Mathematics - 2020-07-06 (page 1 of 2)

Thorgott

00:03

in quantum physics, each particle can be thought of as a wave function, which is a complex-valued function

so I think it's fair to say that complex numbers do exist in nature

obliv

Oh interesting, didn't know that @thor I'll have to learn QM before I can agree or disagree :P

I guess I just got annoyed that $\sqrt{-1}$ doesn't have a real number value

Like, why would we even consider complex numbers as a thing or spend time diddling with something like that :P

but mathematicians have no chill with their curiosity

Thorgott

the imaginary number $i$ appears in the Schrödinger equation

so I think it's pretty uncontested to say that complex numbers are indispensable for doing quantum physics

obliv

Right I just meant I never learned it so I can't say anything about it.

Thorgott

engineers also care a lot about complex numbers

something to do with Fourier transforms

user2103480

00:21

"Complex numbers" is just a bad term that we're stuck with imo. One could've chosen a name like "rotational numbers" or whatever. Anything that gives a hint about the structure

obliv

Yeah that would make more sense actually. Didn't hamilton develop quaternions or whatever right after anyway?

complex numbers on their own in like a college algebra class or whenever people learn them seem very unintuitive

Calvin Khor

00:53

@Thorgott those regularity results on bounded subsets of $\mathbb R^n$ aren't trivial but are incredibly standard, I assume the move to manifolds isn't too hard

@user69608 if you want $A = |v+w|^2 + C$ and you want to know what $C$ is, its of course $C=A - |v+w|^2$? unless you assume $a,b,c$ are orthogonal the answer isn't clean. Since you understand the previous line, then you don't need me to open $|v+w|^2$ for you?

Beliod

02:35

Can someone help me understand why an arithmetic series formula looks like the following?

So, we have the series {3, 5, 7...}

The put it into the form:

a(n) = a(n-1) + 2

But I'm not seeing why it shouldn't be written:

2(n-1) + a

Doing it the first way, and suppose you take the third term, you'd get the following:

3(3-1) + 2

3(2) + 2

6 + 2

= 8. Which is incorrect.

Doing it the second way, one gets:

2(3-1) + 3

2(2) + 3

4 + 3

= 7.

Which is correct.

Calvin Khor

@Beliod when you. write $2(n-1)+a$, what does $a$ mean?

I don't think you understand that $a(n)$ is not "$a$ times $n$"

If you have the sequence $\{ 3,5,7,9,11,\dots\}$, then $a(1)=3,a(2) = 5, a(3) = 7, a(4) = 9,$ and so on

$a(2) = a(1) + 2$, for example

user69608

03:31

@CalvinKhor that "something" is $a^2 +b^2+c^2-((a+b+c)^2/4))$ i am interested in this value.

Calvin Khor

$|a|^2 + |b|^2 + |c|^2 - \frac14|a+b+c|^2 = \frac34 (|a|^2+|b|^2+|c|^2)-\frac12 (a\cdot b + a\cdot c +b\cdot c)$

user69608

@CalvinKhor yeah now what is summation a.b

Calvin Khor

I still don't understand. Summation a.b means? $\text{write something in \LaTeX}$?

user69608

dot product

Calvin Khor

$a\cdot b$ means dot product. There's no summation?

user69608

03:38

i mean value of a.b +b.c+c.a

Calvin Khor

@user69608, I have no idea what you want, because that is my final answer. You asked me about something in the middle of a computation, and didn't provide any context. I can't read minds and I don't know what bigger problem you're trying to solve

Mary Star

06:53

Hello!! Does someone of you have an idea about my question:

0

Q: Characteristic polynomial - map & matrices

Let $1\leq m,n\in \mathbb{N}$ and let $\mathbb{K}$ be a field. For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by $$\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac$$ We have that $trace(\mu_a)=n\cdot trace(a)$ and $\det (\mu_a)=\det (a)^n$. Ho...

?

