Mathematics - 2017-01-07 (page 2 of 4)

Kirill

17:05

Dear people, "if we look at the polynomial $f=x^2 - x$ as at the polynomial with the coefficients in $\mathbb{F}_2$, we see that $f(0)=f(1)=0$, so $f$ must be a Null-Polynomial." What does the man who wrote it meant with this? What does it mean, that coefficients are in $\mathbb{F}_2$? $f(0)=f(1)$ also in $\mathbb{R}$, so why should it be a null-polynomial?

DHMO

@Kirill F_2 means the field {0,1}

Ali Caglayan

The polynomial is over the field F2 not R

Kirill

@DHMO I know the field, I know the traditional definition o a polynomial. But what does it mean, that coefficients are in this one?

DHMO

it means F_2[x]

Kirill

@DHMO ok, that stands on the next page in my book. This sentence is something like an introduction to what a polynomial ring is.

arctic tern

17:10

@MaryStar the cyclic generator of ${\rm Gal}(L/K)$ is $x^{|K|}$. In particular this means $x^{|K|}$ is trivial on $K$. Then we you get to $x^{|L|}$, that's trivial on $L$. You're talking about ${\rm Gal}(\Bbb F_p(a),\Bbb F_{p^n})$ where $\Bbb F_p(a)=\Bbb F_{p^{p^n}}$, so the trivial automorphism will be $x^{p^{p^n}}$.

Kirill

@DHMO but how can $f$ be a null polynomial, if $f=0$ is not $f=x^2-x$?

DHMO

no, f(0) is 0^2-0 = 0

Kirill

@DHMO ok, what is the null polynomial? Is that not $f=0$?

DHMO

Oh, I think your sentence meant that it is not the null polynomial although f(0) = f(1) = 0

arctic tern

@DHMO the quote that Kirill gave said it is a null polynomial

DHMO

17:14

@arctictern the quote said "it must be a null-polynomial"

the "must" should be introducing a false concept

Kirill

@DHMO still, it should be one. The "must" introduces my bad translation :)

DHMO

@Kirill what is the original quote?

arctic tern

maybe. Kirill might not be giving us the whole context. or the author is using null polynomial to mean polynomial that evaluates to zero on every input in the scalar field, instead of zero polynomial. which is the rather obvious alternative from context.

Mary Star

@arctictern By trivial you mean that $\tau (x)=x$ ?

Kirill

@DHMO "Wenn wir das Polynom $f=x^2-x$ als Polynom mit Koeffizienten in $\mathbb{F}_2$ anschauen, sehen wir, dass $f(0)=f(1)=0$, also müsste $f$ nach diesem Polynombegriff das Nullpolynom sein."

arctic tern

17:18

@MaryStar by trivial automorphism of $L$ I mean $\tau(x)=x$ for all $x\in L$

@Kirill and does it define nullpolynom somewhere? I know german uses "null" for "zero" which might throw off my alternative interpretation.

or it might even be using "polynomial" to mean "polynomial function," which would be ick

Mahmoud

Hi.

Mary Star

We have that $\tau$ is a K-automorphism of L, so for eaxh x in K we have that $\tau(x)=x$, right? For $x\in L\setminus K$ we have that $\tau (x)=x^{p^n}$, or not? @arctictern

Kirill

@DHMO I propose that you were right. "Müsste" means something like "ought to"

DHMO

I see

arctic tern

@MaryStar yes

Mary Star

17:21

@arctictern So, to find the order we have to find the $k$ such that $\tau^k(x)=x$ for $x\in L\setminus K$, right?

arctic tern

@MaryStar yes

Kirill

@arctictern Nullpolynom := constant function f(x)=0

Mary Star

So, we have to find k such that $x^{p^{kn}}=x$. Does this mean that $p^{kn}=1$ ? Or do we use the fact that $x\in \mathbb{F}_p(a)$ ? @arctictern

arctic tern

13 mins ago, by arctic tern

@MaryStar the cyclic generator of ${\rm Gal}(L/K)$ is $x^{|K|}$. In particular this means $x^{|K|}$ is trivial on $K$. Then we you get to $x^{|L|}$, that's trivial on $L$. You're talking about ${\rm Gal}(\Bbb F_p(a),\Bbb F_{p^n})$ where $\Bbb F_p(a)=\Bbb F_{p^{p^n}}$, so the trivial automorphism will be $x^{p^{p^n}}$.

