralled the students.

Scope & Scope ambiguity​

Consider this statement:

Linus does not like all children.

Where does the 'not' scope to?

• Interpretation 1: $\exists x(child(x) \land (\sim like(Linus, x)))$
• Interpretation 2: $\sim\exists x(child(x) \land like(Linus, x))$

The scope of the quantifiers is ambiguous.

Feb 23

Semantic types​

Consider the following sentences:

1. Alex ate apples.
2. * Alex did run eat.

Why do some words combine but others don't?

Type e​

Individual entities. Proper nouns, definite nouns.

Alex, Mike, these dogs, the dog.

Type t​

Things that can bear truth values (aka complete propositions).

Alex runs.

Type <e,t>​

The braket notation denotes $\langle \text{input, output} \rangle$. This type contains mainly (intransitive) predicates, like verbs, adjectives or nouns after copula (e.g. to be sth).

runs, (is) a teacher, tall, apple.

Takes a type e element and returns type t.

Saturation​

Consider the following case:

The word "saw" is transitive, so it takes 2 arguments. Such predicates would first take a type e, but continues to return an unsaturated predicate $\langle e, t\rangle$. It needs to take yet another type e to become saturated. The type e is said to be saturating the type <e, t>.

Alex gives a book to Mike. Give is a ditransitive verb, which means it takes 3 arguments. Thus, give is of the type $\left\langle e,\langle e, \langle e, t\rangle \rangle \right\rangle$.

Type <<e, t>, e>​

Determiners. Consider the sentence:

The apple fell.

Fell is an intransitive verb, so it is of type <e, t>. The whole sentence is type t. Therefore, the segment "the apple" must be type e.

However, since nouns like apple are type <e, t> (review type <e, t>), we know that determiners like the must be type <<e, t>, e>. $\Box$

Type <<e, t>, <e, t>>​

Some degree modifiers and adverbs.

very, quite, slowly, happily.

Takes an <e, t> (adjectives, verbs) and outputs an <e, t>.

Type <t, t>​

Negations. Takes a truth value, outputs an (opposite) truth value.

Feb 28

Lambda abstraction​

We remember that a proposition like "Lilly likes Mike" could be abstracted into $LIKES(x,y)$ and substituting $x,y$ by Lilly and Mike. However, this does not give us any information on which argument is saturated first. Thus, we introduce lambda abstraction:

$\lambda y\, \lambda x[LIKES(x,y)](Mike)(Lilly)$

The function takes arguments in the parentheses in the order from left to right. Why do we know $y$ is saturated first? We note that the sentence has the following syntactic structure:

Lambda conversion​

The process of converting the abstracted form into a saturated sentence is as follows:

1. $\lambda x[LIKES(x,Mike)](Lilly)$

2. $LIKES(Lilly, Mike)$

Denotation​

The denotation of a predicate is represented as: $||like||=\lambda y\, \lambda x[LIKES(x,y)]$

Some types of words (e.g. type <<e, t>, <e, t>>) not only takes individual entities (type e) but also predicates. A predicate is conventionally represented by a capital letter like $P$. For instance: $||\text{playful}_{MOD}||=\lambda P\,\lambda x[P(x)\land \text{PLAYFUL}(x)]$ Predicates like this are considered complex predicate that takes other predicates as arguments. In this case, notice that $\text{playful}_{MOD}$ as in "playful dogs" (type <<e, t>, <e, t>>) is a different type from $\text{playful}$ "the dog is playful" (type <e, t>). The change of type for the same lexicon is called a type shift.

Identity map​

The identity map in semantics is the indefinite determiner: $||\text{a(n)}||=\lambda P[P]$

The iota operator​

To understand definite determiners, we need to introduce the iota operator $\iota$. Let $a$ be a specific individual made reference to in the real world. $\iota x(P(x))=a \iff \exists x(P(x) \land \forall y(P(y)\iff y=x)\land x=a)$

$x$ such that $x$ is a $P$ (e.g. dog) is $a$ if and only if there exists a $P$ (e.g. dog), $x$, and any $y$ is a $P$ if and only if $y$ is $x$, and that $x$ is the same entity as $a$.

E.g. "The one", "the dog", etc.

Mar 7

Optionality formalism​

Notations:

• $(x)$: $x$ is optional
• $(*x)$: $x$ must be absent
• $*(x)$: $x$ is obligatory

Polarity sensitive items (NPIs)​

Yet, anymore, in months, until yesterday...

NPI licensing environment​

A semantic/pragmatic environment in which an NPI could survive.

Upward- and downward-entailing​

• Upward-entailing (UE): the subset entails the superset

  • E.g. "Some musicians snore loudly" $\implies$ "Some musicians snore" is UE.

• Downward-entailing (DE): the superset entails the subset

  • E.g. Let the set of red apples be the subset of the set of apples. The statement "John did not eat an apple" $\implies$ "John did not eat a red apple" is DE.
  • NPIs only work in DE environments!

"The Pizza Test"​

To test whether an triggering item survives in UE or DE, try constructing a sentence using "Pizza" and "Pepperoni pizza".

