Everyone

A conjecture I just created out of thin air.

@robjohn Thanks! I understood you correctly. Thanks a lot! That indeed is a big help to me!!!

I wonder if people except for those whose research field is algebra-related use category theory frequently.

Can anyone please help me understand or rather plot the graph of |y|=sin(x) ?

It's a strange graph indeed.

I tried plotting it, in the following way:

I considered 2 cases: y>0 and y<0

If y>0, then y=sinx

If y<0, then -y=sin x implies y=-sinx

Thus, if y>0, then y=sinx and if y<0, then y=-sinx

Now, the graph of sinx and -sin x is:

tiny terminology thing, i wouldn't call that set a 'graph', but OK, yes, you're graphing the (x,y) satisfying either (1) y = sin(x) and sin(x) >= 0, or (2) -y = sin(x) and sin(x) >= 0.

so the chunk of the usual graph of y = sin(x) (which is indeed the graph of a function) that lies on or above the x-axis, together with that chunk reflected about the x-axis

I thought this is the graph of |y|=sinx or rather the above image represents |y|=sinx

But it's not the case

the set of (x,y) satisfying |y| = sin(x) is not going to include any points with x such that sin(x) is negative

|y| = sin(x) tells you, among other things, that sin(x) is the absolute value of something, so is >= 0

@leslietownes Exactly, that was my first approach, I though that the graph,would be sinx, such that sinx\geq 0

But this is the correct graph as it turns out?!

yeah, that corresponds to those (x,y) where -y = sin(x) and sin(x) >= 0

I thought initially the graph would be like this

OK, i mean, i get that a person could have a thousand initial thoughts about what the graph might look like

but |a| = b means either (1) a = b and b >= 0, or (2) -a = b and b >= 0

and vice versa

@leslietownes can you please elaborate about why the graph is like this? Am I missing something 😕

it definitely matters which side the absolute value is on, and all of this other stuff

it's what you get when you look at the points (x,y) where sin(x) >= 0 and y is either sin(x) or -sin(x)

y = |sin(x)| is not that at all, for example, because this latter graph includes points (x,y) where sin(x) is negative (and hence sin(x) is not the absolute value of anything), and does not include any points (x,y) where y is negative (which the plot of |y| = sin(x) certainly does)

maybe forget about sin(x) and just think about, given the graph of y = f(x), can you understand what the picture of the set of points (x,y) satisfying |y| = f(x) will look like

that might make it conceptually clearer. you could make general sketching problems involving absolute values arbitrarily difficult with complicated enough expressions, but the idea behind the set |y| = f(x), where f(x) is a function of x whose graph you know, is pretty simple

relative to the universe of things that might involve absolute values or case analysis

I am trying to prove that f(x) is bijective function.

But because it is set so I don't know how I will prove it like this f(x)=f(y)=>x=y.

@leslietownes I am so much grateful! I get it now. The jist is: the graph of |y|=f(x) while sketching we only consider those points, where f(x) is non-negative and now, conclude, that y can either be f(x) or -f(x) where f(x) is non-negative. This realization makes this utterly simple. Now, we just have to draw the non-negative portion or rather those points such that f(x) is positive and then reflect them to the x-axis, to obtain the required graph of |y|=f(x). Thank you so much!😊

👍👍

hi, does anyone know if this result about smooth submersions of manifolds without boundary extends to manifolds with boundary : $\pi : M \rightarrow N$ is a smooth submersion iff every point of $M$ is in the image of a local section of $\pi$ (and $\pi$ is assumed smooth)

so i want to apply the constant rank theorem in the case when $\partial M \neq \emptyset$ can happen, and when I try to do that what I end up wanting is for the kernel of $d \pi$ at these boundary points to not be the entire tangent space of $M$'s boundary

and i think that happens as long as $N$ has dimension $\geq 2$

so it seems to be fine in that case

is there some easy way to see the statement is true for $N$ of dimension $1$?

ugh, also im assuming here actually that $\partial N = \emptyset$

so what I said above does not resolve what I want to see is true

basically, does this statement hold when $M$ and $N$ can both have nonempty boundary?

