Mathematics - 2019-10-21 (page 1 of 2)

12:29 AM

So, I’ve got this puny little calculator on hand: ernst.mulder.com/calculators/sharp-elsimateel326s-index.html

And I just noticed a little quirk of it

Suppose I enter the ops 1,+,2,=

I get 3, of course

If I now do = repeatedly, I get 5,7,9,11,...

So it’s doing +2 over and over. That’s what I’d expect it to do, and that’s also what it does for division and subtraction

But: if I do 1,x,3 and then repeatedly hit =

I get 3,3,3,3...

So, for multiplication only, it interprets repeated multiplication as -from the left-

Leaky Nun

have you tried 2x3=====?

Semiclassical

Yeah. 6,12,24,48...

Looking up the data sheet for it, the main application they have in mind is for taxes

So I wouldn’t be shocked if that’s somehow behind it

I’m tempted to make an SE question about it. Not sure what site tho

Ted Shifrin

12:50 AM

confuzled

Ted Shifrin

1:02 AM

Hi @loch

Leaky Nun

... a thousand dollars?

Ted Shifrin

I doubt it.

So, Leaky, have you met loch in person?

Leaky Nun

yeah

Ted Shifrin

Good.

Have you introduced yourself to Artin and said hi for me? :P

Leaky Nun

not yet

Ted Shifrin

1:10 AM

I'd tell you other names, but you're less likely to want to talk to them.

Leaky Nun

what are the names?

Ted Shifrin

Mattuck, Guillemin, Melrose, Mrowka

loch

1:52 AM

hi @TedShifrin

Semiclassical

2:06 AM

hi chat

Harry Battersby

hi

Semiclassical

trying to think of how to word something. I've got two version of an inequality, one as $A^2\leq B^2$ and the other as $|A|\leq |B|$. Obviously these are equivalent, but how would I describe the latter in relation to the former?

Leaky Nun

@Semiclassical take the square root of both sides

Semiclassical

Yeah. My inclination was something like "Aside from an overall square root, Eq. (2) is equivalent to Eq. (1)"

Harry Battersby

how should I go about proving that a strictly increasing function is injective? I am not quite sure about it since the definition of a strictly increasing function negates the definition of an injective function.

Semiclassical

2:15 AM

What?

Leaky Nun

@HarryBattersby how does it negate it?

Semiclassical

I mean, strictly increasing means $x>y\implies f(x)>f(y)\implies f(x)\neq f(y)$.

Harry Battersby

Yes, so isn't it the negation of the definition of injectivity which suggests that there exists $x_1$ and $x_2$ such that if $f(x_1)$ = $f(x_2)$ ⟹ $x_1$ = $x_2$?

Semiclassical

No. The negation of $f(x_1)=f(x_2)\implies x_1=x_2$ would be $f(x_1)\neq f(x_2)\implies x_1\neq x_2$.

the contrapositive of injectivity, however, would be $x_1\neq x_2\implies f(x_1)\neq f(x_2)$.

fixed

And contrapositive is equivalent to the original statement, so strictly increasing implies injective (which is what you're trying to show)

Harry Battersby

But the thing that confuses me is that since the function is strictly increasing, then how can two points be equal to each other and produce $x_1$ = $x_2$?

Leaky Nun

2:23 AM

$\lambda \ell$ looks like $\mathcal{M}$

4 mins ago, by Harry Battersby

Yes, so isn't it the negation of the definition of injectivity which suggests that there exists $x_1$ and $x_2$ such that if $f(x_1)$ = $f(x_2)$ ⟹ $x_1$ = $x_2$?

no it doesn't say "there exists"

injectivity means that whenever we have $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$, then $x_1 = x_2$

Harry Battersby

I understand. I totally forgot about the contrapositive. So using the contrapositive we can basically show that since there doesn't exist $x_1$ and $x_2$ such that $x_1$ = $x_2$, therefore $f(x_1)$ won't be equal to $f(x_2)$

which is essentially the contrapositive of the definition.

But I'm still finding it hard to visualize how the function is injective. If the function is always increasing and doesn't have an instantaneous point where the slope is 0, then how can the definition of injectivity hold?

I understand that you can use the contrapositive, but I don't quite get the logic behind it..

