Mathematics - 2018-06-17 (page 3 of 4)

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MatheinBoulomenos

8:00 PM

@Rudi_Birnbaum I think you know some quantum mechanics, right? A Lie algebra is an algebraic structure that has similar properties to the commutator bracket $[-,-]$ or the Poission bracket

math.uconn.edu/~kconrad/blurbs/galoistheory/separable2.pdf

Def 5.3

@MatheinBoulomenos found

On the physical interpretation of the infinite expansion of the solution to $y = y'$, this question has received some positive attention: math.stackexchange.com/q/2822137/338817, but I'm wondering if anyone here has any additional insight.

@MatheinBoulomenos $[-,-]$

MatheinBoulomenos

@LeakyNun Keith Conrad's notes are truly a treasure trove!

yes indeed

8:02 PM

@MatheinBoulomenos: Yes, I am quite familiar with QM. So does it correspond to anything in QM?

MatheinBoulomenos

@Rudi_Birnbaum when you have a space of some operators on a Hilbert space, then taking the commutator of those operators i.e. $[f,g]=f \circ g - g \circ f$ is an example of a Lie bracket, and the space of operators together with that Lie bracket is an example of a Lie algebra

Oh, in physics you probably have some $i\hbar$ in there or something like that

the simplest example is a finite dimensional Hilbert space, i.e. $\Bbb C^n$ and you just take the commutator of matrices

I don't really know quantum mechanics myself, but I've been told that this is important

@MatheinBoulomenos: Yes thats kind of the essential difference to classical mechanics.

@MatheinBoulomenos: The main conclusion to be drawn from that (that the bracket $\ne 0$) is that small particles dont move in trajectories.

If you were studying a group what would you choose to study about it?

@geocalc33: Of course for the group it seems a very natural thing. What I was wondering just is if that would work with rings or fields as well.

MatheinBoulomenos

@Rudi_Birnbaum I've looked at those books you mentioned and they seem to only spend some part on finite groups, i.e. Fulton&Harris is mostly about Lie algebras with only a bit on finite groups in the beginnning and Vinberg has more material on compact groups and Lie groups than on finite groups

But from your questions, I assume that you're really interested in finite groups?

8:12 PM

I'm just asking what are some types of things you can investigate with groups?

@MatheinBoulomenos: Thing is that in practice I will only get in touch with finite groups. But in case I can learn much more for a small price I will prefer the latter.

MatheinBoulomenos

@Rudi_Birnbaum I'd say Fulton&Harris is definitely a good choice, though the chapters on Lie algebras may not be directly related to what you do

@MatheinBoulomenos I'm lucky I did tensor products

so does this look good?

class is_reduced_comm_ring (R : Type u) [comm_ring R] : Prop :=
(reduced : ∀ r : R, ∀ n : ℕ, r ^ n = 0 → r = 0)

class is_separable_intermediate_extension
  (F : Type u) [field F]
  (AC : Type v) [field AC] [is_alg_closed_field AC]
  [field_extension F AC] [is_algebraic_closure F AC]
  (S : set AC) [is_intermediate_field F AC S] : Prop :=
(separable : is_reduced_comm_ring (tensor_a F S AC))

@geocalc33: The thing which determines everything in chemistry and quantum mechanics is the Hamiltonian, an operator in Hilbert space. It commutes with the symmetry group elements of the molecule.

MatheinBoulomenos

@Rudi_Birnbaum Vinberg does compact groups after finite groups, which gets technically really fast because you need Haar measures

8:15 PM

@MatheinBoulomenos: I see, thank you.

MatheinBoulomenos

Fulton&Harris have an early chapter on real representations which seems especially relevant for you

@LeakyNun what is this?

an automated theorem prover?

not automated

it's a proof assistant

MatheinBoulomenos

ah, I see

github.com/kckennylau/local-langlands-abelian/tree/master/src

my repo

MatheinBoulomenos

@LeakyNun oh, a finite extension $L/K$ is separable iff the symmetric trace-bilinear form which sends $(a,b)$ to the determinant of the $K$-linear endomorphism of $L$ given by multiplication with $ab$ is non-degenerate

8:17 PM

structure finite_Galois_intermediate_extension
  (F : Type u) [field F]
  (AC : Type v) [field AC] [is_alg_closed_field AC]
  [field_extension F AC] [is_algebraic_closure F AC] :=
( S : set AC )
( intermediate : is_intermediate_field F AC S )
( finite : vector_space.dim F S < cardinal.omega )
( separable : is_separable_intermediate_extension F AC S )
( proj : Gal F AC → Gal F S )
( proj_commutes : ∀ f : Gal F AC, ∀ x : S, ((proj f).to_equiv x).1 = f.to_equiv x.1 )

MatheinBoulomenos

(at least one direction from that is really important for ANT)

@geocalc33: The eigenvalues of the Hamiltonian tell us about the permissible energy levels of any molecule. So its nice to know group theory.