RScrlli

09:12

Hello everyone, I was wondering, do you know of some extension of Minkowski inequality to the case of negative exponents?

flowian

09:32

A sequence of real numbers converges to a number $s$ relative to the usual topology iff it is eventually in each neighborhood of $s$.

Don't get how it is eventually in *each* neighborhood of $s$.
$S_n$ is eventually in set $U$ if there is an $m \Bbb N$ s.t $n \geq m \Rightarrow \, S_n \in U$.

Calvin Khor

its eventually in $U$ for each nbd $U$, but there is no $n$ such that its in each (i.e. every) neighbourhood

flowian

@CalvinKhor It is not eventually in every nbd $U$ at the same time?

Calvin Khor

yes, as a rough first approximation, think about the sequence 1/n and all the balls around 0

for each $\epsilon$, $1/n$ is eventually in $B(0,\epsilon)$

but that doesnt imply that its also in $B(0,\epsilon/100000)$

flowian

Got it. Thanks

Calvin Khor

you're welcome :)

hasan

10:11

The polar form of a complex number has magnitude and phase. Can I say the magnitude and phase are independent to each other? What is the condition for that?

Thorgott

define "independent"

hasan

Let $x=re^{j\theta}$. Can I say pdf $f(r,\theta)=f(r)f(\theta)$?

Thorgott

what's $f$

hasan

$f$ is the probability distribution function.

Thorgott

of which distribution?

hasan

10:22

I am trying to find the integral of multiplication of two Gaussian Q function $Q(x)$. The conditional pdf is $y(y|r,\theta)=Q(rcos(\theta))*Q(rsin(\theta))$

I am trying to find the marginal pdf of $Z(y)$ where, $z(y) = \int\limits_0^\infinity {\int\limits_0^{2\pi } {Q(rcos(\theta ))*Q(rsin(\theta ))f(r,\theta )} } drd\theta $i

I was thinking can I write $f(r,\theta ) = f(r)f(\theta )$. If I can do that what is the condition?

NoseBleed

11:15

Can't theorem 3 be proved with just the isomorphism?

I mean since it needs to be bijection and that means it needs to be surjection

then both has to have finite order n?

I got bunch of question today

Fargle

They're just being careful because the fact that $f(g)^n = e_H$ does not mean that the order of $f(g)$ is necessarily $n$---it just divides $n$.

But since an isomorphism gives an inverse which is also an isomorphism, you can apply the argument both ways and get equality.

NoseBleed

@Fargle So I am right right?

Fargle

I think I'm unclear on what you're asking.

NoseBleed

I mean proving the theorem 3 by just bijection

which is faster

Fargle

As opposed to...?

NoseBleed

11:19

and efficient

I mean can we proof the theorem with just saying since it is isomorphism then it means it needs to be bijective which needs to be subjective so □

Fargle

No, because you're talking about a single element.

You're saying if $g \in G$ has order $n$, and $f$ is iso, then $f(g)$ has order $n$.

Surjectivity is not enough to require this.

NoseBleed

But for this proof isn't it required to be cyclic?

Fargle

No.

$G$ and $H$ are arbitrary isomorphic groups.

$g$ is an arbitrary finite-order element of $G$.

NoseBleed

because I think only cyclic becomes neutral when they have their order as exponents

Alessandro Codenotti

@NoseBleed What is the definition of order?

Fargle

11:22

A cyclic group is one where every element is a power of some specific element.

NoseBleed

@AlessandroCodenotti size

Alessandro Codenotti

size of what

NoseBleed

@AlessandroCodenotti Size of group I guess or it's elements

Alessandro Codenotti

The order of a group is its size, but what is the order of an element?

NoseBleed

I am confused why elements have orders? It should be 1 for each

Fargle

11:24

Your book should have defined order of an element.

Go back and look

NoseBleed

you mean for example elements of $D_3$ ?

I forgot how to spell it

Alessandro Codenotti

@NoseBleed The theorem you're asking about talks about the order of an element $g\in G$, so your book should have defined what that means at some point before theorem 3

NoseBleed

dihelral...

