If I'm interpreting math.stackexchange.com/a/437873/76284 correctly, it's saying that the axiom of extensionality is not a conservative extension of the other axioms, right? As in it's really necessary to have it as an axiom rather than "just" a definition (as is the case with "subset")?

Question: is there a concept of defining a topology inside of a topology? I had the thought of a topology on the power set of a set, and set in open sets in that topology are equipped with a topology, while the nonopen sets are equipped with another topology.

@Rithaniel I'm not sure if I understand your question fully, but a topology $\mathcal{T}$ is defined for any set $X$. But $\mathcal{T}$ itself is a set too, so you can take a subset $A \subset \mathcal{T}$ and put a topology on $A$, perhaps that's what you meant

The other interpretation of a topology within a topology would certainly be the subspace topology

reddit.com/r/math/comments/agp62o/…

Equality, which is the following inference rule:

is actually a very important logic rule that need to be understood

As is A=B \\ P(B) \\ ----- \\ P(A)

(\\ = new line)

yup

From this, we can see that the syntax of logic, is similar to algebraic structures except that the "axioms" have => instead of =, making them unidirectional most of the time

How does matrix exponentiation work with tensor products?

If $M:V\to V$ and $N:W\to W$ are linear, then $M\otimes N:V\otimes W\to V\otimes W$. What's $\exp(M\otimes N)$?

Is that even simplifyable?

What about $\exp(M\otimes I)$? That's $e^M\otimes I$, right?

(and not $e^M\otimes e^I=e(e^M\otimes I)$)

And $e^M\otimes e^M=\exp(M\otimes I+I\otimes M)$ and not $\exp(M\otimes M)$?

--

On a similar note

if $[M,N]=MN-NM$

then$$[M\otimes I+I\otimes M,N\otimes I+I\otimes N]=\\ [M,N]\otimes I+I\otimes[M,N]$$

but $[M\otimes M,N\otimes N]$ doesn't equal $[M,N]\otimes[M,N]$

chat meta-question: could the description not just link the math scripts themselves so that I don't have to go to the bookmarks?

It's a really big link

ah, it does not allow you to give it a different text, should have imagined!

is it okay to start spoiling a proof of a theorem after some time if you can't solve it or is it better not to read the proof at all and just leave it and go on?

I would read the proof

but you might lose the idea or technique that you could get by proving a theorem later

@AkivaWeinberger You have a map $GL(V) \times GL(W) \to GL(V \otimes W)$. The thing the exponential map does is take as input a matrix and spit out a homomorphism $\phi: \Bbb R \to G$ which that matrix as its derivative at zero. Such homomorphisms are determined by their derivative at zero.

Therefore, if $f$ is a Lie group homomorphism, we have $f(\exp(tz)) = \exp(t df_0(v))$.

For us, the starting tangent vector is $A_1$ and $A_2$, and the derivative of the map $$(I+A, I+B) \mapsto (I + A) \otimes (I+B) = I + A \otimes I + I \otimes B + A \otimes B$

The last term has zero derivative. What you find then is that your desired exponential is $\exp(t(A \otimes I + I \otimes B))$.

"Your desired exponential"? @MikeMiller

@famesyasd Read the start of the proof, see if it gives you a hint on how to finish the proof

I guess that's not what you were asking, sorry. The point is that this $\exp(M \otimes N)$ is unrelated to things coming from the two factors separately.

So you can't *simplify it.

And $e^M\otimes e^N$ is the exponential of $M\otimes I+I\otimes N$.

Right, I missed what you were asking.

So the reason I was asking was because I was reading about representations

of groups and of Lie algebras

and the formula for a tensor product of representations is different for those two things

Sure.

In one you act diagonally and in the other you act once on each factor, no?

As in $Xu\otimes Xv$ versus $Xu\otimes v+u\otimes Xv$?

(So the second one looks like the product rule)

I would hesitate to use X for both a lie algebra element and a group element

OK

Those are different fellas

And the second should for sure look like the product rule

You get it by taking the derivative of the first

In terms of the relationship between them, it's because of the exp thingy

In terms of the definition of a representation of a Lie group, it's because

@Secret yikes

so it plays well with the bracket

Actually I am not well versed enough in tensor algebra to knew whether there exists an analogue to the rule $(e^x)^y=e^{xy}$ for reals

There's no rule A^y

Only e^y

If $e^A=e^B$, does $e^{A\otimes M}=e^{B\otimes M}$?