Leaky Nun

injectivity can also be stated as: whenever we have $x_1$ and $x_2$ that are distinct, i.e. $x_1 \ne x_2$, then $f(x_1) \ne f(x_2)$

@HarryBattersby if it rains, then the grass is wet

if the grass isn't wet, then it isn't raining

(because if it were, the grass would be wet)

Semiclassical

"Distinct inputs give distinct outputs"

Harry Battersby

oh that makes sense! That's quite an abstract idea to think about.

feynhat

Hi

Semiclassical

2:32 AM

By comparison, strictly increasing means: "If one input is bigger than another, then its output is also bigger than the other's output."

Harry Battersby

So would directly proving that a strictly increasing function is injective be possible? (Without using the contrapositive)

Semiclassical

But if one input is bigger than another, then those inputs are distinct

and if one output is bigger than the other, then the outputs are distinct

So strictly increasing implies injective.

Leaky Nun

@HarryBattersby I mean that depends on what you consider to be a direct proof

can I use trichotomy?

feynhat

What is the reason behind the coboundary sign convention in Bredon or May or Milnor&Stasheff?

Leaky Nun

i.e. for any $x_1$ and $x_2$ we have either $x_1 < x_2$ or $x_1 = x_2$ or $x_1 > x_2$

Harry Battersby

2:34 AM

@Semiclassical @LeakyNun Thanks a lot. That really cleared up a lot.

Semiclassical

It's probably a good idea to write this out as an explicit proof, of course.

Harry Battersby

@Leaky I haven't really learned about trichotomy, so I'm not exactly sure how that would work

Leaky Nun

I just stated it

Semiclassical

Namely, show that: If $f$ is strictly increasing and $x\neq y$, then $f(x)\neq f(y)$.

Harry Battersby

@LeakyNun so you basically divided it into different cases?

Leaky Nun

2:35 AM

three cases

Semiclassical

Trichotomy just says: "Given two real numbers, one of three things has to be true: the first number is bigger than the second, the first number is equal to the second, or the first number is smaller than the second."

Leaky Nun

@TedShifrin I'll hand it over to you :P

Semiclassical

In particular, trichotomy implies: If two numbers aren't equal, then one is bigger than the other.

Harry Battersby

Ahh that sounds interesting. So far I have only touched negations, contrapositive and direct proof (Some statement, P that implies some other statement, Q). So, I still have much to cover :p

Semiclassical

Note that trichotomy isn't a property of logic (as negation/contrapositive/converse are)

it's a property of the real numbers

Harry Battersby

2:40 AM

That's really interesting indeed. Are there any specific circumstances under which we can use this strategy?

Semiclassical

If you're dealing with real numbers and their orderings, it's an obvious thing to remember.

so if you've got real numbers $x,y$ which are distinct, then you know that $x>y$ or $y>x$.

Harry Battersby

I see. That actually sounds really interesting and useful. I will definitely check it out at a deeper level.

So, just out of curiosity, would proving that an increasing function is injective using a direct proof (P implies Q) would be possible in this case? Are there any mathematical problems that can only be solved using a certain way due to their complexity?

Leaky Nun

3:07 AM

51

Q: An easy example of a non-constructive proof without an obvious "fix"?

I wanted to give an easy example of a non-constructive proof, or, more precisely, of a proof which states that an object exists, but gives no obvious recipe to create/find it. Euclid's proof of the infinitude of primes came to mind, however there is an obvious way to "fix" it: just try all the n...

@HarryBattersby ^

Leaky Nun

3:30 AM

@loch why isn't H^2(Gal(C/R),C*) = 0?

isn't C* divisible

loch

3:58 AM

dunno
idk galois cohomology

Leaky Nun

just treat it as group cohomology

it's just H^2(Z/2Z, C*)

where the non-trivial element of Z/2Z acts on C* by conjugation

loch

ah
then i can say idk group cohomology instead!

but at least i can confirm that C* is divisible

Leaky Nun

lol

hmm what happened to cyclic cohomology

$\widehat{H}^{n+2}(\Bbb Z/n\Bbb Z,M) = \widehat{H}^n(\Bbb Z/n\Bbb Z,M)$

and $\widehat{H}^0(\Bbb Z/2\Bbb Z,\Bbb C^\times)$ is the cokernel of $\Bbb C^\times/\langle z \div \overline{z} \rangle \to \Bbb R^\times$