Oh now I remember

It is order of a element treated as generator

oh now I see I understand the proof

one more help in this book I got another too but talk it for later

in the first picture after the proof it talks about different between dihedral group and rotations of hexagon

how it theorem 2 is applied

so that they know they are not isomorphic

and there is alternative way which I kinda don't understand

I think they used theorem 2 (iii)

ok I understand it

I just need to understand the one begin with alternativey one

Fargle

What's the alternative way, out of curiosity?

Oh---I see, it was in your last screenshot.

NoseBleed

As I send first picture it says:"Alternatively, bal bla bla"

@Fargle it's first

Fargle

11:35

The punchline is that the rotations of the hexagon contain an element of order 6.

NoseBleed

but $d_3$ also has 6 orders

but $d_3$ also has 6 as order

Fargle

If there was an isomorphism with $D_3$, then that would imply $D_3$ has an element of order $6$---by Theorem 3.

Not just order 6 on the group---order 6 on an element.

Thorgott

you need to differentiate between the order of the group and the order of an element in the group

NoseBleed

oh now I get it

my bad

Thorgott

fun fact: these two groups are the only groups of order 6

2

Sebastiano

11:39

Good morning from Italy everybody into chat. Why my question is yet closed?

NoseBleed

@Thorgott r u sure? I think we can create one

Sebastiano

0

Q: When is it possible to use complex analysis for solve the Riemann integrals?

Supposing to have these commons integrals with $x\in\Bbb R$, for example I have these: $$\int\limits_{-\infty}^{+\infty}\frac{dx}{(x^2+1)^2}=\frac12\left(\arctan x+\frac{x}{1+x^2}\right)\tag 1$$ $$\int\limits_{0}^{+\infty}\frac{x^2}{x^4+1}dx=\sqrt{2}\pi/4 \tag 2$$ $$\int\limits_{0}^{2\pi}\frac{1}...

integration complex-analysis proof-writing proof-explanation improper-integrals

Thorgott

I am sure

the only groups of order 6 up to isomorphism, I should specify

NoseBleed

ok now we move to fibonacci number

Sebastiano

Thank you very much for your suggestions.

Best regards.

NoseBleed

11:40

@Thorgott I hope to find another one

Thorgott

that is to say, any group of order 6 is either isomorphic to $D_3$ or the group of rotations of a hexagon

you can try to find another one

it's oftentimes instructive to fail

NoseBleed

@Thorgott but they are not isomorphic lol

Sebastiano

Thank you very much for reopen my question. THank you with my heart.

Calvin Khor

@Sebastiano you have 4 reopen votes already, you can probably just wait for another. But also, I don't know why it should be reopened. I don't understand the question

Sebastiano

@CalvinKhor What is not clear?

Calvin Khor

11:42

Lol well it has now been opened. But I still don't understand

Thorgott

I'm not saying they are, though

Calvin Khor

"Whether an integral cannot be solved by classical methods when I will have to use to complex analysis and what needs to be done to solve them?" this sentence, I can't understand it

Thorgott

I'm saying that there are two groups of order $6$, $D_3$ and the groups of rotations of a hexagon, which I shall call $C_6$, that these two are not isomorphic, but that any other group of order $6$ is isomorphic to one and exactly one of $D_3$ and $C_6$

Sebastiano

@CalvinKhor I never dealt with complex analysis at university. I did some private tutorials in 2003 but I don't remember anything. The question is: when do I have to use complex analysis if I can't solve it by substitution or by parts, or by artifice?

NoseBleed

@Thorgott paulsible

let's move to geometry of numbers

Sebastiano

11:45

@CalvinKhor Is there a criterion, a clue that makes me think that certain integrals can also be solved through complex analysis and how to solve them?

Calvin Khor

@Sebastiano OK, I understand now, and I don't think you can get a simple answer besides learn more and examples and practice more examples

feynhat

@Thorgott The statement I saw was: $\Omega^k = \Delta\Omega^k \oplus \mathcal{H}^k$. Why is $\Delta\Omega^k = \delta\Omega^{k+1} \oplus d\Omega^{k-1}$?

beta_me me_beta

I just came across an interesting topic, Relation between Kirchhoff's circuital laws and Matrix tree theorem, can someone provide any refrence or insights

NoseBleed

why is path from 1 to $n$ is $f_n$?