5 hours ago, by arctic tern

@MaryStar First off, if $\tau(x)=x^{p^n}$ then $\tau^k(x)=x^{p^{kn}}$. The trivial automorphism is $x^{p^{p^n}}$ so we must solve $kn=p^n$, i.e. $k=p^n/n$. On the other hand, the charactersitic is $p$, so solving $\tau^k(a)=a+k$ with $k$ minimal gives $k=p$.

Kirill

I mean, I know polynomials since the school time and learning rings at the moment. I can imagine the field $\mathbb{F}_2$, and can see that coefficients fo this polynomial are 1 and 1. But I do not get it, how can they be in not be in the $\mathbb{F}_2$ How these things are related? It is something like the following for me: I know what is an egg, I know what is a fridge. But I cannot get it what the egged fridge is...

Secret

17:28

[Random]

arctic tern

@Kirill "how can the be in not be in the F_2 How these things are related" what?

Secret

=>Recipe book

arctic tern

@Secret I did put a line in the chat guidelines ("Don't Hog") about trying to limit space consumption when other people are conversing...

Secret

ok done

arctic tern

@Kirill You know how to write down a polynomial with coefficients in F_2. So you know what a polynomial in F_2[x] is. You know how to add and multiply such polynomials. So you know what the operations in F_2[x] are. You know how to evaluate such polynomials at values in F_2. So you know how to do something with polynomials over F_2. What's the issue?

Kirill

17:32

@arctictern how can coefficients be or not be in $\mathbb{F}_2$? The polynomial is a polynomial, the field is a field. What does is it mean, that these coefficients are in this field?

arctic tern

@Kirill How can something be or not be in a set? Well, apple is in {apple,banana} but it is not in {orange,kiwi}. I think you know how set membership works.

When you write down a polynomial, you have a bunch of powers of x added together with scalars in front of them. Those scalars in this case are just elements of F_2.

Kirill

@arctictern the one of us does not get it right

arctic tern

Your "egged fridge" analogy would be better as a clothesline analogy. You know what a real number clothesline is, but what is a F_2 clothesline? Well, you hang real numbers on a real number clothesline, and elements of F_2 on a F_2 clothesline.

It would be like asking "I know what a clothesline for pants is, but what is a clothesline for shirts?"

Granted, the analogy breaks down because actual clotheslines have multiple types of things on them, whereas polynomials must have all their scalars from a fixed choice of field.

You're probably cool with the idea of polynomials having: (a) rational coefficients, (b) real number cofficients, (c) complex number coefficients. There's no reason to limit ourselves to just Q, R and C though. The reason we can add and multiply polynomials is the scalars out in front exist in a ring, so we can form polynomials over any ring. (For evaluation to make any sense though it should be a commutative ring.)

Ali Caglayan

In this case coefficients are in F2 which gives you a choice of 0 or 1

Note that -1 = 1 in F2

so the -x term becomes x

Kirill

@arctictern The polynomial $f=x^2-x$ has a form $f=a_n \cdot x^n + a_{n-1} \cdot x^{n-1}$. So, $a_n =1$ and $a_{n-1}=1$. Yes, $1 \in \mathbb{F}_2$, but one is also almost in every ring. I do not get why we need rings for this whole story.

arctic tern

17:41

@Kirill You mean almost every ring has an element labelled with the symbol 1. That's not the same thing as different rings sharing the element 1. Now, Q, R and C all share the element 1, but their 1 is different from F_2's 1. Indeed, in F_2 we have 1+1=0, which doesn't happen in Q,R,C.

Alessandro Codenotti

because you need to get the coefficients from somewhere and that "somewhere" is a ring

Kirill

@arctictern *neutral element

@AlessandroCodenotti so I need a set of coefficients to define a polynomial?

arctic tern

@Kirill Since not every polynomial has 0s and 1s as coefficients, you cannot reasonably claim to have told the "whole story" in your example.

Take $\Bbb F_4=\Bbb F_2(a)$. What about $x^2+a^2x+a$? That doesn't even share the symbols $0$ and $1$ that are used in other rings.

