Mathematics - 2020-12-13

12:42 AM

2

Q: What are preferences and preference functions in multi-objective reinforcement learning?

In RL (reinforcement learning) or MARL (multi-agent reinforcement learning), we have the usual tuple: (state, action, transition_probabilities, reward, next_state) In MORL (multi-objective reinforcement learning), we have two more additions to the tuple, namely, "preferences" and "preference f...

copper.hat

12:55 AM

@traducerad still lots of obvious things waiting to be discovered :-).

robjohn

$\left(\!{n\atop k}\!\right)$$\binom{n}{k}$$n\choose k$

The first is a bit smaller, but is a bit more complicated.

porridgemathematics

9:02 AM

does this reasoning make sense? $S^1 x [0,1)$ is homeomorphic to the open unit disk via $(z,r) -> rz$,$[0,1)$ is homeomorphic to $[0,+\infty)$, so $S^1 x [0,1)$ is homeomorphic to $S^1 x [0,+\infty)$, and this is homeomorphic to $C$ via $(z,r) -> rz$, so $C$ is homeomorphic to the open unit disk?

whoops its formatted badly $S^1 \times [0,1)$ is homeomorphic to the open unit disk via $(z,r) -> rz$,$[0,1)$ is homeomorphic to$ [0,+\infty)$, so$ S^1 \times [0,1)$ is homeomorphic to $S^1 x\times [0,+\infty)$, and this is homeomorphic to $\mathbb{C}$ via $(z,r) -> rz$, so $\mathbb{C}$ is homeomorphic to the open unit disk?

*$S^1 \times [0,+\infty)$

Balarka Sen

S^1 x [0, 1) -> D, (z, r) |-> zr is not a homeomorphism. For example, (z, 0) and (w, 0) are mapped to the same point, 0.

porridgemathematics

ah crap

is it even homeomorphic to the open unit disk?

Balarka Sen

Nope :)

Astyx

it has a hole

porridgemathematics

but i think the $[0,1) \leftrightarrow [0,+\infty)$ perhaps is useful ?

to find a homeomorphism

or maybe not

Balarka Sen

9:09 AM

Your ideas are sound, it needs a little polishing.

porridgemathematics

ok ill try to rethink how i put this together, thank you for spotting this!

Astyx

What are you trying to find/prove ?

porridgemathematics

that the open unit disk is homeomorphic to the plane

would $D \rightarrow \mathbb{C}$ given by $r exp(i \theta) \rightarrow h(r) exp(i \theta)$ work? where $h : [0,1) \rightarrow [0,\infty)$ is some homeomorphism?

the original idea i was going for was 'scale things near the boundary more'

Balarka Sen

Yes, that works, @porridgemathematics.

porridgemathematics

oh cool

lyme

10:03 AM

hello, is it true to write $\int_0^x\int_0^x\int_0^x(x-t)^3f(t)dtdtdt$= $\frac12\int_0^x(x-t)^5f(t)dt$?

Leonhard Euler

11:41 AM

which mathematical journal has the highest impact factor?

mathguy

12:22 PM

Can someone answer this:

0

Q: Closed form of the sum $\sum_{n=1}^{\infty}\frac{H_n}{n^x}$

Some days ago I derived the identity $$\sum_{n=1}^{\infty}\frac{H_n}{n^2}=2\zeta(3)$$ where $H_n$ is the $n$th Harmonic number. Other related identities include $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}$$ $$\sum_{n=1}^{\infty}\frac{H_n}{n^4}=\frac{-1}{6}\pi^2\zeta(3)+3\zeta(5)$$ $$\su...

sequences-and-series summation closed-form harmonic-numbers digamma-function

123

12:38 PM

Hi All...

Tobias Kildetoft

1:32 PM

@LeonhardEuler I think either annals or JAMS

RewCie

Hey Russian

Georg Friedrich Bernhard Riema

2:30 PM

Riemann is here

Had to name myself Riema because of name copyright issues

2

nbro

3:41 PM

1

Q: How are these equations of SGD with momentum equivalent?

I know this question may be so silly, but I can not prove it. In Stanford slide (page 17), they define the formula of SGD with momentum like this: $$ v_{t}=\rho v_{t-1}+\nabla f(x_{t-1}) \\ x_{t}=x_{t-1}-\alpha v_{t}, $$ where: $v_{t+1}$ is the momentum value $\rho$ is a friction, let say it's...

deep-learning comparison optimization stochastic-gradient-descent momentum

schn

4:03 PM

I have an index $kk'$, and I'd like to specify that it corresponds to the sets "$n$ choose $k$" from $\{1,...,n\}$. Is there a compact way of expressing this, in words or in symbols?

Also that $k<k'$.

"$n$ choose 2" of course.