I cannot see the pattern

what is the reason behind it which is not mentioned

Sebastiano

@CalvinKhor Can you help me, please? Would you kindly edit my question? I don't know how many times I've changed it. There's three wholemeal, two impropres and one definite.

NoseBleed

11:46

I tried drawing the rabbit breeding chart

still can't get it

beta_me me_beta

0

Q: Relation between Kirchhoff's Circuital law and Matrix tree Theorem

I'm not a professional mathematician, just an undergraduate student. I was reading Introduction to Graph Theory by West, I came over the topic which discuses the methods to find the spanning trees in a general graph. Kirchhoff's Matrix tree theorem was there. Out of curiosity, I thought, could th...

graph-theory spectral-graph-theory trees spanning-tree

feynhat

Oh okay. It may not be. But $\Omega^k = \Delta\Omega^k \oplus \mathcal{H}^k$ implies $\Omega^k = \delta\Omega^{k+1} \oplus d\Omega^{k-1} \oplus \mathcal{H}^k$.

Its just set theoretic

NoseBleed

It there any way we can see the pattern?

Sebastiano

@CalvinKhor I hope I haven't disturbed anyone. Thank you so much for your comments.

Thorgott

@feynhat It is, but it's not entirely trivial. In fact, you can go like $\Delta\Omega^k=d\delta\Omega^k\oplus\delta d\Omega^k=d\Omega^{k-1}\oplus\delta\Omega^{k+1}$.

It boils down to a handful of adjointness computations. I'm in a seminar rn, but I can try telling you the proof later if you want.

Calvin Khor

11:50

@Sebastiano In my opinion, there can't be a good answer, but I would remove most of the information, especially about the integrals you can already solve, and add this which you said to me: "Is there a criterion, a clue that makes me think that certain integrals can also be solved through complex analysis and how to solve them?"

also you don't need to say Riemann integrals, "solve the integrals" is fine

feynhat

I am doubtful about the second equality.

NoseBleed

is it called fibonacci zigzag path counting or there is any resource online to see this

You guys seem busy so I will question it later

Sebastiano

12:16

@CalvinKhor Done!! Thank you very muchhhhhhhhhhhhhh.

Calvin Khor

@Sebastiano you're welcome. Hope I'm wrong and someone gives a great answer

NoseBleed

12:47

no worries I found the pattern by matching with points going through (n-1) and (n-2) then I found that it is $f_n$ and is the same one as fibonacci number

I should had done it before

I appreciate your helps guys! Thanks for involving in discussion!

NoseBleed

13:28

quick question again :p why does $\int_a^{x}(f'(a)+\int_a^{x_1}f''(x_2)dx_2)dx_1$=$f'(a)(x-a)+\int_a^{x}\int_a^{x_1}f''(x_2)dx_2dx_1$?

It says it is integration by parts

but I think this is mean value theorem used here but instead of $c$ it is $a$

trying to review Taylor theorem derivation

feynhat

huh? Its just linearity of $\int$.

NoseBleed

?

Calvin Khor

feynhat is right

NoseBleed

What does it means

$[f(a)]_a^b$?

Calvin Khor

it means ur overthinking it, theres no fancy calculation, just $\int 3f+g dx = 3\int f dx+\int g dx$

NoseBleed

13:38

I know it

but

I actually mean why

wait lemme type

user69608

how to integrate $\frac{1}{(a+bcosx)^2}$ where a greater than b

NoseBleed

why does $\int_a^{x}f'(a)dx_1$=$f'(a)(x-a)$?

I think I knew it in past but now I forgot

Calvin Khor

why does $\int_a^x 5 dx = 5(x-a)$?

Fargle

Yeah, the ticket here is that $a$ is constant

user69608

@CalvinKhor ?