Alessandro Codenotti

you said a polynomial has the form $f=a_nx^n+a_{n-1}x^{n-1}...+a_1x+a_0$ but you need to specify what the $a_i$ are

also since you want to add and multiply polynomials you need to have some operations of addition and multiplication defined for the $a_i$, so you get them from a ring

Kirill

@arctictern yes, neither German or English are my native languages. I meant this example as a whole story, not all the possible examples you can imagine.

arctic tern

17:45

@Kirill if it doesn't cover all possible examples then it's not the whole story

(the story of polynomials over a given field, to be specific)

Kirill

@arctictern yes, thank you for the advice, now I know that the "whole story" can be misinterpreted in English.

@AlessandroCodenotti So, is there any ring where $f(1) \ne f(0)$ for this one?

arctic tern

@Kirill The expression $x^2-x$ can be interpreted as a polynomial over any ring, since any ring has a $0$ and $1$ element. And in every single one of those cases, $f(0)=f(1)=0$. But when you're working with other domains that have more elements than just the ones called $0$ and $1$, the polynomial $f$ will not evaluate to $0$ at any of the other values.

(I said domain instead of ring for a reason. The polynomial $f(x)=x^2-x$ still evaluates to $0$ at all elements of $\Bbb F_2[\varepsilon]/(\varepsilon^2)$ for instance.)

Kirill

@arctictern than, according to the sentence, why do we need to take the coefficients from the $\mathbb{F}_2$ to see that $f(1)=f(0)=0$, if they are 0 in every ring?

arctic tern

@Kirill Did that sentence claim we "needed" to take coefficients from F_2 in order for f(1)=f(0)=0?

In any case, you are also ignoring the important thing that in F_2, the only elements are 0 and 1, which means f() evaluates to 0 at every element of F_2. Even if f(0)=f(1)=0 in other rings, f() will not always be 0 in other rings because there are other elements that do not evaluate to 0.

Example: in $\Bbb F_3$ the polynomial $f(x)=x^2-x$ evaluates as $f(0)=0$, $f(1)=0$, $f(2)=2$.

Kirill

@arctictern the logic of the sentence: "if we look at this polynomial and take coefficients in F2, so we see that ... 0". Yes, this sentence looses meaning for me after the explanation you gave - that $f(1)=f(0)$ for every ring.

arctic tern

17:57

@Kirill That sentence mentioned how f "ought" to be a "nullpolynomial." Surely that "nullpolynomial" means "evaluates to 0 at every scalar." This is true if the field is F_2. It is not true if the field is not F_2. That means it was necessary to take F_2. Not because it's necessary for f(1)=f(0)=0 (which is true in any ring), but because that's when {0,1} is every element.

Kirill

@arctictern I think I am getting it

@arctictern Ill try it again

arctic tern

I don't understand why it would say "we need a different polynomial definition."

Oh, I think I get it.

That is my guess. Hard to say with only translations of snippets.

Kirill

@arctictern "If we look at the polynomial ... shouls be a zero-polynomial. But we want to use elements from bigger rings, and it can happen that the values will differ from zero. So, we need a different definition of polynomials. The idea is to define them as a sequence of their coefficients..."

arctic tern

Okay. Now I assume the author was first offering a definition of polynomial as a type of function. But since $x^2-x$ being the zero polynomial depends on which ring you work over and the author doesn't like this, the author then needs a different definition of polynomials.

Kirill

@arctictern can I ask this in this way?: what it the problem with the old definition? Is that not well-defined, or?

wyattbergeron1

18:10

"Rarely if ever expressible as a ratio of integers"

That's great

arctic tern

@Kirill The problem is we don't have a good way of talking about "zero polynomials."

Over F_2 we don't want x^2-x and 0 to be the same polynomial, even though they define the same function.

So we can't simply define polynomials as a type of function.

Semiclassical

Hi chat

arctic tern

(I mean you could, if you really wanted to, by restricting the domain to be algebraically closed. But then you'd need to prove the equivalence of definitions. And the point is polynomials should be an algebraic gadget, of which our intuition is they are just these things with powers of variables that are added and multiplied a certain way.)

