Mathematics - 2015-08-31 (page 2 of 3)

12:08 PM

@r9m Faà di Bruno's formula reduces it to a combinatorial problem. Not sure whether that is easy to get a grip on, but may be worth a try.

@DanielFischer Hey

hello @BenLim

hello

hi

@DanielFischer with/without faa di bruno we get an awkward recursion .. $a_n = \sum\limits_{k=0}^{[n/2]} \frac{a_k}{(n- 2k)!}$

Balarka Sen

12:12 PM

@BenLim how's your preparation for quals going?

Ben Lim

@BalarkaSen dunno....

@DanielFischer I was wondering if I can run an argument by you.

Daniel Fischer

You can run, but you cannot hide.

Ben Lim

@DanielFischer Ok, I have the following problem.

r9m

@DanielFischer lool :P

Ben Lim

Let $G$ be a finite group, $G_p$ a $p$-Sylow subgroup, $V$ an irreducible linear representation of $G$ over $\overline{F_p}$. Then $\dim V \leq [G: G_p] - 1$.

By some previous problem, we know that G_p, being a p-group itself fixes a line pointwise.

Let $v$ be the vector that spans that line. Let $G_p, g_2G_p,\ldots, g_nG_p$ be left cosets that partition $G$.

Since $V$ is irreducible, we know first of all that the set $\{gv\}_{g \in G}$ spans all of $V$. However, I claim more, namely that $v, g_2v,\ldots,g_nv$ spans $V$. @DanielFischer

Indeed, suppose we have a vector $gv$ with $g$ belonging to some left coset $g_iG_p$. We can write $g = g_ih$ for some $h\in G_p$. Then $gv = g_ihv = g_iv$ since $G_p$ fixes $v$ pointwise, and so the claim follows.

Daniel Fischer

12:17 PM

Yes.

Ben Lim

So this immediately shows that $\dim V \leq [G: G_p]$.

Now I also want to claim that $v + g_2v + \ldots + g_nv = 0$.

This should be clear no? It's fixed by all of $G$, and the representation is irreducible, so it has to be the zero vector. @DanielFischer

If I know this, it would follow immediately that $\dim V \leq [G: G_p] - 1$.

Daniel Fischer

@BenLim I guess so. Can't remember what the definition of an irreducible representation was. No nontrivial invariant subspaces?

Ben Lim

yea.

Daniel Fischer

Then you're home and dry.

Ben Lim

@DanielFischer I just have one doubt. Where did I use the fact that $G_p$ was a $p$-Sylow subgroup? I only used the fact that it was a $p$-subgroup of $G$.

Khallil

12:21 PM

Can anybody offer a helping hand here with a question on uniform continuity?

in Metric Spaces, yesterday, by Rupsa

i have a question if f :R^n to R defined by f($\x_1$,..., $\x_n$)= max{|$\x_1$|,..,|$\x_n$|} then how can i prove that f(x) is uniform continuous?

Ben Lim

@DanielFischer Maybe I should ask on main.

Daniel Fischer

@BenLim Well, if $G_p$ is not a Sylow subgroup, it's still true - then you claim strictly less, since $[G : G_p] > [G : \operatorname{Syl}_p(G)]$.

@BenLim Ah, no, by $\operatorname{Syl}_p(G)$ I meant some Sylow $p$-subgroup of $G$.

Not the number of such subgroups.

Ben Lim

oh really is that true? Hmm I didn't know that

Daniel Fischer

@BenLim Every $p$-subgroup $G_p$ is contained in some Sylow $p$-subgroup $S$. Then $[G : G_p] = [G : S]\cdot [S : G_p]$.

Ben Lim

Oh pff of course.

@DanielFischer I only came to the solution of the problem while rolling around in bed at 5am

Daniel Fischer

12:30 PM

@BenLim That's not a good time. Too late to go to bed, and waaaaaaaaaayyyyyyyyyyyyy too early to wake up.

Remember me

2

Q: If $|H|=112$ then $A_7\cap H \lhd H$?

I posted this because Alex Clark asked in chat and I'm not sure how to proceed. Let $G$ be a group such that it has a fixed subgroup isomorphic to $A_7$, which we denote simply by $A_7$. Let $H$ be a subgroup of $G$ that has order $112$. Prove that $H\cap A_7 \lhd H$. Since $|H|=2^47$ and $|A_7|...

Ben Lim

@DanielFischer I don't see why the vector $v + g_2v + \ldots +g_nv$ is fixed by $G$.

Remember me

@Dream You are there?

Daniel Fischer

@BenLim What is the natural operation of $G$ on the set of (left) cosets of $G_p$?

Ben Lim

left multiplication

Remember me

12:40 PM

Hello@Balarka

Daniel Fischer

And what does that do with the set, @BenLim?

Balarka Sen

grm, i can't seem to be able to find a flaw in an obviously false argument.

hi @Remember

Ben Lim

the problem is this. If we choose any $g \in G$, we write it as $g = g_ih$ for some $h \in G_p$. Then $g \cdot (v + g_2v + \ldots + g_nv) = g_iv + g_ig_2v + \ldots + g_ig_nv$. Now why is this equal to $v + g_2v + \ldots + g_nv$ @DanielFischer?

Daniel Fischer

@BenLim You are overcomplicating it. What does $aG_p \mapsto (ga)G_p$ do with the set of cosets of $G_p$?

Ben Lim

it sends it to another left coset.

Daniel Fischer

12:43 PM

Every single one. What does it do with the set?

Ben Lim

it permutes the left cosets.

Daniel Fischer

Aha. It permutes.

anon

your sum is a sum over the orbit of v right?

Ben Lim

@anon chat.stackexchange.com/transcript/message/23778643#23778643

Balarka Sen

hey @anon. wanna help me with something?

Ben Lim

12:44 PM

@anon No it's not. It's a sum over the coset representatives.

