Mathematics - 2017-11-20 (page 1 of 3)

@usukidoll \|

Liam

This question claims that $(w, Av) = (A^H w, v)$ for normal matrix $A$ with eigenvectors $v,w$ associated with different eigenvalues. I'm trying to understand why that is true.

@Liam that is the definition of the Hermitian isn't it

@Adeek sick

Liam

12:14 AM

Maybe it follows immediately from the definition and I'm just being slow, but the definition I've always seen is A^H is A conjugate transpose. @LeakyNun

oh the definition I see is $\langle Av, w \rangle = \langle v, A^*w \rangle$

I suppose you can prove the equivalence by... expanding the matrix

Liam

Yeah I guess it is listed as a basic property on the conjugate transpose page of wikipedia...

So that doesn't even require that $A$ be normal. It's just a property of the conjugate transpose.

Thanks for pointing that out

12:26 AM

[Malicious plan] One idea is to force the Burali Forti paradox to trigger not only for the class of all ordinals $\Bbb{On}$, but as early as $\omega_1$. If successful, either we will be able to show explicitly that $\omega_1$ is uncomputable unless you have uncountably many oracles, or that more than half of ZF will come crashing down because without $\omega_1$, Borel algebra and hence all nice topologies, and hence analysis, does not exist

@LeakyNun

@Daminark As I was reading complex manifolds check this out

So one way of definining the tangent bundle of a manifold X is as follows

Actually, not necessary, we just don't end up with the Borel algebra forming a set, but becomes a proper class

because in such scenario, all ordinals are countable, and there is a proper class of all ordinals

If X is covered by $\{U_i\}$ then we have that $T_X$ is covered by $U_i \times R^m$ where the gluing occours as follows

$U_i \cap U_j \cap R^m \rightarrow U_i \cap U_j \cap R^m$ given by $(u,v) \mapsto (u,\phi_{*}(u))$

the jacobian at the point u of the map $\phi_{ij}$

recall $\phi_{ij} := \phi_i \circ \phi_j^{-1} : U_i \cap U_j \rightarrow U_i \cap U_j$

@LeakyNun

@Daminark if you think about it this is actually the same way we define the topology on the tangent bundle by transfering the topology

in regular way from undergrad diff topology

When I get on my computer and will be able to see the TeX I'll check this out

bump me when you go on your computer I would like to discuss few things

when discussing things I can clear my head on ideas.

12:39 AM

> So in particular there are uncountable ordinalities. There is therefore a least un- countable ordinality, traditionally denoted ω1.
This least uncountable ordinality is a truly remarkable mathematical object: mere contemplation of it is fascinating and a little dizzying. For instance, the minimality property implies that all of its initial segments are countable, so it is not only very large as a set, but it is ex- tremely difficult to traverse: for any point x ∈ ω1, the set of elements less than x is countable whereas the set of elements greater than x is uncountable! (This makes Zeno’s …

math.uga.edu/~pete/settheorypart3.pdf

Not to mention, it almost surely does not exist

@LeakyNun

@Adeek I know s*** about topology

I was just messing with you :P

...

what for

bored that is all sorry

12:56 AM

> Every computable function has a finite procedure giving explicit, unambiguous instructions on how to compute it. Furthermore, this procedure has to be encoded in the finite alphabet used by the computational model

Ugh, I want Countable alphabet!

@Adeek absolutely disgusting behavior

Yeah I know

You should be executed for this

@Daminark are you back home yet ?

yeah

Probabilistic Turing machine

In computability theory, a probabilistic Turing machine is a non-deterministic Turing machine which chooses between the available transitions at each point according to some probability distribution. In the case of equal probabilities for the transitions, it can be defined as a deterministic Turing machine having an additional "write" instruction where the value of the write is uniformly distributed in the Turing Machine's alphabet (generally, an equal likelihood of writing a '1' or a '0' on to the tape.) Another common reformulation is simply a deterministic Turing machine with an added tape full...

Hypercomputing $\omega_1$ using a probabilistic turing machine with uncountable steps:

Take any string as the starting string e.g. 0000000000...

