Mathematics - 2020-09-14

Jeff

03:03

I am told that $\lim_{x \to 6} \frac{f(x)-15}{x-4} = 8$ and $\lim_{x \to 4}\frac{f(x)-15}{x-4} = 3$. I've easily determined that $\lim_{x \to 6} f(x) = 31$. How do I find the $\lim_{x \to 4} f(x)$?

cubesnyc

i am looking over a research paper on poker suit isomorphism, and one of hte introductory paragraphs is rather cryptic to me

the paragraph reads as follows:

Our first building block will be a procedure for indexing
M-rank sets, which are sets of M cards of the same suit.
With out loss of generality, let us consider the ranks to be in
{0, 1, . . . , N} and consider a set in decreasing order.
The colex function (Bollobas 1986), provides precisely
this indexing. As much of the indexing scheme is based off
the same principles, we describe the construction here.
Without loss of generality, we order the sets lexographically. For example if M = 2, h4, 1i > h3, 0i > h2, 1i. Given

the actual paper is here

cs.cmu.edu/~kwaugh/publications/isomorphism13.pdf

i am having a hard time following his illustration that <4,1> > <3,0> > <2,1> or what that even means since it seems like it wasnt explained anywhere?

Jeff

04:16

@Jeff Is it as simple as the same method for the first one, but realizing that since $\lim\limits_{x \to 4}\frac{-15}{x-4}$ DNE then the whole limit can't exist? Also, what happened to the "Reply" feature in this chat?

Ted Shifrin

05:36

@Jeff The reply did just disappear. Dunno! If the limit the quotient exists and the denominator goes to $0$, what must the numerator go to?

Astyx

10:25

Why does the Dolbeault cohomology use the $\bar \partial$ operator and not the $\partial$ one ?

Mike Miller

10:47

@Astyx You want the 0th cohomology to be holomorphic forms

Astyx

Oh that makes sense

What happens if you take $\partial$ ? I expect you get something isomorphic ?

Mike Miller

I don't really remember! I think that might interchange (p,q) and (q,p) cohomology

Yeah seems right

Balarka Sen

Scary

Astyx

I think the isomorphism is just conjugation actually

Mike Miller

Yeah

Knight

12:41

Can someone tell me where is Ron Maimon these days?

Mike Miller

Hasn't he been banned for years

Knight

I thought someone might now where is he now

Mike Miller

At home or work, most likely

Googling his name in quotes quickly shows he doesn't seem to be active much anywhere on the internet under his own name

Knight

Okay

adamaero

14:06

What's the name for the curve plotting points produces? It's used in controls engineering...

tip of my tongue..

With Root Locus...

A term to describe the plotting of a shape?

Or the path..

Lol, I think it's just the locus.

Leaky Nun

zero-locus?

Astyx

14:29

Is the point of the Dolbeault cohomology that it's a finer one than de Rham cohomology because it imposes holomorphic forms ?

But it's still the same as any complex cohomology no ?

I've heard all cohomology theories were "isomorphic"

Mike Miller

14:42

@Astyx I don't understand this statement

@Astyx Any collection of groups $E^\bullet(X,A)$ whose input is a pair of a topological space $X$ and a subspace $A$, and which satisfies the Eilenberg-Steenrod axioms, is canonically isomorphic to singular cohomology $H^*\bullet(X,A;G)$ with coefficients in some Abelian group $G$

The input for Dolbeault cohomology is a complex manifold, which is more structure than a topological space and for which that structure exists only on very few topological spaces

It's completely incomparable, at least from abstract principles like you're talking about

If $M$ happens to be a closed Kahler manifold it follows from Hodge theory that $H^n_{dR}(M; \Bbb C) = \bigoplus_{p+q=n} H^{p,q}_{Dolbeault}(M)$

But this requires work and is again nothing to do with general principles

Astyx

@MikeMiller Oh ok, I didn't know the details of this

Hence my confusion

Mike Miller

Nobody should have told you that without telling you what it means :(

Astyx

Many people have haha

Mike Miller

There are far more things that people call cohomology than things satisfying the Eilenberg-Steenrod axioms

Astyx

So homology of complex manifolds is "richer" than homology of topological manifolds

Mike Miller

14:52

It's unrelated

They measure different things

That there is a relationship at all is miraculous and what gives rise to the modern subject of Hodge theory

Astyx

Cool, thanks !