@wyattbergeron1 thanks

Kirill

@arctictern aha, so he needs a definition that works independently from the domain?

arctic tern

yes

Kirill

18:15

@arctictern thank you! That was a great help, even if the question was too obvious for you.

arctic tern

some polynomial rings you don't even care about evaluation at all. for instance in H[x] where H is the quaternions (which is a noncommutative ring), it is not at all meaningful to evaluate polynomials at elements of H.

wyattbergeron1

If you have a zero term polynomial it isn't even a polynomial, is it?

arctic tern

yes it is.

consider the polynomials f(x)=x and g(x)=-x. you want f(x)+g(x) to be a polynomial too don't you?

wyattbergeron1

But it doesn't have terms

arctic tern

what's your point

wyattbergeron1

18:17

idk

arctic tern

"x" only has one (nonzero) term, even though "poly" means multiple

wyattbergeron1

It was just a thought

Kirill

it is still one

arctic tern

we should just interpret the "poly" in "polynomial" not as "more than one" but instead as "some number of" terms.

Kirill

btw it has an infinite number of terms, but allmost all of them have zero as coefficients.

SteamyRoot

18:37

That kind of vagueness is very dangerous

Whenever you say something like that, you should clarify what "almost all" means.

If "almost all" means all but finitely many, then you have a polynomial. If not, you have a power series.

arctic tern

dangerous for technical talk, not for informal as much

Kirill

18:55

@SteamyRoot translation question. German "fast all" means "all but finitely many". We have a polynomial, if the sequence $f: \mathbb{N} \to R, i \mapsto a_i$ defined on the commutative ring $R$ has all but finitely many $a_i$-s that differ from zero.

was that ok?

SteamyRoot

Looks good :)

Fruitful Approach

19:21

What is a sufficient condition on $n \in \Bbb{Z}, |G|$ such that there exists a nontrivial subgroup of $G$ solving $X^n = 1$ in the ring $R$ in which $G$ is a multiplicative group?

:math.stackexchange.com/questions/2087845/…

Jessy Cat

2

Q: Direct product of two nilpotent groups is nilpotent and direct product of two solvable groups is solvable

Let $G = G_{1} \times G_{2}$. I need to prove the following two things: If $G_{i}$ is nilpotent of degree $n_{i}$, $i = 1, 2$, then $G$ is nilpotent of degree $n = \max \{ n_{1}, n_{2} \}$. If $G_{i}$ is solvable (some people call them soluble) of degree $n_{i}$, $i = 1,2$, then $G$ is solvabl...

Still looking to award that bounty.

Tobias Kildetoft

@FruitfulApproach Well, if $G$ is finite then this is pretty straightforward. Otherwise I don't think much can be said

Fruitful Approach

Yes, but I have a ring involved

Confusing to me...

Tobias Kildetoft

@JessyCat I really don't feel like providing additional details on that. If the current ones are not enough, you should spend some more time on this yourself.

@FruitfulApproach If you know $|G|$ and this is finite, then the ring is irrelevant

Fruitful Approach

Yes, in my case $G$ is multiplicative group of $\Bbb{Z}/(m)$

Tobias Kildetoft

19:26

If the characteristic of the ring is non-zero that helps too

Fruitful Approach

No it's not, it's $R = \Bbb{Z}$.

Tobias Kildetoft

@FruitfulApproach Ahh, that is a very easy case to deal with

Fruitful Approach

$G \neq \Bbb{Z}/m$

Tobias Kildetoft

what? Which is it?

arctic tern

@FruitfulApproach I don't see R=Z/m anywhere in your question.

Fruitful Approach

19:28

$G $ the ring's multiplicative group which I understand can be "weird" like $D_{2n}$ and stuff

arctic tern

@FruitfulApproach erm, how can G be the multiplicative group of R=Z/m and also R=Z?

Fruitful Approach

Oh, I was generalizing my post. But my case is ring $Z/m$

arctic tern

you said commutative ring in your question

Fruitful Approach

Sure, that's ok

$R = \Bbb[Z}/(m)$

arctic tern

@JessyCat The answer there covers everything. Is there something about the answer you don't understand?

Tobias Kildetoft

19:32

@FruitfulApproach Do you know the order of the group of units of that ring?

mick

19:50

1

Q: How to show that the fabius function is nowhere analytic?