Let the probabilistic turing machine be T, assigned with a probability distribution P on a linearly ordered uncountable set S

Then perform the following procedures:

1. Plug the starting string 0 into T, obtain T(0) after perhaps uncountable number of steps

2. Compute S1=S-T(0). Update the probability distribution on S1

3. Repeat steps 1,2 uncountably many times

4. As the computation near its end after uncountably many iterations, the probability distribution will become so peaked at finite number of elements in Sn, that eventually, all will be exhausted

5. At the end of the procedure is a well ordering produced by T(x) where x in S

So in theory, running a quantum computer for an uncountable amount of time should give you $\omega_1$

Real computation

In computability theory, the theory of real computation deals with hypothetical computing machines using infinite-precision real numbers. They are given this name because they operate on the set of real numbers. Within this theory, it is possible to prove interesting statements such as "The complement of the Mandelbrot set is only partially decidable." These hypothetical computing machines can be viewed as idealised analog computers which operate on real numbers, whereas digital computers are limited to computable numbers. They may be further subdivided into differential and algebraic models (digital...

Bekenstein bound

In physics, the Bekenstein bound is an upper limit on the entropy S, or information I, that can be contained within a given finite region of space which has a finite amount of energy—or conversely, the maximum amount of information required to perfectly describe a given physical system down to the quantum level. It implies that the information of a physical system, or the information necessary to perfectly describe that system, must be finite if the region of space and the energy is finite. In computer science, this implies that there is a maximum information-processing rate (Bremermann's limit...

grrrrrrrr

> Similarly, a neural net that somehow had Chaitin's constant exactly embedded in its weight function would be able to solve the halting problem,[5] though constructing such an infinitely precise neural net, even if you somehow know Chaitin's constant beforehand, is impossible under the laws of quantum mechanics.

"Infinite computational steps" models

And now, something more interesting...

> According to Shagrir, Zeno machines introduce physical paradoxes and its state is logically undefined outside of one-side open period of [0, 2), thus undefined exactly at 2 minutes after beginning of the computation.[13]

Jenna Sloan

1:16 AM

@Secret Nice wall of text

@Adeek yeah I am

What's the difference between differential geometry and differential topology?

Okay cool

@Daminark okay let us go over things

I was wondering if there is any intuitive difference between

real vector bundles and complex vector bundles

@AkivaWeinberger I think differential geometry you also have some kinda of metric

Q: Can a Hamkins infinite time Turing Machine with infinite Super Turing jumps (from higher type oracles) get the power to decide $\Sigma_1^2$ sets?

1:35 AM

8

Hamkins showed that his infinite time Turing machine has the power to decide some $\Delta_2^1$ sets. I wonder if some modifications of the machine could be made to reach level $\Sigma_1^2$ sets, or, if no modifications on sight, if the power of his machine plus infinitely iterated super jumps fro...

Yes, the ITTM model has only ordinary tapes with $\omega$ many cells. I believe our oracle concept for sets of reals is the right one to consider, but I suppose I can imagine stronger ones. If you want to use uncountably many cells, then if the cells are to be well-ordered (in order to mak sense of traversing them), then you will not just be putting the set on the oracle tape, but the set in some order, which would carry more information. — Joel David Hamkins Mar 26 '13 at 11:34

I brb

So combining the above with:

> This least uncountable ordinality is a truly remarkable mathematical object: mere contemplation of it is fascinating and a little dizzying. For instance, the minimality property implies that all of its initial segments are countable, so it is not only very large as a set, but it is ex- tremely difficult to traverse: for any point x ∈ ω1, the set of elements less than x is countable whereas the set of elements greater than x is uncountable!

and:

Forever Mozart

whence

and thus

$\square$

> Now it's worth noting that there are many results in set theory which indicate that there can be no "concrete" picture of $\omega_1$, in various senses. For example, it is consistent with ZF that there is no injection from $\omega_1$ to $\mathbb{R}$ or from $\mathbb{R}$ to $\omega_1$. In ZFC of course there will be an injection from $\omega_1$ to $\mathbb{R}$, but it can still be very complicated: any such injection $i$ induces a relation $\trianglelefteq_i$

> on a subset $S_i$ of $\mathbb{R}$, and any such pair $(S_i, \trianglelefteq_i)$ must be extremely complicated - certainly not Borel, and under additional ("large cardinal") hypotheses not even projective.

