Mathematics - 2018-06-04 (page 2 of 3)

Secret

08:00

[RNdom]

Consider RNS. Define the topohology of the grid f to be as follows:

{d.,df.,.s,a,EE,(fade)

Secret

08:16

Topolography of g is given my: x |_|c^v ~ rf#Iop(a&a)

Leaky Nun

08:34

@Silent $U = \{ (a,b,c,d) \mid a+b=c+d \} = \{ (a,b,c,d) \mid a+b-c-d = 0 \} = \ker f$ where $f : \Bbb R^4 \to \Bbb R : (a,b,c,d) \mapsto a+b-c-d$

@abenthy welcome

@loch will you be there?

Forever Mozart

@Daminark

loch

@LeakyNun I am here

Leaky Nun

ok

loch

@LeakyNun are you in common room?

Leaky Nun

@loch no, I'm at home

it will take me an hour to get there

I'll tell you when I arrive

loch

08:49

I see

Leaky Nun

10:05

@loch I’m in the common room now

loch

@LeakyNun cool let me come in 5 minutes

Leaky Nun

11:02

62

A: Motivating Lubin-Tate theory

Sorry I didn’t see this earlier. My memory is vague, and probably colored by subsequent events and results, but here’s how I recall things happening. Since I had read and enjoyed Lazard’s paper on one-dimensional formal group (laws), which dealt with the case of a base field of characteristic $p...

It’s true that it was Lubin who found the formal groups that do everything for you, but it was Tate who understood all the implications of their existence, and put everything together in the paper you’re referring to. — Lubin Oct 16 '15 at 0:59

“it was Lubin”

said by Lubin himself lol

MatheinBoulomenos

lol

Secret

12:09

Took me almost a DECADE to break into That Barrier

The Barrier that separates Real life and Fantasy

Now where is my Order...

BeginningMath

13:09

anyone here is good with prenex?

Mancala

13:43

What is a surface of revolution in $\mathbb{H} ^ 3$?

GFauxPas

14:03

is a set with exactly one mapping partially ordered by inclusion? asking for a proof

what is $\mathbb H^3$?

Akiva Weinberger

Hyperbolic space?

GFauxPas

oh I don't f*** with those

lol

Akiva Weinberger

I mean, wouldn't it be the same thing as Euclidean space?

You can rotate around a line just fine

Mancala

Yes, hyperbolic space with dim 3

Leaky Nun

@loch i’m in the common room now

Mancala

14:16

I also figured this out, but what would a reflection isometric across cycle be? math.stackexchange.com/questions/2807101/…

loch

@LeakyNun im coming in 20 sec

Akiva Weinberger

14:33

@Mancala They're reflecting across a plane

It works similarly to reflecting across a plane in Euclidean space

I'd assume the book dealt with isometric transformations in hyperbolic space already

Oh, it's not a book.

Secret

I am starting to wonder whether these handshake problems is basically to find the number of edges in a complete graph of N vertices

mathworld.wolfram.com/HandshakeProblem.html

GFauxPas

hmm, I'm trying to use Zorn's Lemma on the following set but I need to show there's an upper bound, any ideas? here's the set

Secret

5

Q: Handshake problem

I was given the following math puzzle which (I thought) has an interesting solution. A mathematician and her husband attended a party with $n-1$ other couples. As is normal at parties, handshaking took place. Of course, no one shook their own hand or the hand of the person they came with. And no...

GFauxPas

$S, T$ are well ordered. $S' = S \cup $ initial segments of $S$, $T' = T \cup $ initial segments of $T$, $\{ f : S' \to T' : f \text{ is an order isomorphism} \}$

I hope it does have an upper bound

ordered by inclusion

if it doesnt have an upper bound my book is wrong, but the book doesnt say why it has an upper bound

GFauxPas

15:06

oh of course I see it

Semiclassical

@Secret yeah, that’s how I think of it

That or just “n choose 2”.