Consider the fabius function https://en.m.wikipedia.org/wiki/Fabius_function https://people.math.osu.edu/edgar.2/selfdiff/ How does one show that this function is nowhere analytic ? Probably related , Maybe even a step in the answer : how to evaluate this function for nonreals ? Is it defined...

complex-analysis proof-writing implicit-differentiation

s.harp

@mick math.stackexchange.com/questions/12989/…

Krijn

@TobiasKildetoft If you have time I have a small question for you: If we have $\mathfrak{g} = \mathfrak{z} \oplus \mathfrak{g}'$, where $\mathfrak{z}$ is the center, and we have a irrep $V$, where we know how it acts irreducibly on $\mathfrak{z}$ by $\rho$

Can we then somehow find the right irrep on $\mathfrak{g}'$ to make the outer tensor product work?

Tobias Kildetoft

@Krijn I am not quite sure what you mean by knowing how it acts on the center

Do you mean you know how the center acts?

Krijn

Yeah

Let me try to rephrase it correctly

We know that there is some $\lambda$ in the dual of the center such that $Z(v) = \lambda(Z)v$ for all $Z$ in the center

Ted Shifrin

Heya @Tobias @Krijn

Tobias Kildetoft

20:04

@TedShifrin Hi

Ted Shifrin

ohhh, there's @Kaj ... you buried under snowdrifts yet?

Krijn

Hey @Ted

Kaj Hansen

There was absolutely no snow on the ground outside my house. None. Not even ice.

We were supposed to get "3 inches"

Tobias Kildetoft

How would 3 inches make you "buried"? :)

Ted Shifrin

I of course wasted tons of worry, spent $175 to change my airplane reservation, and escaped yesterday without seeing my friends in Athens. I just can't count on weather forecasters anywhere near GA. I didn't want to be stuck inside for 5 days and miss everything I had appointments for back in San Diego. :(

Kaj Hansen

20:05

@TobiasKildetoft, people in the South freak out and overhype every time winter weather is possible

Ted Shifrin

@Tobias: That's buried in Georgia.

Tobias Kildetoft

@TedShifrin Ahh, right

Kaj Hansen

You should see grocery store shelves. It's honestly ridiculous

Ted Shifrin

Well, the truth is that things ice up and there's nothing to do.

I drove through feet of snow and blizzards in New England for many years.

@Kaj: I will NEVER again visit Georgia except fall or spring.

Kaj Hansen

The 2014 scenario was kind of bad, but only because we don't ice our roads. It was all better 48 hours later

Ted Shifrin

20:06

@Kaj: No, if that's the year I'm remembering, I was stuck in my house for 5 days straight.

As it turns out, I couldn't have gotten to school to teach even though I wanted to. I hate being prisoner to a tiny bit of weather.

Kaj Hansen

That was the year I had dif. geo from you. I might not be remembering quite well

Ted Shifrin

Yeah, we did exams at night to make up two of the missed classes.

Kaj Hansen

Doug showed me pictures of ATL. Even there they barely got dusted. This was the most overhyped winter weather scenario ever.

Ted Shifrin

So I'm getting various sarcastic texts and pictures from the people I canceled on in Athens.

I'm over it.

I wonder what's become of @Danu. I haven't seen him in weeks.

Kaj Hansen

I see him log in on chess.com a good bit

Definitely in the last 3 days

Ted Shifrin

20:10

@Kaj: There was a cool geometry question yesterday. Recall that in hyperbolic geometry there are infinitely many parallels to a given line through a given point (not on the line). The OP asked how to see that on a pseudosphere.

Hmm, well, I guess he's mad at math (or at us) for now.

Kaj Hansen

That is a cool question. I like thinking about the revolved tractrix instead of the hyperbolic plane just because I think it gives me better intuition about what's possible on a surface in $\mathbb{R}^3$ with the usual metric, instead of passing to a harder-to-intuition space.

I get that there's advantages to doing that though

Ted Shifrin

Well, you lose completeness, which is a big deal.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

But it's a good local model, but much harder to see the intrinsic geometry.

Heya @Alessandro :)

Kaj Hansen

I should re-read your geometry notes. I'm sure I'm very rusty

Ted Shifrin

20:14

Anyhow, @Kaj, one of my exercises in the geodesics/Clairaut relation section was to find the geodesics on a pseudosphere. You can figure out pretty clearly that given any meridian (profile curve), we can find lots of geodesics through a given point not meeting it.

Well, I'm rusty, too :)

Krijn

@TedShifrin You are Iron Man?

Ted Shifrin

And tin man, too.