So that means, in order to express $\omega_1$, not only one needs an infinite binary string, in fact, they need a string with $\aleph_1$ entries. So $\omega_1$ is uncomputable even in an infinite time turing machine

:Leaky

@Adeek: Not necessarily a metric, but a connection on a vector bundle (a way of differentiating sections). A Riemannian manifold has a unique (torsion-free, for the experts) connection on its tangent bundle that is compatible with the metric, but diff geo deals with lots of other vector bundles.

And in things like symplectic geometry there is no metric, but a symplectic form that provides the structure.

1:48 AM

@Adeek re real and complex vector bundles, the first thing that comes to mind is that a vector bundle homomorphism between complex bundles will have to respect a higher amount of structure than if you consider them just as real vector bundles. Beyond that I dunno anything. Though @Ted is here so yeah halp

Ted is about to pack for his trip, however. :)

Where are you heading?

Georgia for 9 days.

I hope I don't get shot.

I have to cook Thanksgiving dinner for 9 1/2 people.

Why would you get shot?

Oh lord

Because students can have guns on campus now.

And I'll be at UGA for parts of two days.

1:50 AM

Oh you should've seen this one time last year we had a game called "Humans vs zombies" going, so a bunch of us came to analysis with nerf guns and our prof had a field day

Nerf guns won't upset me. This country is in serious s**t./

"I know there's all this stuff about campus carry but I at least want to see your gun. At least with him I know I have to give him a good grade, if I tell you that your work is shit I'll think I'm safe and then bam"

But yeah, there are different (holomorphic) complex vector bundles with the same underlying real bundles ... e.g., different complex structures on a complex manifold.

No way in hell I would have compromised on my grading standards ... but I'm just saying I'm glad I am done.

I mean he was joking since it was clearly just nerf guns. But yeah I mean campus carry business is strange but hopefully won't change much day to day?

The country has definitely changed radically in terms of the outright hate being shown and acted out.

heya DogAteMy

1:56 AM

Heya

I just had a performance with a Broadway actress

Yeah, that's never a fun time but, in the words of overly optimistic parents about their kids, hopefully it's just a phase

Did you outshine her?

Lol no

Damn, DogAteMy ... you let me down :D

but it was chamber choir (~16 kids) and her

1:57 AM

Cool :) I'm sure it was very exciting for you.

There’s a Place for us

►

I realize you probably don't know which one is me

but that's the video taken by my mom

Hmm, it blurs when I make it full-screen. Did one of the girls step in front of you? :P

Yeah my face is a tiny bit above and to the right of the girl sitting down holding the violin

Alternatively: Rightmost boy wearing glasses

LOL

Got it. :)

You look like DogAteMy :P

We also performed at a bunch of synagogues yesterday and Friday night (sans Broadway actress)

2:06 AM

Aren't you the busy beaver ... :)

(Also sans microphones)

It's nice to be multi-talented :)

DogAteMy: Did you do some songs in Hebrew?

Yeah

Very cool.

I hope everyone has a good T-day holiday. I'll be gone mostly for 9 days.

wtff

while I was reading Huybrechts I noticed the following lemma

It can be shown that if a continous function on an open set $U \subset C^n$ is holomorphic in one variable it is holomorphic in every variable

complex analysis is some strong ***

2:13 AM

you're misreading

haha

holomorphic in each variable separately ... then jointly holomorphic

yeah that is what I meant but not wrote carefully

but, yes, that's one of Hartogs' theorems and it's surprising

yeah very

@AkivaWeinberger nice :)

2:15 AM

the other Hartogs' Theorem is more interesting to me ... that in several variables if you're holomorphic outside a compact set, then you actually extend holomorphically.

What does jointly homomorphic mean?

oh wow

intuitively it make sense ... you have to blow up on something of (complex) codimension 1, which is not inside a compact set.