E.g.:

• every: "Every [pizza] is good" $\implies$ "Every [pepperoni pizza] is good" (DE)
• A believes: "A believes [pizza] is good" $\implies$ "A believes [pepperoni pizza] is good" (DE)
• Never: "Some [pizza] is good" $\impliedby$ "Some [pepperoni pizza] is good" (UE)

As a math(s) person, I also re-worded some definitions of terms in standard logic operators or the typical way of framing a mathematical definition. They might be easier to understand if you are more used to those.

Jan 19

Sense vs. reference​

Consider the phrase the same sandwich.

• Sense: a sandwich made from the same ingredients
• Reference: the EXACT SAME piece of bread and filling etc.; a.k.a. THAT exact sandwich.

Specific vs. non-specific indefinite​

Consider "a dog", an indefinite noun phrase.

• Specific: "there is a dog in my bathroom" which is a specific dog.
• Non-specific: "I want a dog" which does not point to a specific dog, but rather everything that has a sense of a dog (their properties make them qualify as a dog).

Quantifiers​

Consider " Alex loves themself." It means "Alex loves Alex".

However, "Everyone loves themself" does not mean "# Everyone loves everyone"!

Proposition​

A potential fact that could be true or false.

• E.g. "This is a blue marker" (it has a truth value)
• Non-e.g. "Blue marker" (does not have a truth value)

Entailment​

Let p1, p2 be propositions.

p1 entails p2 $\iff$ if p1 is true, p2 is true.

The same as the logic "imply", or $\implies$.

• E.g. "Beyoncé is a famous woman" $\implies$ "Beyoncé is a woman"

• Non-e.g. "Maisha hit the window" $\nRightarrow$ "The window broke"

Denial / contradiction​

Let p1, p2 be propositions and p1 entails p2. Then "p1, but not p2" is a contradiction.

• E.g. "Beyoncé is a famous woman, # but she's not a woman"

• Non-e.g. "Maisha lengthened the rope, but the rope is still short" (p1 does not entail p2!)

Common ground (CG)​

Set of propositions that participants in a conversation accept as true for the purpose of this conversation.

• E.g. "the person I am conversing with is called Amelia"

The common ground would have more elements added as the conversation progresses. These elements are called assertion.

Assertion​

Proposed addition to the CG.

And conversely...

Presupposition​

Propositions that are already assumed to be in the CG. [THEY DO NOT ACTUALLY HAVE TO BE IN THE CG, just ASSUMED to be]

• E.g. Personal pronouns like "my" presupposes the ownership and automatically is a presupposition.

Consider the sentence "Priya, have you stopped drinking wine for breakfast?"

The presupposition is "Priya used to drink wine for breakfast". The presupposition trigger is the verb phrase "stopped", which asserts that the action following it used to happen.

Priya, who does not drink wine, probably would neither want to say yes or no. Instead, she would say "hey wait a minute..."

Presupposition holes​

Putting the sentence containing the presupposition into a question, negation, conditional, modal does not eliminate the presupposition.

For instance:

• Question: Has Priya stopped drinking wine for breakfast?
• Negation: Priya did not stop drinking wine for breakfast.
• Conditional: If Priya stops drinling wine for breakfast, I would give her a cookie.
• Modal (might, could, should): Priya might stop drinking wine for breakfast.

In all these sentences, the presupposition that "Priya used to drink wine for breakfast" still holds.

These holes could be use to diagnose the presuppositions of a sentence that might not be obvious sometimes.

Accomodation​

The act to accept a new proposition into the CG.

• E.g. Hearing "My sister is in New York" from others, the listener would accomodate

Presupposition failure​

Failure to make the other listener accomodate the presupposition the speaker assumed. (The assumed presupposition does not get added to CG)

• E.g. "I parked my spaceship over there"
  • Presupposition: The speaker owns a spaceship"
• Reaction: "What? Spaceship?"

Entailments and presuppositions are both implicative relations between propositions. However, they are very different in nature. Entailment is a logical relation in which the truth of one or more propositions guarantees the truth of another proposition. Presupposition is a relation in which one proposition takes for granted the truth of another proposition (or that it is uncontroversial).¹

Jan 24

Vagueness​

Indeterminacy​

Is this a cup or a mug?

A paper cup with a foldable handle²

The words themselves are indeterminant with respect to a lot of features that exist in the real world.

For instance, factors that makes a mug a (prototypical) mug includes:

• being ceramic
• being cylindrical
• having a handle Whereas for prototypical cups:
• being plastic / paper / etc.
• no handles

However, human languages do not have enough words for the amount of features existing on their referents. The boundaries are fuzzy.

Indeterminacy depends on specific languages. When language A might have more precise words for some features, language B might only have broader vocabularies that are indeterminant with respect to language A.

Ambiguity​

Lexical ambiguity​

Consider the following statement (phonologically realized):

Atheism is a non-prophet organisation.

The homophones 'prophet' and 'profit' result in a lexical ambiguity, as the meaning of the word resulted in an ambiguous sentence.

Structural ambiguity​

The chicken is ready to eat and the child is too.

The syntactical structures resulted in the ambiguous sentence.

Referential ambiguity​

Two Soviet ships collide, one dies.

It is unclear what the pronoun 'one' is pointing to: person or ship?