Daniel Fischer

@BenLim It's more obvious if you write it as $$G_p v + g_2 G_p v + \dotsc + g_n G_p v.$$

Ben Lim

Ok, if we write it that way, then $g \cdot (G_pv + \ldots + g_n G_p v) = ....$

@DanielFischer I think I see it now!

anon

@BenLim well, I suppose the stabilizer of $v$ could be bigger than $G_p$, but then it's just a sum of v's orbit times [Stab:G_p] methinks

Balarka Sen

$X$ be a simply connected space. $\Bbb RP^2 \to X$ be a map. this is zero on $\pi_1$, so image of $\Bbb RP^1$ under this map is a loop in $X$ which is homotopic to the zero loop. this implies that the map extends to a map $\Bbb RP^2 \cup C(\Bbb RP^1) \to X$ modulo homotopy. that is, every map $\Bbb RP^2 \to X$ (is homotopic to a map that) factors through the quotient map $\Bbb RP^2 \to \Bbb RP^2/\Bbb RP^1$.

Ben Lim

fuuuu that was trippy.

anon

12:47 PM

if you could divide by p it'd be $|G_p|^{-1}\sum gv$ of course

Balarka Sen

this gives me a bijection between homotopy classes of based maps $[\Bbb RP^2, X]$ and $[S^2, X]$.

what's the flaw in this argument?

Ben Lim

@anon I'm studying for quals now.

Remember me

@Khallil What have you tried till now?

I guess it is just the kind of $\epsilon$-$\delta$ proofs which we used to do in analysis@Khallil

Khallil

Yep, that's what I thought.

Just gotta find a suitable $r$ and $e$.

I'm notoriously bad at that though. ^_^"

Remember me

Yes.

Balarka Sen

12:52 PM

@anon Are you having a look at what I asked?

Daniel Fischer

@Khallil $\lVert x\rVert_\infty \leqslant \lVert x\rVert_2$, so $f$ is Lipschitz with Lipschitz constant $1$.

Khallil

in Metric Spaces, 15 hours ago, by Khallil

The $|\dots|$ on the inequality on the RHS is equivalent to $\sqrt{(\dots)^2}$, @Rupsa.

anon

@BalarkaSen nope. dunno enough to.

Balarka Sen

oh, i thought you knew about $\pi_1$'s and homotopies.

anon

dunno why the image of RP^1 would be a loop for instance

Balarka Sen

12:55 PM

RP^1 is a circle. image of a circle is precisely what we call a loop.

Daniel Fischer

@anon That's because the projective line is an $S^1$.

anon

woops confusing it with RP^2

k

Ben Lim

@DanielFischer I'm off now.

anon

what's C(RP^1)?

Remember me

@Khallil What daniel told is a very good idea .

Balarka Sen

12:56 PM

cone over RP^1

anon

@BenLim seeya

Daniel Fischer

@BalarkaSen Why would the fact that the image of the line is nullhomotopic imply that the full map is homotopic to a map that is constant on the line?

Balarka Sen

that is, RP^1 \times [0, 1] with RP^1 \times {0} pinched to a point

anon

so, putting RP^1 around an open disk?

Balarka Sen

yeah

anon

12:57 PM

well, I have to be getting ready for work, but looks like something I could actually read

Balarka Sen

@DanielFischer image being nullhomotopic means you can extend it to RP^2 \cup C(RP^1). this is homeomorphic to RP^2/RP^1, as (RP^2, RP^1) is a CW-pair

@anon yeah, that's why i asked you

Remember me

Now since every Lipschitz continuous map between two metric spaces is uniformly continuous the map is uniformly continuous

Well this method is comparatively better because it will be tedious to go with the epsilon delta's especially when you have $\Bbb{R}^n$@Khallil

Daniel Fischer

@BalarkaSen Why should it be homeomorphic? If you glue a cone to a surface, in general you don't get a surface.

Balarka Sen

sorry, homotopy equivalent

X \cup CA is homotopy equiv to X/A in general, whenever (X, A) is a CW-pair

Khallil

Thanks, @DanielFischer and @Rememberme. :-)

Daniel Fischer

1:00 PM

@BalarkaSen Okay, that may be true.

Balarka Sen

it is true :)

if you can find a flaw in that argument, it'd be of great help. i wasted a whole afternoon on trying to do that.

Daniel Fischer

@BalarkaSen How do you know the conclusion is wrong?

Balarka Sen

@DanielFischer Take X = CP^\infty. [RP^2, X] is in bijective correspondence with H^2(RP^2; Z), which is Z/2. However, if every map did factor modulo homotopy into a map S^2 --> X, then we'd have a bijection to [S^2, X] = $\pi_2(\Bbb CP^\infty) = \Bbb Z$.

Daniel Fischer

Okay, that would indeed be a problem.

Balarka Sen

I'm totally confused about why the argument is false, though.

Remember me

1:23 PM

Hello@Huy

Khallil

Can I ask what the symbol for a sphere is in $\mathbb{R}^3$? It has an $S$ or something.

Remember me

$S^2$ If I am not wrong@Khallil

Khallil

Thanks, @Rememberme.

Balarka Sen

@Rememberme $S^2$ is not a symbol for a "sphere inside $\Bbb R^3$". It denotes there 2-sphere regardless of any space it's embedded in.

But it doesn't hurt to use it for the subspace of $\Bbb R^3$ homeomorphic to $S^2$ either.

Fixed.

Aditya Agarwal

1:47 PM

Hello all.

0

Q: Find the value of $f(0)$, where $F'(a)+2$ is the area bounded by...

Let $$F(x)=\int_{x}^{x^2+\frac{\pi}{6}}2\cos^2tdt$$ for all $x\ \epsilon \ \mathbb {R}$ and $f:[0,\frac12]\to[0,\infty)$ be a continuous function. For $a \ \epsilon \ [0,\frac12],$ if $F'(a)+2$ is the area of the region bounded by $x=0;y=0;y=f(x);x=a$, then what is the value of $f(0)$? So we fi...

functions definite-integrals contest-math area

Please this is urgent.