$z \div \overline{z}$ has magnitude $1$

it looks like the circle

so it's the cokernel of $\Bbb R_{>0} \to \Bbb R^\times$

where $r \mapsto r \overline{r} = r^2$

aha so it is $\Bbb R^\times / \Bbb R_{>0} = \Bbb Z/2\Bbb Z$

and therefore $H^2(\operatorname{Gal}(\Bbb C/\Bbb R),\Bbb C^\times) = \Bbb Z/2\Bbb Z$

@MatheinBoulomenos hej

northerner

4:34 AM

Hello

Leaky Nun

hi

Leaky Nun

5:29 AM

oh wow it follows from L C F T that $H^2(\operatorname{Gal}(\Bbb C/\Bbb R),\Bbb C^\times) = \Bbb Z/2\Bbb Z$

MatheinBoulomenos

5:50 AM

@LeakyNun LCFT in the Archimedean case is obvious

Leaky Nun

"obvious"

ÍgjøgnumMeg

8:26 AM

Morning y’all

Mikhail

We didn't die!

Rithaniel

Well maybe you didn't

Slereah

10:28 AM

If I have a connected manifold $M$, are all manifolds of the form $M \setminus U$ for some open set $U$ diffeomorphic?

Do they need to beeeee... homogeneous?

Mike Miller

No and you can easily come up with counterexamples

Slereah

Can I

Hm

lemme think

Mike Miller

You're literally just asking if every closed submanifold of a connected manifold is diffeomorphic, and connectedness is more or less irrelevant here

Slereah

Ah I think I can see one

Double torus minus $S \times [0,1]$

Mike Miller

Not sure I understand tbh, that's not an open set

Slereah

10:33 AM

$(0,1)$ yes

Mike Miller

I guess I was expecting you wanted the complement to be a manifold without boundary but if you allow boundary it's even easier to come by yeah

Slereah

Boundaries are fine

Hm

Mike Miller

So for this example the complement is a disjoint union of two punctured tori and you just need to come up with absolutely any open set with non-diffeo complement

Slereah

Is this true if $U$ is a disk tho?

Mike Miller

lol yes but that changes the question dramatically, and you should probably assume U is the interior of a closed disc

Slereah

10:35 AM

Just trying to prove things relating to a connected sum of a manifold to itself

So it will involve a lot of disks

Mike Miller

at which point you prove this by showing that any two discs oriented the same in a connected manifold are isotopic (this is a lemma of Palais I think but it's straightforward)

It's also true topologically but much harder

Slereah

Fortunately all my manifolds are smooth manifolds

I am pretty sure that the connected sum of a manifold to itself is equivalent to the connected sum of that manifold to an $(n-1)$-sphere bundle over $S^1$

or something close to that

Mike Miller

Language way too fancy. It's either the connected sum with $S^{n-1} \times S^1$ or with the $n$-dimensional Klein bottle, where you take $S^{n-1} \times I /(v, 0) \sim (r(v), 1)$, $r$ being some reflection

Depending on whether or not you do the gluing in an or-preserving or or-reversing fashion

References for the claim about discs being isotopic

6

A: smooth homotopy 4-balls with sphere boundary in dimension 4

Yes, you can perform that ambient isotopy: any oriented embedding $i: B^n \to M^n$ is isotopic to any other. (This is a lemma proven independently by Cerf and Palais1, but the idea is quite clear: shrink the image of $i$ until it's contained in the chart, then take the limit that defines the deri...

Slereah

I once discussed this with a math people who told me that things could get fancy due to the whole spheres that could be homeomorphic but not diffeomorphic

Not sure if this is correct or relevent

Or if it doesn't matter if I assume everything smooth

Mike Miller

Yeah they're wrong

It's not that it's irrelevant they're just making mistakes because they don't know the definition of connected sum

Slereah

10:46 AM

Well to be fair I don't use a typical connected sum

It's more of a general cut and paste procedure

Mike Miller

Define your terms dude this is getting frustrating

Slereah

Sorry

Take a manifold $M$, remove two open disks from it, quotient the resulting manifold with boundaries along some gluing function $h$

So $$M' = (M \setminus (S_1 \cup S_2)) / \sim_h $$

Not assuming orientation-preserving for $h$ though it is smooth at least

Mike Miller

Yeah that's indeed not a connected sum and indeed you can't say anything better about this than "connected sum with a sphere bundle" like you did above

Slereah

Alas

This is physics-related so I'm not 100% sure what the exotic smooth structure business would be like

Exotic smooth structures are rather poorly studied in general

Mike Miller

The connected sum procedure is carried out as follows. Choose an oriented embedding $i_1, i_2$ of the disc into a manifold $M$, $M'$. Delete the interior of the images. Then the connected sum is what you get when you identify $i_1(x) = i_2(x)$ for points $x$ in the unit sphere.