More examples containing more than one source of ambiguity​

Collected from the LGCS106 class worksheets.

Time flies like an arrow. Fruit flies like a banana.

Lexical & structural: like (prep.) vs. like (v.), flies (n.) vs. flies (v.).

• How do you make a turtle fast?
• Take away his food.

Lexical & structural: fast (v.) vs. fast (adv.).

Jan 26

Cooperative principle and Gricean maxims​

A family picture with four signs reading 'Excited, engaged, expecting, Emily'³

What could be inferred from this picture?

Cooperative principle​

Make your contribution such as is required, at the stage at which it occurs, by the accepted purpose or direction of the talk exchange in which you are engaged.

Cooperative principle is usually assumed at the start of the conversation for other speakers, especially under the assumption that the speakers are neurotypical.

Maxims​

• Quality: always say true information to the best possible extent
• Quantity: adequate amount of information (enough but not TMI)
• Relevance / relation: correct logic relationships and relevance to the topic
• Manner: orderly speech without ambiguity

For instance:

Q: What time is it?

A: The dining hall just opened.

First, assume that the answerer is coorporative. By Maxim of Quantity, the speaker does NOT know what the exact time it is, but gave out the best possible guess by referring to the opening time of the dining hall. The dining hall must have a fixed opening time, as otherwise the answer would not be relevant.

Conversational implicature​

Cancellable without contradiction​

Conversational implicatures are cancellable without contradiction. For example:

His sister is often complaining.

The "often" implies not always. But if there is a follow up:

His sister is often complaining. In fact, she's always complaining.

This does not sound like a contradiction, but rather like a clarification. The implicature 'not always complaining' is therefore cancelled without a blatant contradiction. A non-implicature meaning can incur contradictions:

#Mike regrets that he touched that car. However, he did not touch that car.

Reinforceable without redundancy​

Consider:

His sister is often complaining. However, she is not always complaining.

The implicature 'not always complaining' is reinforced here without giving a sense of redundancy. On the contrary, non-implicature meanings would deliver redundancy:

#Maisha broke the window. The window broke.

Calculable​

You can always describe why you reached the conclusion you reached with one or more of the Gricean Maxims.

Jan 31

Face, face wants and face loss⁴​

Erving Goffman’s notion of face is our social persona adopted into politeness theory.

Face wants are the desire to protect our positive face and negative face from threat or damage.

The two types of faces are:

• Negative Face (i.e. desire for autonomy): The want of every competent adult member of a community that their actions be unimpeded by others.
• Positive Face (i.e. desire for approval): The want of every competent adult member of a community that their wants be desirable to at least some others. ‘Love me, love my dog.’

In a conversation, we would like to protect both our own face and the listener's face. Therefore, there are different politeness decisions to make.

Face-threatening acts (FTAs)​

In a conversation, FTAs are inevitable components that necessarily threaten the speaker’s and/or hearer’s positive face and/or negativeface.

• E.g. "Can I use your car?" "Could you tell me the time?"

Examples of FTAs.

In Brown and Levinson’s framework, they require the speaker to decide whether or not to mitigate the threat and which politeness strategies to use.

Politeness strategies​

The decision tree of politness strategies.

• Negative politeness strategies = An action, phrase or utterance that indicates attention is being paid to the negative face wants of an interlocutor. Often achieved through shows of deference.

  • E.g. "Could you by any chance bring that box here?"
  • 

• Positive politeness strategies = An action, phrase or utterance that indicates attention is being paid to the positive face wants of an interlocutor. Often achieved through shows of friendliness.

  • E.g. "Hey sweetheart, can you get me that box?"
  • 

• Bald, on record = A technical term in Brown and Levinson’s theory of politeness. Refers to an inherently face-threatening act made without any softening through positive or negative politeness strategies.

  • E.g. "Get the box here!"

• Off the record = basically, hinting at something

  • E.g. "Huh, I wonder how good it would be if the box were here!" (no direct request made)

Feb 2

Use-conditional meaning​

aka conventional implicatures.

Speaker-oriented​

Appositives​

Who said the statement after 'aka'?

Alex is house-sitting for Derek, aka McDreamy

Maisha said that Alex is house-sitting for Derek, aka McDreamy

Maisha did not say the aka no matter what. Only the speaker of the utterence said it!

Chuck said I could have his lovely vase!

Chuck did not say it's lovely. The speaker said it!

Epithets​

Every Democrat advocating a proposal for reform said this stupid thing is worthwhile.

Who think that it's stupid?

Independent from truth valaue​

Obama, a Texan, was president for 8 years.

#No that's not true!

What the 'not true' is targeting? Not the claim that he is a texan.

A: Alex is dating Sheila, my nemesis.

B: That’s not true!

1. it’s not true that Alex is dating Sheila
2. #It’s not true that Sheila is the speaker’s nemesis.

Conversely,

A: Alex is dating Sheila, my nemesis.

B: Oh, come on, Sheila is not your nemesis.

1. Challenges the content of the appositive
2. Makes no comment re: Alex and Sheila dating

Anti-backgrounding​

Lance survived cancer. #Lance, a cancer survivor, was interviewed by the press.

Redundant background info!