Please explain someone or some people.

Thanks in advance (y).

morphic

(y)

Aditya Agarwal

Is someone from you all downvoting the question.

morphic

I upvoted but I guess someone downvoted again

And again

Aditya Agarwal

Whats happening?

Ohk leave it plz answer my query.

morphic

dunno, i don't even know how to do the question

Aditya Agarwal

1:54 PM

Oh. (y)

morphic

What is (y)

Moses

@DanielFischer I am busy studying the proof of Riesz Decomposition which is: Let $a \in \mathcal{A}$. Suppose that $\sigma(a) = \sigma_{1} \cup \sigma_{2}$ where $\sigma_{1} \cap \sigma_{2} = \emptyset$ and $\sigma_{1}, \sigma_{2} \neq \emptyset$.

Then there exists non-trivial idempotents $E_{1}, E_{2} \in \overline{\text{Alg}(1,a)}$ such that $E_{1} + E_{2} = 1$.

If $\mathcal{A} \subset \mathcal{L}(X)$ then $E_{1}X,~ E_{2}X$ are closed invariant subspaces under $a$ and $E_{1}X \vee E_{2}X = X$.

Aditya Agarwal

(y)

Daniel Fischer

@Moses Sounds right so far.

Moses

@DanielFischer Do you maybe know why it follows that $E_{1}X = \text{ker}(1 - E_{1})$?

Daniel Fischer

2:04 PM

@Moses That's always the case for idempotents. $(1 - E_1)(E_1X) = (E_1 - E_1^2)(X) = (E_1 - E_1)(X) = 0(X) = \{0\}$. And if $x \in \ker (1 - E_1)$, then $x = E_1 x + (1 - E_1)x = E_1 x + 0 = E_1 x \in E_1(X)$.

Moses

@DanielFischer Yeah I understand. One more thing, it states that $E_{1}X$ and $E_{2}X$ are complimentary i.e. $E_{1}X \vee E_{2}X = X$ and $E_{1}X \cap E_{2}X = \{ 0 \}$. Are you familiar with the notation $E_{1}X \vee E_{2}X = X$? Is '$\vee$' just the union of the images of $E_{1}$ and $E_{2}$?

Daniel Fischer

@Moses Not the union, a vector space is never the union of two proper subspaces. I guess your $X$ is a vector lattice? Then $A \vee B$ would be the subspace generated by $\{ a \vee b : a \in A, b\in B\}$, I expect. Or it could be an unusual notation for the direct sum.

Chris's sis the artist

2:21 PM

0

Q: A very tough integral $\int_0^{\pi} \arctan^3\left(\frac{\sin (x)}{2 \sqrt{2}}\right)\csc ( x) \, dx$

My research shows that $$\int_0^{\pi} \arctan^3\left(\frac{\sin (x)}{2 \sqrt{2}}\right)\csc ( x) \, dx$$ $$=\frac{3}{16} \pi \sinh ^{-1}(1) \log ^2(2)-\frac{1}{96} 85 \pi \log ^3(2)+\frac{61}{16} \pi \log (3) \log ^2(2)-\frac{13}{32} \pi \log (4) \log ^2(2)+\frac{9}{16} \pi \log (6) \log ^2(2)+...

calculus real-analysis integration definite-integrals

I've decided to post on main the crazy cubic version. Some that like much polylogarithms might come up with great simplifications.

Moses

@DanielFischer It is just given as a Banach space. But in the proof it also states that $X = E_{1}X + E_{2}X$.

Daniel Fischer

@Moses Then it's probably an odd notation for the subspace sum - actually, more likely sum than direct sum, I now think.

Moses

@DanielFischer Yeah but if it was the direct sum, why would they add $E_{1}X \cap E_{2}X = \emptyset$ as a separate condition. Is this not implied my direct sum.

Daniel Fischer

@Moses Yes, that's why I think it's more likely that they mean sum than direct sum now.

Moses

@DanielFischer Oh yeah I misread.

Daniel Fischer

2:27 PM

Makes kind of sense, since the sum of two subspaces is the supremum in the lattice of linear subspaces.

Moses

@DanielFischer Yeah.

Mary Star

2:47 PM

Hello!! A general formula of a polynomial in one variable is $\sum_{i=0}^n a_i x^i$. Which is the general formula of a polynomial in two variables? Is the general formula maybe the binomial theorem, $(x+y)^n=\sum_{i=0}^n \binom{n}{i}x^iy^{n-i}$ ?

@DanielFischer @robjohn @TobiasKildetoft do you have an idea?

Daniel Fischer

$$\sum_{k = 0}^K \sum_{n = 0}^N a_{kn}x^ky^n$$

Owatch

Hello!

Got a question

I've got $\int{sin^{4}(3x)}dx$

I'm told I can do the following:

> If the powers of both sine and cosine are even, use the half-angle identities

The other recommendation are for solving situations where either sin or cos have odd exponents.

But this is just $sin$, with an even exponent. Should I go right for half angle?

Mary Star

@DanielFischer I see... Can we represent each element of $F[x,\sqrt{x^2-1}]$ as $(x+\sqrt{x^2-1})^n, n \in \mathbb{Z}$ ?

Daniel Fischer

@Owatch It's not the only good way, but one good way.

Owatch

Okay, I'm already down that path, so I'll continue

r9m

3:01 PM

@Chris'ssistheartist INSANE!!!

Chris's sis the artist

@r9m :-))))))))

r9m

@Chris'ssistheartist is the 4th power even longer?

Daniel Fischer

@MaryStar No, for example $0$ is not a power of $(x + \sqrt{x^2-1})$. If you mean as a sum $$\sum_{n = -N}^N a_n (x+\sqrt{x^2-1})^n,$$ that's different.