@geocalc33: And the eigenvectors tell us where the electrons "are".

MatheinBoulomenos

@LeakyNun now about Galois: $L/K$ is Galois iff elements of $K$ are the only ones fixed by every $K$-automorphism of $L$

@geocalc33: The permissibilty of any molecular process (light absorption, magnetic property) finally in described by an "integral" between <eigenvector|Operator|eigenvector>, then is good to be able to deduce from symmetry if they possibly can be \ne 0

@MatheinBoulomenos nice

oh and if I have the topology on Gal(F-bar/F), do I have a topology on Gal(F-sep/F) for free?

MatheinBoulomenos

8:22 PM

these sets are the same

oh right

which direction is easier?

@loch no algebraic topologist specializes in homotopy of spheres that topics was big in the 70s now it’s just a part of topos theory

MatheinBoulomenos

You always have surjection Aut(F-bar/F) -> Aut(L/F) for every algebraic extension L/K (I prefer to write Aut instead of Gal if I don't know if the extensions are algebraic)

@Zee Is there any way to explain to a layman what a "topos" is?

MatheinBoulomenos

The kernel of that is Aut(F-bar/F-sep). To see that this is the trivial group, note that this is purely inseparable and the automorphism group of a purely inseparable extension is always trivial (think about the minimal polynomial of an element, it has only one root, so permuting those is not very interesting)

Hi @Arrow

8:25 PM

Hello @MatheinBoulomenos :) How are you?

@MatheinBoulomenos don't you need L to be normal

MatheinBoulomenos

@Arrow I'm doing fine, thanks. How are you?

@LeakyNun F-sep is normal, but yeah, it's better to assume that

@Rudi_Birnbaum you could say an elementary topos is a context in which you can use much of ordinary mathematical language, such that its meaning will depend on the topos.

MatheinBoulomenos

you can still lift automorphisms, but it's not so clear what the kernel might be

@MatheinBoulomenos I'm good. Your answer was very enlightening but also raised some more questions for me. I finally feel I'm getting the bigger picture though.

MatheinBoulomenos

8:28 PM

@Arrow that's nice to hear

@Arrow about your suggestion with considering $R[G]^G$-modules, I don't see what that gives you over for example taking the ideal-submodule product $R[G]^GV$, this will always be contained in $V^G$ and if $|G|$ is invertible in $R$, then it is equal to $V^G$

@Arrow: So that might hold true for any category, I guess. In how far does it generalize the topological group?

@Rudi_Birnbaum they are categories of sheaves

@MatheinBoulomenos yes, I let that go. Here's my latest question.

@Zee: And the sheave is some "linear" approximation of a general space in one point, right?

@Rudi_Birnbaum I am not sure where you're coming from. Any category generalizes a group (=one object groupoid). A Grothendieck topos can be viewed as a sort of categorification of the notion of locale.

8:32 PM

@Fargle: I see you're taking my name in vain again!!

@TedShifrin D:

hi @Mathein, @Arrow, @Leaky

MatheinBoulomenos

Hi@Ted

hi @Ted

Hello @TedShifrin, how are you?

8:33 PM

Heya @EricSilva.

yoyo how goes it @Ted

@Rudi_Birnbaum I don't think there's anything inherently linear about the notion of sheaf. Sheaves can take values in many different sorts of categories.

Just gave and graded my AoPS final exams. Meh.

oh did the kiddos not do so hot or something

Hello @MikeMiller, how are you?

8:35 PM

Well, my superstar kid (who won the junior AMO) didn't take it. My second best student did OK. But one problem with having only on-line homework is that they are soooo sloppy and haven't been trained to write anything clearly. Next year I'm determined to give some hand-graded homework in calculus (whether they do it is another matter).

hi

heya Mike

MatheinBoulomenos

Hi @Mike

@TedShifrin and the writing prompt graders are extremely inconsistent

Hey @Mathein @Leaky @Ted usw.