Balarka Sen

15:29

The Dolbeaut double complex filters the de Rham complex, it's a structure coming out of the complex structure - this is the best you can say

The double complex cohomology of del + delbar is de Rham, but how is it related to del and delbar cohomology? I guess is unclear

Mike's statement is saying if M is Kahler then the spectral sequence degenerates at E^2

Astyx

Do you know JP Demailly's Complex Analytic and Differential Geometry ? Do you recommend it ? @MikeMiller

Mike Miller

I know nothing, especially about complex geometry books

Astyx

www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf

Mike Miller

Seems very specialized to me. I dunno what your goal is

I've personally never heard Lelong numbers much less read 70 pages about them

Astyx

I'm having a course that covers roughly the first chapter this week

Balarka Sen

15:36

Yikes

Very scary

Mike Miller

Sounds like an SCV course

Balarka Sen

Whoa wait

Pseudoconvexity

Great stuff!!

I will read

Mike Miller

Tfw

Astyx

It looks cool if you ask me

Alessandro Codenotti

That's a scary looking book

Adeek

15:48

hi @BalarkaSen @MikeMiller

Astyx

hi

Adeek

Hi @Astyx

I want to discuss a theory I am developing with someone interested in algebraic geometry.

anyone wants to discuss

Mike Miller

Hi; I've got nothing to contribute about that, sorry

Adeek

it is a theory that would connect geometry and algebra

@MikeMiller fair.

@Astyx would you like to discuss ?

does Ted still come here ?

Astyx

I'm not very knowledgeable in algebraic geometry

I doubt I could contribute anything useful

Adeek

15:54

I see.

@TedShifrin can you check my email in regards to a theory that in development by me.

@TedShifrin Thank you so much.

@MikeMiller how are you ? Are you a postdoc now ?

Tobias Kildetoft

Isn't algebraic geometry precisely a theory that connects geometry and algebra?

Adeek

Sure I mean I guess you could call Cohomological analysis

my theory

It would connect a lot of areas not just algebra and geometry.

but also analysis as well.

@TobiasKildetoft

Balarka Sen

Seems useful

Adeek

@BalarkaSen very useful yeah.

I emailed Ted the main idea just want some critique towards it.

I am planning to publish it during my PhD.

Balarka Sen

Do you use higher category theory

Adeek

16:02

yes

@BalarkaSen can I email you the idea

Balarka Sen

I wouldn't understand it

Adeek

I put it in layman's term

The main idea for people that don't know any higher categories

Balarka Sen

Oh but I'm just not familiar with much of algebraic geometry, or algebra or geometry for that matter. I think Ted will have useful feedback

Just cool to hear you used higher category theory in algebraic geometry to connected algebra and geometry

Adeek

yeah and I will generalize K-theory and regulators and other stuff as a way to link different cohomologies generated by my theory and provide a way to measure perspective.

I call it cohomology of perspective

my theory.

Mike Miller

I'm technically a postdoc, yeah. I'm not paying much attention now because I'm preparing some calculus lectures

Adeek

16:07

oh I see @MikeMiller very cool.

brb

Adeek

16:48

Hi Ted. "Sorry left my computer unattended"

Knight

16:58

@MikeMiller What level?

Adeek

Ted if you could reply back from email that would be nice :)

Ted Shifrin

I just got it. Honestly, I know nothing about this stuff, so I cannot say anything useful.

Adeek

@TedShifrin I see no problem :). I hope everything is ok with you.

Ted Shifrin

Pandemics, fires, dangerous incompetent president ... yup, just fine.

Adeek

yes that is why I asked :S

I hope things goes towards the better soon

Ted Shifrin

17:10

We all do ... :)

Silent

Mike Miller

@Knight multivariable integration

Silent

Since above $y$ is transcendental over $F$, we have that $F(y)$ is isomorphic to $F(x)$. But, then again, in the above result we claim that $[F(x) : F(y)]$ may be greater than one! I am not able to digest this.

Leaky Nun

$[F(x) : F(x^2)] = 2$

Knight

If you don’t mind terribly I would like to have your lecture (whether it is video or PDF).

Silent

17:15

Sure!