Secret

yeah pretty much

GFauxPas

wait no i dont see it

every initial segment has an upper bound but why does all of $S'$ have an upper bound

Secret

All initial segments of $S$ are nested subsets of $S$, thus $S$ will be the upper bound since there is no set in $S'$ that contains $S$ except $S$ itself

Likewise for $T$

GFauxPas

nice

wait but why is that an upper bound in $S'$

$S \subseteq S'$ but $S \notin S'$

is it?

Secret

15:21

For each initial segment $S_{\alpha}$ of $S$ ordered by inclusion, there is always initial segments of $S$ including $S$ itself that contains them and any $S_{\beta}$ where $\beta < \alpha$,thus they are greater than or equal to $S_{\alpha}$ under inclusion

Also $S'$ is not a set of sets thus $S$ or any of its initial segments cannot be elements of $S'$

GFauxPas

so should it be initial segments $\cup \{S\}$?

rather than initial segments $\cup S$?

Secret

initial segments $\cup \{S\}$ will produce $\{\text{initial segments, S}\}$?

GFauxPas

yes

Secret

but you are given initial segment $\cup S$ which is different

GFauxPas

right

okay so

any initial segment has an upper bound by definition

every element $s \in S$ has an initial segment $S_s$ so that has an upper bound?

thus every chain in $S'$ has an upper bound, namely, the initial segment defined by $S_s \in S'$?

im a bit confused

i think I get it

oh no Dami is here, hide the roulette tables

Secret

15:37

I cannot think of an intuitive and elementary example without using ordinals to show that every initial segment of a well ordered set has an upper bound

GFauxPas

that's just the definition of an initial segment

wosets aren't intuitive anyway

Secret

I see, $S_s = \{x \in S : x < s\}$ thus we always have $s$ as an upper bound

GFauxPas

yup

Akiva Weinberger

Suppose $\phi>\varphi>0$

Nah I'm kidding

GFauxPas

the results are up on ProofWiki for ordinals but I want these results to not depend on the existence of ordinals because No Proper Classes in my BackYard

Secret

15:41

yeah, I also want to deal with wosets directly without ordinals, cause I rely on them too much

Nûr

45

Secret

Hmm... does $S$ always have a upper bound? Depending on what $S$ is, it may not have an element larger than or equal to all other elements

For example, the naturals

GFauxPas

it does not. consider the naturals, for example

Secret

I mean, every initial segment of the naturals have an upper bound (in fact it has a supremum, namely the successor element), but the naturals in its entirely does not have an upper bound since any such is not a natural

GFauxPas

right

GFauxPas

15:59

proofwiki.org/wiki/User:GFauxPas/Sandbox whew, I need a break

thanks for the help Secret

Mancala

@AkivaWeinberger I wanted to see a drawing in the $\mathbb{H}^2$ case to understand better.

Akiva Weinberger

16:19

The curves forming boundaries of triangles are actually lines in the hyperbolic plane

Reflecting a triangle across one of those lines turns it into another triangle

(of opposite color)

Abr001am

triangles in noneuclidean round planes (spherical or ovalic ... etc) have >$\pi$ sum of inner angles.

Mancala

@AkivaWeinberger Planes here would be union of straight lines, right? What would a reflection across totally geodesic plane? I'm having trouble understanding this specific case, it seems to me to be something that comes out from the inside out, is that it?

Secret

16:38

I found a good way to not mix up meet and join:

Let up > down, then observe $\lor$, the vertex of this symbol is the supremum as everything else are upper bound that are larger than it

Likewise for $\land$, the tip is the infimum since everything else is smaller in the cone defined by the shape of the symbol

Abr001am

here i was trying to understand why the area of a circle drawn upon a sphere is bigger same as is but as a cap in an euclidean isometry. It was long time ago.

Secret

And when one put $\land, \lor$ together, they get an X, which looks just like the intuitive lattice when duplicated many times

Mancala

I'll check it out, but I'll be right back!

Secret

and thus we have the notion of the algebraic structure of a lattice $(L,\land,\lor)$ being encoded by the shape of the binary operators

I wonder if we can do similar things for rings...