Not like conversational implicatures​

Not cancellable w/o contradiction​

I sometimes get in really bad fights with Max, my nemesis … #but they’re not really a nemesis.

Compare conversational implicatures:

I sometimes get in really bad fights with Max … but they’re not really a nemesis.

Cancellable!

Not reinforceable w/o redundancy​

I sometimes get in fights with Max, my nemesis… #they really are my nemesis.

Redundant when reinforced.

Compare conversational implicatures:

I sometimes get in really bad fights with Max… they really are my nemesis.

Reinforceable w/o redundancy.

Persists under presupposition holes​

Alex is house-sitting for Derek, aka McDreamy.

Inference: Derek is also known as McDreamy.

• Negation: Alex is not house-sitting for Derek, aka McDreamy.
• Question: Is Alex house-sitting for Derek, aka McDreamy?
• Conditional: If Alex is house-sitting for Derek, aka McDreamy, I will lose it.
• Modal: Alex might be house-sitting for Derek, aka McDreamy.

Inference remains: Derek is also known as McDreamy.

Persists under presupposition plugs​

Baseline facts (non-plugs): cancellation is a contradiction. For instance:

• Sam’s kangaroo is sick.
  • Presupposition: Sam owns a kangaroo.
• Sam’s kangaroo is sick… #but that’s ridiculous–Sam doesn’t even own a kangaroo.
• Sam, owner of a kangaroo, is sick … #but Sam doesn't own a kangaroo.

Usage-based meaning or presupposition cannot be cancelled by the speaker without it being a contradiction.

However, there are plugs...

Attitude predicates are presupposition plugs​

Jay believes that Sam’s kangaroo is sick, but that’s ridiculous — Sam doesn’t own a kangaroo.

• Presupposition is plugged by “believe”
• Presupposition is cancellable without contradiction

However,

Jay believes that Sam, owner of a kangaroo, is marriage material … #but Sam doesn't own a kangaroo.

• Usage-based meaning is not plugged by “believe”
• Usage-based meaning still cannot be cancelled without a contradiction

Why? Because UBM are speaker-oriented and are not scoped by the attitude predicates.

Verbs of saying / performatives are plugs too​

Harry promised to introduce Bill to the current king of France.

Speaker does not presuppose that a king of France exists.

McConnell promised that the damn Republicans would be less partisan.

Meaning conveyed by ‘damn’ persists under “promise".

Feb 9

Speech acts & Grammatical forms​

Some speech acts are canonically attached to different grammatical forms.

Declarative​

Usually used to make an assertion.

I made you a tea.

Interrogative​

Usually used to ask a question.

Do you like to make tea?

Imperative​

Usually used to make an order.

Make a tea please!

Other speech acts​

Other speech acts, such as:

• Requests
• Promises
• Conferrals
• Pronouncements / declaration

are not particularly attached to a specific grammatical form.

Felicity conditions​

An appropriate request / promise / etc. is called a felicitous one. There are different felicity conditions for different kinds of speech acts:

Summary of meaning diagnosis​

Note: I am just a note taker and organiser and therefore did not produce the content or the proofs myself. All rights and wisdom belong to Prof Goins!

Assumed knowledge and quick references​

The references of concepts that might not be familiar to everyone would be included in toggle boxes. I might not include the concepts that would not significantly hinder the understanding of this document!

There would also be references and further readings included in these boxes. Check them out if you are interested.

What are Ramanujan's conjectures on tau functions?​

The tau functions​

In 1916, an Indian mathematician S. Ramanujan was interested in a sequence of integers $\tau (1), \tau (2)... \tau (n), ...$ defined via the Taylor Series

where $|q| < 1$ (this makes sense since you would want $1-q^n$ to converge).

The conjectures​

Here are four of the simpler conjectures Ramanujan has made about $\tau$ functions:

1. $\tau (mn) = \tau (m) \tau (n)$ when $\gcd (m,n) = 1$.

We could actually conclude more generally following from here. That is, if $n=p_1^{e_1}...p_r^{e_r}$, then $\tau(n)=\tau\left(p_1^{e_1}\right) \tau\left(p_2^{e_2}\right) \cdots \tau\left(p_r^{e_r}\right)$ where $p_i$ is prime for all $i=1,2,...,n$. Note that this makes sense since prime numbers would always have $\gcd (p_i,p_j) = 1 \space (i\neq j)$, so as their powers.

2. $\tau\left(p_{e+1}\right)=\tau(p) \tau\left(p^e\right)+p^{11} \tau\left(p^{e-1}\right)$ for all prime $p$.

3. $|\tau(p)| \leq 2 p^{11 / 2}$ for all prime $p$. A stronger statement would be that $|\tau(n)| \leq \sigma_{11}(n)$, where:$\sigma_\alpha(n)=\sum_{d \mid n} d^\alpha$ This is saying that the $\sigma_\alpha(n)$ is the sum of $d^\alpha$ for each $d$, where it is a divisor of $n$. Have not done the proof...

   Give me an example

   For instance, let's consider when $n$ is a prime number. Then the divisors would only be 1 and itself. Thus we could say in this case that $\sigma_\alpha(n)=1+n^\alpha$.