Leaky Nun

put $y = x^2$ in your query

user193319

Given a ring $(R,+,\cdot)$, let $(R^{op},+,\star)$ be the opposite ring formed from it. I am trying to show that $Hom_{R}(R,R) \cong R^{op}$ as rings. What are the ring operations on $Hom_{R}(R,R)$? Usual function addition and composition?

Leaky Nun

yes

note that the Hom is R-module Hom, i.e. R-linear maps

you can't add two ring homs together

user193319

You can't add two ring homomorphisms? Why can't I define $f + g$ as $(f+g)(r) := f(r)+g(r)$? Why doesn't that work?

Thorgott

of course you can add them, but the resulting map is usually not a ring hom anymore

user193319

17:25

Oh, I am confused...so then what are the ring operations on $Hom_{R}(R,R)$?

Mike Miller

I'd rather not share lectures with people not taking the class, sorry.

I'm certain that there are very polished lectures on the same material you can find elsewhere

Knight

@MikeMiller I understand the reason, there are chances of stealing, eh?

Silent

@LeakyNun Thanks :)

Thorgott

pointwise addition as addition and composition as multiplication @user193319

Knight

I shouldn’t have asked for it in first place, I never thought how corrupt someone might estimate me.

Ted Shifrin

17:31

It's not a matter of corruption, @Knight.

Knight

Is it about fees?

Ted Shifrin

But there's an implicit contract with the students that this is a course for them, not to have their names and faces made public. When my lectures were videoed and put on YouTube, my students agreed to it.

Mike Miller

There's nothing interesting enough to steal from a calculus lecture, I assure you

Ted Shifrin

I've had sleepless nights wondering if people might steal my computation of gas mileage as an application of linear approximation for functions of two variables. :P

Mike Miller

I would just want them to be much more polished were they to be for "public consumption". Part of Ted's implicit contract (to me!) is that I am in communication with students afterwards --- if something is confusing, I can clarify more or less immediately

Making something public extends that privilege to others; but I only get paid for answering some of those questions ;)

Ted Shifrin

17:39

You mean I get paid for answering questions? Who knew!

Oh, you do :D

Mike Miller

For now, at least

Balarka Sen

academia destroyed not anymore mwahaha

Ted Shifrin

Howdy, a @Balarka.

Balarka Sen

all the math you have learnt will be forgotten

4

im not shocked

@TedShifrin Hi!

Ted Shifrin

This is sounding like Hippa's meme.

Balarka Sen

17:41

:D

Ramanewbie

@TedShifrin kind of

Ted Shifrin

You were here years ago for Hippa's meme?

Damn, that was something like 4 years ago.

Ramanewbie

Sure
I can't get if off my mind

Balarka Sen

Has to be at least 6 years

Ramanewbie

You don't remember me @TedShifrin :o

Ted Shifrin

17:43

Well, those videos were made spring 2014, so I suppose 6 years is barely possible.

Balarka Sen

Oh ok

I have been a member of MSE for 6 years and I feel like I have seen that meme at the very beginning

Time flies

DarkRunner

18:03

Hi all, I'm confused about the partial derivative of a cross product

I understand that $\frac{d}{d t}(\mathbf{u}(t) \times \mathbf{v}(t))=\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)+\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)$

I'm trying to use that to prove $v \times E = r \times \frac{\partial E}{\partial t}$

If anyone can help, that would be great

Mike Miller

That's not a partial derivative --- partial derivatives are things you take of multivariable functions, and this is a 1-variable function (of t!)

You seem to be using facts from physics here that you're not telling us

What are r, v, E?

DarkRunner

@MikeMiller Sorry, let me clarify: $E$ is the electric field of a charge. $v$ is the velocity of that charge, and $r$ is the position function

Balarka Sen

$E$ is always in the direction of $r$, so $r \times E = 0$

Use that plus the cross product rule you just wrote.

Knight

@BalarkaSen $r$ is the position of source particle

Mike Miller

There's a sign error then

DarkRunner

18:11

@BalarkaSen OK, so here's what I have:

I know by the Cross Product rule that $\frac{\partial}{\partial t}{r \times E} = \frac{\partial r}{\partial t} \times E + \frac{\partial E}{\partial t} \times r$

Balarka Sen

@Knight The position of the source particle is the origin. $r$ is the position vector from the origin to the test charge, or the other way around - I don't keep track of sign.