NoseBleed

13:43

@CalvinKhor hmm... $\int a dx=ax+c$

Calvin Khor

@user69608 dont know

NoseBleed

I don't know hehe

user69608

@CalvinKhor u must be joking :)

NoseBleed

@CalvinKhor Why. No riddles please ;_;

Calvin Khor

its not a riddle idk what else to say

if you can't integrate a constant then review that before taylor's theorem

user69608

13:45

any hint?

NoseBleed

you r right may be need a serious review of integration lol

no I don't need review my first step is correct then I perform $|_a^b$

lmao wth did I ask... really embarrassing

so it means that $f'(@)$ is constant in order to be treated like this

Fargle

Yep---$f'(a)$ doesn't vary as you vary $x_1$.

It might in principle vary if you varied $a$, but integration only cares about whether the inside looks constant to the variable you're integrating on.

NoseBleed

oh cr@p u r right

user69608

can anyone help in that integral?

NoseBleed

thank u guys

en.m.wikipedia.org/wiki/Double-precision_floating-point_format

ok enough for today

Knight

14:26

Hello everyone! I got this question

satan 29

hi

@user69608 which one?

Knight

Find all real solutions $(x,y)$ such that

$$sin^4 ⁡x=y^4+x^2 y^2−4y^2+4 \\cos^4 ⁡x=x^4+x^2y^2−4x^2+1$$

Here is my attempt:

$$
\cos^4 x - \sin^4 x = x^4 - y^4 + 4y^2 -4x^2 - 3 \\
(cos^2 x - \sin^2 x )(\cos^2 + \sin^2 x) = x^4 - y^4 + 4y^2 -4x^2 - 3 \\
\cos 2x = x^4 - y^4 + 4y^2 -4x^2 - 3
$$

satan 29

thats what i did too.

now you might wanna try: -1<= cos(2x) <= 1

Knight

$$
-1 \leq \cos 2x \leq 1 \\
-1 \leq x^4 -y^4 + 4y^2 - 4x^2 -3 \leq 1 \\
2 \leq x^4 - 4x^2 +4 -(y^4 -4y^2 +4) \leq 4 \\
2 \leq (x^2 -2)^2 - (y^2 -2)^2 \leq 4
$$

abhas_RewCie

@Knight I want to ask for clarifications.

Knight

14:31

You can

Getting that last inequality it is possible to find as many real $x$ and $y$ as we wish.

abhas_RewCie

@Knight If your equation for $\sin^4 x$, LHS is a function of $x$ only and RHS is a function of $x,y$ i.e. 2 variables, did you notice that?

5 mins ago, by Knight

Find all real solutions $(x,y)$ such that

$$sin^4 ⁡x=y^4+x^2 y^2−4y^2+4 \\cos^4 ⁡x=x^4+x^2y^2−4x^2+1$$

Knight

Yes Abhas why I wouldn't notice that

abhas_RewCie

Do you know what that means?

Knight

no

satan 29

um, why is that an issue?

Knight

14:33

it means nothing substantial

abhas_RewCie

It means, take all $x$ to one side and $y$ to other, the thing you get on both sides, corresponds to a constant.

Now you can easily find all such $(x,y)$ pairs

Knight

Moving on we have

satan 29

@Knight i also noticed something else

Knight

Solution for such system of equation we have
$$
S = \{ (x,y) | 2 \leq (x^2 -2)^2 - (y^2 -2)^2 \leq 4 \}
$$

@satan29 YES?

abhas_RewCie

@Knight No, that's silly thing.

You say $\cos 2x = x^4 - y^4 + 4y^2 - 4x^2 - 3$

Actually solve it like this:

$$ cos 2x - x^4 + 4x^2 = -y^4 + 4y^2 - 3 = C $$

satan 29

14:37

@Knight The first equation gives sin^4(x)= y^2( x^2 + y^2-4) +4

and the next one gives cos^4(x)= x^2(x^2 + y^2-4) + 1

abhas_RewCie

That expression gives bounds for $y$, solve it, then solve possible values of $x$

satan 29

add them

abhas_RewCie

I've already given enough hints...