8:36 PM

Hey guys!

MatheinBoulomenos

@ÍgjøgnumMeg @Daminark hi

Hello @everyone

@POTUS

I don't have the energy to rewrite their answers

Oh, the on-line kids have written stuff, @MikeM? None for the brick-'n-mortar. I wrote a bunch of my own stuff for them to work on, though, but not part of homework.

8:37 PM

yeah, there are writing prompts once per week.

Different for me.

I am going to try to deviate from the calculus course some since there are hardly any students anyhow. I'd rather cover less, do a few proofs, and make them learn some math better.

I think the online mechanism (the way the classroom is set up) is quite good but I do not have necessarily exceptional feelings about the scripts or graders or

I'm a heretic, I realize.

MatheinBoulomenos

@Arrow about your recent question, I'm not sure how to answer that, but note that the isomorphism between induction and coinduction does not depend on $|G|$ or $(G:H)$ being invertible in $R$, in fact, in group cohomology, the isomorphism of induction and coinduction for $\Bbb{Z}[H] \to \Bbb{Z}[G]$ where $H$ has finite index in $G$ is frequently used

I have complained a lot about the precalculus course.

8:38 PM

@TedShifrin hello, please can you give me the expression of the complementary of the following set :$$\{(x,y):x^2+y^2\le1\}\cup\\\{(x,y):-1\le x\le1\text{ and }y\le0\}$$

i found : $]1,+\infty[\times]1,+\infty[\cup ]-\infty,-1[\times]1,+\infty[$ but i think that is not true

I guess I don't want to leave that public

I probably shouldn't, either. But oh well.

I need a picture, @Poline. Have you drawn one?

@TedShifrin what's wrong with it

I am taking on 3 more courses for them in the next quarter and then I think I'm done

It stresses me out when the kids get lost bc bad example and I am not sure how to get them back on track

the green picture: i.imgur.com/o4wLImc.png

8:40 PM

It's not precalculus after the first half, Eric. It's a linear algebra course that I could improve on a lot.

oh odd

@MatheinBoulomenos but it seems dependent on the invertibility of the order of $H$ in the ring. And the explicit isomorphism in Qiaochu's answer seems mysterious in general. Is there a canonical arrow in the other direction perhaps?

And they totally went off the deep end doing some linear fractional transformation stuff.

seems an odd thing to do for little kiddos

OK, @Poline.

Eric, I know you'll find it hard to believe, knowing me as you do, but I was vocal with criticisms.

8:41 PM

lolol

btw @Ted gonna eat some Venezuelan food tonight

Oooh, nice, Eric.

MatheinBoulomenos

@Arrow, no the order of $|H|$ doesn't have to be invertible in $R$, see for example Prop. 7.5.4 on p.133 here: math.ucla.edu/~sharifi/algnum.pdf

v excited

@Poline: So shouldn't we have $]1,\infty[ \times ]-\infty,\infty[$, for example? Spacing doesn't work write with European intervals.

MatheinBoulomenos

that's for $R=\Bbb Z$, but it probably works in general

8:43 PM

@EricSilva assuming there aren't private parties taking up the restaurant space all week

i hope not :(

being turned away a second time would be sad

@MatheinBoulomenos ah, perhaps the key that we want invertibility in order to have some splitting, rather than "merely" an isomorphism.

@TedShifrin i don't understand

MatheinBoulomenos

@Arrow yes, that sounds plausible

@MatheinBoulomenos because I bug you with details, perhaps I should first ask: what is the significance - in the representation theory of finite groups - of the fact induction and coinduction coincide?

8:45 PM

we have the complement of the ball with $|x|>1$ or $y>0$

@TedShifrin

Where is the right adjointness of induction to restriction of scalars actually used?

@MatheinBoulomenos i'm an idiot

I just need to talk to the second years about lean

that's what my professor always told me

somehow I always forget

@Poline: So any point with $|x|>1$ is in the complement. Now what about when $|x|\le 1$?

Ah, @Arrow and @Mathein and @Leaky can have their own kindred spirit club now ...

that's why my professor wanted me to do it in Lean

@Ted Thanks for having G-P errata, by the way. I was stuck on a problem for a while until I saw it.

8:48 PM

if $|x|<1$ we see if $y>0$ then it is in the complement if $y<0$ then $(x,y)$ is not in the complement @TedShifrin

Yeah, Griffiths/Harris also have tons of errors. I kept handwritten notes and never typed them all up. I don't think they corrected them, either.