From there, we could see a quite remarkable congruence...​

4. $\tau(n) \equiv \sigma_{11}(n) \bmod 691$

   What does this mean?

   This means that $\tau$ and $\sigma_{11}(n)$ produce the same reminder when they divide 691.

   Read more in Modular arithmetic.

Proofs (not really, it's just a walk-through)​

Well you might now be thinking why these numbers appear to be this random, so here is a sketch about how these congruences worked out. They are not formal proofs since I kind of didn't understand the whole thing myself but I would try comb through the process!

The proofs largely follow the ones written in Daniel Bump's Automorphic Forms and Representations.

Lemma 1: The Eisenstein Series​

A proposition...​

Let's first introduce a function $\Delta (z)$ as follows:

Where $f$ is the Taylor Series as conjectured by S. Ramanujan. Therefore,

Observe that $\Delta (z)$ is a modular form for the special linear group

which is a 2x2 integer matrix with a $\det$ of 1.

Then, let's define the Eisenstein Series.​

For all $k\in \mathbb Z$ is even and $k>4$, define the Eisenstein series to be:

Observe that $E_k\left(\frac{a z+b}{c z+d}\right)=(c z+d)^k E_k(z)$ is true. (Took me a while to realize!)

Lemma 2​

By a Fourier Transform that Prof Goins did not go into since it was too complicated, you could derive the following from Lemma 1: $\sum_{n\in\mathbb{Z}}{\frac{1}{(n-\tau)^{k}}}={\frac{(2\,\pi\,i)^{k}}{(k-1)!}}\sum_{n=1}^{\infty}n^{k-1}\,e^{2\pi i n\tau}$ where $\tau \in \mathbb C$.

Lemma 3​

Upon substituting $\tau = mz$ into the Eisenstein Series $\forall \space m \in \mathbb Z$, we get: $E_{k}(z)=\zeta(k)+\frac{(2\pi i)^{k}}{(k-1)!}\sum_{n=1}^{\infty}\sigma_{k-1}(n)\,q^{n}$ where $q=e^{2\pi i z}$, $\sigma_\alpha(n)=\sum_{d \mid n} d^\alpha$, and $\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^k}$ is the Riemann zeta function for any $k\in \mathbb Z$ is even and $k>4$ as defined before.

Note that $\zeta(k)$ here is simply a constant, and it would be used in the next lemma to cancel out the $\pi$. Not much more of it would be discussed here.

Lemma 4​

Now that we have the Eisenstein series in lemma 3, let's focus on the cases $k=4,6,12$:

• $k=4$: we have $\zeta (4)=\frac{\pi ^4}{90}$. Then we define $G_4(z)$ to be: $G_{4}(z)={\frac{E_{4}(z)}{\zeta(4)}}=1+240\sum_{n=1}^{\infty}\sigma_{3}(n)\,q^{n}$ Note that the $\zeta (4)$ here really just gets rid of the $\pi ^4$ that the coefficients would otherwise have.
• $k=6$: similarly, we have $\zeta (6)=\frac{\pi ^6}{945}$. Thus:$G_{6}(z)={\frac{E_{6}(z)}{\zeta(6)}}=1-504\sum_{n=1}^{\infty}\sigma_{3}(n)\,q^{n}$
• $k=12$: we have $\zeta(12)={\frac{691\,\pi^{12}}{638512875}}$. Thus:$G_{12}(z)=\frac{E_{12}(z)}{\zeta(12)}=1+\frac{65520}{691}\sum_{n=1}^{\infty}\sigma_{11}(n)\,q^{n}$

Lemma 5​

Now we have some very exciting coincidences congruences! Namely:

1. $1728\,\Delta(z)= G_{4}(z)^{3}-G_{6}(z)^{2}$.

   Remember that we defined $\Delta (z) = f(e^{2\pi i z})$ in the proposition in lemma 1.

2. $691\,G_{12}(z)=441\,G_{4}(z)^{3}+250\,G_{6}(z)^{2}$

Since $G_k$ results from $G_{k}(z)={\frac{E_{k}(z)}{\zeta(k)}}$ of the Eisenstein series $E_k$ defined in lemma 1, we could conclude: $G_k\left(\frac{a z+b}{c z+d}\right)=(c z+d)^k G_k(z)$ holds true too like the Eisenstein series.

And by Lemma 5-1, surprisingly we can see how

For all $\left(\begin{array}{ll} a & b \\c & d \end{array}\right) \in S L_2(\mathbb{Z})$.

This says that $\Delta (z)$ is indeed a modular form for

which is the set of 2x2 integer matrices with a $\det$ of 1, as we have proposed in lemma 1.

Now we are finally ready for...

Proof for congruence 4​

Using lemma 4-3 timing both sides by 691, we have:

Since $691\,G_{12}(z)=441\,G_{4}(z)^{3}+250\,G_{6}(z)^{2}$ by lemma 5-2, observe that $441+250=691$. Whoa!

Combining lemma 5-1, we have:

Remember that we defined $\Delta (z) = f(e^{2\pi i z})=\sum_{n=1}^{\infty} \tau(n) (e^{2\pi i z})^n$ in the proposition in lemma 1. Since we already defined $q=e^{2\pi i z}$ in lemma 3, we could say that $\Delta (z) = \sum_{n=1}^{\infty}\tau(n)\,q^{n}$.