Chris's sis the artist

@r9m It's shorter! But it is still unsimplified. I need to find ways of doing that.

Daniel Fischer

@Owatch What's your first step?

r9m

3:02 PM

@Chris'ssistheartist 'kay!

Chris's sis the artist

@r9m The MOST amazing thing is that it can be done for all powers!!!

Owatch

@DanielFischer I changed it to two double angle identities

r9m

@Chris'ssistheartist I see!!

Daniel Fischer

@Owatch Namely?

Mary Star

@DanielFischer So, can we represent each element of $F[x,\sqrt{x^2-1}]$ as $\sum_{n = -N}^N a_n (x+\sqrt{x^2-1})^n$ ?

Owatch

3:05 PM

$\int{[\sin(3x)]^{2} * [\sin(3x)]^{2}}\, dx$ to $\int{ [ \frac{1}{2}(1-\cos(6x))] *[ \frac{1}{2}(1-\cos(6x))]}\,dx$

Daniel Fischer

@MaryStar Well, go and look. Start with something simple, like $x$.

@Owatch Good. Then multiply, and one more half/double angle identity.

Mike Miller

Morning.

Owatch

And one more?

I have this now

Balarka Sen

Morning, @MikeMiller.

I asked a question above, have you seen it?

Owatch

$\frac{1}{4}* \int{dx} - \frac{1}{2} * \int{cos(6x)}dx + \frac{1}{4} * \int{ [cos(6x)]^{2}}dx$

Mike Miller

3:09 PM

You said you resolved it, I thought.

Balarka Sen

I thought I did, but I really didn't.

Mike Miller

OK, so precisely what's the question?

robjohn

@Owatch $\sin^4(3x)=\frac18\cos(12x)-\frac12\cos(6x)+\frac38$

Owatch

That isn't the answer

Until I'm done with mine.

robjohn

@Owatch That allows you to integrate it easily

Balarka Sen

3:13 PM

$X$ be a simply connected space. $\Bbb RP^2 \to X$ be a map. It's zero on $\pi_1$, so the copy of $\Bbb RP^1$ maps to a loop in $X$ homotopic to $0$. So $\Bbb RP^2 \to X$ extends to a map $\Bbb RP^2 \cup C(\Bbb RP^1) \to X$. So the map is homotopic to a map $\Bbb RP^2 \to X$ which factors as $\Bbb RP^2 \to\Bbb RP^2/\Bbb RP^1 \to X$.

This implies $[\Bbb RP^2, X]$ is in bijection with $[S^2, X]$

Now, this is a flawed argument. My question is : where is the flaw?

Mike Miller

I don't see the "This implies"...

Balarka Sen

Well, $\Bbb RP^2/\Bbb RP^1 \cong S^2$.

Mike Miller

So?

It's not even clear that you've written down a well-defined map $[\Bbb{RP}^2, X] \to [S^2, X]$. Why doesn't this depend on the choice of homotopic map that factors?

robjohn

$\int\sin^4(3x)\,\mathrm{d}x=\frac1{96}\sin(12x)-\frac1{12}\sin(6x)+\frac38x+C$
I just checked, and that is the same answer that Mathematica gives

Balarka Sen

@MikeMiller Let $\Bbb RP^2 \to S^2 \stackrel{f}{\to} X$ and $\Bbb RP^2 \to S^2 \stackrel{g}{\to} X$ be two homotopically maps. Assume $f, g$ are homotopic. Then extend the homotopy to the whole map.

Mike Miller

3:16 PM

Not what I said. Pick two choices $\Bbb{RP}^2 \to S^2 \to X$ that are homotopic. Why can you extract the homotopy type of $S^2 \to X$?

Why can't you pick $f$ and $g$ to not be homotopic, but the whole map to be?

Balarka Sen

$\{F_i : \Bbb RP^2 \to X\}$ be the intermediates of the homotopy between the two maps. Then you can factor each $F_i$ into a map of the form $\Bbb RP^2 \to S^2 \to X$. Pick the maps $S^2 \to X$ to get a homotopy between $f, g$.

Why can't that work?

Mike Miller

Why should you be able to factor each map??

Mary Star

@DanielFischer So, we have to find a N such that $x=\sum_{n = -N}^N a_n \left [ \sum_{i=0}^n \binom{n}{i}x^i(\sqrt{x^2-1})^{n-i}\right ]$, right?

Balarka Sen

6 mins ago, by Balarka Sen

$X$ be a simply connected space. $\Bbb RP^2 \to X$ be a map. It's zero on $\pi_1$, so the copy of $\Bbb RP^1$ maps to a loop in $X$ homotopic to $0$. So $\Bbb RP^2 \to X$ extends to a map $\Bbb RP^2 \cup C(\Bbb RP^1) \to X$. So the map is homotopic to a map $\Bbb RP^2 \to X$ which factors as $\Bbb RP^2 \to\Bbb RP^2/\Bbb RP^1 \to X$.

Mike Miller

and even then, why in such a way that depends continuously on $t$?

It's homotopic to one. Come on.

Balarka Sen

3:20 PM

yeah, of that I am not sure anymore.

@MikeMiller ok, good point.

Daniel Fischer

@MaryStar The binomial expansion looks different for $n < 0$. However, it's better to start with looking intensely at $(x + \sqrt{x^2-1})^{-1}$.

Owatch

Well, I fucked it up somewhere.

I hate that, so much paper wasted.

Balarka Sen

thanks for that, @MikeMiller.

I was pretty confused about this argument.