Oh wait is that like

that's like actually differentiable

2:16 AM

the derivative is some matrix and it exists

in the sense of best linear approximation (with complex limits), yeah

I see

but matrix of certain type though

Mhrm. Very not true in $\Bbb R$.

yeah

$\Bbb R^n$, very.

2:17 AM

I guess complex differentiability is very strong property

Well, you should already have known that from one variable.

I can't wait to learn more of complex geometry and complex analysis

And yet so many things satisfy it!

One derivative implies infinitely many and convergent Taylor series.

@TedShifrin I didn't take one variable

2:18 AM

You need to learn it, Karim!

Even annoying things like $\zeta(s)$ are analytic!

My university sucked, so most of the background that I needed I covered by myself from books

Yeah I learned it @TedShifrin

I learned it by myself though

(Erm, $\zeta(s)(s-1)$, to get rid of the pole?)

Not everywhere, DogAteMy.

OK, I need to eat dinner and pack for my trip. Y'all misbehave without me.

sigh... the flags

2:25 AM

@TedShifrin One thing I found interesting is the following problem

what is flagged?

Classification of Calabi Yau manifolds is still open problem

it would be cool to work on that after I get a lot of knowledge in complex geometry

@Secret this lol

I see

2:48 AM

we have based calculus on functions that behave well although in the real world few functions do...

what is the point?

@TedShifrin

"we have based calculus on functions that behave well although in the real world few functions do..." - what?

what?

how have we 'based calculus' on those functions?

aren't functions studied in calculus the ones who behave well?

that's cause the functions that show up in the real world are the ones that behave well

2:54 AM

really?

definitely

I mean are we sure that the real world exists? Real numbers don't so that doesn't bode well for the world

@EricSilva like the stock market?

it's the canonical example of continuous and non-differentiable function

@Leaky I mean not every function that comes up in the real world is analytic or something

solutions to SDEs are usually fairly tame compared to dumb shit you come up with as counter examples

2:57 AM

But you can find quite a lot of behavior in physics that is modeled quite nicely by many nice functions

just their regularity tends to be "in average" rather than pointwise

so i would say that stuff that shows up in finance is a bad example of what youre trying to say @Leaky

hmm

surely there are no cantor function in the real world?

@Secret cantor function is quite nice in some aspects

I think I'm stepping into chaos theory with this line of thought

double rod pendulum

right, that should be what I'm looking for

the tools of calculus still help to understand those things though

3:02 AM

for example?

have you ever studied dynamics??

could you do a brief introduction?

calculus shows up all the time

like how?

I think chaos theory being deterministic means that let t be time and x be initial conditions, there is a f such that f(x,t) always exist

3:04 AM

ODEs are the historical root of dynamics

if you wanna study fractals for ex. a lot of the time you have to use tools from geometric measure theory that came about from people trying to use calculus for rough things, and it usually works

that is, take a region of spacetime with chaos. Then the trajectory of all particles are well defined curves such that the curves are always continuous in x,t, and there is only one such configuration of curves for each x

Meanwhile, if the system is indeterministic, there are more than one possible configurations of curves in the region of spacetime for some given x

So in theory, if we can see both the past and the future at the same time, all chaotic process will be predictable because you are effectively reading the history of the whole system

but it would be a computational nightmare

yeah, it will be a jumbled mess

that doesnt strike me as relevant to what you were saying

uh, the jumbled mess is responding to Leaky, and Leaky is responding to my wall of text, probably

3:17 AM

$f(x)=x^2+x+1$ has no positive root by Descartes rule of signs. but $f(-x)=x^2-x+1$ has two negative roots by Descarte's rule of signs. But when I solve it, I found two complex roots. Why do this phenomenon occur? Please help me. Please point out mistake. where do I go wrong?