im trying to think about it in terms of embedding both of them into their doubles

but im not sure if that really helps here

here smooth submersion just means smooth + $d \pi$ is onto everywhere

no guarantee of surjectivity

im thinking maybe if I say since $N$ can be viewed as a regular domain in its double which is without boundary, $\pi : M \rightarrow N \hookrightarrow D(N)$ is still a smooth submersion, so if I can prove it in this case with $D(N)$ replacing $N$, a local section defined in an open nbhood of $\partial N \subset D(N)$ restricted to an open neighbourhood of a point of $\partial N$ only in $N$ should still be a local section

so maybe wlog $\partial N = \emptyset$ is fine

(by only in $N$ i mean restricted to a relatively open neighbourhood of a point of $\partial N \subset D(N)$, relative to $N \subset D(N)$)

which is what we want anyway

i mean it will still be smooth

okay so i guess that maybe reduces the question to prove it for $N$ boundaryless of dimension one

assuming connected... this should be doable without too much heavy machinery like constant rank theorem, unless its just false

hmm, in the edge case I mention, I get the impression that we can extend $\pi$ to a smooth submersion $D(M) \rightarrow N$

($\partial N = \emptyset$ still)

yeah, it ought to be the case ,because there is no spillling

at the boudnary points in this edge case

yeah, so I guess in the bad case, what we could do is since $d \pi = 0$ on $\partial M$, basically use the schwarz reflection principle to define $\pi$ on the other copy

so that should give us a smooth extension of $\pi$ to $D(M)$ which is intuitively what we want

i.e. near $\partial M \subset D(M)$ , in coordinates, set $\pi(x) = - \pi(-x)$

then by gluing this gives us a smooth submersion $ \pi : D(M) \rightarrow N $, and now both $D(M)$, $N$ have non-empty boundary, so the result holds in this case, and you get a local section $\sigma : (U,u_0) \rightarrow (D(M), m_0)$ for $m_0 \in \partial M$ for $\pi$

then since $N$ is one dimensional and boundaryless we may as well take $U$ to be an open interval, so that there is a first point at which $\sigma(t_0)$ touches the boundary

finally $\sigma \restriction [0,t_0)$ will be the required local section for $\pi$, defined on $M$

errr, whoops, I mean $\sigma \restriction (0,t_0]$

but that isn't defined on an open subset of $N$, hmm

here $t_0$ is $u_0$ in coordinates

i dont think this is avoidable then, since a local section is going to be an injective immersion, so at $t_0$ I will spill over to the other copy..

sorry if im flooding the chat, is anything im saying making sense or is there something im stupidly missing

btw the ''constant rank theorem'' is the version where the target manifold is boundaryless, the base manifold has nonempty boundary, and the statement is the same as the usual constant rank theorem (with boundary charts on the base manifold at boundary points), but it only goes through if the map is an immersion, or if its kernel at boundary points isnt a subspace of the tangent space of the base manifold

its some exercise in lee

(if the map is an immersion it goes through even if the target has boundary, but that isnt our situation anyway)

okay i realized i said something really dumb, the case of dim $N \geq 2 $ is even worse

afk for a while sorry

i think it should work out since $\pi$ has full rank, so whatever extension we choose for it on $D(M)$ will still be a submersionn

although apriori without heavier machinery we cant control the behaviour of that extension much

I was solving a basic equation : cos 3x=cosx. However, if we solve this equation by using cos^{-1} function then, we get 3x+2npi=x or x=-npi( such n\in \Bbb Z) which can also be written as x=npi. But, if we solve it in this way: cos3x-cosx =0 or 2cos(2x)sin(-x)=0 which implies either sinx =0 or cos2x=0.

In the former case, x=npi (same as earlier) and in the later case 2x=2n\pi +pi/2, where n is an integer. Thus we see by solving in this method, we are getting two different solutions contrary to the previous method where we got only one set of solutions. Why is this anomaly? I can't find the explicit reason for this.

Strange!!!!

@Franklin that's a really big anomaly

I myself have no idea bout it

😂😂😂

@Arthur Yup! That's what confuses me !!!!!!!!!

Maybe it's something about trig functions have to essentially fa torized into an expression containing two products of derived expressions which should be evaluated! Not just directly doing inverse trig functions, like you did in ur former case? Maybe this is an explanation but I would suggest verifying this with other community members as I myself not so sure about this....

@robjohn sir if u can reactivate my old room please?

Can anyone help me in guess ing the pattern in this sequence 20,10,10,15,30,75,x

X is the next number in the sequence I wish to find

I tried subtract addition but division but no general pattern is being observed

Hello! Quick question, If i calculate the fourier transform twiceof x(t), with the difference being the sign in the exponential of the fourier transform integral, will the two outputs be X(omega) and X*(-omega) ?

@Astyx Maybe you are aware of. There was a course at PCMI: youtube.com/playlist?list=PLldN_DpkXL3Y-ktK5Nq5IYgRbNjYLMNhF

@Arthur It might be so, but I need a complete conclusion.