I assume you mean they're poorly studied in physics which fine

Arrow

10:54 AM

Hello. Is the submersion $\mathbb R^2\to \mathbb R$ given by $(x,y)\mapsto xe^y$ a fiber bundle? It seems the gradient flow is complete and gives parallel transport. On the other hand, the Palais-Smale condition is not satisfied, so maybe I'm missing something...

Slereah

@MikeMiller mostly yes

There's some papers on the topic tho

unfortunate that our universe happens to be exactly the correct dimension for that to matter

Mike Miller

For that to matter to someone pathological in a particular way, sure

Slereah

heh

You never know!

The Terrible Puddle

What's the difference between a knot and a loop?

Mike Miller

What is a knot and what is a loop?

The Terrible Puddle

11:02 AM

I'd say a knot is a simple closed curve

Mike Miller

And you're looking to know what a loop is?

The Terrible Puddle

Yeah

Mike Miller

Drop the simple condition

The Terrible Puddle

Oh

Arrow

@MikeMiller hey, could you help me with some confusion about complete connections?

Mike Miller

11:17 AM

@Arrow Not right now sorry. I think your thing is a fiber bundle by direct calculation ie write down local trivializations

Arrow

@MikeMiller (I also think so. My question is about the Reeb foliation $(x^2-1)e^y$.) If you happen to have a couple of minutes later, I'd love to bug you. Thanks anyway!

1010011010

12:15 PM

Who can point me to good material on Matsubara summation?

Actually no, let's break it down further

So I have a summation that i'm representing as a sum of residues of a function

How does the contour deformation work precisely?

There is no pole at infinity, it's a zero instead

So if I have a sum of poles on the real axis

And two poles in the lower half plane

(say)

I'm saying my summation is over the real axis poles, can I deform the clockwise contours around these real axis poles to a contour enclosing the complete lower half plane only ?

Or option 2: do I need to also have a contour in the upper half plane?

The imaginary part of the weighting function is positive in the upper half plane

The imaginary part of the two lower half plane poles is negative after the transformation x -> x+iepsilon

Secret

2:23 PM

The power set operator is injective, but not surjective

Any inverse of power set operator is not total

e.g. there are no sets such that its power set is the natural numbers

anakhro

2:41 PM

@Secret to speak of "surjective" you must first define its codomain.

Secret

well, the problem here is the "codomain" is the proper class, V

I don't know how functions that maps to proper class works

The powerset operator can be described as mapping between one set $S$ to another of cardinality $2^{|S|}$ along with the subset properties this new set has to obey

anakhro

@Secret en.wikipedia.org/wiki/Ordered_pair#Morse_definition

Secret

yeah, I think that will work

so $\mathcal{P}$ is not surjective because e.g. no set can map by a power set to the naturals

and $\mathcal{P} : V \to V$, as a definable function on classes

Leaky Nun

3:03 PM

$\bigcup$ is a left inverse of $\mathcal P$

Secret

$\bigcup 16 = 15 \neq 4$?

Leaky Nun

16 isn't the power set of 4

Secret

ok nvm

Shine On You Crazy Diamond

Every commutative boolean monoid has an associated commutative boolean ring

:math.stackexchange.com/questions/3402876/…

The finite monoid I'm working with is the set of subintervals in $\Bbb{N}$ of $[0, |s|)$ where $s$ is a finite string.

You can also limit the length of the subintervals and get a subring of that.

I'm actually interested in subintervals $[i,j)$ only up to length $j - i = \lfloor |s|/2\rfloor$.

Since any larger interval cannot result in a compression of $s$.

Shine On You Crazy Diamond

3:53 PM

@Ultradark

Rithaniel

4:15 PM

So, how would you find the automorphism group of the semidirect product $Q_8\rtimes Q_8$?

Leaky Nun

I wouldn't

Rithaniel

(Where the homomorphic image from $Q_8$ to $\text{Aut}(Q_8)$ is the klein four group)
Also, that is fair, Leaky

Mike Miller

4:40 PM

@Arrow I can't stay consistently but if you ask your question I might see it later

Arrow

Dear @MikeMiller, here is my question(s).