I'm guessing no

but I'm not sure

Wow that was foolish

I wasn't looking when you wrote that so I don't know what you typed

The trick should be to choose one so that both of those are zero

A = 0 and B not zero

So that the B o M exponential is nonzero

So you want $e^B=I$ and $e^{B\otimes M}\ne I$?

Oh you know what

Right

Let $i$ be that 2x2 matrix that squares to $-1$

$e^0=e^{2\pi i}$ but $e^{0\otimes\frac12I}\ne e^{2\pi i\otimes\frac12I}$

Ah sure

because the latter is $-I\otimes I$

I was thinking this but worried it went badly because everything commutes

But no, we are fine

And instead of the dual being the inverse transpose (as it is for groups), it's the negative transpose @MikeMiller

@AkivaWeinberger Right, with the moral being to always take derivatives.

Wait, how do you get that out of derivatives?

I was just thinking of it as coming from $(e^A)^{-1}=e^{-A}$

Wrong answer, but there's going to be a correct way to say this, because the concepts of (simply connected Lie group reps) and (Lie algebra reps) are the same, with the equivalence passing through "take the derivative at 0"

Well at 1

Oh I got it

The derivative of $\dfrac1{I+tA}$ is $\dfrac{-A}{(1+tA)^2}$

Set $t=0$ on the right

Hm. The derivative of $(x+y)^n$ with respect to $x$ is $n(x+y)^{n-1}$. Is this visible from the binomial expansion?

$(x+y)^n=\sum_k\binom nkx^ky^{n-k}$

$D_x(x+y)^n=\sum_kk\binom nkx^{k-1}y^{n-k}$

$n(x+y)^{n-1}=\sum_k n\binom{n-1}kx^ky^{n-k-1}$

${}=\sum_k n\binom{n-1}{k-1}x^{k-1}y^{n-k}$

Thus $k\binom nk=n\binom{n-1}{k-1}$

This is in fact true

and, in fact, $kB_{n,k}=nB_{n-1,k-1}$ and $B_{n,0}=1$ uniquely determine the binomial coefficient fairly easily (you get $B_{n,k}=\frac nk\frac{n-1}{k-1}\frac{n-2}{k-2}\dotsb\frac{n-k+1}1$)

so if you didn't know the formula for the binomial coefficients, you could derive them from differentiation

What's your view on the prevalence of mental illness among PhD students and academics, in Mathematics in particular?

I'm not either of those things so I can't really comment

If I had to guess, though, there's probably a correlation between autism and interest in science/math

That's the stereotype, anyway

I dunno about other disorders like OCD, AD(H)D, or anxiety though

And I think I've heard from people that student pressures can induce depression or burnout

Hey hey

I got a quick question about notation of functions containing mixed dterministic and stochastic components

Say that I have a function f such that f(x) = b*x^2

such that each time f is evaluated, b should be drawn randomly, say from a bernoulli distribution Ber(q)

whats the correct notation for this?

just f(x) = b*x^2, b ~ Ber(q) ?

or do I have to set up b as an instance of a stochastic process

which i would rather avoid

How does a polynomial in a polynomial ring of countably many variables look like? Will it be a polynomial in finite variables?

Has anyone here ever taken real analysis before?

@ThomasShelby yes

@AlessandroCodenotti Thanks a lot.

Okay, how do I prove the Chinese Remainder Theorem.

What is wrong with my rationale for integrating the volume of a cylinder in rectangular coordinates? Say we have the function z = -2, and the domain x^2 + y^2 = 7. This should be solvable with ∫∫ -2 dx dy = ∫ [-2x] dy. The integration seems to be from x = -√(7 - y^2) to √(7 - y^2), but this yields 0, which means the second integration will yield 0, which is not the volume of the cylinder.

I'm thinking some kind of induction with a base case where I have two n

I think you might be dropping a minus sign, @user10478

thanks

@Rithaniel you have a ring homomorphism $\varphi : R \to R/I_1 \times \cdots \times R/I_n$, show it's a surjection and find the kernel of it

(you'll need the pairwise coprimality of the $I_i$ to show that the product of the $I_i$ is equal to their intersection)

(which can be done inductively)

(parentheses)

Okay, so I need to be thinking of the system of congruences as the space $\mathbb{Z}_{n_1}\times\mathbb{Z}_{n_2}\times\mathbb{Z}_{n_3}\times\cdots\times \mathbb{Z}_{n_m}$?