Astyx

I don't understand what you mean

schn

For $\{1,2,3\}$, one has $\{1,2\}, \{1,3\}$ and $\{2,3\}$. Then $k\in\{1,2\}$ and $k'\in\{2,3\}$.

I'd like the index $kk'$ denote the elements of the sets, in order.

Astyx

Somethin like $\sum_{k=1}^n\sum_{k'=k+1}^n \dots$

Mike Miller

I'm not sure what you're indexing but depending on context I would probably write something like $(a_{ij})_{1 \leq i < j \leq n}$. I understand that's ordered indices but it's cleaner than $(a_{\{i,j\}})_{1\leq i < j \leq n}$.

You could of course write $a_{\{i,j\}}$ if you really wanted to.

schn

I have a collection of matrices $R_{ij}^{x1x2}$, where the upper index denotes the $x1x2$-plane.

Mike Miller

4:14 PM

The nested indices are pretty nasty notation, I'd do $R^{mn}_{ij}$ or something like that, where the upper indices indicate you're working in the plane spanned by $e_m, e_n$.

If $n$ is taken make that $R^{km}_{ij}$ or something

schn

Right. $x_k$ is always the dependent variables of $x_1,x_2,...,x_{k-1}$, so one is never interested in the plane $x_kx_{k-1}$, although it's the same as the $x_{k-1}x_k$-plane.

For instance, $xy$-plane kind of implies one studies $y(x)$.

Mike Miller

I probably don't agree, but I can believe that's convention in your application.

Balarka Sen

@MikeMiller Literally the capital of Iceland

schn

@MikeMiller How would one then specify that there are "$n$ choose 2" of $e_m, e_n$?

Edward Evans

lool

Astyx

4:21 PM

haha

Edward Evans

pick one

Astyx

Three Lemmas for the Elven-kings under the sky,
Seven for the Dwarf-lords in their halls of stone,
Nine for Mortal Men doomed to die,
One for the Dark Lord on his dark throne
In the Land of Mordor where the Shadows lie.
One Lemma to rule them all, One Lemma to find them,
One Lemma to bring them all and in the darkness bind them
In the Land of Mordor where the Shadows lie.

Mike Miller

@schn There are as many 2-elements subsets of [n] as there are ordered pairs (i,j) with i < j.

So specifying that 1 <= k < m <= n says that there are C(n,2) choices of index k,m

Balarka Sen

this is random but why are there so many dipshits who are really really into black metal, Tolkien and Dungeons&Dragons

Edward Evans

don't you mean "kings"

Astyx

4:30 PM

I think he means dipshits

Edward Evans

@Astyx stronk

nah he means kings and if you say otherwise I'll get my medieval flail and swing it at you in the cold north of Norway while wearing my leather pants and bullet belt

Balarka Sen

i have neo Nazis in mind tbh lmao

schn

@MikeMiller Thanks a bunch!

Edward Evans

NSBM könige

Balarka Sen

yeah lol

Edward Evans

4:32 PM

nah that's bullshit

I was at a Mayhem gig in Mannheim and I felt somewhat shunned because of my leather jacket and bald white head

Balarka Sen

huh

Edward Evans

as in, NSBM is bullshit and NSBM fans should be shunned, but I was shunned for my appearance

which is ironically quite an NSBM attitude

Balarka Sen

lol

one of my favorite bands, Drudkh, i found out much later, is borderline NSBM

Edward Evans

Oh really

I like Drudkh too

Balarka Sen

yeah

Edward Evans

4:35 PM

I'm always wary when listening to German BM

Balarka Sen

otoh there's a lecture by Fenriz in youtube where he frequently says Latin American BM has been a strong influence for the Norwegian scene

Edward Evans

Fenriz is a king

Balarka Sen

'gree

Astyx

Con Fenriz y darkthrone, norwegian reggaeton

Edward Evans

Fenriz Navidad

I love him in Until the light takes us getting busted on a train for having tear gas or smth in his bag

Balarka Sen

4:43 PM

lool

schn

@MikeMiller Maybe a super basic question, but given the $n\times n$ matrix $R_{ij}$, how does one compactly specify the indices $i,j$, where each can range from $1,...,n$.

Is there a way to store a collection of matrices, like one stores a collection of vectors in a matrix?

geocalc33

5:03 PM

what did the igneous rock say to the granite Silverstone slab?

"I'm a rock"

schn

What kind of tensor is a vector with $> 3$ components?

geocalc33

why can't anyone invent an anti-microwave?

user 6232128

5:21 PM

A macro-wave?

geocalc33

a microwave=a thing to heat food up

user 6232128

so a freezer is an anti-microwave

geocalc33

but if I have a hot plate of food, say I want to cool it down in 30 seconds or less

a freezer can't do that

say I want to cool it to 40 degrees in 30 seconds

and say it initially was at 80 degrees

robjohn

@geocalc33 complain to the Law of Thermodynamics Department.