More generally, how many mathematical symbols actually have literal interpretations and how large is this set of algebraic structure that has this property

Akiva Weinberger

@Mancala You see how, in that image, the hyperbolic plane is a disk, and the hyperbolic lines are circles that meet the edge of the disk at right angles?

In the 3D case, hyperbolic space is a sphere, and hyperbolic planes are spheres that meet the edge of the space at right angles.

Unfortunately, images of this are less illuminating

GFauxPas

16:52

kinda pretty tho

Akiva Weinberger

(This is a tiling of hyperbolic space with icosahedra; it shows the edges of the icosahedra, which are lines, as well as one icosahedron in the middle shown in blue)

Mike Miller

@AkivaWeinberger It's one of those remarkable cases where we see that our mind's eye is more powerful than any depiction we could create

Akiva Weinberger

I mean I really only need a single plane for what I'm trying to explain

so the whole tiling is kinda overkill

A single plane, an object on one side of it, and its reflection on the other side

but hyperbolic

Mike Miller

True, but I think it's much easier to create a 3-dimensional image internally than externally

anakhronizein

@MikeMiller any idea of how to make RP^3 obviously look like a circle bundle over RP^2?

Akiva Weinberger

17:05

Hopf?

Mike Miller

@anakhronizein $S^3 \to S^2$ factors through $\Bbb{RP}^3$ (actually, the latter is the Euler class 2 circle bundle over $S^2$). Then just compose with the quotient map $\Bbb{RP}^3 \to S^2 \to \Bbb{RP}^2$.

You need to check that fibers are connected but I believe the yare

Akiva Weinberger

/pf/ is a pretty rad affricate

"Hopf"

Pferfect

anakhronizein

What do you mean it factors through RP^3, exactly?

Akiva Weinberger

There's a map $f:{\rm\Bbb RP^3}$ that makes $S^3\to{\rm\Bbb RP^3}\xrightarrow f S^2$ into the Hopf map, I think

anakhronizein

I don't quite know what it means to by "Hopf".

Akiva Weinberger

17:13

Actually, that might not be right

Hold on

So let's say $H:S^3\to S^2$ is the Hopf map

and let $x$ be in $\rm\Bbb RP^3$

then we can choose a $\tilde x\in S^3$ such that its projection in $\rm\Bbb RP^3$ is $x$

We have two choices 'cause $S^3\to{\rm\Bbb RP^3}$ is a double cover

Take $H(\tilde x)\in S^2$

I think this does depend on the choice of $\tilde x$

ÍgjøgnumMeg

@AkivaWeinberger pfanne!

Akiva Weinberger

Or, does it?

Ugh, how is Hopf defined algebraically

GFauxPas

anyone here know German

Akiva Weinberger

@ÍgjøgnumMeg, Ted, Mathein

ÍgjøgnumMeg

Yes?

oh

GFauxPas

17:17

you know German?

ÍgjøgnumMeg

@GFauxPas yeah

Akiva Weinberger

I think it doesn't depend on the choice of $\tilde x$ actually

Is $H$ even?

In any case, $x\mapsto H(\tilde x)$ is the ${\rm\Bbb RP}^3\to S^2$ thingy

GFauxPas

I want to add as a historical note to ProofWiki that Zermelo thought the axiom of choice was obvious. But I don't want to use a published translation for copyright reasons. Can you translate the sentence please he says right before the exposition of the axiom?

Akiva Weinberger

and then we can concatenate with $S^2\mapsto{\rm\Bbb RP}^2$.

@anakhronizein

anakhronizein

Hmmm, I will check the details on this Akiva, thanks!

GFauxPas

17:20

"so läßt sich das allgemeine „Prinzip der Auswahl" auf das folgende Axiom zurückführen, dessen rein objektiver Charakter unmittelbar einleuchtet."

Akiva Weinberger

Have you tried Gurgle Translat?

