Mathematics - 2018-05-31 (page 5 of 5)

MatheinBoulomenos

23:00

So basically: for any adjunction we get a monad for free

Leaky Nun

I see

MatheinBoulomenos

and when we have a monad on a category, then we can look at the category of algebras over that monad (I think defining morphisms of algebras over a monad is not necessary at this point)

but in our example something remarkable happened: the category of algebras over the monad was just the other category from the adjunction

but all these parts that define a monad live just in the category where the monad lives

so in this case $\mathbf{Set}$

As it turns out, this is not always the case

for an adjunction with a left adjoint $F:C \to D$, $D$ might be a different category than the category of algebras over the monad induced by the adjunction

you can check that this stuff works if we replace free monoids with free groups, free modules, etc. but not with topological spaces

So this is a special situation

we call this a "monadic adjunction"

But when we have a monadic adjunction, then this is tells us quite a lot

GFauxPas

@B.Mehta @Daminark did you figure out the thing about $L^1$ and $L^\infty$

MatheinBoulomenos

so when the left adjoint is $F:C \to D$ and this is part of a monadic adjunction, then we can completely describe the category $D$ as objects in $C$ together with some additional structure (given by the algebra structure over the monad from the adjunction)

And I think this is also a bit like your result looks like

and faithfully flat descent is also a special case of this

Leaky Nun

in what sense?

MatheinBoulomenos

23:08

I think it's probably a monadic adjunction + some other duality

the dual of a torus is a finitely generated free abelian group, right?

Leaky Nun

dual?

MatheinBoulomenos

character group

Leaky Nun

the character of a torus is a f.g.f.a.g

right

of a split torus

MatheinBoulomenos

okay so in faithfully flat descent you basically prove that if $f: R \to S$ is a faithfully flat ring homomorphism, then restriction of scalars along that functor (for algebras, modules, schemes over Spec(R), module-sheaves etc.) part of a monadic adjunction

you always have the restriction of scalars and extension of scalars adjunction of course

but when you have a faithfully flat ring homomophism, it's actually a monadic adjunction

so algebraic geometers call the algebra structure associated to the algebra for that monad "descent datum"

but for Galois descent, i.e. $f: K \to L$ is a Galois extension, then this descent datum/ this algebra structure over the monad from the adjunction is just an action from the Galois group

Leaky Nun

could you state the theorem?

MatheinBoulomenos

23:14

yeah that's the problem I don't know the exact statements of the concrete results

Leaky Nun

I see

thanks

MatheinBoulomenos

But I thought this might still help you get what Galois descent is about

Leaky Nun

dankschee

MatheinBoulomenos

galois descent

►

user193319

How does one prove that $\Bbb{R}$ are $\Bbb{R}^n$ are isomorphic as $\Bbb{Q}$-vector spaces if it has NOT yet been proven that two vector spaces are isomorphic if they have the same dimension?

Leaky Nun

23:16

may Galois ascend to heaven

@user193319 it starts with "choose a basis..."

what you said is very easy to prove

MatheinBoulomenos

@user193319 I don't think that you can prove that without basically writing down the proof for that statement

Leaky Nun

you literally build an isomorphism based on the fact that they have the same dimension

user193319

Okay.

MatheinBoulomenos

you take a bijection between bases and extend linearly

@LeakyNun that video basically describes what the algebra structure over the monad for the adjunction is

Leaky Nun

thanks

MatheinBoulomenos

23:20

(I should note that geometers who are less fond of category probably have another PoV on this stuff)

Leaky Nun

they don't exist

ok i'm joking their PoV might be useful

user193319

So the irrationals form a linearly independent set in $\Bbb{R}$, right?

Leaky Nun

nah

MatheinBoulomenos

you won't be able to write down a basis explicitly

Leaky Nun

@MatheinBoulomenos is the analogy in the beginning basically talking about sheaves?

MatheinBoulomenos

23:22

@LeakyNun you mean with covering spaces?

I'd say no

Leaky Nun

yes

user193319

I'm not trying to write down a basis; I'm just trying to show that $\Bbb{R}$ is infinite dimensional.

Leaky Nun

that isn't enough

you need to argue on cardinality

ok that's enough because (infinite cardinal)^n = (same infinite cardinal)

user193319

An infinite linearly independent set isn't enough to show infinite dimensionality?