@Arthur problem resolved! I posted that as a question. I found out the reason. You might want to check it!

@DakkVader you ca compute this mechanically for yourself: Let $f$ be some nice enough function and take its fourier transform, so you will get $\hat{f}(\eta) = \int f (x)\exp(2 \pi i \langle x , -\eta) dx$, then if you take the inverse fourier transform (what you refer to as changing the sign in the exponential), of this object, you will get $f(x) = \int \hat{f}(\eta) \exp(2 \pi i \langle x , \eta) d(\eta) = \int \int f(x) \exp(2 \pi i \langle x, -\eta) dx \exp(2 \pi i \langle x, \eta) d(\eta) $

when $f$ is sufficiently nice, you can swap the order of integration, so that the exponential factors in the integral cancel to $1$

which means you get back the original function you started with

if you play around in basically the same fashion with the definitions of the fourier/inverse fourier transform you will quickly see the answer to your question and know in future how to figure it out for yourself

its also a good idea when you're learning this stuff for applications to just assume you can always swap the order of integrals/sums, because the formalization is really just a tool that in some sense comes later and tells you what the minimal conditions you need are to perform these swaps

lehigh.edu/~yuu221/files/teaching/complex/lecture9.pdf

Thereom 19, second line in the proof: C hat is compact? C hat is the extended complex plane.

why does f have finitely many poles?

Anyone have prove bijection of of a function that maps to set?

I wonder how it is done.

@NotTfue what does that mean?

functions are always between sets

and they always take as arguments, sets

everything is a set

@porridgemathematics I mean this function.

I want to prove if f(x)=f(y) then x=y

I wonder if it can be done

may be I will ask it. Few downvotes won't matter lol.

@Koro because $f$ is first and foremost a continuous function from $\hat{\mathbb{C}}$ to itself, and $f$'s poles as a set is equal to the preimage with respect to $f$ of a single point, namely $\infty$, this is a closed subset of $\hat{\mathbb{C}}$ which is compact itself, and is thus also a compact set

@Koro $\hat{\mathbb{C}}$ is compact because it is the one point compactification of the plane $\mathbb{C}$

or in other words, its the "smallest" compact Hausdorff space containing $\mathbb{C}$ as an open dense subspace

that its compact should have already been covered in your notes

Intuitively this must be bijective.

now why does $f^{-1}(\infty)$ ($f$'s poles) have to be a finite set? This is where you need to use that $f$ is more than just a continuous function

$f$ is a holomorphic function between complex manifolds, in this case your complex manifold is the next simplest after $\mathbb{C}$ itself, and $f$ is a holomorphic self map of it

so holomorphic functions between complex manifolds are just continuous functions between the underlying topological spaces, which are holomorphic functions from open subsets of $\mathbb{C}$ to $\mathbb{C}$ when written in local coordinate charts

so when you work with holomorphic maps between compact complex manifolds, you still retain the identity principle from standard complex analysis

that is to say, in the classical case, if you have a hol. function $f : U \rightarrow \mathbb{C}$ hol., where $U$ is an open connected subset of $\mathbb{C}$ , then its zeros are necessarily isolated, in fact even more is true, if there is any sequence of (Distnct) points $u_n \in U$ s.t. $f(u_n) = 0$ for all $n \geq 1$, and the $u_n$ converge in $U$ to a point of $U$, say $u_0$, then $f$ is identically equal to zero on all of $U$

so, in your situation, (and in the classical case), the same is true for the points of $U$ at which $f$ takes any possible value in $\mathbb{C}$

@porridgemathematics ah I see. Thank you very much!

I was thinking 'closed and bounded'.

you just take that finite value $a$ and look at $f - a$, and then apply the identity principle

so you have that $f^{-1}(\infty)$ is a compact set, plus you also know that it inherits the discrete topology

this is actually important to understand

for the above case, I know a result that discrete subspace of a topological group is closed.

it inherits the discrete topology precisely because $f$ is holomorphic

so you have the identity principle + continuity

to get that $f$'s poles form a compact set

you ONLY need that $f$ is continuous

in the space you're working in (riemann sphere) or any self map of compact topological hausdorff spaces

to get that preimages of $f$ are always finite

you need that preimages are discrete

meaning that they inherit the discrete topology as a subspace of $\hat{\mathbb{C}}$

and if you have a discrete + compact set, then this set must be finite

can you say why? its important to know

this is a purely topological question

any topological space that is both discrete and compact, must be finite

(and vice-versa, as long as you always endow finite sets with the discrete topology)

@porridgemathematics or we can say that since f is holomorphic for large z, then for some R>0, for all z, |z|<R, poles lie within this circle. So if they were infinitely many, then by Bolzano Weierstrass, they will have a convergent subsequence converging to some p say.

dont use the closed + bounded def. here

its simpler than that\

what are the open subsets of a discrete space

But this violates the definition of meromorphic functions (in which we want, set of poles to be discrete).