MatheinBoulomenos

@Arrow I'm not that familiar with representation theory beyond the basics, but for group cohomology it is quite relevant

it gives you a stronger version of Schapiro's lemma, among other things

@Poline: No, remember that if $|x|\le 1$, we have to be outside the disk $x^2+y^2\le 1$.

@Fargle: At least with my diff geo text, I've corrected all the errors instantaneously as soon as I became aware of 'em :P

@MatheinBoulomenos I don't know anything about group cohomology. All I want is to understand the basics conceptually :D

And I do appreciate it.

8:50 PM

so $y$ must be greater then 1 @TedShifrin

@EricSilva i dont kjnow what venezuelan food is like

Not quite, @Poline. Only when $x=0$ is that so.

MatheinBoulomenos

@Arrow a lot of group cohomology is just about understanding $\Bbb{Z}[G]$-modules

of course the focus is on some Ext and Tor functors

why ? the point (1/2, 2) is in the complement

Sure, but what about the point $(.5,.9)$?

8:53 PM

@MikeMiller it's sick, arepas are big, and they're great, if u have ever had any south american food there are big similarities, i like pabellon venezolano a lot, that's like a particularly venezuelan dish

@EricSilva: You need to come cook a few meals at my house and we'll invite Mike.

im down

@MatheinBoulomenos ah! If you like derived functors perhaps you'll find this answer satisfying. It relates universal delta functors with Kan extensions along localizations.

MatheinBoulomenos

@Arrow since you seem very interested in functors between categories of modules over say $R[G]$ and $R[H]$ where $H$ is some subgroup of $G$, it might be worth to look at some introductory notes on group cohomology, one usually takes $R=\Bbb Z$, but most stuff works more generally I think. We have restriction, corestriction, induction, coinduction, inflation and coinflation in group cohomology :)

@TedShifrin $(1/2,2)$ is in the complement

8:56 PM

@MatheinBoulomenos I am actually not very interested in such functors. I am just taking a course on representation theory and trying to understand the magic.

MatheinBoulomenos

ah, okay

@Poline: I answered you (without a ping) earlier. Sure, but what about $(0.5, 0.9)$?

@MatheinBoulomenos I don't really know any math :D I can't understand much without categorical language

And I can't understand much at all with categorical language.

Good thing I'm retired.

Alessandro Codenotti

Hi everyone

8:57 PM

hi demonic @Alessandro

Hi @Alessandro

MatheinBoulomenos

Hi @AlessandroCodenotti

no that's why we need y>1

@Poline: Try again.

$y\geq 1$

8:59 PM

@MatheinBoulomenos perhaps I can bug you about something small from these short notes? Certainly there's the canonical map $X^G\to X_G$ from the limit of a functor to its colimit. However, what can be said about all maps in the opposite direction?

No, @Poline.

is there a way to count the number of vertical asymptotes of a complex function

You should take my suggestions seriously.

@geocalc33: I'm not sure what that means. But only around a pole does $|f(z)|\to\infty$. Around essential singularities the function takes on almost every value.

(0.5,0.9) is under the ball $\{(x,y), x^2+y^2<1\}$

Did you compute?

9:01 PM

@MatheinBoulomenos in your answer you explained that natural arrows $X\to X^G$ are easy to define representably via Yoneda, but if we're being pedantic $X,X^G$ do not live in the same category - the former is an $R[G]$-module while the latter is an $R$-module.

ohh you are right

Cela se passe de temps en temps :P

MatheinBoulomenos

@Arrow you can think of $X^G$ as living in $R$-mod or living in $R[G]$-mod

it's the largest $R[G]$-submodule of $X$ such that the action of $R[G]$ factors over the counit $R[G] \to R$

for adjunction purposes, it's of course nicer to say that the functor $X \to X^G$ takes values in $R$-mod

so i don't know how to do , we must consider $x\leq0.5,$ $x\geq0.5$

Yes, that's my choice of pedantry

Living in $R$-modules

Are natural arrows $X_G\to X^G$ possible to find as easily?

Would the sum of elements of $G$ appear canonically again?

9:04 PM

@TedShifrin i was primed to read portuguese so when i tabbed in and saw this comment my brain exploded from the romance language switch misfiring in my brain

@MatheinBoulomenos ah clearly im not very good with the internet

o. .o

LOL @EricSilva: It shows to go you shouldn't pay detention.