Knight

@satan29 Please do it :-)

satan 29

after Adding, On the RHS, let x^2 + y^2=t

Knight

14:39

please use latex

satan 29

so you get sin^4(x) + cos^4(x) = t(t-4)+5

$sin^4(x)+cos^4(x)=t(t-4)+5$

Knight

okay

keep going

user69608

@satan29 integrate$\frac {1}{(a+bcosx)^2}$ where a greater than b

satan 29

This might make things simpler. Since The LHS can be maximized/minimized easily

@Knight we can then

Knight

@satan29 Brother we got to find all the real solutions

satan 29

14:43

write the LHS as $(sin^2(x) + cos^2(x))^2 - 2sin^2(x)cos^2(x)$

$=1-(sin^2(2x)/2)$

Knight

okay keep going

satan 29

now, we can write $sin^(2x)= (1-cos(4x))/2$

Knight

Brother can you take a moment to read the question?

satan 29

which simplifies to $1/4 (3+cos(4x))$

Knight

okay

satan 29

14:47

which is bounded: it has a maximum value of 1 and a minimum value of 1/2

which means the RHS must be between 1/2 and 1

Knight

did you check with calculus?

satan 29

uh cos(4x) can be 1 at max

and -1 at min

and the value of that expression depends only on cos(4x)

The RHS was $t^2-4t+5$

Knight

okay

user69608

when anyone gets any hint to the integral please ping me.

satan 29

so have $1/2 <= t^2-4t+5<=1$

@user69608 i have a nice idea, ill ping in a moment

@Knight

subtracting one from both sides, we get

-1/2 <= t^2-4t+4<=0

-1/2<= (t-2)^2 <=0

Knight

14:51

yes

satan 29

which means the only solution is t=2

Knight

yes, but we have $t=x^2 +y^2 = 2$ and hence infinite solutions

satan 29

thats not the only equation though,brother

Knight

(And I really just accepted your calculations without a check)

@satan29 Thats why I said read the question, we are asked to find all solutions

satan 29

remember, we had $1/4(3+cos(4x))= t^2-4t+5$

when we put t=2

Knight

14:54

So, what's your solution set?

satan 29

cos(4x) has to be 1

i.e $4x=0, 2\pi$

i.e x=0 or pi/2

Knight

@satan29 Do you see this step?

satan 29

yes

Knight

You got a square less than 0, that means no real solutions

abhas_RewCie

kar di bejjati

satan 29

14:57

less than or EQUAL to zero @knight

abhas_RewCie

OK

meri galti thi

satan 29

@Knight and thats exactly why t=2 is the only solution

Knight

okay

satan 29

so 0, sqrt(2) is a solution for x,y

but we need to check once

and yes, it indeed is a solution

Knight

Thanks

abhas_RewCie

15:00

correkt!

satan 29

and we need to do it for x=pi/2 and y=sqrt(2-pi^2/4) also

but its unlikely that thats a solution.

and so i believe we have found all the solutions!

(x,y)=(0,sqrt(2))

Knight

okay

satan 29

@Knight :)

@user69608

I have got an idea

we need to integrate $1/ (a+bcos(x))^2$

equivalently stated, we need to find a function whose derivative is $1/(a+bcosx)^2$

the square in the denominator reminds me of the quotient rule.

so i guess its reasonable to assume that the required function is of the form $f(x)/(a+bcos(x))$

abhas_RewCie

Just find $p(x)$ for which $p'(x)(a+b\cos x) - (a + b\cos x)'p(x) = 1 $

satan 29

when we differentiate this, we get f'(x)(a+bcos(x)) - (-bsin(x))(f(x)) / (a+bcos(x))^2

and we want the numerator to be 1

abhas_RewCie

15:06

But, that's impossible (I guess, not sure), because that's unintelligible class of function that's why limits of integration is given... (Not sure, I'm damn weak with integration)

satan 29

and its not hard to see that f(x) = sin(x) does the job.

abhas_RewCie

@satan29 What?

cool observation!