@ManeeshNarayanan I thought the rule says "at most"

@LeakyNun, there's a reason calculus was fine without epsilon-delta for almost hundreds of years

Descartes' rule of signs

In mathematics, Descartes' rule of signs, first described by René Descartes in his work La Géométrie, is a technique for determining an upper bound on the number of positive or negative real roots of a polynomial. It is not a complete criterion, because it does not provide the exact number of positive or negative roots. The rule is applied by counting the number of sign changes in the sequence formed by the polynomial's coefficients. If a coefficient is zero, that term is simply omitted from the sequence. == Descartes' rule of signs == === Positive roots === The rule states that if the terms of a...

the pathological examples for which one needed rigour didn't show up that often

I got the solution. Thank you.

Forever Mozart

3:27 AM

After much work, I have refined my question to something quite interesting

I am now past the "bumping around in the dark" stage :)

now it's time to prove something

I love it

One needs an $\alpha$ th oracle to make $\omega_{\alpha}^{CK}$. It follows that anything beyond the Church Kleene Ordinal is impredicative in the general sense in that its definition is self referential

Forever Mozart

umm... that's not math

proof by cult

3:41 AM

hey @Daminark

want to discuss something small

Yo

Sure

No larger than 76.2 epsilon though :V

$\epsilon^{\epsilon^{\epsilon^{\epsilon^{\cdots}}}}$ should be small enough

Suppose you have a real vector space V. We introduce a complex vector space by the following Operator $J : V \rightarrow V$ such that $J^{2} = -id$. Then we introduce the complexified vector space $V_{\mathbb{C}} = V \otimes \mathbb{C}$.

we can extend $J$ to a map between $V_{\mathbb{C}}$ as follows

$J : V_{\mathbb{C}} \rightarrow V_{\mathbb{C}}$ given by $u \otimes (x + iy) \mapsto J(u) \otimes x - y$ ?

is that the extension ?

lol epsilon joke

or is it just $J(u) \otimes (x + iy) $?

oh yeah it is just that

because $J^2 u \otimes x + iy \mapsto - u \otimes x + iy$

$= - (u \otimes (x + iy)$

Yeah that would preserve the property that J^2 = -1

yeah

3:58 AM

@AsafKaragila Yes it clicks now. I wondered for many years why infinite sums are not commutative because I expected you can prove by induction that they are commutative. I suspect I somehow mixed potential with actual infinity. — Trismegistos Aug 7 '14 at 14:34

hmm

and these two behave very differently, the latter may not actually exist

there is something I am wondering about

So my book defines the following

Oh no

i had a similar thought as well

but you cant get to infinity by induction

@Daminark if we have X being complex manifold then it says that for a chart $\phi_i : U_i \rightarrow C^n$ we have that this gives rise to the push forward $(\phi_i)_*$ which identifies the tangent bundle with $U_i \times C^n$

Q: Why are induction proofs so challenging for students?

4:01 AM

56

This forum already has many good, simple examples of induction proofs, a great resource. As I am soon to teach induction for the $n^\textrm{th}$ time—this time to some perhaps under-prepared college students (in the US)—I would like to understand why induction proofs are often challenging for stu...

undergraduate-education proofs induction

I mean I understand how we normally have the real tangent bundle as $U_i \times C^n$ but why does the coordinate chart gives rise to that identification ?

Indeed, the limit case is separate from the inductive case

which was a little surprising when i first learned aobu tit

That I'm not sure about. Keep in mind I only know the very basics of manifold theory and nothing about vector bundles

My example for the induction thing is this: Consider $x_n = 1 - \frac{1}{10}^n$. We can show by induction that $\forall n\in \mathbb{N} $ we have $x_n > 0$, but we know $1 = 0.999...$. Hence, we are done.

that should be $(\frac{1}{10})^n$

@Secret

4:12 AM

@Daminark nvm I see it

it is very weird

okay here is how it works

So given $\phi : U_i \rightarrow C^n$ we get an induced map $\phi_{*,p} : T_p \rightarrow C^n$ and as p varries through $U_i$ we get that we get indeed

that through $\phi_*$ we get the result above.

@orbit-stabilizer yeah, I think the 1=0.9999... case need to be proved separately, e.g. via cauchy sequences

Right, but if we have that, then we're good

I like using the fact that the reals are dense to prove 1 = 0.999...