Poline Sandra

5:20 PM

hello, if $M_n=\max(u_n, v_n)$ why $u_n^3+v_3\geq M_n^3$

Poline Sandra

5:33 PM

have you an idea ? because by definition $u_n\leq M_n$

Thorgott

Take $u=1,v=2$, then $M=\max(u,v)=2$, but $u^3+v=1+2=3\not\ge 8=2^3=M^3$.

Poline Sandra

$u_n^3+v_n^3\geq M_n^3$

@Thorgott

Thorgott

ah, that is indeed true as long as $u,v\ge0$, because $M^3$ is either $u^3$ or $v^3$

Shine On You Crazy Diamond

5:49 PM

@Ultradark you could probably grasp my post, since you know what a ring is

I haven't brought in any theory yet, but i'll most likely be using basic properties of rings

Ultradark

okay I'll take a look!

@ShineOnYouCrazyDiamond can you have a colored (morphisms) category theory diagram

Shine On You Crazy Diamond

Yes, I do in BananaCats

though the color is just a user preference

Though it's not ready for use yet

Ultradark

oh cool

Mike Miller

6:04 PM

@Arrow I don't really have anything to say, sorry

Arrow

@MikeMiller no problem. Thanks anyway!

Shine On You Crazy Diamond

6:23 PM

-1

Q: Associative $+$ operation for a certain multiplicative subsets of a finite, commutative, boolean monoid?

Let $X$ be a finite, commutative, boolean monoid with a zero element $0$. Define the set $C$ of subsets of $X$ to be $C = \{ x \subset X : 0 \in x, $ and $ \forall a,b \in x, \ ab \in \{a, b, 0\}\}$. Define for $x, y \in C$: $$ x + y = \{0\}\cup (x \Delta y) \setminus \{u \in x \Delta y: \exis...

abstract-algebra elementary-set-theory associativity boolean-ring

anyone good at associativity proofs?

Jake Rose

6:49 PM

Can anybody explain the sub manifold bit?

Ted Shifrin

7:38 PM

@JakeRose If you're trying to figure out what a manifold (in Euclidean space) is, you might want to take a look at my YouTube lectures (linked in my profile) 3500 day 40 and 3510 day 23. The parametric definition you copied above appears as one of three equivalent ways of understanding manifolds.

Leaky Nun

a manifold is just [insert abstract and convoluted definition]

T_01

How can i determine $\lim\limits_{t \rightarrow 0}{\frac{v_1^2 \sin(tv_2)}{t^3v_1^4+tv_2}}$ where $||(v_1, v_2)|| = 1$ ?

Leaky Nun

Taylor is your friend

T_01

In what way could I use Taylor here?

MatheinBoulomenos

@LeakyNun the category of smooth manifolds embeds fully faithfully into the category locally ringed spaces over $\mathrm{Spec}(\Bbb R)$

Leaky Nun

7:44 PM

@MatheinBoulomenos can you classify all the subspaces $V$ of $\Bbb R^\Bbb N$ closed under $\delta$ satisfying the property that there is a linear map $f: V \to \Bbb R$ such that $f(v) = v(0) + f(\delta(v))$ for all $v$?

$\Bbb R^\infty$ is such a subspace

MatheinBoulomenos

what is $\delta$?

Leaky Nun

and no such subspaces can contain $(1,1,1,\cdots)$

the shifting operator

MatheinBoulomenos

no idea

Leaky Nun

ok thanks

T_01

@LeakyNun I guess you did not meant me, right? Uh, maybe L'Hospital works?

Leaky Nun

7:49 PM

@T_01 you would probably need to case on whether $v_2 = 0$

or $v_1$

T_01

and if none of both is zero?

Leaky Nun

@MatheinBoulomenos how about $C^k$ manifolds

@T_01 sure

T_01

@LeakyNun sure what?

Leaky Nun

maybe it's just $v_1^2$ afterall

MatheinBoulomenos

@LeakyNun same argument should work

Leaky Nun

7:50 PM

unless $v_2=0$ in which case it is $0$

@MatheinBoulomenos how about $C^0$ manifolds

MatheinBoulomenos

yeah, should work as well

Leaky Nun

and PL manifolds?