I just showed, while working on another problem, that if we define $U(\mathbb{Z}_n)=\{[a]\vert \text{gcd}(a,n)=1\}$, then $|U(\mathbb{Z}_{mn})|=|U(\mathbb{Z}_m)||U(\mathbb{Z}_n)|$ if we have that $\text{gcd}(m,n)=1$. Would this be similar to what you're saying with showing that the product is equal to the intersection?

Well $\Bbb Z/(mn) \cong \Bbb Z/(m) \times \Bbb Z/(n)$ when $(m, n) = 1$ is the statement of the chinese remainder theorem for $\Bbb Z$

that the unit group of the first is equal to the product of the unit groups of the latter two is a statement for rings

though I guess this is not particularly helpful

"unit group" is the name of that $U$ group? I'll need to remember that.

Yes, the group of units of a ring is the group of invertible elements of the ring

(in $\Bbb Z/(n)$ these are the elements that are coprime to $n$, since in this case $ax \equiv 1 \bmod n$ has a solution in this case)

@Rithaniel alternatively you can just construct a solution and then show uniqueness

I believe ring homomorphisms is going to be something we study later in the semester, so it would be good to familiarize myself with this sort of stuff, though I'm not going to immediately follow it.

For a more constructive approach, if you have coprime integers $n_1, \dots, n_\ell$, you can write down a solution modulo $n_i$ for each $i$ by defining $N_i = N/n_i$ and then building a solution to your system of congruences mod the $n_i$ using these

and Bézout

(I mean pairwise coprime of course)

Reading Bézout's Identity now.

Ah, I know this.

@Rithaniel Bézout is fundamental so I'd get familiar with this

Yeah, I proved this exact statement earlier this week. I just didn't know it's name.

Ah okay, good!

Alright, after working through it, I think I see how you approach the problem. You start with a base case of two $n_i$, and demonstrate the existence of a solution by picking a value and showing that this particular value works for the system. Then, inductively, you just add another $n_i$ and multiply the previous $n_i$ together. The only thing that I think is missing is that the solution is unique for $0\leq x<N$ where $N$ is the product of all $n_i$.

If you have two solutions mod $n_i$ for each $n_i$ then $n_i \mid x - y$ for all the $n_i$. The coprimality means that $N \mid x - y$ (because this is the product of the $n_i$). Since $0 \leq x, y < N$, you have $x = y$

Hi there, chat

Hullo

@ÍgjøgnumMeg Hi !

Hi @Jacksoja

is Q[ a,b ] = Q[ a+b] ? @ÍgjøgnumMeg

a= sqrt 2 , b = sqrt3

i think no

It is in this case

let me show you my argument

Q[a,b] has all rationals and sqrt 2 and sqrt 3 ie x+y sqrt2 +z sqrt 3

when Q[a+b] has this form, x' + y' ( sqrt 2+ sqrt 3 ]

the degree of $\Bbb Q(a, b)/\Bbb Q$ is $4$ so you're missing one basis vector

@ÍgjøgnumMeg this question is on ring theory, so before field theory ^^

artin book, rings first chapter

$\Bbb Q[\alpha] = \Bbb Q(\alpha)$ when $\alpha$ is algebraic over $\Bbb Q$

@ÍgjøgnumMeg what is the difference btween parenteses and brackets?

Does anyone have a good resource on how to do mod calculations with a negative exponent for RSA, for example 7^-1 mod 80

by hand

@Jacksoja well if you have rings $R \subseteq S$ and $\alpha \in S$ then $R[\alpha]$ is the smallest subring of $S$ containing $R$ and $\alpha$, while $R(\alpha)$ is the smallest subfield of $S$ containing $R$ and $\alpha$

(Like for polynomial rings you have $\Bbb Q[X]$ and the field of rational functions $\Bbb Q(X)$)

(In this case $\Bbb Q(X)$ is the field of fractions of $\Bbb Q[X]$, though this is only true because $\Bbb Q$ is a field)

@ÍgjøgnumMeg okay thank you !

in the case of your question, it's easier to think of $\Bbb Q(a, b)$ as a $\Bbb Q$-vector space and then make a change of basis to show that $\Bbb Q(a, b)$ and $\Bbb Q(a + b)$ are isomorphic as $\Bbb Q$-vector spaces

Note also that $\Bbb Q(\sqrt{2}, \sqrt{3})$ and $\Bbb Q(\sqrt{5}, \sqrt{7})$ (for example) are isomorphic as $\Bbb Q$-vector spaces (because they are finite dimensional $\Bbb Q$-vector spaces of the same dimension) so you'll also want to show the equality as sets

or you can just show the set equality probably idk

bit of a time-waster

Hey, I am new at real analysis and I have been practicing Calculus the limits, sequences series, and convergence tests. Overall, what is the best way to study for real Analysis? I am reading Real Analysis and Foundations by Steven Krantz. But I feel like I am reading it passively and not really understanding the proofs in the book. What is the best way to prepare for real analysis so that one can really understand the proofs. For example the proofs about the Number Systems.