@porridgemathematics any subset

honestly you shouldn't think of 'meromorphic' as being different from holomorphic, it isnt your fault, there are few books that treat them as instances of the same thing

@porridgemathematics yes, I know this result.

there are only complex analytic/holomorphic maps of complex manifolds

@porridgemathematics sure. As you asked above: the open sets are singletons also so I take an open cover consisting of singletons.

By compactness, we get the finiteness.

yup, so what you're doing here is applying that result to the discrete (because $f$ is holomorphic and you are taking the preimage of $f$ at a point working inside a connected space, which also happens to be compact) subspace $f^{-1}(\infty)$

yup

This is the outline of the proof.

@porridgemathematics ohh

honestly the $\infty$ symbol there can be useful for visualizing the riemann sphere

but it doesn't really mean anything

you can replace it with whatever

it will help you think of how to metrize it

I recognise that as 1 pt. compactification ...

(The riemann sphere)

yeah just saying, this kinda stuff sometimes gets forgotten

even by instructors teaching it lol

@porridgemathematics never really understood the purpose of Riemann sphere.

basically the purpose is to eliminate the word 'meromorphic'

there is a section on it, but then it is not talked about later in chapters unless I am mistaken, of course.

from being used again

because you are taught about meromorphic functions defined on the complex plane

with 'singularities

these are one and the same as holomorphic functions defined on the complex plane, except there are no singularities

the target is just the Riemann sphere

and the ''poles" (non-essential singularities) become the point at infinity

but they ought to at least sketch or tbh give you first the full definition of riemann surface

without that, its bad pedagogy

and it should click much easier if they bothered

I'm studying from Stein Shakarchi complex analysis right now.

try forster

Ohh

Ok I'll try. Thanks for the reference.

Everyone I got that it is 0.5 for first and 0.5 difference so last is 2.5

Could you please also suggest me where I can learn CW complexes and chain complexes considering that I am a beginner at those?

but let me rephrase/correct one thing before i try and answer your next question, it isnt true that every function with singularities is meromorphic

the purpose of the riemann sphere is to shove under the rug meromorphic functions

so the ones that explode near the singularities in size

and they do after you see a bunch of definitions, become holomorphic

but places where there are singularities, but no explosion

and also there isnt boundedness

(i.e. it isnt removable)

these are true singularities

called '"essential singularities""

where the behaviour near them is truly, bizzare

for holomorphic functions on the riemann sphere, these are the only singularities possible

(since poles aren't anymore once you add infinity)

(on open subsets of the Riemann sphere*)

for your next question, it depends what kind of course you are learning it for

is it a algebraic topology course that is going to get into the details of proofs

or is it a course that will basically serve geometers

'the latter you will see a lot of pictures

in any case, if you want a mixture of both flavors, try reading the actual chapter 0 of hatcher

plus the appendix section on CW complexes

and google the definition of what a 'diagram, cocone,cone,colimit,limit' is

@JackRod Great Vivekananda avatar!

then prove for yourself what the colimit of the diagram that you see when people define what a CW complex is is a colimit in the category of topological spaces

it took me some time to get used to identifying and using the same symbol post identifying

this happens a lot in other books too

@porridgemathematics yes

but its necessary because otherwise things get so notationally bothersome noone would bother to write them

but its bad if you're trying to feed a computer your supposped proof to check

*supposed

(and there are weirdly many incorrect proofs in big named titles where errors are committed because identifications are misused)

@porridgemathematics actually I tried reading that chapter, I understand how cell-attachment works and all but still I'm not able to solve exercises.

then you need to start drawing pictures

you should do both

if you understand formally what is permissible

you know what is allowed to be drawn roughly better

than if you dont at all

(i cite the thing that tripped you up before and a likely very exhausted ted)

For example: proving $S^n$/{p,q} $\simeq S^n$ \/ $S^1$

\/ is wedge sum

that is a homotopy equivalence

and the following one:

not a homeomorphism, btw

Give an example of two actions of C_2 on S^n such that orbit spaces are not homotopy equivalent.