Owatch

$\frac{1}{4}\int{dx} - \frac{1}{2}\int{cos(6x)dx} + \frac{1}{4}\int{(cos(6x))^{2}dx}$

Became $\frac{1}{4}x - \frac{1}{12}sin(6x) + \frac{1}{96}(x + sin(12x))$

robjohn

@Owatch remember that $\sin^2(x)=\frac{1-\cos(2x)}2$ and $\cos^2(x)=\frac{1+\cos(2x)}2$

Owatch

3:29 PM

That last piece, was converted from $\frac{1}{4}\int{(cos(6x))^{2}}dx$ to $\frac{1}{4}\int{\frac{1}{2}[1 + cos(12x)]dx}$

Which the became $\frac{1}{8}\int{1+cos(12x)}dx$

Which then became what I finally had above, from the substitution of $u = 12x$, $dx = \frac{du}{12}$ in $\frac{1}{8}\int{1 + cos(u)\frac{du}{12}}$

Mary Star

$x$ can be written as followed:

$$x=\frac{1}{2}(x + \sqrt{x^2-1})^{-1}+0(x + \sqrt{x^2-1})^{0}+\frac{1}{2}(x + \sqrt{x^2-1})^{1}$$

right?

Daniel Fischer

@MaryStar Right. Since $(x+\sqrt{x^2-1})^{-1} = x - \sqrt{x^2-1}$. Now it should be easy to write $\sqrt{x^2-1}$ in that form too. And then you can see that you can write every polynomial in $x$ and $\sqrt{x^2-1}$ in that form.

Owatch

So uh, is my substitution with double angles right?

Daniel Fischer

3:50 PM

@Owatch You shouldn't have included the constant $1$ when substituting. Or you have forgotten the $12$ when substituting back, $\frac{1}{96} u \leadsto \frac{1}{96}(12 x) = \frac{1}{8}x$.

Owatch

Oh

I see

I moved 1/12 out

But I can't just do that with the 1. Damn

Balarka Sen

$[\Bbb RP^2, S^2]$ is $\Bbb Z_2$, isn't it? 'Cause this is in bijection with $[\Bbb RP^2, \Bbb CP^\infty]$ (by restricting to $2$-skeleton) which is bijective to $\Bbb Z_2$

Mike Miller

Yes, though again you need to be more careful about "this is in bijection with..."

That those are bijective needs proof.

Balarka Sen

ok, fair enough.

Well, the map $[\Bbb RP^2, S^2] \to [\Bbb RP^2, \Bbb CP^\infty]$ is defined by composing with homeomorphisms with $\Bbb CP^1 \subset \Bbb CP^\infty$. the other way map is defined by restriction to $2$-skeleton.

These are apparently inverses to each other, so you're done.

Owatch

@DanielFischer Wait, is it $\frac{1}{8}\int{(1 + cos(u))*\frac{1}{12}*du}$ or $\frac{1}{8}\int{ 1 + \frac{cos(u)}{12}*du}$?

Nevermind

Stupid question

Mike Miller

3:59 PM

@BalarkaSen No. What does "restricting to the 2-skeleton" even mean? Why should the image of $\Bbb{RP}^2$ be in the 2-skeleton?

Moses

@DanielFischer Riesz Decomposition is given by: Let $a \in \mathcal{A}$. Suppose that $\sigma(a) = \sigma_{1} \cup \sigma_{2}$ where $\sigma_{1} \cap \sigma_{2} = \emptyset$ and $\sigma_{1}, \sigma_{2} \neq \emptyset$.

Then there exists non-trivial idempotents $E_{1}, E_{2} \in \overline{\text{Alg}(1,a)}$ such that $E_{1} + E_{2} = 1$.

If $\mathcal{A} \subseteq \mathcal{L}(X)$ then $E_{1}X,~ E_{2}X$ are closed invariant subspaces under $a$ and $E_{1}X \vee E_{2}X = X$.

Does it follow that $X = (E_{1} + E_{2})X$ by noting that $(E_{1} + E_{2})(x) = I(x) = x$ for all $x \in X$. We get this…

Balarka Sen

@MikeMiller I mean, any map $\Bbb RP^2 \to X$ is homotopic to map that has image inside the $2$-skeleton, right?

Mike Miller

OK, and why is this choice of map well-defined?

(Also, why is that true?)

Mary Star

@DanielFischer I see... Thank you!! :-)

Daniel Fischer

@Moses It does. But note that generally $(S+T)(X)$ can be a proper subspace of $S(X) + T(X)$.

Owatch

4:06 PM

Yes, got it!

Moses

4:17 PM

@DanielFischer Why is it equal in this case. I thought it follows from linearity?

Daniel Fischer

@Moses You always have the inclusion $(S+T)(X) \subset S(X) + T(X)$. So if $(S+T)(X) = X$, it is necessarily equality.

Balarka Sen

Grr, I am keep getting disconnected.

@MikeMiller Good question! I recall there was something called the cellular approximation theorem, but I don't now how to prove it.

Mike Miller

Don't worry about proving it. What's the statement?

Balarka Sen

$X, Y$ be finite CW-complexes, $f : X \to Y$ be a map between them. Then $f$ is homotopic to a cellular map.

A cellular map, iirc, is a map which takes the $n$-skeleton of $X$ to $n$-skeleton of $Y$

Mike Miller

OK, good. That's what you need. There's also a relative version, mind.

Balarka Sen

4:23 PM

(I don't know if there were any other assumption, though. Maybe there was)

Mike Miller

No.

Also, 'finite' is unnecessary.

Balarka Sen

ok, that's nice.

@MikeMiller really?

Mike Miller

Why would it be?

Balarka Sen

Take a map between CW-complexes which has uncountably many cells in each dimensions. Then is it still homotopic to a cellular map?

Mike Miller

I guess maybe because you want to make it cellular 'one cell at a time'.

Yeah, sure, why not?

Balarka Sen

4:24 PM

@MikeMiller ah, yeah, that's what I was thinking of.