MatheinBoulomenos

but it's a $R[G]$-submodule of $X$! you can choose your pedantry as you like, but I don't see anything wrong with saying that it is an $R[G]$-module

i actually went cross eyed for half a second lol

9:05 PM

Can I get my identicon back if I changed my profile picture ?

Well, I tried to misfire you in English as well, @Eric :P

lolol

@MatheinBoulomenos because if I decide it's an $R$-module then it can only become an $R[G]$-module upon restricting scalars along the counit, and that's something I cannot omit without being confused.

@MatheinBoulomenos I try to be very careful with notation because I get lost easily with all the bimodule structures

@Adam: Those are unrelated.

what do u mean

MatheinBoulomenos

9:06 PM

@Arrow, oh you wouldn't like how I write stuff. I like to omit forgetful functors if it's clear e.g. from the index in the Hom set

@MatheinBoulomenos it's probably convenient when you really understand things, but for me the notation is very helpful. It's easy to see where to use adjunctions, etc

@Adam: Perhaps I misunderstood. Never mind.

okay

if you have a function where the complex part goes to positive infinity and the real part goes to negative infinity what kind of singularity is that?

That makes no sense, @geocalc.

9:09 PM

why

What example are you thinking of?

just wondering in general

Give me an example.

ted: if |x|<1/2, then y must be greater then 1/2 ?

@Poline: No, you're falling in a trap of trying to oversimplify, I think. If $|x|\le 1$, you need to say $y>\sqrt{1-x^2}$.

MatheinBoulomenos

9:14 PM

@Arrow if we consider $R$ as a $R[G],R[G]$-bimodule via the counit (or maybe some people prefer to call that augmentation map), then we have for the coinvariants $X_G \cong R \otimes_{R[G]} X$, so Hom-tensor adjunction gives us that for any $R[G]$-module $Y$, we have $\operatorname{Hom}_{R[G]}(X_G,Y) \cong \operatorname{Hom}_{R[G]}(X,Y^G)$

So canonical maps $X_G \to X^G$ are the same as canonical maps $X \to X^G$

which we already covered

Ah, fantastic. Yoneda ensures sections downstairs corresponds to natural sections too, it seems. Great!

This gives a simple case of isomorphy of induction and coinduction (along a terminal arrow from a group). It seems the general case should be very similar, but I still can't find my footing. I have no intuition for the adjoint triple induced by a ring arrow. Maybe I should look for some geometric intuition from schemes

coinduction would be some sort of "co-pullback".. dafuq

MatheinBoulomenos

@Arrow I don't have much intuition for that myself, I only started thinking about it due to your questions

@TedShifrin if you have a group what are some things you could study about it?

@MatheinBoulomenos if $A$ and $B$ are $R$-algebras, then are $(A \otimes_R B) \otimes_A (A \otimes_R B)$ and $A \otimes_R (B \otimes_R B)$ isomorphic?

If so, in which category?

prof ted: so we have $]-\infty,-1]\times]-\infty,+\infty[\cup [1,\infty[\times]-\infty,\infty[\cup ]-1,1[\times [\sqrt{1-x^2},+\infty[$

9:23 PM

@geocalc33: I'm not an algebraist. I'm interested in how the group acts on spaces. But since you're beginning, you might ask things like: If the group isn't abelian, can more than half the elements be their own inverses? You could play around with that fraction.

@Poline: You'd better pay attention to the direction of the brackets. And pay attention to $x=1$.

It is abelian but I'm interested in how the group acts on spaces

Most interesting groups aren't abelian, though.

MatheinBoulomenos

@LeakyNun that's just associativity of the tensor product

woah

shots fired

@MatheinBoulomenos regarding Maschke's theorem: here's a neat viewpoint. A functor $\mathsf C\overset{F}{\longrightarrow}\mathsf D$ induces a natural transformation $\mathsf C(-,=)\Rightarrow \mathsf D(F-,F=)$. If $F$ is a left adjoint, this natural transformation has a retraction iff the unit of adjunction has a retraction. If $F$ is a right adjoint, this natural transformation has a retraction iff the counit has a section.

But having a retraction to $\mathsf C(-,=)\Rightarrow \mathsf D(F-,F=)$ implies that $F$ reflects left/right/two-sided inverses, so in particular splitting of short exact sequences - the content of Maschke's theorem.