I'm not asking for solution right now.

All I am asking is I have no idea how to do these.

I know homotopy equivalence but still.

for sure that holds when $n=2$

yes, it is S^1

I don't want a solution right now.

:)

The point is I have no idea how to do any one of the two above.

ah yes my bad

in general you can say $S^n / S^i \equiv S^n \lor S^{i+1}$

your case is $i = 0$

here the equiv is homotopy equivalence

okay, think about a CW structure for $S^n$ in general first

(for the idea on how to approach this using what you've read in the appendix)

there isn't just one

this is a somewhat mean fact, I suggest adding another space to the equation, namely $S^n$ with an interval attached at the points $p$ and $q$

@porridgemathematics oh I'm afraid. I haven't yet studied the appendix.

yeah, thats more accessible @Thorgott

I'll go through that, if that's needed. Thanks.

basically the intuition as to what you do to solve this stuff is in Ch0

hatcher does exactly what @Thorgott just suggested

indeed it would be more directly useful for you rn

you can also do it without a drawing if you use a CW decomposition of $S^n$ and see what you're collapsing

there's still justification after that but Ch0 is honestly abysmal

formally speaking

there is too loose of an idea (next to none) given to the reader of whats allowed and whats not because he draws pictures and justifies them using things like being able to collapse a closed subcomplex of a CW complex

the whole good pair business

so you are left trying to figure out which way you are 'supposed' to do it

and honestly since he even needs to justify he is using a special fact about collapsing subcomplexes

i think the answer is clear, its just not doable in a one semester course

he does to be fair always refer you to the appendix

its all in there, its just not obvious how to read the book best

and the ordering is kinda weird

but the homology cohomology stuff is great

anyway , I would say if you actually understand the appendix, and then every picture and accompanying result justifying it in Ch0

you will be able to eat hatcher

(and do all the exercises in both the appendix and ch0)

@porridgemathematics Hmm, I haven't read your rant (sorry), but what about the projection $S^1\rightarrow[-1,1],\,(x,y)\mapsto x$? It's not a submersion because the tangents at $(-1,0)$ and $(1,0)$ get collapsed, but there's still obvious local sections in both the clockwise and counter-clockwise direction at both of these.

you're right

i think what i said earlier does work for ''if submersion -> exist local section at every pt''

i didnt actually remark on the other direction

thats a good point

(which you've disproven)

this example also can be used to answer one of Koro's earlier questions, but that was actually accidental

hmm, I guess the forward implication might make sense

a submersion $M\rightarrow N$ induces a submersion $f^{-1}(\partial N)\rightarrow\partial N$, which has local sections and then you can restrict to adapted charts and extend these horizontally or something?

i was in the process of writing the forward down before rant

yeah basically

so what i was staring at was lee theorem 4.25

*4.26

he proves it easily using constant rank thm

for manifolds without* boundary

and nowhere else in the book is it remarked whether this or one direction of theorem 4.26 holds

for general manifolds with possibly nonempty boundary

so my first idea was just use constant rank

and try to do so even though we arent dealing with manifolds without boundary

the first reduction for that was to prove the forward direction (submersion implies local sections), embed $N$ as a regular domain in its double

reason being because then there is hope of applying a constant rank thm

so prove it wlog $N$ having empty boundary

(here $N$ is the target, $\pi : M \rightarrow N$) in lees notation

assumed a smooth sub

now the annoying thing is to use const. rank in this situation you need a special condition on $Ker d \pi$ at boundary points

I had an argument with Yves Daoust (who seems to be making new accounts all the time, keeping rep low) on a recent question

so to avoid that I figured just embed $M$ in its double too, so that $M$ is a regular domain of that, and extend $\pi$ to an open subset of $M$ in $D(M)$

where it remains a smooth submersion

this should be possible but its going to require whitney embedding and tubular neighbourhood theorem

Formally the integrand in the question is wrong, and I acknowledge that

because its easy enough if you are extending a smooth function $f : A \subset M \rightarrow \mathbb{R}^n$ where $A$ is closed

without anything like tubular neighbourhoods and whitney embedding

But Yves is saying that makes the given formula totally invalid, while I disagree

just use the usual partition of unity thing

Which one of us is right?