@ÍgjøgnumMeg can you please tell me some reading material about this ? I did not get there yet in my book

@ÍgjøgnumMeg is this subject of field extension ?

@Jacksoja I used Dummit and Foote's abstract algebra

also, being more careful i'd say $R(\alpha)$ would be the smallest subfield of $\operatorname{Quot}(S)$, though even in this case it probably only makes sense when $R$ is an integral domain or smth

Well $\Bbb Q(\alpha)$ is a field extension yes

there are also ring extensions

though field extensions are easier to read about to begin with because fields are simple objects

or rather, simpler than rings, in some sense or another

@ÍgjøgnumMeg thank you so much ! I have that book , should I complete the chapters o n rings first then read chap 13 ?

Well it's a good idea to be familiar with rings before you start reading on fields

I think 13 is fields and then14 is Galois theory right?

If these were given on a true/false test, many students would mark "5+2=7+3=10" as "true" and "5+2=4+3=7" as "false"

Similarly, many students would be unable to correctly answer the question "Complete the blank, 5+2=__+3" on a test

especially if it's multiple choice and 7 is one of the choices

(I'm talking young, like grades 1-4)

Many of my classmates would have a similar issue ngl

There's a fundamental misunderstanding of what the equals sign means, and if they don't get corrected, they won't understand algebra

@ÍgjøgnumMeg correct

because "fill in the blank" is basically an algebra question (replace the blank with an x)

@ÍgjøgnumMeg but I do not need to read about modules and vector spaces, right?

after the 3 chapters on ring , i can move to fields @ÍgjøgnumMeg

it's a good idea to know about vector spaces when studying fields (a field extension is a vector space over the base field)

Incidentally, there isn't a good way to write "5+2=7 and 7+3=10" without repeating yourself

what mathematical operation would be the same as shifting the binary representation of an int by one to the right >> 1 for both positive and negative numbers?

Could you give an example? @Rick

one sec

Like, 1011 (binary for 11) turning into 101 (binary for 5)?

Or is it 0001011 turning into 1000101?

Like, we know what length the thing is supposed to be, and the rightmost bit wraps around to the left

for example -3 would be -2 and 3 would be 2

or 1011 $\mapsto 0111$

dunno why I mixed text and mathjax

but it's hurting my soul

@Rick I don't see… Are you doing something to the binary representation?

@ÍgjøgnumMeg You can edit messages

that's a lot more effort than my soul is worth

well usually, I would just floor. but for negatives, I would need to round up, I think

So wouldn't 3 turn into 1?

@ÍgjøgnumMeg thanks I shall prepare those then, taking Galois theory this semester

@Rick There's an operation that takes the floor for positive numbers and ceiling for negative numbers, but I forget what it's called

Like, a named operation

@Jacksoja you'll have a hard time understanding Galois theory (or what the point of Galois theory even is) if you don't know about fields

@ÍgjøgnumMeg I shall do fields properly then ! :)

I found it

Wikipedia calls it (drum roll) "round-towards-zero"

round-toward-zero sound anti-climactic

@Jacksoja enjoy!

thanks :) :)

Or maybe "truncation"?

no truncation won't work

It looks like there's a programming language that calls it "trunc", which is probably short for truncation

i tried that

but truncation does not produce the correct result

Well you gotta divide by two first, no?

What do you mean you tried it? Like, in a programming language?

ya it just removes the trailing decimal

Isn't that what you want? 3 -> 1.5 -> 1, and -3 -> -1.5 -> -1

('Cause in binary 11 turns into 1)

it does not yield the same result as the >> 1 operator

What does 3 turn into when you do that operator?

let me double check, to be sure one sec

Why? I don't understand

sorry should I get 1 for 3 and -2 for -3

Why -2?

It looks like it's "divide by 2 and then round down" in that case

correct i think

How are negative numbers represented? 2's complement?

well it's ((-2 +-1)/2)>>1

Math.floor will work for negative numbers but it will be wrong for positive numbers

This is Python?

no Javascript

you have Math.round() Math.floor() Math.trunc() Math.ceil()

they should mirror for both the right and left hands sides of the zero

sorry for the shift operation you don't divide by 2

What do you get when you do it with 13, for example?