Leaky Nun

but that isn't trivial

MatheinBoulomenos

23:23

@LeakyNun Basically taking a pullback along a covering is like extending scalars along your field extension

that's the analogy

Leaky Nun

you need to show that the cardinality is equal @user193319

ok he's now invoking the Galois correspondence of covering spaces

@MatheinBoulomenos oh right it's fibre product in the category of schemes

MatheinBoulomenos

exactly

user193319

Why do I need to show equality between cardinalities? I thought that sameness of dimension was what I needed to show they are isomorphic.

Leaky Nun

"sameness" = "equality"

user193319

Yes

Leaky Nun

23:27

dimension = cardinality of basis

user193319

But that's an extra thing to argue; I don't have that theorem yet.

Leaky Nun

@MatheinBoulomenos $E/F$ field extension with Galois group $\Gamma$, $A$ is $F$-algebra, why $(A \otimes_F E)^\Gamma \cong A$?

(fixed points)

hmm, covering space analogy is extensively used in the video

MatheinBoulomenos

@LeakyNun not sure

Leaky Nun

interesting

the theorem is very easy to state

and for me it is too good (to be true)

@MatheinBoulomenos so this is just a problem in Galois theory now

MatheinBoulomenos

It would be enough to prove this for vector spaces actually

Leaky Nun

23:35

Does this work: $(A \otimes_F E)^\Gamma \cong A \otimes_F E^\Gamma \cong A \otimes_F F \cong A$

or is it bollocks?

MatheinBoulomenos

the first equality is kind of mysterious

you have finite sums of elementary tensors

Leaky Nun

the Galois group is $\Gamma = \operatorname{Gal}(E/F)$

the action is just changing the $E$ part

finite sums?

hmm I think my argument works

GFauxPas

Do I need AoC to arbitarily choose a countably infinite subset of an uncountable set?

Leaky Nun

@GFauxPas define uncountable

GFauxPas

no injection into the naturals

countable: yes injection into the naturals

Leaky Nun

23:39

so we just want to construct an injection $\Bbb N \to S$ right

given that no injection $S \to \Bbb N$ exists

GFauxPas

ya

Leaky Nun

I think you can use the UMP of $\Bbb N$ to construct such a map

MatheinBoulomenos

@LeakyNun I guess it works, but I'd argue with choosing a basis for $A$ as $k$-vector space to be safe

GFauxPas

dont know what that is

Leaky Nun

(i.e. induction / recursion)

GFauxPas

23:40

ump?

Leaky Nun

that's category-theoretic language :P

universal mapping property

GFauxPas

oh

Leaky Nun

a function $\Bbb N \to S$ is the same as the following data

1. a point in $S$
2. a successor function $S \to S$

eh... I may have stated the theorem wrongly

@MatheinBoulomenos can correct me

@MatheinBoulomenos and then it works, and we can throw away the basis

GFauxPas

okay proofwiki says you need at least the axiom of dependent choice

Leaky Nun

I see

GFauxPas

23:41

or axiom of countable choice

Leaky Nun

I suppose you need that

I can prove that there is injection $n \to S$ for each $n \in \Bbb N$ that satisfies the obvious diagrams

then I suppose I need countable choice to glue them together to get a global function $\Bbb N \to S$

category theory has polluted my mind. it's too late now.

MatheinBoulomenos

@LeakyNun yes

Leaky Nun

cool

MatheinBoulomenos

@LeakyNun I'm not sure whether you need countable choice or dependent choice, tbh

GFauxPas

"The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?" - Jerry Bona

MatheinBoulomenos

23:43

I'd say dependent choice

Leaky Nun

I'm lost beyond that point

@MatheinBoulomenos let's play the noetherian ring game

MatheinBoulomenos

No it works with countable choice

Axiom of countable choice

The axiom of countable choice or axiom of denumerable choice, denoted ACω, is an axiom of set theory that states that every countable collection of non-empty sets must have a choice function. I.e., given a function A with domain N (where N denotes the set of natural numbers) such that A(n) is a non-empty set for every n ∈ N, then there exists a function f with domain N such that f(n) ∈ A(n) for every n ∈ N. == Overview == The axiom of countable choice (ACω) is strictly weaker than the axiom of dependent choice (DC), (Jech 1973) which in turn is weaker than the axiom of choice (AC). Paul C...

@LeakyNun I want to know how IGN would rate the Noetherian ring game

Leaky Nun

lmao

MatheinBoulomenos

@LeakyNun I should sleep now

Leaky Nun

gute nacht

bona nox

MatheinBoulomenos

23:49

gute Nacht und viel Erfolg mit dem Projekt

Leaky Nun

dank'schee

[Galois Descent (for Algebras)] [only statement]

Let $F/E$ be a Galois extension with Galois group $\Gamma := \operatorname{Gal}(F/E)$.

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