Kaj Hansen

So the revolved tractrix can be thought of as a union of uncountably many individual tractrices. Any individual one should also be a geodesic I'd imagine

(That's an awkward way of saying that, but I can't think of a better way)

Ted Shifrin

Of course. So, taking a point off a given one of those, you have one line parallel to the original. But we want infinitely (uncountably) many of 'em through the fixed point.

Kaj Hansen

I'm guessing these can be found with Clairaut's relation. Can one simply choose a point and an angle w.r.t. the meridians and find a geodesic that goes through that point at that angle?

Well, not "found" explicitly, but qualitatively

Ted Shifrin

20:20

But you have to be careful to make sure that geodesic will never wrap around and hit the original line (which I'm choosing to be one of your tractrices).

Kaj Hansen

Oh yes, I'm not even asking with that problem in mind. I'm just doing memory-refreshing on geodesics in general

One can find a geodesic at any angle through any given point, right?

Ted Shifrin

Yuppers.

Alessandro Codenotti

Is there a nice way to generalize the idea of measure from $\Bbb R^n$ to a smooth manifold? I ask because of Sard's theorem, in G&P they only define what it means for a subset of a manifold to have measure $0$, but that doesn't seem to work as nicely if I want to measure stuff with positive measure?

Ted Shifrin

I'm puzzled by this question. I think it's wrong, but I'm playing with it.

Kaj Hansen

@AlessandroCodenotti, and then there's $0^\circ$ K

Ted Shifrin

20:22

Measure 0 is well-defined independent of any choices, @Alessandro.

But the easiest construction is to have a Riemannian metric (e.g., induced from $\Bbb R^n$ in the cases you're looking at), and then the volume form gives you a standard measure.

Mike Miller

I just had that discussion with a colleague.

Ted Shifrin

Otherwise, you can glue together local measures with a partition of unity and get something totally uncanonical.

Heya @MikeM

Mike Miller

@TedShifrin Equivalently described via densities, of course.

Well, if you want your measure to be smooth.

Ted Shifrin

Well, the only advantage of a density in this discussion is that it applies to a non-orientable creature.

Alessandro Codenotti

Hm, I know only a few of those terms, I guess my curiosity will have to wait

Mike Miller

20:25

@TedShifrin I agree. On the other hand, it's equivalent to choosing a (positive) volume form on an oriented creature.

Ted Shifrin

No argument there. (No modulus either.)

Mike Miller

@AlessandroCodenotti I frequently use the induced measure on a smooth manifold, but the induced measure is coming from a Riemannian metric.

Alessandro Codenotti

I don't know what a Riemannian metric is

Mike Miller

Whereas you need less than a metric to define a measure, for a lot of other constructions you might like to do, you need a metric

One day you will.

Ted Shifrin

@Alessandro: It's just a smoothly varying dot product on each tangent space. :)

For the manifolds you're playing with, just use the restriction of the usual Euclidean dot product.

Have you encountered the Gram determinant?

Alessandro Codenotti

20:32

nope

Ted Shifrin

So here's a great linear algebra exercise for you. You up for it?

Alessandro Codenotti

I should be studying numerical analysis but I'm very curious now

Ted Shifrin

This discussion is your fault, you know. But this is an important result.

I hand you $k$ vectors $v_1,\dots,v_k\in\Bbb R^n$. I ask you to compute the $k$-dimensional volume of the parallelepiped they span. Give me a reasonable formula for it.

(E.g., start with $k=2$ and $n=3$.)

Martin Sleziak

Seeing that Ted Shifrin is here, perhaps I might mention that after a long time somebody posted a question in Differential geometry chat room.

Ted Shifrin

LOL, oops. Thanks, @Martin. I'm working on a recently-posted question which I totally do not believe.

That question you linked is on the borderline with physics. Interestingly, I was recently asked to review a book by an applied mathematician on "shape geometry." But I'll look when I'm done with this other query.

Mike Miller

20:39

Which I assume is totally unrelated to shape theory.

Ted Shifrin

@MikeM: Are you familiar with the notion of asymptotic curves on a surface in $\Bbb R^3$?

Mike Miller

Yes

Ted Shifrin

Take a gander.

Kirill

What is a zero sequence in $R[x]$, R is a commutative ring?

Ted Shifrin

I've never heard of such a thing.

Mike Miller

20:42

I'm working, but maybe Balarka can do that.