GFauxPas

no because I'd rather it be an actual person :P

ÍgjøgnumMeg

@GFauxPas I think perhaps @Mathein would be a better person to ask, my German only extends to non-technical useage

lol

GFauxPas

-_-

ÍgjøgnumMeg

Translation: I haven't done any maths in German

anakhronizein

17:21

"the general "principle of choice" can be traced back to the following axiom, whose purely objective character is immediately apparent.

@GFauxPas

ÍgjøgnumMeg

nise

Akiva Weinberger

Ah, you can Germ?

GFauxPas

thanks!

not the quote I wanted then

i wanted him saying it's obviously true

lemme see

anakhronizein

"whose purely objective character is immediately apparent." is pretty close to saying "whose truth is obvious"

GFauxPas

but he was objecting to other formulations that people objected were too vague, from what I gather

he formulated something about choice but people complained the exposition was too vague

maybe this

‘wird aber in der mathematischen Deduktion überall unbedenklich angewendet’

?

anakhronizein

17:27

"but in mathematical deduction it is used everywhere in a harmless way."

GFauxPas

ah yes

cartesian product of non empty is non empty

:)

obvious

thanks anak

anakhronizein

No problemo.

Akiva Weinberger

Give me an element of $\prod\mathcal P(\Bbb R)$

anakhronizein

$(\emptyset,\emptyset,\emptyset,\emptyset,\emptyset,\emptyset,\emptyset ,...)$

Szabolcs

Is every biconnected graph homeomorphic to some triconnected graph? In other words, will I always get a triconnected graph if I "smooth out" the degree-2 vertices of a biconnected graph?

No, it's not. I'm stupid.

GFauxPas

17:41

okay found the whole quote

want me to paste it or nah?

anakhronizein

If you wanted.

GFauxPas

Der vorliegende Beweis beruht auf der Voraussetzung, daß Belegungen $\gamma$ überhaupt existieren, also auf dem Prinzip, daß es auch für eine unendliche Gesamtheit von Mengen immer Zuordnungen gibt, bei denen jeder Menge eines ihrer Elemente entspricht, oder formal ausgedrückt,

daß das Produkt einer unendlichen Gesamtheit von Mengen, deren jede mindestens ein Element enthält, selbst von Null verschieden ist. Dieses logische Prinzip läßt sich zwar nicht auf ein noch einfacheres zurückführen, wird aber in der mathematischen Deduktion überall unbedenklich angewendet.

So kann z. B. die Allgemeingültigkeit des Satzes, daß die Anzahl der Teile, in die eine Menge zerfällt, kleiner oder gleich ist der Anzahl aller ihrer Elemente, nicht anders bewiesen werden, als indem man sich jedem der betrachteten Teile eines seiner Elemente zugeordnet denkt.

i put it in google translate so you dont have to translate that (unless you want to)

but what I needed help with was narrowing ti down and you did that so thanks :)

Akiva Weinberger

> daß das

GFauxPas

that's in the original

anakhronizein

Cute!

GFauxPas

17:46

i'm guessing it's "that that" which appears in English

that that is concerned with, e.g.

or something

ÍgjøgnumMeg

@GFauxPas right, except nobody writes "daß" anymore because of a spelling reform, it's just "dass" now

lol

although actually here it's just "that the"

GFauxPas

well it's from 1904 lol

ÍgjøgnumMeg

yeah fair, I'm surprised it's even as modern sounding as it is hahah

I have something from Ernst Kummer where often you see "Theile" instead of "Teile"

etc.

GFauxPas

"the proposition that the number of parts into which a set disintegrates is less than or equal to the number of all its elements can not be proved otherwise than by thinking of each of the considered parts of one of its elements."

hmm, not sure what theorem he's talking about here

that the number of subsets cant be more than the cardinality of the set?

number of distinct subsets

mercio

he is saying that you need a representative for each set

oh hmm

Daminark

17:58

Yo

mercio

if you replaced the last "of" with a "by"

or a "as"