Knight

@BalarkaSen particle is moving with $v$

Balarka Sen

What

DarkRunner

@Knight That's right

Balarka Sen

How does that contradict what I said

DarkRunner

18:13

Simplifying, I have $\frac{\partial}{\partial t}{r \times E} = v \times E + \frac{\partial E}{\partial t} \times r$

Knight

@BalarkaSen You said particle is at the origin

Mike Miller

@DarkRunner You swapped the order here on that second term... Should be r x E_t

ABHIJIT BHATTACHARJEE

Hello everyone i want to ask a question about composition series of finite groups. We usually construct a composition in a finite group by a sequence of maximal normal subgroups. Can we construct a composition series in a finite group by a sequence of maximal normal subgroups ?

DarkRunner

@MikeMiller OK, got it

I think I messed up my $LaTeX$ by mistake

Amin Idelhaj

@ABHIJITBHATTACHARJEE so I am not the most accurate reader in general but I think you said the same thing twice, did you mean "minimal" the second time?

Balarka Sen

18:16

Lol

Hi @Amin

gian

Hi all! If I have a diffeomorphism $\phi$ and I'm only given its matrix representation $M$, how do I compute $\phi '$?

Amin Idelhaj

Also hey everyone, what's up!

ABHIJIT BHATTACHARJEE

Oh!! I should be minimal normal subgroup

Balarka Sen

The next question would be what is a minimal normal subgroup

DarkRunner

If anyone can help, that would be great ...

Mike Miller

18:18

what is the matrix representation of a diffeomorphism?

ABHIJIT BHATTACHARJEE

Can we construct a composition in finite group by a sequence of minimal normal subgroup

Amin Idelhaj

So I think the answer is no Abhijit, the reason maximal normal subgroups work nicely is that if $N$ is normal in $G$, then normal subgroups between $N$ and $G$ correspond to subgroups $G/N$

DarkRunner

@BalarkaSen Yes! Yes! You're right!

Thank you!

Balarka Sen

@DarkRunner Whew!

Amin Idelhaj

But normal subgroups of $N$ need not be normal in $G$

Balarka Sen

18:19

I don't know physics, just to clarify.

Amin Idelhaj

Or wait hold on I might be BSing you lemme think for a bit

Balarka Sen

But Abhijit, what is a minimal normal subgroup of a group?

Would that be the trivial group?

Amin Idelhaj

I think he's saying among non-trivial groups

So $N$ is a minimal normal subgroup of $G$ if it's non-trivial and any proper subgroup of $N$ which is normal in $G$ is trivial

Tobias Kildetoft

@BalarkaSen Whenever we use the term minimal or maximal, we always add the requirement that it be either non-trivial or proper depending on which one. But only if the context makes those uninteresting examples

Balarka Sen

OK that makes sense.

Tobias Kildetoft

18:23

So maximal abelian subgroup could be the entire group (since not all groups are abelian). But maximal normal cannot be the entire group

Minimal normal subgroups are characteristically simple, which basically classifies them for finite groups

ABHIJIT BHATTACHARJEE

We take a finite G if it is simple then we are done. If it is not simple we get a minimal normal subgroup. Now we take the quotient group by the minimal normal subgroup. If the quotient group is simple then we are done otherwise we take a minimal normal subgroup of that. We will repeat the process and at some point of time it will stop since G is finite.

Amin Idelhaj

Yeah okay so in a composition series you only need $H_i$ to be normal in $H_{i+1}$

skullpatrol

Tobias Kildetoft

@ABHIJITBHATTACHARJEE That will not produce a composition series on its own

Ted Shifrin

Howdy, @Tobias, Demonark, skull.

Amin Idelhaj

18:28

So in principle you can say, whenever a quotient isn't simple you can take an intermediate normal subgroup

skullpatrol

Hi professor

Amin Idelhaj

If you want to take the minimal one, maximal one, etc. But as Tobias said you can't just take minimal then ignore what's happening underneath

ABHIJIT BHATTACHARJEE

Why that will not work ?

Tobias Kildetoft

@TedShifrin Hi

Because a minimal normal subgroup need not be simple

Amin Idelhaj

The minimal normal subgroup of $G$ might not be simple. Now you can say, okay take a minimal normal subgroup in there, continue until you can't, repeat that whole process over and over

Hey Ted and Tobias!