Idk, how rigerous it is

I wish I had more grad student in working in my area it is easy to clarify ideas by talking about them

Two real numbers are distinct iff there exists another real number in between them. Suppose there exists such a number between 1 and 0.999.... But then, it can't be just anything less than 9 in the tenths place, it can't be less than 9 in the hundreths place, and so on.

hmmm, but this seems to fall prey to the induction thing earlier, right?

Quick quiz:

About which invention was the following said, in 1958?:

> It is with understandable enthusiasm that I give you today an exclusive report on this news: A 'zipperless zipper' has been invented – finally. The new fastening device is in many ways potentially more revolutionary than was the zipper a quarter century ago.

4:23 AM

hmmm

yeah, I think the fact that the reals are dense and complete (whose proof might had something along the lines of dedekind cuts or cauchy sequences) is what really establish there is no real number between 1 and 0.99999....

@LeakyNun "$X$ is a sphere bundle over $Y$"

@Secret, right but my 'proof' wouldn't work right

Which is equivalent to what anon suggested anyway

@AkivaWeinberger, Ziploc!!

ziploc bags

4:24 AM

yeah, cause your induction proof only checked any arbitrary finite number of cases, not the limiting case

@orbit-stabilizer Nope

Good guess though, I hadn't thought about that

Oh, wait, no. What's an example of a sphere bundle which is not a linear sphere bundle? (By that I mean, cannot be given a linear structure?)

@Secret darn. it was my go to proof to explain to ppl why 1 = 0.999...

it does build intuition tho

@AkivaWeinberger I don't know...

Want the answer?

yes

4:26 AM

Actual infinities are weird, they cannot be reached from anything below unless you move as fast as the actual infinity itself

are inaccessible cardinals numbers so large that they can't be reached by taking power sets?

(Also, apparently Ziploc was invented in 1968, ten years after the above quote)

@orbit-stabilizer Velcro.

that's what i think of when i hear inaccessible cardinals

@orbit-stabilizer Yup, they are (along with some extra property that I forgot)

@AkivaWeinberger interesting! turns out it wasn't a big deal?

@Secret are there cardinals greater than inaccessible cardinals?

4:28 AM

@orbit-stabilizer What do you mean? Velcro was pretty influential

@AkivaWeinberger I'm just thinking about shoes :D

Is the inclusion $SO(n) \to Homeo^+(S^n)$ a homotopy equivalence? I think so

Hm, what about fucked homeomorphisms which are not diffeomorphisms

@orbit-stabilizer There are many, Cantor's attic give some information about them. You can go all the way up to the Kunen inconsistency where after that point, the notion of cardinality breaks down because nothing is consistent anymore

Fun fact, Velcro wasn't popular until NASA used it in their space missions

idt those can be homotoped off to a linear diffeomorphism.

4:30 AM

set theory/model theory/logic is so cool

Mind you, that inconsistency is more like a receding goalpost, in that you can never actually reach it, let alone specify exactly where it is

space travel has been good to us

I'mma look up the history behind Ziploc now because that might be interesting as well

I am so sad that uncountable well orderings cannot constructively exist

Ah, yeah, that's exactly it. Any exotic $S^7$ can be built off as two 7-disks glued along $S^6$ by a homeomorphism. If that was isotopic to a linear diffeomorphism then this would be the standard $S^7$ by isotopy extension

So $n = 6$ is a counterexample

well rip me

4:32 AM

(Don't tell me about $\omega_1$. This thing cannot even exist without an oracle of the same degree as itself)

I guess $SO(n+1) \to Diff^+(S^n_{standard})$ is a homotopy equivalence anyway. For $n = 3$ that's Hatcher's theorem IIRC.