GFauxPas

I mean this is google translate so

mercio

yeah

GFauxPas

i guess its something like

a set is non empty iff it has an element, so any time you have a disjoint collection of non-empty sets, you just pick an element from each set

something like that

mercio

uh

GFauxPas

18:01

so the number of singletons is the number of elements

im not sure tbqh

@anakhronizein can you help us

mercio

I think he is saying that if you have a partition of a set, the partition can't be larger that the set, but to prove that you need to assume the axiom of choice

GFauxPas

oh that makes sense

mercio

(you can actually get partitions strictly larger than the set if you don't have the axiom of choice)

GFauxPas

?!?!

mercio

one of the pigeons is trying to claim my room as his territory

GFauxPas

18:03

how does not having the axiom of choice allow such a monstrosity to be considered

mercio

because you need the axiom of choice to prove that a quotient set cant be larger than the set

iirc one example is the set of finitely supported functions from Z to .. say .. N

which is countable

and then you consider the quotient of that set by translations

wait uuuuh

scratch that

maybe just functions from Z to N

or was it quotienting by changing a finite number of elements ?

i don't remember

also cardinality is a bit fuzzier without the axiom of choice

I don't remember if he proved that there couldn't be injections one way or surjections the other way

115

A: Why worry about the axiom of choice?

How I Learned to Stop Worrying and Love the Axiom of Choice The universe can be very a strange place without choice. One consequence of the Axiom of Choice is that when you partition a set into disjoint nonempty parts, then the number of parts does not exceed the number of elements of the set be...

:(

no example there

maybe in the comments

GFauxPas

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

mercio

but i remember seeing something a lot more specific

GFauxPas

18:27

hmm, Zermelo writes about disjoint sets $A, B, C, \ldots$. Does that mean he only needed choice for sets of sets up to cardinality 26?

that is so weird that a quotient set can be bigger than a set without choice

good thing I'm a choicean

mercio

18:41

it's not a bug it's a feature

Nûr

19:04

hello

mercio

hello

Eulb

19:28

hello

can someone recommend a good point set topology reference for just lots of examples to supplement munkres?

GFauxPas

Any argument where one supposes an arbitrary choice to be made a nondenumerably
infinite number of times [ . . . ] is outside the domain of mathematics.- Borel lol

Vulthuryol

@Eulb try Janich(Topology) or Simmons(Topology and modern analysis)

GFauxPas

lol this is so wild, these mathematicians snarking at each other over choicew

Vulthuryol

I am interested in learning representation theory, I was browsing the wiki on it. There are so many areas to read in. Can anyone suggest where should an undergraduate begin?

GFauxPas

people are objecting to Zermelo not proving it, and he says he knows he didn't prove it, but people use it all the time without saying so explicitly GOSH

he says that denying choice is like saying that "we can't prove the parallel postulate therefore we won't study geometry"

lul

so sassy I love it

The relatively large number of criticisms directed against my small note
testifies to the fact that, apparently, strong prejudices stand in the way
of the theorem that any arbitrary set can be well-ordered. But the fact
that in spite of a searching examination, for which I am indebted to all
the critics, no mathematical error could be demonstrated in my proof and
the objections raised against my principles are mutually contradictory and
thus in a sense cancel each other, allows me to hope that in time all of this

Zermelo ... has 8 pages of polemics with
Poincar´e, Bernstein, Jourdain, Peano, Hardy, Schoenflies etc., which are
very funny and hit the logicists hard. He stresses the synthetic character of
mathematics everywhere and accuses Poincar´e of confusing set theory and
logicism." - Hessenberg

HIT THE LOGICISTS HARD

Frankel, on Zermelo defending himself: “a treatise which has in respect to its sarcasm nothing like it in mathematical literature.”

Waiting

19:49

I wonder if I should improve the solution given here: math.stackexchange.com/questions/2789752/…. What does chat think? Do I need a cleverer solution?

@anon haven't you been made yet an emeritus professor?

@Abr001am !!! What complicated integral do you try to calculate now? :-)

loch

@Vulthuryol there are some notes on introduction to representation theory by pavel etingof and some of his students which are pretty nice

Abr001am

Waiting waiting

anon

@Waiting your solution looks fine. Tolaso's profile says he/she hasn't been on MSE since 3 hours before you posted your answer, so it's a bit premature to ask if OP is confused.