@satan29 great dude :)

satan 29

no wait, does it?

abhas_RewCie

@satan29 don't confuse me!

satan 29

@user69608 were you sure it was one in the numerator?

@abhas_RewCie lol sorry

abhas_RewCie

15:10

lol

@user69608 Are you sure denominator is squared? otherwise, the integration is simple and straightforward!

that square hurts

satan 29

bruh

Yeah @user69608 i dont think the numerator is 1

abhas_RewCie

@satan29 why?

WRA solved it with one...

satan 29

if its some f(sin(x),cos(x)) , then the comparing thing works out quite nicely i suppose

in the numerator

abhas_RewCie

WRA solved it with one...

I still don't understand why the limit of the integration was 1? It doesn't appear good even after putting the limits $a$ and $b$... or does it?

Solve this integration $\int \frac {\sin x} x dx $

and this $\int e^{-x^2} dx$

atleast, they are less headache!

bye

satan 29

@abhas_RewCie limit?

idk man

maybe @robjohn has an insight...

Thorgott

15:38

@feynhat Ok, let's do this. First of all, for any two $k$-forms, $\omega_1,\omega_2$, we have $\langle d\delta\omega_1,\delta d\omega_2\rangle=\langle\delta\omega_1,\delta\delta d\omega_2\rangle=0$, so $d\delta\Omega^k$ and $\delta d\Omega^k$ are orthogonal. Next assume $\omega\in d\delta\Omega^k\cap\delta d\Omega^k$, so that there are $\omega_1,\omega_2\in\Omega^k$ with $\omega=d\delta\omega_1=\delta d\omega_2$. On one hand, $d\omega=dd\delta\omega_1=0$ and $\delta\omega=\delta\delta d\omega_2=0$, so $\omega\in\mathcal{H}^k$. On the other hand, $\langle\omega,\gamma\rangle=\langle d\delta\…

Onto the second part: For $\omega_1\in\Omega^{k-1},\omega_2\in\Omega^{k+1}$, we have $\langle d\omega_1,\delta\omega_2\rangle=\langle\omega_1,\delta\delta\omega_2\rangle=0$, so $d\Omega^{k-1}$ and $\delta\Omega^{k+1}$ are orthogonal. Assume $\omega\in d\Omega^{k-1}\cap\delta\Omega^{k+1}$, so that there are $\omega_1\in\Omega^{k-1},\omega_2\in\Omega^{k+1}$, such that $\omega=d\omega_1=\delta\omega_2$. We have $d\omega=dd\omega_1=0$ and $\delta\omega=\delta\delta\omega_2=0$, so $\omega\in\mathcal{H}^k$. We also have $\langle\omega,\gamma\rangle=\langle d\omega_1,\gamma\rangle=\langle\omega_1,…

you could probably structure this better, but it's not like this is sophisticated argument

feynhat

@Thorgott What. The first part is the definition of $\Delta$, right?

Thorgott

no

the reverse inclusion is non-obvious

you need to show that something of the form $d\delta\omega_1+\delta d\omega_2$ is actually of the form $d\delta\omega+\delta d\omega$

user69608

@satan29 c'mon man wolfram alfa most of the times gives complex wierd answers .

@satan29 this question is from "resonance high level problems"

satan 29

15:53

@user69608 see what i tried to do earlier.

It has to be along the lines of this only....and i say this because I have previously solved a similar question before, (very similar) using the same procedure.

Thorgott

I mean, in general you won't have $(f+g)V=fV+gV$ for linear operators on vector spaces

feynhat

Oh ofcourse lol. take $f = I, g= -I$.

user69608

@satan29 should i send the answer?

feynhat

So yeah. Its not that obvious.

Thorgott

ye

it looks tautological but isn't

still, the actual work is only to show that $\left(\mathcal{H}^k\right)^{\perp}=\Delta\Omega^k$ and the rest is more of a linear algebra exercise

feynhat

15:59

I mean, for me the statement is what I wrote. So I wouldn't need that LA exercise.

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