9:25 PM

@Ted I sent you a thing, for when you get around to it.

prof ted: the point (1,0) is not in the ball

@TedShifrin $\mathbf{Z}$ is p cool tho

@MatheinBoulomenos why?

Your original problem had the closed ball, @Poline.

MatheinBoulomenos

$(A \otimes_R B) \otimes_A (A \otimes_R B) \cong A \otimes_R ((B \otimes_A A) \otimes_R B) \cong A \otimes_R (B \otimes_R B)$

9:26 PM

OK, @Fargle. I'll look. I'm about to do an hour-long web-class with two high school kids, so I have to think a bit about what we're gonna do.

@geocalc: Talk to other people about this. I don't have time right now.

MatheinBoulomenos

@LeakyNun it works in whatevery category you likely need: $(A,B)$-bimodules or $R$-algebras for example

yes i'm sorry

MatheinBoulomenos

Or maybe I should say $A \otimes_R B^{op}$-modules that's a bit stronger than $(A,B)$-bimodules

@MatheinBoulomenos how does the first step work? in which category? the bases of the tensoring are different...

@Daminark

9:28 PM

$]-\infty,-1[\times]-\infty,+\infty[\cup ]1,\infty[\times]-\infty,\infty[\cup [-1,1]\times ]\sqrt{1-x^2},+\infty[$

MatheinBoulomenos

@LeakyNun you prove associativity of tensor products with different bases if you do stuff right

is it right prof ted?
$]-\infty,-1[\times]-\infty,+\infty[\cup ]1,\infty[\times]-\infty,\infty[\cup [-1,1]\times ]\sqrt{1-x^2},+\infty[$

MatheinBoulomenos

@LeakyNun wait, are $A$ and $B$ commutative?

sure

@MatheinBoulomenos I think it's something like $(A \otimes_R B) \otimes_A (A \otimes_R B) = X \otimes_A (A \otimes_R B) = X \otimes_R B$

that I have seen before, I think

actually is it true

MatheinBoulomenos

@LeakyNun that is true yeah

9:31 PM

why?

MatheinBoulomenos

but if everything is commutative, then I would just write it as $(B \otimes_R A) \otimes _A (A \otimes_R B)$

sure

Yes, @Poline. I have trouble reading European brackets, but I think it's right.

MatheinBoulomenos

associativity of tensor products as I said. If you do stuff right, then you prove associativity for different bases

could you state the general theorem for associativity?

9:33 PM

$(-\infty,-1)\times(-\infty,+\infty)\cup (1,\infty)\times(-\infty,\infty)\cup [-1,1]\times (\sqrt{1-x^2},+\infty)$ like this prof ted

Yup. It's fine :)

MatheinBoulomenos

If $A,B,C,D$ are rings (we don't need commutative and noncommutative is instructive to make sure we don't write stuff that is unnatural) and $M$ is a $(A,B)$-bimodule and $N$ is a $(B,C)$-bimodule and $K$ is a $(C,D)$-bimodule, then we have a natural isomorphism of $(A,D)$-bimodules
$M \otimes_B (N \otimes_C K) \cong (M \otimes_B N) \otimes_C K$

Oh, no, the second one isn't

@MatheinBoulomenos thanks

@Poline: You have to write the second explicitly as $\{(x,y): x\in [-1,1], y>\sqrt{1-x^2}\}$.

9:35 PM

ok thank you very much

@MatheinBoulomenos can I do it for algebras?

MatheinBoulomenos

@LeakyNun I'm not sure about the most general form for algebras

ok

@MatheinBoulomenos I have a group that is abelian and another group that is a symmetry of it. would these be two distinct groups or just considered the same group?

MatheinBoulomenos

@geocalc33 these are two different groups

9:38 PM

okay

MatheinBoulomenos

@LeakyNun but if you actually need it, you can just take the isomorphism and check if it respects multiplication

I see

MatheinBoulomenos

the forgetful functor from $R$-algebras to $R$-modules (or even Sets) is conservative

so the first one is abelian, and the second one has inverses that do not form a group

but the second group still has inverse elements, they just don't form a group

@MatheinBoulomenos I'm trying to follow this carefully but I'm confused. For me $(-)_G\cong \varepsilon_!R[G]\otimes_{R[G]}-$ where $\varepsilon _!R[G]= \!\!\; _{R}R_{R[G]}$. I am not sure about the second isomorphism or how it gives the conclusion. (Sorry!)