Waiting

@anon look at that before drawing more conclusions: math.stackexchange.com/questions/2805416/…

He keeps a completely wrong result despite all the explanations given to him.

Abr001am

4 integrals ? lol that's enough for me to blowout of overthinking.

anon

19:58

Weird

Waiting

@Abr001am A piece of cake.

anon

@Abr001am it's really just "two"ish integrals before you split apart the three

xyz is treated as one thing at first

Abr001am

yes there is 3 variables, i noticed now :D

Evinda

Hello

We have that $\int_{-\infty}^{+\infty} x^{2n} e^{-2 \lambda x^2} dx \leq \frac{n!}{\lambda^n} \int_{-\infty}^{+\infty} e^{-\lambda x^2} dx=\sqrt{\frac{\pi}{\lambda}} \frac{n!}{\lambda^n}$

Why is the last quantity finite?

Abr001am

i'm coping with telescoping series and i often come accross a circular definition.

how to avoid, or atleast forecast something like that ?

anon

20:03

@Evinda the last quantity is obviously finite. do you mean the last integral?

Faust

DRUGS

anon

@Abr001am you haven't been sufficiently specific about what kind of situation you're talking about, so I can't see how to answer that question

Faust

are good for you

but only if taken in large doses

MatheinBoulomenos

@Faust that's a healthy attitude

Evinda

@anon No I meant $\sqrt{\frac{\pi}{\lambda}} \frac{n!}{\lambda^n}$ .

Is it finite because it does not depend on x? Or do we check what happens when $n \to +\infty$ ?

Faust

20:07

@MatheinBoulomenos math is the same way, bad for you in small doses good in large ones, i think its cause it messes up your brain chemistry

anon

@Evinda I really don't understand your question. It's like asking why $n\lambda$ is finite. If you plug numbers in, it becomes a real number. That's how algebraic expressions work.

Faust

thats not really a helpful answer...

for a given n and a given $\lambda$ the RHS is finite

Abr001am

@anon well it's almost like wheras a system of equations is solvable or not, but in this case there is a determinant.

Lozansky

$\lambda =0$?

MatheinBoulomenos

But $\pi$ is infinite, there are infintely many digits after the period

Faust

20:10

well the statement doesnt even make sense for lambda less than 1

@Lozansky

anon

@Abr001am I thought you were talking about telescoping series, but now you're talking about systems and determinants. What?

Faust

@MatheinBoulomenos its finite and transcendetal stop muddying the waters

Abr001am

@anon yes well, i gave you an assimilation.

anon

convince me you're not a bot

Faust

lmao

Evinda

20:12

Ah I thought that it wouldn't be finite, because when $n \to +\infty$ the quantity tends to $+\infty$. So don't we check what happens when $n \to +\infty$ @anon

Faust

it depends what the question is...

and your statement is nonsense

your implying the limit existing means something is finite

MatheinBoulomenos

there's no limit or sup or anything, so no. You also wouldn't say that $n^2$ is infinite where $n \in \Bbb N$ because it goes to infinity for $n \to \infty$

Faust

or vice versa

anon

@Evinda You can say $\lim_{n\to\infty}(\rm stuff)$ isn't finite. Or you can say $(\rm stuff)$ is an unbounded function of $n$. Perhaps you mean one of these two statements.

Faust

the limit is not finte

but the RHS is finite

this is why math has very specific definitions O.o

Evinda

20:16

@Faust Ah because we consider the RHS as a function of x, right?

anon

no, because when we call it finite, we consider it a number, not a function

Evinda

Ah I see, because we consider n to be fixed, right?

anon

the word "finite" applies to "number" or "value" (in particular, it applies to a "limit") while the word "bounded" applies to "function"

Faust

what you need to do is look up the definitions of a limit existing and the definitions of finite

@Evinda

if you use those it will become clear and smarter people then me came up with them and explain them better

i cant belive i have turned into a native speaker of mathimatics...