Now that we know $691+65520\left[\sum_{n=1}^{\infty}\sigma_{11}(n)\,q^{n}\right] \equiv 65520\left[\sum_{n=1}^{\infty}\tau(n)\,q^{n}\right]\,\mathrm{mod}\ 691$:

$\sum_{n=1}^{\infty}\sigma_{11}(n)\,q^{n} \equiv \sum_{n=1}^{\infty}\tau(n)\,q^{n}\quad \mathrm{mod}\ 691$

Thus, we finally arrive at congruence 4: $\tau(n) \equiv \sigma_{11}(n) \bmod 691$

□

Some remarks at this point​

So currently here is a kind of humanly understandable walk-through for the proof for the congruence 4. To prove 1-3, there have to be some fancier stuff like Euler product, L-series for $\Delta (z)$, cusp forms, etc. Prof Goins talked about all of them but I would probably have to organise them again when I have more time to think. Until then, check out Daniel Bump's Automorphic Forms and Representations if you are interested!

[TO BE CONTINUED]

Latest update:

• Derived closed form L-seminorm on operator norm for 2x2 matrices.

• Derived closed form L-seminorm on Forbenius norm.

• Rederived the closed form for operator norm for 2x2 matrices for the complex field.

Intro​

• Metrics: measures of distances

• How about distances between subsets? Hausdorff distance

Hausdorff distance​

E.g. Let $A$, $B$ be two subsets in $\mathbb R$, and $a \in A, b \in B$.

Then:​

• Say A={1, 2} and B={4, 9}. Then the Hausdorff distance would be $\max \{4-1, 9-2\} = 7$.

Quantum Gromov-Hausdorff distance​

If $A= \overline{\underset{n \in \mathbb{N}}{\bigcup} A_n}$ where $A_1 \subseteq A_2 \subseteq ...$,

then the quantum Gromov-Hausdorff distance is:​

or more precisely, $\text{dist}_Q ((A, L), (A_n, L_n))$ since an extra structure L-seminorm is defined on the two spaces.

Norms​

There could be different types of norms defined on the same spaces. E.g. consider $\mathbb R^2$, define $\lvert \cdot \rvert_2$ and $\lvert \cdot \rvert_1$. Thus the spaces ($\mathbb R^2$, $\lVert \cdot \rVert_2$) and ($\mathbb R^2$, $\lvert \cdot \rvert_1$) are different as normed vector spaces.

• As a result, different norms have different geometries;

• $\underset{n\to \infty}{\lim} \lVert (x, y)\rVert_n = \lVert (x, y)\rVert_{\infty}$

Operator norm​

Let $a\in M_n(\mathbb C)$.

The operator norm $\lVert a\rVert_{op}$ is defined as:​

$\lVert a\rVert_{op} = \sup\{\frac{\lVert a \textbf{x} \rVert_{2} }{\lVert \textbf{x} \rVert_{2} }: \textbf{x}\in \mathbb C^n, \textbf{x}\neq \textbf{0} \}$

where the Euclidian norm is $\lVert \textbf{x} \rVert_2 = \sqrt{\sum_{k=1}^n \lvert \textbf{x}_k \rvert ^2 }$.

Closed form of the operator norm for 2x2 matrices​

Take as a fact that $\lVert A\rVert_{op} = \sqrt{\lambda_{\max} (A^* A)}$ since the operator norm above is defined by the Euclidian norm ¹.

Let $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in M_2(\mathbb C)$. Therefore it could be computed that:

Since it is a 2x2 matrix, the characteristic polynomial of it would be: $\lambda^2 - (|a|^2+|b|^2+|c|^2+|d|^2)\lambda + (|ad-cb|)^2=0$

From the characteristic polynomial, the $\lambda$ could be calculated:

$\lambda = \frac{1}{2}\left(|a|^2+|b|^2+|c|^2+|d|^2\pm \sqrt{\left(|a|^2+|b|^2+|c|^2+|d|^2\right)^2-4(|a d-c b|)^2}\right)$

Taking $\lVert A\rVert_{op} = \sqrt{\lambda_{\max} (A^* A)}$, $\left(|a|^2+|b|^2+|c|^2+|d|^2\right)^2-4(|a d-c b|)^2$ is real and that $A^*A$ is positive semidefinite which only has nonnegative eigenvalues, the squareroot must be nonnegative.

$\lambda _{\max} = \frac{1}{2}\left(|a|^2+|b|^2+|c|^2+|d|^2+ \sqrt{\left(|a|^2+|b|^2+|c|^2+|d|^2\right)^2-4(|a d-c b|)^2}\right)$

Therefore, the operator norm in its closed from would be:​

$\lVert A\rVert_{op} = \sqrt{\frac{1}{2}\left(|a|^2+|b|^2+|c|^2+|d|^2+ \sqrt{\left(|a|^2+|b|^2+|c|^2+|d|^2\right)^2-4(|a d-c b|)^2}\right)}$

Seminorms​

Seminorms satisfy the properties of norms, with the exception that a non-zero object could have a seminorm of zero.