Ted Shifrin

18:30

@Ramanewbie Hey, it didn't click until now. You are "little Hippa." :D Where have you been for years?

Balarka Sen

Normal subgroups are confusing

Nothing very normal about the notion

Alessandro Codenotti

They should be called quotientable subgroups clearly

Balarka Sen

Agreed

That's perfect, I love it.

Amin Idelhaj

Group ideals

jk

Tobias Kildetoft

ideals should be called quotientable subrngs

Balarka Sen

18:37

Ugh

No

Ideals are much better than normal subgroups

Tobias Kildetoft

really, they are all just kernels

Alessandro Codenotti

Coalgebras are a terrible offender here. There are subcoalgebras, left coideals, right coideals and two sided coideals. You can't quotient by a subcoalgebra, you need a two sided coideal to take a quotient. By the way a two sided coideal is not the same as a left coideal which is also a right coideal

Balarka Sen

coquotient lmfao

Alessandro Codenotti

Who came up with those names? It's terrible

Balarka Sen

you take the dual algebra, quotient by the ideal, then dualize

ABHIJIT BHATTACHARJEE

18:40

Maximal normal subgroups in the group may not be simple as well but the quotient group formed by the maximal normal subgroup must be simple. For example in finite p group of order p^n any subgroup of order p^(n-1) is maximal normal. To construct a composition series in a finite group the most important part is to look for the quotient group

Alessandro Codenotti

No when you dualize the second time you're not guaranteed to get a coalgebra in infinite dimension

Amin Idelhaj

F

Alessandro Codenotti

Dual of a coalgebra is always an algebra, dual of an algebra is a coalgebra in finite dimension, but messed up in general

Amin Idelhaj

18:56

Yo @astyx

Astyx

Oh howdy

Amin Idelhaj

Happy day after your birthday!

Astyx

long time no see !

Thank you !! :D

How are you ?

Amin Idelhaj

Been doing alright, how about you?

Astyx

Good as well, I started my second year of master's last week, and so far thecourse have been very interresting

What are you doing these days ?

Amin Idelhaj

19:00

Oh nice! Same except PhD lol

What kinds of stuff are you doing?

Astyx

I had one week of differential geometry and this week I have complex analysis

What's your PhD about ?

Amin Idelhaj

So here it's a 5 year PhD, basically first 2 years are equivalent to a masters. So I'm still in background. General direction I'm going is number theory and harmonic analysis

Astyx

oh spooky

I have done very little number theory

I'm hesitant to take one course this year

Amin Idelhaj

doit doit doit

Astyx

The "issue" is I can take only two course per semester

And I'd like to do some algebraic geometry as well

Amin Idelhaj

19:09

Ah gotcha

Astyx

I don't know if it's reasonable to take additional courses, they said they calibrate the amount of work required so that two is plenty

Amin Idelhaj

That makes sense lol, immersion is probably good anyway

Astyx

That's true as well

Amin Idelhaj

Almost as good as embedding

hehehe

Astyx

Ok you can leave now

Safdar

19:53

How would you prove this?

$$N=\sum^\infty_{a=1}\sum^\infty_{b=1}\sum^\infty_{c=1}\mathrm{\frac{ab(3a+c)}{4^{a+b+c}(a+b)(b+c)(c+a)}}=\frac{1}{2}\sum^\infty_{a=1}\sum^\infty_{b=1}\sum^\infty_{c=1}\frac{1}{4^{a+b+c}}$$

Astyx

20:03

Sum throughout all the permutations of a, b, c

This gives $3\sum\sum\sum {1\over 4^{a+b+c}}$

Then divide by the number of such permutations to get the result

@Safdar

Safdar

@Astyx you can do this because $a,b,c$ are independant of each other?

Astyx

You can do this because a,b and c are mute variables

For a simpler example

$$2\sum_{x=1}^k\sum_{y=1}^k{x\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x\over x+y} +\sum_{x=1}^k\sum_{y=1}^k{x\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x\over x+y} + \sum_{y=1}^k\sum_{x=1}^k{y\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x\over x+y} + \sum_{x=1}^k\sum_{y=1}^k{y\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x+y\over x+y} = k$$

Safdar

@Astyx so you are saying that $x, y$ are interchangable here.. Similarly in the above case, the same applies for $a,b,c$.. Have I understood you correctly?