Evinda

Ok, I will.. thank you both @anon @Faust

Faust

20:21

sorry i cant be better help but i promise if you read those two definitions it will become clear

GFauxPas

20:41

In b4 numbers are just constant maps

Faust

O.o

Mancala

@AkivaWeinberger Thank! I understood my mistake now!

Faust

21:04

@MatheinBoulomenos do you really need the axiom of choice to pick an element from a nonempty set? surely not.

peano axioms should give you that much

MatheinBoulomenos

@Faust I was joking

Alessandro Codenotti

You need the axiom of choice to choose whether you believe in the axiom of choice

mercio

oh my

Faust

@MatheinBoulomenos axiom of choice makes me head hurt shouldnt say confusing things like that

Maximus

$Is ||(\vect{x} + \vect{u})/r||^2 = ||(\vect{x} + \vect{u})||^2 * 1/r^2 ?$

that by what property are we allowed to take constants in and out of norms?

$||(\vec{x}+\vec{u})/r||^2=||(\vec{x}+\vec{u})||2∗1/r^2$

MatheinBoulomenos

21:17

I think it's sometimes called "absolute homogenity"

$\| \lambda v\|=|\lambda| \|v\|$

Maximus

but this property

is true correcT?

MatheinBoulomenos

yes

it's part of the definition of a norm

Maximus

Thanks MatheinBoul

MatheinBoulomenos

it tells you intuitively that if you scale a vector, the length gets scaled in the same way

Maximus

oh

you're right its on wikipedia too

p(av) = |a| p(v) (being absolutely homogeneous or absolutely scalable).

Leaky Nun

21:19

how dare you doubt @MatheinBoulomenos's teaching

Maximus

haha

user193319

If $a$ is an element in some unital $C^*$-algebra, is it true that $a-||a||1$ is a positive element?

Mike Miller

@user193319 Is a self-adjoint?

Presumably you need that for this to be true

Then your new thing would be a negative operator becauss the spectral radius is bounded by the norm

Maximus

21:44

can anyone give me some tips on how to separate/solve this integral?

$\int e^{-\pi ||(\vec{x} + \vec{u})/r||^2} d \vec{u}$

where r is just a constant

Lozansky

@Maximus Limits?

Maximus

limits are over a closed ball

$\int_{||\vec{u}|| \le n 2^n \lambda_n(L)} e^{-\pi ||(\vec{x} + \vec{u})/r||^2} d \vec{u}$

where lambda is just a constant

MatheinBoulomenos

@MikeMiller this is probably a pretty simple question, but since we know that homotopy classes of maps $X \to \operatorname{Gr}(k, \Bbb R^\infty)$ correspond to rank $k$ vector bundles on $X$, when we set $X=\operatorname{Gr}(k, \Bbb R^\infty)$, we get an element corresponding to the identity.
Intuitively, this should be the tautological bundle on the Grassmanian, right?

user193319

@MikeMiller Yes.

Lozansky

@Maximus Is $\vec{x}$ somehow related to $\vec{u}$?

Maximus

21:48

no

I need some kind of a bount

bound

Lozansky

What course is this?

Maximus

I also know that $||\vec{x}|| \ge \sqrt{n} r$

and $r > 2^{2n} \lambda_n(L)$

No course, some paper I'm working on

MatheinBoulomenos

@Maximus I don't know if that bound is good for you, since this estimation seems pretty rough, but anyway: you cann pull the constant $e^{-\pi/r^2}$ out of the integrand, so we just work with $e^{-\|\vec{x} + \vec{u}\|}$.
Since $e^x$ is always positive, you can bound the integral with your specific area of integratation by the same integral, but over the whole of $\Bbb R^n$ (this may not even as bad as an estimate as it sounds, as $e^{-x^2}$ falls off pretty quickly.)
But then we can just do the substitution $\vec{y}=\vec{x}+\vec{u}$, so that the $u$ term vanishes. Then we're left with

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