MatheinBoulomenos

idk

the argument for $C^k$ $k \in [0,\infty]$ does not work for PL manifolds (because I don't think PL manifolds admit PL partitions of unity)

Leaky Nun

I'm not sure whether for my presentation tomorrow I should cite the Albert--Brauer--Hasse--Noether--theorem without proof or define $\Bbb C \otimes_\Bbb R V$ as $V + iV$

MatheinBoulomenos

why do you need Albert-Brauer-Hasse-Noether?

Leaky Nun

7:53 PM

maybe I don't

doesn't it talk about Br(C/R)

or is that just LCFT

T_01

@LeakyNun sorry, i am a bit confused - so is L'Hospital the right way to go (as i can interprete enumator and denominator as functions in $t$) , or not?

Leaky Nun

@T_01 yes, but I like Taylor more than L'Hopital

MatheinBoulomenos

that's just LCFT and in the Archimedean case, it's just obvious

Leaky Nun

how about the A-structure thing

it can be generalized

MatheinBoulomenos

Albert-Brauer-Hasse-Noether is about global Brauer group

T_01

7:54 PM

@LeakyNun Could you explain, how i could use the taylor series to get the limes? I really do not now much about analysis, sorry

MatheinBoulomenos

@LeakyNun sure include that if you want, you don't need to cite me, I'm sure that's well-known to people who thought about it

Leaky Nun

I'm saying, I want to generalize things, but I don't know if I should

MatheinBoulomenos

you should

Leaky Nun

it hurts me to state things that I know can be generalized :P

but it hurts the audience to generalize

MatheinBoulomenos

no pain no gain

Leaky Nun

7:57 PM

I'm failing to figure out which language does mathematics in Latin

MatheinBoulomenos

I don't understand that sentence

Leaky Nun

Serbo-Croatian :o

Limes (matematika)

Limes je jedan od osnovnih pojmova u matematičkoj analizi. == Limes niza == Neka je ( a n ) {\displaystyle (a_{n})} niz realnih ili kompleksnih brojeva. Reći ćemo da niz ( a n ) {\displaystyle (a_{n})} konvergira broju L (realan ili kompleksan broj) ako vrijedi ( ∀ ϵ > 0 ( ∃ n…

Croatian:

Limes (matematika)

Limes je jedan od osnovnih pojmova u matematičkoj analizi. == Limes niza == Neka je ( a n ) {\displaystyle (a_{n})} niz realnih ili kompleksnih brojeva. Reći ćemo da niz ( a n ) {\displaystyle (a_{n})} konvergira broju L (realan ili kompleksan broj) ako vrijedi ( ∀ ϵ > 0 ( ∃ n...

anyway @T_01 taylor series gives you sin(tv2) = tv2 + t^3 v2^3/6 + ...

MatheinBoulomenos

Limes (mathematica)

Limes, in mathematica, est quantitas ad quam alia quantitas adpropinquit. Definitio haec est: lim x → a f ( x ) = b {\displaystyle \lim _{x\rightarrow a}f(x)=b} significat ∀ ϵ > 0 ∃ δ > 0 ut | x − a | < δ ⇒...

Leaky Nun

well that's Latin

did Euler use the word limes

@MatheinBoulomenos $\Bbb C \otimes_\Bbb R V = V + \overline{V}$

so in general $L \otimes_K V = \sum_{\sigma \in G} \sigma(V)$?

and this is the content of Galois descent?

MatheinBoulomenos

@LeakyNun it doesn't appear in his "introductio in analysin infinitorum"

Leaky Nun

8:11 PM

yeah he just summed a series by writing 1+1/1+1/1.2+1/1.2.3+&c.

MatheinBoulomenos

@LeakyNun yes

well, that follows from the normal basis theorem

Leaky Nun

well Galois descent also needs normal basis theorem oder

MatheinBoulomenos

no, I don't think you need it

Leaky Nun

what's the content of Galois descent?