Ted Shifrin

Balarka hasn't learned enough geometry. But I truly think something's wrong, as I posted.

Martin Sleziak

@Kirill This probably depends on the context. But I would guess zero polynomial; polynomials are often represented of sequences of elements from R which have only finitely many non-zero terms (i.e., sequences with finite support).

Mike Miller

I agree with your suggested counterexample but I'm not comfortable enough with this to try to fix the result.

Ted Shifrin

I think it's totally wrong. I'm about to have a counterexample. Go do your work :)

Ah, of course, @Martin. Thanks ;)

Kirill

@MartinSleziak true. if $f: \mathbb{N} \to R, i \mapsto a_i$ and $g: \mathbb{N} \to R, i \mapsto b_i$ and $f+g: \mathbb{N} \to R, i \mapsto a_i + b_i,$ so I need to find a zero polynomial to say that R[x] is an abelian group, but I am confused about which exactly is a zero polynomial

Martin Sleziak

20:49

Well, zero polynomial is the sequences consisting only of zeros: $(0,0,0,\dots)$.

I.e. the function $z$ such that $z(i)=0$ for each $i$.

Ted Shifrin

@MikeM: I think the only thing to say is that $\langle\vec x_{uu},\vec n\rangle = \langle\vec x_{vv},\vec n\rangle = 0$. :)

Kirill

@MartinSleziak or, is that a sequence that has 0 as limit?

Mike Miller

Fair enough.

Tobias Kildetoft

@Krijn Did we ever get to what your question was?

Martin Sleziak

@Kirill Every sequence you're dealing with here has zero as a limit. (Since you have only finitely many non-zero terms.)

Krijn

20:50

@TobiasKildetoft Yeah, it was how it could get a irrep on $\mathfrak{g}'$ so that the outer tensor product would be $V$

Ted Shifrin

@Martin: The paper that guy in the chatroom refers to needs money to download, so I'm not going to do it. :)

Tobias Kildetoft

@Krijn Ahh, that is just the restriction of $V$ to the semisimple part

Martin Sleziak

@TedShifrin I can email you the paper. Will I find your address somewhere on your website?

Kirill

@MartinSleziak so, really just $z: \mathbb{N} \to R, i \mapsto 0 \quad \forall \quad i$?

Martin Sleziak

I see it is almost at the top. I hope it is ok to email you.

Ted Shifrin

20:53

Sure thing. Thanks. I'm already answering him in the chatroom, though :)

Martin Sleziak

@Kirill Yes.

Mike Miller

@TedShifrin Finally found the definition of shape I was looking for. A space has trivial shape if every map to a CW complex is null-homotopic.

Kirill

@MartinSleziak thank you!

Ted Shifrin

Not a geometer's notion of shape @MikeM. See — I was right! :)

Alessandro Codenotti

@Ted If I have $n$ linearly independent vectors in $\Bbb R^n$ the volume of the spanned parallelepipid is the determinant of the matrix with those vectors on the columns (maybe there's some sign problem)

Mike Miller

20:54

Just call me a differential geometer.

Ted Shifrin

You're different, all right @MikeM.

Sure, @Alessandro. We're trying to generalize that.

Mike Miller

Anyway, this is a good refinement of "weakly contractible". The topologist's sine curve does not have trivial shape.

Ted Shifrin

@MikeM: and you've never been deferential.

Mike Miller

Maybe in the early days.

Ted Shifrin

What's a space that's weakly contractible but not contractible?

Mike Miller

20:56

Topologist's sine curve.

Ted Shifrin

@Martin: Thanks. I got the paper. I'll look.

Martin Sleziak

Thanks for doing that.

Mike Miller

This is a notion folks like Bob Edwards were working with in their fundamental work on topology... A "cell-like map", which featured prominently, is one with fibers of trivial shape.

Ted Shifrin

Right — I remember lots of talks about cell-like maps at topology conferences of yore.

Mike Miller

You might need some conditions like compact fibers or something.

Alessandro Codenotti

20:58

I suppose that a reasonable formula won't involve a change of basis to have all of the coordinates after the first $k$ equal to $0$

Mike Miller

I knew all of this for about two weeks then forgot it.

Ted Shifrin

@Alessandro: Correct. And it would have to be a correct sort of basis change to make sure that "volume" is preserved.

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$Mathematics$ Mathematics