$\omega_1^{CK}$ is perhaps only slightly better in that it is only impredicative, but not nonconstructive

math.toronto.edu/weiss/model_theory.pdf

this is greek to me

Ziploc's history is not interesting

Anyway, a brief summary of ordinals in predicative set theory (which means throwing away powerset axiom and axiom of replacement):
---The proper class of ordinals is given by $[0,\omega 2)$
For a less predicative set theory where the axiom of replacement is retained:
---The proper class of ordinals is given by $[0,\omega_1^{CK})$
For an even less predicative set theory where finite n instance of oracles are permitted:
---The proper class of ordinals is given by $[0,\omega_{n+1}^{CK})$

Q: Which of the following statement is correct[CSIR-2017]

4:39 AM

0

Let $f=u+iv$ be an entire function where $u,v$ are the real and imaginary parts of $f$ respectively. If the Jacobian $$ J_a= \left[ {\begin{array}{cc} u_x(a) & u_y(a) \\ v_x(a) & v_y(a) \\ \end{array} } \right] $$ is symmetric $\forall a\in \mathbb C$, then (1) $f$ is a p...

complex-analysis harmonic-functions holomorphic-functions

Please help me.

and for a finitist (throwing away axiom of infinity)
--- The proper class of ordinals is given by $[0,\omega)$

@Balarka so at some point in life I'm gonna ask you how spectral sequences work because... what the fuck?

and for an ultrafinitist (where induction is restricted to some fixed finite number)
--- The set of ordinals is [0,n]

Not now because I have to write a bio assignment

hey, @BalarkaSen do you want to discuss few complex manifold stuff? Suppose you have a complex manifold X. Then we would like to establish a relationship between $T_{X,R}$ and $T_X$ i.e the real tangent bundle and the complex tangent bundle. We natrually have $T_X \subset T_{X,R} \otimes C$. We also have that the subbundle $T^{(1,0)}_{X}$ is the subbundle generated by the bundle of eigenvectors of I for eigenvalue i.

Here $I$ is the natural complex operator

I don't understand how is $T_X^{(1,0)}$ naturally isomorphic to $T_{X,R}$ ?

4:56 AM

@orbit-stabilizer Also it took years for the inventor to figure out a way to efficiently manufacture Velcro

'cause it's not obvious how you make those things

Lots of tiny thingies

Q: Any weird 'modular like mathematical space' that behaves like as if it is infinite until a threshold is reached?

6

(Using the advice from Mathoverflow, I have rephrased and splitted up the question) (Might be a bit layman because I don't have rigorous math term to describe the concept) Generalize it to mathematical spaces, are there spaces which are sort of like a. Having multiple $\mathbb{R}^2$ spaces th...

calculus geometry

Lagranian

0

Q: What is the maximal value of $|A\cup B|$?

For $A$ and $B$ sets that is different from null, $4. |A-B| = 3 . |A' \cap B|$ $3. |A| - |B| = 20$ What is the maximal value of $|A\cup B|$? It seems so confused right now. Can you take a look?

elementary-set-theory

Can you take a look?

I'm so confused atm

hmm.. now that I look at this question again, it appears I am basically asking what happens if there are finite sets that behave like infinite sets (the converse is easy, and are given by the infinite dedekind finite sets) and that n+n<m for any n<m finite

anon

@LeakyNun the formula you gave for $S\to S^3$ seems to be some kind of augmentation of the identification $S^1\star S^1\simeq S^3$ (where $\star$ is join), which would have just been $(\varphi,\theta,\omega)\mapsto(\sin\varphi\exp(i\omega),\cos\varphi\exp(i\theta‌))$. not sure how to interpret the extra "twisting" factor you have geometrically or how you derived it.

Lagranian

I've tried to apply $|A\cup B| = |A| + |B| - |A\cap B|$

5:07 AM

@Lagranian make venn diagram, note that for $|A \cup B|$ to be max we must have $|A \cap B| = 0$. Use given equations to get the answer.

Hi i

hi

I saw today some good properties of some probability distributions

nice

i will have statistics next sem

then we will talk :P

Like a rv which has same mean,median,mode is a normal random variable

Andrv ehich has. Same mean and variance then it is a poisson rv

So i was interested to see many more hidden properties of other distributions or some nice properties ... So asking in the main site by tagging it to be a soft question will be a good idea??

5:12 AM

depends on the people that view it.

@NV-US there is also a probability statistics chat room, i admit tht it is not so active with users...

discuss here, this is a very active room