6

but I think I'm doing something wrong

I am adding two numbers before the shift.

but 6 seems like the correct result

(a+b)>>1 is the operation the works for all my test cases

Math.floor ( (a+b)/2 ) works for negative numbers

and Math.trunc((a+b)/2) seems to work for all positive numbers

I think there might be some sort of glitch. Math floor works when I remove the >> operation from the code entirely

but truncation will still fail, but floor seems to works

I dunno how much I can help you, tbh, I don't know any JavaScript

where 0 is treated as an index on the right-hand side of the numberline

I know some Java but I'm told that that's useless for JS

You helped a lot, I was building a segmented tree where I wanted the segments can have negative and postive intervals

removing >>1 allowed Math.floor to work

so rounding toward 0 was key, truncation would not have worked, which was what I was using before.

thx :)

Interesting picture I found

The description suggested there was neat math/topology about it but I didn't understand it

From what I can tell, each face of the cube is replaced by two parallel squares of wood, and the squares are joined together in such a way that the lower square of one face is joined to the upper square of the adjacent face

hi @AkivaWeinberger

The shapes on the faces are chiral, and the shapes of the upper squares is the mirror image of the shapes of the lower squares, if that makes sense

Cool arrangement

@LeakyNun Hi

how's it going

Good

What's the topology of that image, viewed as a 2D surface with boundary?

How many boundaries does it have, first of all

I don't even know what that thing is...

Oh wow it looks like six

One per vertex, and each boundary is a trefoil knot

Whoa

@LeakyNun It's some wood arranged in a weird shape

hmm I see it

interesting

but it still breaks my mind :P

Yeah there's a full boundary visible in the photo

I think there's 8 of them though

Trefoil

@AlessandroCodenotti Oh right yeah

A cube has 8 vertices

Derp

Yes that's the one I was talking about in the photo

We can also think of it as a polygonal surface

with twelve faces

I'm just wondering what it's homology would be

So that's like a Seifert surface for 8 unlinked treefoils

96 edges?

Ah no wait, they are actually linked

And 144 vertices

@AlessandroCodenotti Oh yeah good catch

hi there , can anyone recommend me a textbook where transferring norms and metrics from R^(mn) to L(Rn,Rm) ( vector space of linear maps ) for differentiability of higher order is well-explained ?

OK so V-E+F=144-96+12=60

Hm wait how do you find the genus of a surface with boundary

No idea

I feel like I counted it wrong

Oh yeah I messed up on the edges

120 edges

And actually only 96 vertices

so V-E+F=96-120+12=-12

The Euler characteristic just drops by 1 for each boundary component you add, so take your Euler characteristic, add # of boundary components, and that's 2-2g for the desired g

and then add the eight boundary components ('cause I think that's what you do) and I end up with $2-2g=-12+8=-4$

Thanks Mike

and thus $g=3$

So this is topologically a genus-3 surface with eight holes punched out

Nice

What does that mean for its homology? ('Cause Leaky asked)

Homology of something with boundary is zero in top degree so there's only something to say about $H_1$

Then using a presentation of a surface as the gluing of sides of a polygon, you see that in fact $\pi_1 \Sigma_{g,n} = F_{2g+n-1}$ for $n > 0$ the number of punctures.

F being the free group

@MikeMiller does Stokes theorem have any implication regarding the de Rham cohomology?

$H_1$ is then clearly free of rank $2g+n-1$

Yes, Stokes is what says the de Rham isomorphism between differential form cohomology and singular cohomology is true

At least, that there is a chain-level pairing

aha

So $\Bbb Z^{13}$

so we want $\Omega^k(M) \times C_k(M;\Bbb R) \to \Bbb R$ sending $(\omega, f) \mapsto \int_{f(\Delta^n)} \omega$?

You could probably repeat this to make a three-layered version

The faces have the shape they do 'cause there's only two levels. If there were three levels, the middle shape would look like a hash symbol

("Lower" = "inner", "higher" = "outer")

I'm not actually sure that this wouldn't intersect itself

Nah it won't

It'll be fine

Oh wow there's more

That's the same surface but it's clearer what's going on

Seifert surfaces are weird

(Surfaces with knotted and/or linked boundaries like that one)

I remember the first time someone told me about Seifert surfaces (either Mike or Balarka) it took me a while to convince myself the Hopf link can be the boundary of a surface without self-intersections