L-seminorm on a matrix using operator norms​

L-seminorms are used in equation [eq:quantum-g-h].

Let $M_n (\mathbb C)$ and let $\beta > 0$. Let $a\in M_n (\mathbb C)$.

Then:​

$L_{\beta}^{op}(a) = \frac{1}{\beta}\cdot \lVert a-\frac{1}{n}\cdot tr{(a)} I_n\rVert_{op}$

The operator norm is defined in equation [eq:operator-norm].

If it uses the Forbenius norm instead of operator norm, it might be able to measure its distance from an nxn block-diagonal matrix.

L-seminorms​

Closed form L-seminorm on operator norm for 2x2 matrices​

In the previous section, the closed form of the operator norm for $M_2 (\mathbb C)$ had been derived in equation \ref{eq:op-norm-2x2}. Since the L-seminorm on a matrix using operator norms is $L_{\beta}^{op}(a) = \frac{1}{\beta}\cdot \lVert a-\frac{1}{n}\cdot tr{(a)} I_n\rVert_{op}$, the operator norm could be substituted in.

Let $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in M_2(\mathbb C)$. Therefore,

Let $\frac{a-d}{2}=k$. Thus,

Then we could derive:

Substituting the $k$ back in: $\begin{aligned} &\lVert A-\frac{1}{n}\cdot tr{(A)} I_n\rVert_{op}\\ = &\sqrt{\frac{1}{2}\left(2\left| \frac{a-d}{2}\right|^2+|b|^2+|c|^2+ \sqrt{\left(2\left| \frac{a-d}{2} \right|^2+|b|^2+|c|^2\right)^2-4\left| \left(\frac{a-d}{2}\right)^2+bc\right|^2}\right)} \end{aligned}$

Therefore, the closed form formula of the $\beta$ L-seminorm for 2x2 matrices using the operator norm could be expressed:

Closed form L-seminorm on Forbenius norm​

The Forbenius norm of matrices, unlike the operator norm, has a closed form formula for all $M_n(\mathbb C): n\in \mathbb N$ and not just on $M_2(\mathbb C)$. Therefore, it is possible to generalize the L-seminorm for square matrices of all dimensions as well.

The Forbenius norm of a matrix itself $\lVert b \rVert_{F}$ could be expressed as: $$

Since the summation would give rise to $n^2$ terms, this norm is analogous to the Euclidean norm of $\textbf{b} \in \mathbb C^{n^2}$.

Based on this definition of the Forbenius norm, the $\beta$ L-seminorm using Forbenius norm could be defined: $$

From these definitions, the closed form could also be derived.

First let $A\in M_n(\mathbb C)$ have entries $a_{ij}$. Thus, $tr(A)=\sum_{i=1}^n a_{ii}$. Let $\frac{1}{n}tr(A) = k$ be a constant.

Then, let matrix $B=A-\frac{1}{n}\cdot tr{(A)} I_n$. Therefore, matrix $B$ would be in the form:

Therefore, its Forbenius norm could be derived:

by the definition of the Forbenius norm.

In addition, since $|a_{ii}-k|^2 = (a_{ii}-k)\overline{(a_{ii}-k)}$, the terms could be further simplified:

Now that the term $| a_{ii}|^{2}$ has been cancelled out, the $k$ could be substituted back in:

Therefore, the L-seminorm using Forbenius norm would be: $L_{\beta}^{F, n}(A) = \frac{1}{\beta} \left( \left( \frac{1}{n}-2 \right)|tr(A)|^2+\sum_{i,j=1}^{n}|a_{ij}|^2 \right)$

for all $A\in M_n(\mathbb C): n\in \mathbb N$.

Suomeksi​

Rovaniemen Ruokasenkadulla ei ole ollenkaan ruokaa (vaan hautauspalvelua, tietenkin), ja kaaduin kadulla jäässä 5 kertaa päivässä.

Hautauspalvelut Ruokasenkadulla.

Mutta mielestäni mukaan, Rovaniemi oli muutakin kuin tylsyyttä. Ainakin ei kukaan enää vastannut englanniksi kun aloitin vuoropuheluja suomeksi. Sen takia olen oppinut yhtä paljon suomea kuin olen kaatunut lumessa. Mutta, kaikki puhuikin suomea kaverini kanssa, ja he eivät pystynyt ymmärtämään mitään kuitenkin.

Joulupukkia tai revontulia ei ollenkaan näy. Vain käyn taidemuseossa katsomassa, koska yksi kadulla olevien matkustajista suositteli. Mutta mieleni harhaili kuin kuvassa (tulossa löytyy).

Sitten menimme mäkeen, jolla oli paljon lunta. Allen teki lumiukon, ja sitten vahvasti potkii sen päähän. En tiedä mitä saan ajatella tästä.

Allen potkaisi lumiukon.

En tiedä miksi maksoin melkein puoli budjetistani käydä tässä. Se sai minut kysymään itseltä elämästäni. Nyt odotan lennon takaisin Helsinkiin, ja ajattelen, että eilen Rovaniemellä vain oli totaalista outoa unta. No, ehkä on myös elämystä. En ole varma.