to extend $f$ to a smooth map in an open nbhood of $A$

but it should hold in general too, after whitney embedding the target

and then using the retraction of the tubular neighbourhood in post composition

and I think since you start at full rank, this should retain full rank

hmm, I'm not sure doubling a map works that easily

points of $\partial M$ mapping to $\mathrm{int}(N)$ seem worrying

i think at least we can assume $N$ has no boundary wlog

for the forward direction

oh wait, I just realized my "counter-example" was nonsense

the backwards direction is true and trivial

same thing as Lee says in the boundaryless case goes through verbatim

uh thats kinda what I thought before but i departed that line of thought because I forgot i was looking at theorem 4.26

as in what the reverse statement precisely reads

the sections of $S^1\rightarrow[-1,1]$ at the boundary points are not smooth (it's e.g. $\pm\sqrt{1-t^2}$)

haha

ah yes

they are not smooth in the sense of maps between manifolds with boundary

can we say that f is holomorphic if f^5 and f^6 are holomorphic?

hmm nvm

anyway sorry for the rant, if you're interested @Thorgott and if i end up finishing what seems like it goes through ill keep you posted in a bit

may be false still, your example gave me some doubts

I have some confusion regarding Riemann theorem on removable singularities.

The theorem says that if f is a function on $\Omega$, open and connected , which is holomorphic on $\Omega$ except possibly at $z_0\in \Omega$, and is bounded on $\Omega -\{z_0\}$, then f has holomorphic extension to all of $\Omega$.

Suppose that f is holomorphic on $\mathbb C-\{a,b,c\}$. Can I still use this to extend f to all of C analytically?

Let's think of the theorem as a 'local' theorem: Suppose that f is bounded in some nbds. of a,b and that of c. In those nbds, I can extend f analytically.

This brings the question: is the resultant function obtained by pasting together these holomorphic maps still holomorphic?

Or I think the following should be correct: By identity theorem, the answer to my last question is correct.

That means that Riemann theorem on removable singularities is applicable in this case.

@porridge it's true that $f^{-1}(\partial N)$ is an open in $\partial M$ and the restriction $f^{-1}(\partial N)\rightarrow\partial N$ is a submersion, but I realized that doesn't help much. to extend a local section of this restriction of $f$ to a local section of $f$ still requires an appropriate normal form for $\pi$.

@Koro being holomorphic is a local property

so yes, but you don't need to quote a fancy theorem

I wonder if I am getting trolled in comment section.

your statement is too weak, btw, Koro

you only want to require $f$ to be bounded in a deleted neighborhood of $z_0$

@CowperKettle thanks..

@robjohn sir please reactivate my account

@Thorgott ohh, you mean that instead of {a,b,c}, I can take any discrete set as well?

@Thorgott thanks :-).

@Koro that will be a corollary, yes

@JackRod I am not sure which account has been inactivated.

It has a couple of downvotes already. I don't understand . . .

probably cause people don't wanna waste their time checking some garbled nonsense an AI wrote up

it's an interesting experiment, but probably not what most people think MSE is for

But is it really nonsense, though? It seems coherent enough to me.

Hi can you please help me with a matrix question? I have that $\lambda>0$, $\sum_{i,j=1}^n a_{ij}\xi_i\xi_j>\lambda |\xi|^2$ that is $A$ is positive definite. I want to find $\sum_{i,j,k,l=1}^n a_{ij}a_{kl}\eta_{ik}\eta_{jl}$.

the first sentence sounds entirely wrong

it seems to conflate connectedness and path-connectedness

$\eta$ is an $n\times n$ matrix. The answer is given to be $\sum_{i,j,k,l=1}^na_{ij}a_{kl}\eta_{ik}\eta_{jl}>\lambda^2 |\eta|^2$.

The sum is $\text{trace}(A^t\eta A\eta^t)$.

Hence my disclaimer, @Thorgott; I'm not very good at topology. I'd accept an answer that debunks it.

Doesn't path-connecedness imply connectedness anyway?

if it's true, yes

but it's certainly not necessary, so "we need to show that" is not true

Sometimes chatgpt gives wrong proofs. When I say that a statement is wrong then it'll say something like You are right, sorry for the confusion and negation of the statement.

is there an entire function f such that f(1)=2f(0) and satisfying 1) f has no zero in C, 2) For every e>0, there exists a z in C such that |f(z)|<e ?

f(0)= f(1)=2f(0), hence f(0)=0, which is a contradiction. So 2) holds.

Q.E.D.

Any objections?

Is finding the number of domino cards one case of 12 fold way?

Which one is the case of finding the number of domino cards?

Can someone please help me understand how he got FDR1 from equation 9 (I have dervided up to 9): arxiv.org/pdf/1810.00004.pdf ? I would really appreciate it.