Astyx

kind of

interchangeable is not the word

Safdar

@Astyx dummy variables?

Astyx

20:11

yep

Safdar

@Astyx Thanks a lot.. I've been stuck for the past hour trying to solve this..

Astyx

No problem

Whenever you have such a symmetry you should try to use it

Mike Miller

20:49

@AlessandroCodenotti Why are you doing this coalgebra cocrap

Mathphile

Can anyone find a counterexample to this statement:
"If $a^b$ is transcendental then $b^a$ is also transcendental"

Mike Miller

I can yeah

I'll leave it to you to find them though

Thorgott

yeah

Mathphile

21:05

Ah I got one

$e^{\ln 2}$ and $(\ln 2)^e$

user2103480

@MikeMiller that would interest me as well

Mathphile

oh wait do we know if $(\ln 2)^e$ is transcendental?

wolfram says no :(

@MikeMiller Now I am curious to what your counter-example is

geocalc33

If all the quadrants of $\Bbb R^2$ were superimposed would it be possible to write down a metric?

e^1 and 1^e

Mathphile

@geocalc33 lol

what if $a,b \ne 1$

Thorgott

try something with a square root instead of a logarithm

Mathphile

21:13

hmm

ah

$2^{\sqrt{2}}$ and $\sqrt{2}^2$

Thorgott

ye

Mathphile

Alright what about this

"If $a^b$ is transcendental then $a+b$ is also transcendental"

Thorgott

same example still works

Mathphile

oh yeah I forgot $\sqrt{2}$ isn't transcendental

man that was embarassing

Alessandro Codenotti

21:39

@MikeMiller I'm not, but my girlfriend took a Hopf algebras course, and I came up with a few doubts as I was helping her with some of that stuff

Mike Miller

21:59

Poor lass

orientablesurface

22:38

Let $(M,f)$ and $(M,g)$ be global charts on a topological manifold $M$. Then are the unique smooth maximal atlases they generate equal?

Mike Miller

are f and g equal?

orientablesurface

no

not necessarily

Mike Miller

Then no, you know counterexamples

Take M = R

Jeff

Sounds like it's as simple as I thought. But since something has disappeared, here's the question again:
I am told that $\lim_{x \to 6} \frac{f(x)-15}{x-4} = 8$ and $\lim_{x \to 4}\frac{f(x)-15}{x-4} = 3$. I've easily determined that $\lim_{x \to 6} f(x) = 31$. How do I find the $\lim_{x \to 4} f(x)$?

(Once again, I can't reply to myself) But the right limit $-15/(x-4)$ as $x \to 4$ DNE because $\pm \infty$, therefore the whole limit DNE.

Ted Shifrin

I answered this last night, @Jeff.

orientablesurface

22:52

@MikeMiller what if we include that $M$ and $M$ is diffeomorphic to itself wrt to the different unique smooth maximal atlases

Mike Miller

@orientablesurface That's the case for M = R, because any two smooth structures are diffeomorphic on R.

The only answer you'll get is "iff fg^{-1} and gf^{-1} are smooth", which is almost tautological

Ted Shifrin

(On the respective domains with the appropriate differentiable structure.)

Mike Miller

I think these are defined as maps between Euclidean space with standard smooth structure, so I don't think one needs to worry about the parenthetical.

orientablesurface

ahh I see, but if I include the condition that the union of two smooth atlases is a smooth atlas on the manifold, then they determine the same smooth structure

Mike Miller

23:13

Right --- and that's equivalent to the condition I gave above

orientablesurface

yup

user2103480

23:50

@AlessandroCodenotti Any mention of hopf algebras gives me flashbacks to the one time that I looked at Martin Hairer's recent articles on SPDE solutions. That sh*t is wild

Like, I am really unsure if I've ever seen mathematics like that

arxiv.org/abs/1612.08138

and this is the hopf algebra version link.springer.com/article/10.1007/s00222-018-0841-x#Equ7

Jeff

@TedShifrin Yes, but it was kind of vague. I'm just confirming my understanding

Mike Miller

@user2103480 ...

Transcript for

$Mathematics$ Mathematics