Mike Miller

@BalarkaSen: The point here is that $n$-cells, except for their boundaries, are disjoint. So if you're modifying the map on the $n$-skeleton (while keeping the $(n-1)$-skeleton fixed) you can modify all the cells at once without worrying about dealing with overlap.

The relative version of cellular approximation says that if $f$ is cellular on a subcomplex $A$, the homotopy may be taken to be constant on $A$. OK, now use this.

Moses

@DanielFischer We assume that Banach algebra $\mathcal{A} \subseteq \mathcal{L}(X)$. Do you know why it follows that $\mathcal{A}$ has the same unit (i.e. $I(x) = x$) as $\mathcal{L}(X)$? What if the unit of $\mathcal{L}(X)$ is not in $\mathcal{A}$ or $\mathcal{A}$ has it's own unit different from $I$?

Balarka Sen

@MikeMiller OK, I'll try, but I have to vanish for a few hours. I'll come back and do it.

Moses

@DanielFischer But since we are assuming that $\mathcal{A}$ is a unital Banach algebra at the start, I guess the question is if $\mathcal{A} \subseteq \mathcal{L}(X)$ then does $\mathcal{A}$ immediately have the unit of $\mathcal{L}(X)$?

Mike Miller

5:25 PM

Hi @PVAL. Did you get a chance to ask Dan about that question?

PVAL

@MikeMiller No that will be this wednesday.

Mike Miller

Okie

Daniel Fischer

@Moses Not necessarily. You could for example have a decomposition $X = Y \oplus Z$, and have $\mathcal{A}$ correspond to $\mathcal{L}(Y)$, then $\mathcal{A}$ is a unital Banach algebra, but its unit is not the unit of $\mathcal{L}(X)$. But in fact, if you have a unital Banach algebra $\mathcal{A} \subset \mathcal{L}(X)$, then the unit $u$ of $\mathcal{A}$ is idempotent and thus a projection, hence induces a decomposition $X = u(X) \oplus (1-u)(X)$, and you are in the situation described.

Mike Miller

5:41 PM

@BalarkaSen: Welcome back.

Moses

@DanielFischer In that case that you mentioned last, does $\mathcal{A}$ and $\mathcal{L}(X)$ have the same unit?

Balarka Sen

rehi, @MikeMiller.

I was just looking at the statement of the problem.

Moses

@DanielFischer I assume yes, since for any $f \in \mathcal{L}(X)$ it follows that $f(x)(u(x)+(1-u)(x)) = f(x) = (u(x)+(1-u)(x))f(x)$.

Daniel Fischer

@Moses If $\mathcal{A}$ contains the unit of $\mathcal{L}(X)$, that is necessarily also the unit of $\mathcal{A}$ and the decomposition is the trivial $X = X \oplus \{0\}$. If the unit of $\mathcal{A}$ is not the unit of $\mathcal{L}(X)$ and not $0$ (which may be excluded by the definition of a unital Banach algebra but is not necessarily excluded), you have a nontrivial decomposition.

Then $\mathcal{A}$ is a unital Banach algebra in its own right, but it is not a "unital subalgebra" of $\mathcal{L}(X)$.

@Moses But to be the unit of $\mathcal{A}$, we only need that $u\circ f = f\circ u = f$ for all $f\in \mathcal{A}$, not for all $f \in \mathcal{L}(X)$.

Balarka Sen

@MikeMiller OK, consider a based map $\Bbb RP^2 \to \Bbb CP^\infty$. We can take the basepoint to be the $0$-cell on each complex. This map is then homotopic rel basept to a cellular map.

The cellular map takes all of $\Bbb RP^2$ to the $2$-skeleton of $\Bbb CP^\infty$. What else there is to prove?

Mike Miller

5:56 PM

Again, well-definedness. You have not at all convinced me this is a well-defined map $[\Bbb{RP}^2,\Bbb{CP}^\infty] \to [\Bbb{RP}^2,S^2]$.

Balarka Sen

Ah, right.

I'm going to prove that after getting dinner.

Moses

@Danielfischer Okay so in this case where they just state that $\mathcal{A} \subseteq \mathcal{L}(X)$ it could be either case that you mentioned above since they don't specify whether $\mathcal{A}$ is a unital subalgebra or not. Hence the decomposition could either be trivial or not. If trivial then $E_{2}(x) = 0$ and $E_{1}(x) = I(x) = x$. Is this the right idea?

Mike Miller

I prefer to think about this just in terms of the fact that the (obviously well-defined!) map $[\Bbb{RP}^2,S^2] \to [\Bbb{RP}^2,\Bbb{CP}^\infty]$ is injective and surjective, but these are really the same thing, of course.

Daniel Fischer

@Moses That's yet another thing. Here we're talking about the subalgebra $\overline{\operatorname{Alg}(1,a)}$ of $\mathcal{A}$, where $1$ is the unit of $\mathcal{A}$. Whether that is the unit of $\mathcal{L}(X)$ is a different question. The point of the decomposition above is that you can always assume that, since if not, everything happens in $\mathcal{L}(Y)$, and the unit of $\mathcal{A}$ is (by definition of $Y$) just $I_Y$ [well, $I_Y \oplus 0$].

Mike Miller

6:15 PM

@DanielF: Do you get a new operator algebra buddy every year? I think that was me a little over a year ago

Daniel Fischer

@MikeMiller It would not be a good sign if somebody asked operator algebra questions I can answer for much longer than a year, so there has to be some flux.

Mike Miller

True, true.

Moses

6:30 PM

@Danielfischer Yes I understand that the unit of $\overline{\operatorname{Alg}(1,a)} \subset \mathcal{A}$ need not be the same as $\mathcal{L}(X)$. But what I'm saying is that given the statement $\mathcal{A} \subseteq \mathcal{L}(X)$ where it is given that $\mathcal{A}$ is unital but without stating if it has the same unit as $\mathcal{L}(X)$ or not, we get that $X = E_{1}X + E_{2}X$ for some $E_{1},~E_{2} \in \mathcal{A}$ and $E_{1} + E_{2} = 1 \in \mathcal{A}$.