In English​

Diary in Rovaniemi. 🙄

There is no food at all on Ruokasenkatu (‘Foody street’ if I literally interpret it) in Rovaniemi (but only a funeral service, of course), and I fell on icy streets 5 times a day.

But I think Rovaniemi wasn’t quite just boring. At least nobody answered in English anymore when I started dialogues in Finnish. As a result, I have learned as much Finnish as I have fallen in the snow. But everyone spoke Finnish with my friends too, and they couldn't understand anything anyway. There was no Santa Claus or northern lights at all. I only went to check out the art museum because one of the random passengers on the street recommended it. But my mind wandered as in the picture.

Mieleni harhaili taidemuseossa.

Then we went to a hill with a lot of snow on it. Allen made a snowman, and then brutally kicked it in the head. I don't know what to make of this.

Allen kicking the snowman.

I don't know why I spent almost half my budget to visit this place. It made me question myself about my life. Now waiting for my flight back to Helsinki, I'm thinking that yesterday in Rovaniemi was just an absolutely weird dream. Well, maybe there is an experience too. I'm not sure.

Recommended text: G.F Simmons Differential Equations

This document might display better in light mode. I am working on the colors right now...

Intro​

When do we need differential equations?

• Consider Newton's Law of Cooling:

  • Rate of cooling $\propto \Delta T$ between the body and the surroundings

  • Variables:

    1. T(t): dependent variable, temperature
    2. t: independent variable, time
    3. $T_E$: constant, surrounding temperature

  • Shorthand: $-\frac{dT}{dt} \propto T-T_E$

  • Differential equation: $-\frac{dT}{dt} = R \cdot (T-T_E)$, where $R \in \mathbb{R}$.

First order ODEs (Ordinary differential Eqtns)​

Simple ODEs​

$\displaystyle \frac{dy}{dx} = f(x)$

Separable ODEs​

$\displaystyle g(y) \, \mathrm{d}y=f(x)\,\mathrm{d}x$

$\frac{dy}{dx}=\frac{f(x)}{g(y)}$

If no specific points are given, the solutions to ODEs would be a FAMILY of functions instead of a single one.

The order of the diff. eqtn is the order of the highest derivative in the equation.

Linear ODEs​

$\displaystyle \frac{dy}{dx}+ p(x)\, y=f(x)$

A diff.eqtn is linear if and only if the dependent variable and its derivatives are only raised to the power of 1.

Note that $\displaystyle \frac{d^2y}{dx^2}$ is still first order as the operator d/dx is squared but the dependent variable y isn't. However, $\displaystyle (\frac{dy}{dx})^2$ is second order.

• Solve linear ODEs by the Integrating Factor

  • The Integrating Factor $\mu (x)$ makes the equation simple.

  • $\displaystyle \mu (x) \frac{dy}{dx} + \mu (x) \, p(x)\, y=\mu(x)\,f(x)$

  • Consider $\mu(x)\cdot y(x)$ and its derivative:

  • $\displaystyle \frac{d}{dx} (\mu(x)\cdot y(x)) =\mu \frac{dy}{dx}+y\frac{du}{dx}$ (product rule)

  • This equals to the LHS of the ODE provided that:

    $$\begin{aligned} \frac{du}{dx} &=\mu p \\ \int{\frac{1}{\mu}} \, \mathrm{d}u&= \int{p}\, \mathrm{d}x \\ \therefore \textcolor{teal}\mu&{=A\cdot e^{\int{p(x)}\, \mathrm{d}x}} \end{aligned}$$
    • Since the $\mu$ times every term in the ODE, the constant $A$ could be cancelled out.

  • Then: $\begin{aligned} \frac{d}{dx}(\mu\,y) &=\mu\cdot f(x) \\ \implies \mu y&=\int{\mu f(x) \, \mathrm{d}x}+C \textcolor{green}* \\ \implies y&= \frac{1}{\mu} \int{\mu f \mathrm{d}x}+ \frac{C}{\textcolor{teal}\mu} \end{aligned}$

    *Note that $y \neq \frac{1}{\mu} \int{\mu \cdot f(x) \, \mathrm{d}x} + \textcolor{green}C \leftarrow$ The constant of integration shouldn't be an afterthought; it should be added once the $\bf \frac{d}{dy}$ is inverted, in step labelled as $\textcolor{green}{\ast}$.

• E.g. of linear differential equations:

  1. $x\frac{dy}{dx}+(1+2x^2)y=x$, where $y(0) \text{ is finite}$. $\begin{aligned} \therefore \frac{dy}{dx}+(\frac{1}{x}+2x^2)y&=1 \\\therefore p(x)&=\frac{1}{x}+2x^2\\ \therefore \mu&=e^{\int{\frac{1}{x}+2x^2}\,dx}\\&=x\cdot e^{x^2}\\\implies \frac{d}{dx}(x\cdot e^{x^2}y)&=x\cdot e^{x^2}\\\implies y&=\frac{1}{2x}+\frac{C}{x}e^{-x^2}\\y&=\frac{1+2Ce^{-x^2}}{2x}\\\\ \text{To keep y finite at x->0}&\text{, must have:}\\ \lim_{x\to 0}(1+2Ce^{-x^2})&=0 \implies 1+2C=0\\ C&=-\frac{1}{2}\end{aligned}$