MatheinBoulomenos

the version for finite Galois extension is: if $L/K$ is a finite Galois extension with Galois group $G$, then if we consider the category $L^*[G]$-Mod which consists of $L$-vector spaces together with a semilinear $G$-action, then $L^*[G]$-Mod is equivalent to k-Mod via thefunctors $V \mapsto V^G$ and $W \mapsto L\otimes_K W$

that's actually an equivalence of symmetric monoidal categories if we equip $L^*[G]$-Mod with the tensor product over $L$ and $K$-Mod with the tensor product over $K$

thus by abstract nonsense, you also get an equivalence of the corresponding categories of monoid objects and commutative monoid objects

Poline Sandra

8:17 PM

@Thorgott thank you

MatheinBoulomenos

which are $K$-algebras and $L$-algebras with a semilinear $G$-action by ring automorphisms (respectively the commutative ones for commutative monoid objects)

and then continuing by abstract nonsense, you get a category equivalence of the corresponding cogroup objects in commutative $K$-algebras which are dual to affine algebraic groups

there's also some Galois descent for projective varieties, but that's harder

Leaky Nun

I'm still trying to prove $L \otimes_K V = \sum \sigma(V)$ lol

MatheinBoulomenos

wlog $V=K$, then apply the normal basis theorem

Leaky Nun

no $V$ is an $L$-module and $=$ is an L-linear isomorphism

so ok wlog $V=L$

MatheinBoulomenos

oh sorry

yeah that's what I wanted

Leaky Nun

8:20 PM

$L \otimes_K L$

MatheinBoulomenos

you don't even need the normal basis theorem

primitive element is enough

Leaky Nun

$L \otimes_K L = L \otimes_K K[x]/(f) = L[x]/(f)$

MatheinBoulomenos

yeah

$f$ splits over $L$, Chinese remainder theorem, Galois group acts transitively on the roots

Leaky Nun

where's the $\sigma$

MatheinBoulomenos

Let $\alpha$ be a root of $f$, then $\prod(x-\sigma(\alpha))=f$

Leaky Nun

8:24 PM

wait

aha

no we have too little structure

I want $V$ to be a $L[H]$-module

where $H$ is an arbitrary group

MatheinBoulomenos

you probably want to replace the left factor in the tensor product by $K[x]/(f)$

Leaky Nun

now $V$ and $\sigma(V)$ are not isomorphic

I thought it would work by naturality

afterall a $L[H]$-module is a functor from $H$ to $L$-Mod

MatheinBoulomenos

why should it not work out?

Leaky Nun

well i can't even tell the difference between $L$ and $\sigma(L)$

ok so you say we replace the left factor instead of the right factor

MatheinBoulomenos

you have a $\mathrm{Gal}(L/K)$ action on $L$ (the left factor)

you can actually chase the Chinese remainder theorem isomorphism to see that is just permutes the factors

Leaky Nun

8:31 PM

$L \otimes_K V = K[x]/(f) \otimes_K V = \bigoplus K[x]/(x-\sigma(\alpha)) \otimes_K V$

MatheinBoulomenos

according to the action on the roots of $f$

Leaky Nun

ok let's chase the CRT

$K[x]/(f) \to \bigoplus K[x]/(x-\sigma(a))$

we send $p$ to $(p(\sigma(a)))_\sigma$

$L \to K[x]/(f)$ with $\alpha \mapsto x$

$L \to K[x]/(f) \to \bigoplus_\sigma K[x]/(x-\sigma(a))$

$\alpha \mapsto x \mapsto (\sigma(\alpha))_\sigma$

wait this is nonsense lemme start again

MatheinBoulomenos

you want $L[x]/(f)$

Leaky Nun

$L \otimes_K L \to K[x]/(f) \otimes_K L \to L[x]/(f) \to \bigoplus_\sigma L$

$\alpha \otimes 1 \mapsto x \otimes 1 \mapsto x \mapsto (\sigma(\alpha))_\sigma$

suppose $p(\alpha) = \sigma(\alpha)$ for some $p \in K[x]$

then $\sigma(\alpha) \otimes 1 \mapsto p \otimes 1 \mapsto p \mapsto (p(\tau(\alpha)))_\tau = \tau(p(\alpha)) = \tau(\sigma(\alpha))$??

the composition is reversed

@MatheinBoulomenos I'm confused

MatheinBoulomenos

hmm

Leaky Nun

8:46 PM

let $f : V \to \Bbb C \otimes_\Bbb R V$ be given by $f(v) := i \otimes v - 1 \otimes iv$

then $f(iv) = i \otimes iv + 1 \otimes v = i(1 \otimes iv - i \otimes v) = -if(v)$

so $f(zv) = \overline{z} f(v)$

@loch help

$g(v) := i \otimes v + 1 \otimes iv$

$g(iv) = i \otimes iv - 1 \otimes v = i(i \otimes v + 1 \otimes iv) = ig(v)$

$g(zv) = zg(v)$

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$Mathematics$ Mathematics