If $1$ is also the unit of $\mathcal{L}(X)$ (which is $I(x) = x$) then by what I understand you said that we get the trivial decomposition $X = X \oplus 0$. Hence I was asking if that means that in this trivial case $E_{1}(x) = 1$ and $E_{2}(x) = 0$? Since then $X = X \oplus 0 = E_{1}(X) \oplus E_{2}(X)$.

@DanielFischer Is the reasoning fine?

Daniel Fischer

@Moses We can get that only if $\mathcal{A}$ does contain the unit of $\mathcal{L}(X)$, which I assumed as given until you explicitly raised the question. Since $E_1$ and $E_2$ belong to $\mathcal{A}$, you would otherwise only have $E_1 X + E_2 X = Y$, where $Y$ is the subspace of $X$ that the unit of $\mathcal{A}$ projects on. That's an unnatural and inconvenient situation, however, so usually one doesn't consider that situation.

Balarka Sen

@MikeMiller Yeah, that is an easier thing to do, as it's clear that it is well defined.

I don't know how to prove that the map in the other direction is well-defined directly.

Mike Miller

You'll see by the time you prove this map is bijective.

Moses

@DanielFischer When you say we can get 'that', what are you referring to?

Daniel Fischer

@Moses $X = E_1 X + E_2 X$. That requires that all of $X$ is in the range of elements of $\mathcal{A}$, hence in the range of the unit of $\mathcal{A}$, and that means it's the unit of $\mathcal{L}(X)$.

Moses

6:52 PM

@DanielFischer Okay but then what you said earlier almost implies that whenever that is the case (which is that $\mathcal{A}$ and $\mathcal{L}(X)$ have the same unit, then the decomposition of $X$ is the trivial $X = X \oplus \{0\}$). Your comment was : @Moses If $\mathcal{A}$ contains the unit of $\mathcal{L}(X)$, that is necessarily also the unit of $\mathcal{A}$ and the decomposition is the trivial $X = X \oplus \{0\}$. If the unit of $\mathcal{A}$ is not the unit of $\mathcal{L}(X)$ and not $0$ (which may be excluded by the definition of a unital Banach algebra but is not necessarily e…

Alessandro

good evening

Balarka Sen

@MikeMiller It doesn't seem clear that it's injective. $f, g : \Bbb RP^2 \to S^2$ be two homotopically distinct map. Compose with the inclusion $S^2 \cong \Bbb CP^1 \hookrightarrow \Bbb CP^\infty$. Let the resulting maps be $f', g'$. Why can't it be possible that $f'$ can be homotoped to $g'$ by sliding through the higher cells in $\Bbb CP^\infty$?

I am probably sleepy.

Daniel Fischer

@Moses We have two different decompositions. Let's denote the unit of $\mathcal{A}$ by $u$. That may or may not be the unit $I_X$ of $\mathcal{L}(X)$. In any case, $u$ is idempotent in $\mathcal{L}(X)$, hence a projection. Let's call $Y$ the range of $u$ and $Z$ the kernel. Then you have the decomposition $X = Y \oplus Z$. That is the trivial decomposition $X = X \oplus \{0\}$ if and only if $u = I_X$. Then, we look at an element $a \in \mathcal{A}$,

and from a decomposition of $\sigma(a)$ - which I assume means the spectrum with respect to $\mathcal{A}$; if it means the spectrum with respect to $\mathcal{L}(X)$, the functional calculus doesn't stay in $\mathcal{A}$ if $u\neq I_X$ - we get a decomposition $u = E_1 + E_2$, and that gives you a decomposition $Y = E_1 Y \oplus E_2 Y$ [and we have $E_k X = E_k Y$, so we can also write that as $Y = E_1 X \oplus E_2 X$, but that's less interesting].

The "outer" decomposition is however boring, and if $u\neq I_X$, then it makes much more sense to view $\mathcal{A}$ as a subalgebra of $\mathcal{L}(Y)$.

Karim Mansour

Hey @MikeMiller I am just having little trouble with notation lets say I want to represent the topology generated by sub-bases A. Then $\tau = \bigcup_{i \in I: B_i \in A} \ \bigcap_{k = 1}^n \ B_{i,k}$ where $B_i,k \in A$?

Mike Miller

7:14 PM

@BalarkaSen: It's not clear that it's injective. You have to think a little harder.

Indeed, prove to me that $[S^2,\Bbb{RP}^2] \neq [S^2,\Bbb{RP}^3]$.

Balarka Sen

hrm.

Mike Miller

@KarimMansour: I would probably never write such a thing like that at all.

I would use words.

Karim Mansour

oh I see

I guess I think to fix it I would need to include { } for the $B_i,k$ since topology is not a set but collection of sets

I like to do stuff in full details when I am learning a subject since I will learn it more that way.

I hope my class covers up until algebraic topology not only point set topology

Moses

@DanielFischer By range of $u$ do you mean range for which it is a unit?

Daniel Fischer

@Moses I mean $\operatorname{im} u$, $\mathcal{R}(u)$, $u(X)$, whatever is your favourite notation.

Moses

7:23 PM

@DanielFischer I probaably should have asked this earlier but what do you mean by since $u$ is idempotent it is a projection? What definition of projection are using?

Owatch

I'm trying to help my brother do math, and he asks questions I just don't know.

Like for instance, why do I substitute to solve integrals?

I don't know, because you have to?

Why do you write dx inside an integral, why do you write du after differentiating u?

Mike Miller

@BalarkaSen: Got an answer for me?

Balarka Sen

I'm thinking.

You want an answer for the second question, or for the original one?

Mike Miller

Second one, since it's easier.

First one too.

but after.