9:41 PM

@ÍgjøgnumMeg

MatheinBoulomenos

i'm not sure about the $!$ notation

Why? lower shriek for the left-most adjoint of an adjoint triple

MatheinBoulomenos

it depends if you want to consider $X_G$ as a $R$-module or $R[G]$-module

Definitely an $R$-module

(took the notation from this answer)

MatheinBoulomenos

then you can compose with restrictions along the counit as you wish

9:45 PM

@MatheinBoulomenos what if the group has inverse elements that don't form a group

MatheinBoulomenos

$\operatorname{Hom}_{R[G]}(X_G,X^G) = \operatorname{Hom}_R(X_G,X^G)$ (you won't like that notation ;-) )

@geocalc33 what do you mean by "inverse elements that don't form a group"? This doesn't make much sense to me

MatheinBoulomenos

@geocalc33 I don't think that question makes sense

@ÍgjøgnumMeg the groups inverse elements don't form a group

The inverse elements are just the same as all the elements.

If you have a group.

You need to learn definitions.

9:47 PM

Every element in a group is the inverse of some element in the same group so the question doesn't make sense

oh

I'm dealing with functions thogh

That doesn't matter

so my inverse elements are all reflections over the line y=x

I think perhaps you should take @Ted's advice

okay i figured it out

9:50 PM

Here's all I see: $\begin{aligned} \!\!\;_{R}\mathsf{Mod}(X_{G},Y) & \cong \!\!\;_{R}\mathsf{Mod}(\varepsilon_{!}R[G]\otimes_{R[G]}X,Y)\\
& \cong \!\!\;_{R[G]}\mathsf{Mod}(X, \!\!\;_{R}\mathsf{Mod}(\varepsilon_{!}R[G],Y))\\
\\
\end{aligned}$

MatheinBoulomenos

@Arrow I don't think Hom-Tensor applied to $R$ as a $(R,R[G])$-module gives you the same result

the second argument in the last Hom functor is $Y$ equipped with a trivial action

that's the adjunction from my answer on your first question about that

@ÍgjøgnumMeg @MatheinBoulomenos

MatheinBoulomenos

@Arrow maybe this is even an important point? The trivial actions from the left and right are compatible

user image

Is that right?

MatheinBoulomenos

I don't understand what this means

9:55 PM

there's no context

$ G_1 $ is group 1

actually it's one element in group 1

and the -1 means inverse

@MatheinBoulomenos maybe. I am not following so quickly. I agree we have $\begin{aligned} \!\!\;_{R}\mathsf{Mod}(X_{G},Y) & \cong \!\!\;_{R}\mathsf{Mod}(\varepsilon_{!}R[G]\otimes_{R[G]}X,Y)\\ & \cong \!\!\;_{R[G]}\mathsf{Mod}(X, \!\!\;_{R}\mathsf{Mod}(\varepsilon_{!}R[G],Y))\\ & \cong \!\!\;_{R[G]}\mathsf{Mod}(X, \varepsilon ^\ast R\otimes _R Y)) \\ \end{aligned}$ where $\varepsilon ^\ast R= \!\!\;_{R[G]}R_R$.

MatheinBoulomenos

@Arrow this was not my argument

I considered $\operatorname{Hom}_{R[G]}(X_G,Y)$ where $Y$ is a $R[G]$-module

@geocalc33 can you tell me what a group is?

Slightly confused about your questions

a set of elements with inverses, closure and a binary operation

MatheinBoulomenos

9:58 PM

@Arrow note that in this answer, $S$ is viewed both as a $(R,S)$ and a $(S,R)$-module to construct the adjoint triple. You seem to want to avoid the action on one side as it seems (is that right?)

Bleh. I have to go figure out what adjoint triple we're using then. Because it won't be the one induced by the ring arrow given by the counit

MatheinBoulomenos

yeah, I'm viewing $R$ as a $R[G],R[G]$-module as I said

I don't think so: my adjoint triple makes use of both. I have $\varepsilon ^\ast R= \!\!\;_{R[G]}R_R$ and $\varepsilon _!R[G]=\!\!_RR_{R[G]}$. So both sides are used.

@geocalc33 and it should have an identity element!

@geocalc33 in any case, perhaps you could formulate your questions with a little more context because, as it stands, I can't make much sense of anything

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