Daniel Fischer

@Moses Heh, should have phrased that better.At the end of the day, a projection is an idempotent linear map. [In the context here. Of course we have e.g. coordinate projections in situations where linearity isn't defined. And we sometimes call a linear map $p \colon E \to F$ a projection when $F$ isn't even a subspace of $E$.]

Owatch

7:28 PM

I solve math problems with a large amount of abstraction from the actual basic logic. I think I could phrase that better, but I really just follow rules and try not to make mistakes. Things make sense to me, and I can see how to solve certain problems, but beyond the rules and whatnot, I don't exactly know how or why what I'm doing is giving me a solution. I guess that is a intelligence barrier that only some people can cross.

Balarka Sen

Well, bleh, that's easy.

$[S^2, \Bbb RP^2]$ is the same as $\pi_2(\Bbb RP^2) = \pi_2(S^2) = \Bbb Z$, as $S^2$ is universal cover of $\Bbb RP^2$.

But $\pi_2(\Bbb RP^3) = \pi_2(S^3) = 0$.

Mike Miller

Yes.

Balarka Sen

Done.

Daniel Fischer

@Owatch You don't have to. It just makes things simpler sometimes. In principle, you could evaluate every integral that you can evaluate with substitution also without. It just becomes messy quite often.

Mike Miller

Nonetheless it is still true that the map $[\Bbb{RP}^2, S^2] \to [\Bbb{RP}^2, \Bbb{CP}^\infty]$ is a bijection.

So I completely agree that it's not obvious. ;)

Balarka Sen

7:31 PM

Yes, which is why I was kind of boggled about why it should be injective above.

Mike Miller

What have you already proved about it?

Balarka Sen

I know it's a well-defined map.

Mike Miller

OK, so you need injectivity and surjectivity. Can you do either?

Balarka Sen

I am trying to see it's injective. Give me some time.

Chris's sis the artist

I was wondering if I should add another name to my existing ones, that is Ramanujan (but it is not usual at all in my country).

(just kidding a bit)

;)

10

Q: A very tough integral $\int_0^{\pi} \arctan^3\left(\frac{\sin (x)}{2 \sqrt{2}}\right)\csc ( x) \, dx$

My research shows that $$\int_0^{\pi} \arctan^3\left(\frac{\sin (x)}{2 \sqrt{2}}\right)\csc ( x) \, dx$$ $$=\frac{3}{16} \pi \sinh ^{-1}(1) \log ^2(2)-\frac{1}{96} 85 \pi \log ^3(2)+\frac{61}{16} \pi \log (3) \log ^2(2)-\frac{13}{32} \pi \log (4) \log ^2(2)+\frac{9}{16} \pi \log (6) \log ^2(2)+...

calculus real-analysis integration definite-integrals polylogarithm

Moses

7:36 PM

@Danielfischer Okay so it is clear that the proof in the text I am using it is referring to the second decomposition which involves $\sigma(a)$. In your explanation of the second type of decomposition are you assuming that $u \neq I_{X}$? If $u = I_{X}$ then $X = E_{1}X + E_{2}X$?

Daniel Fischer

@Moses Yes and no. If $u = I_X$, then we have $X = Y$ (and hence $X = E_1 X + E_2 X$), so I need not assume that $u \neq I_X$, I only say $u$ may be different from $I_X$ in that. Since typically one is looking at $u = I_X$, it is best to suppose that unless one has reason not to.

Balarka Sen

@MikeMiller $f, g : \Bbb RP^2 \to \Bbb CP^1$ be the two homotopically distinct maps. Compose with inclusion into $\Bbb CP^\infty$ to get the maps $f', g'$. Assume these are homotopic. Then we have a homotopy $F : \Bbb RP^2 \times [0, 1] \to \Bbb CP^\infty$ between the two. Maybe I am thinking too hard, but what if I cellular approximate each map $F(-, t)$ to get a bunch of maps $\tilde{F}(-, t) : \Bbb RP^2 \to S^2$? The top and bottom maps are of course the same as $f, g$.

I am wondering if the intermediates form a continuous family. There's not really much reason that they will.

Mike Miller

You're on the right track but the $\tilde F(-,t)$ idea doesn't make sense and will not work.

Balarka Sen

Right. I guess I should cellular approximate the whole of $F$ instead.

Mike Miller

Go on.

Balarka Sen

7:54 PM

Right, so $\Bbb RP^2 \times [0, 1]$ is a CW-complex. So $F$ is homotopic (rel the maps on the top and bottom, where $F$ is already cellular) to a cellular map $F : \Bbb RP^2 \times [0, 1] \to \Bbb CP^\infty$ the image of which sits inside the $3$-skeleton, which is just the same as $\Bbb CP^1 \cong S^2$.

So this map $\widetilde{F} : \Bbb RP^2 \times [0, 1] \to S^2$ is precisely the homotopy between $f, g$ (note that the top and bottom maps are unchanged during the whole process)

That gives you a contradiction.

Mike Miller

Sure, no need to phrase it in terms of contradiction. You just proved injectivity.

(Note that this is the exact same arhument you would use to show that your invwese map was well-defined.)

Balarka Sen

@MikeMiller fair enough.

oh, yeah.

Mike Miller

OK, surjectivity?

Balarka Sen

Who cares? We have two well-defined maps which are clearly inverse to each other. By general nonsense, we are done :P

Mike Miller

I care.

Balarka Sen

7:58 PM

ok :(

Mike Miller

You should know precisely why that map is surjective. I want an answer within two minutes.

Moses

@DanielFischer In the text I am using it states that $\mathcal{A} \subseteq \mathcal{L}(X)$ and that $X = (E_{1} + E_{2})X = E_{1}X + E_{2}X$ where $u = E_{1} + E_{2}$ where $E_{1}$ and $E_{2}$ are non trivial idempotents. Is is clear then that in this case $u = I_{X}$?

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