Astyx

put it in a very cold freezer?

schn

5:38 PM

What's the notation of the elementwise difference of the components of a vector?

Sha Vuklia

Guys, for $M$ a smooth manifold, an inward pointing vector $v\in T_p M$, where $p\in\partial M$ is defined to be one where there exists a smooth curve $\gamma\colon [0,\epsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$. An outward point vector is one where we have $\gamma\colon (-espilon,0]\to M$ with these properties.

Now, I can see that any vector in $v\in T_p M\setminus T_p\partial M$ has at least one of two curves (by considering boundary coordinates charts), but I don't see why a vector can't have both. So, why can't there be coordinate charts $(x^i)$ where the $n$-th component of $v$ is positive, and other charts $(y^i)$ where the $n$-th component is negative?

Astyx

If you take a coordinate chart surrounding your point, the manifold there looks like half of R^n

And in those coordinates, an inward/outward vector is a vector in that/out of that half of R^n

If you're both, it means you're in the separating R^{n-1} hyperplane, ie the boundary

Sha Vuklia

Ye, I think I see ittt

Mike Miller

@ShaVuklia In a boundary chart $p \in U \cong [0,\infty) \times \Bbb R^{n-1}$ (with $p$ send to $0$), $T_p \partial M \cong 0 \times \Bbb R^{n-1}\subset \Bbb R^n$. Then if you have a smooth curve $\gamma: [0, \epsilon) \to [0,\infty) \times \Bbb R^{n-1}$ with $\gamma(0) = 0$, writting $\gamma_1(t)$ for the first component, we have $\gamma_1(t) \geq \gamma_1(0) = 0$ for all $t \geq 0$. Now taking a derivative from above we get $\gamma_1'(0) \geq 0$

Sha Vuklia

If we were to have $\gamma_1\colon [0,\epsilon)\to M$ and $\gamma_2\colon (-\epsilon,0]\to M$, then we could choose the same boundary charts for both $\gamma_1$ and $\gamma_2$. Since $\gamma_1'^n(0)=v_n=\gamma_2'^n(0)$, we would get a contradiction with the upper half plane.

Yea, thanks. Sorry, I was too tired to realise that giving this one more sec would yield me the solution.

Mike Miller

5:50 PM

Right. This is a nice description of the "tangent half-space", and the fact that the tangent half-space isn't the whole tangent space gives a proof that a manifold-with-boundary is not diffeomorphic to a manifold-without-boundary

Sha Vuklia

I can't seem to think of a definition for the tangent half-space x')

Everything minus the outward pointing vectors?

@MikeMiller I feel like this can be argued easily without talking about inward/outward pointing vectors, since we can't have a diffeo between $\mathbb H^n$ and $\mathbb R^n$ anyways.

user 6232128

@robjohn your webinar link to the UCLA law department discussion inspired my avatar.

Mike Miller

Why can't we?

Thorgott

If I have a 1-manifold (potentially with boundary) and a connected chart thereof, why does the boundary of the chart contain at most two points? This is seemingly self-evident and I believe I do have a proof, but my proof is so ugly that I feel like I'm missing an easy argument.

Sha Vuklia

@MikeMiller Oh oops... wait, let me check my statement that I have absolutely no justification for atm.

Mike Miller

6:00 PM

@Thorgott Connected subsets of R are intervals?

Thorgott

yeah, that should be the idea, but how to turn this into a quick argument?

Mike Miller

I don't understand your question. That's a proof

Thorgott

I don't see the proof

Mike Miller

The intermediate value theorem says that a connected subset of $\Bbb R$ containing $a$ and $b$ also contains $[a,b]$

Open intervals are the union of their closed subintervals

Thorgott

I know that connected subsets of R are intervals, I don't see how that proves my inquiry

Mike Miller

6:04 PM

I'm so lost. What does "boundary of the chart" mean that this isn't self-evident?

Thorgott

I have a 1-manifold $M$ and an open connected subset $I\subseteq M$ homeomorphic to an interval in $\mathbb{R}$. Why does $\partial I=\operatorname{cl}(I)\setminus I$, the closure taken in $M$, contain at most two points?

Mike Miller

OK, a chart is a map (so I was imagining you were asking about a subset of $\Bbb R$)

I get your question now

You have to use Hausdorffness here in an essential way

Thorgott

It's true for intervals in $\mathbb{R}$, but the chart doesn't necessarily extend to a homeo between the ambient closures

geocalc33

@robjohn can you help me with the asymptotic expansion of $H(x)=\int_0^x \exp\big(\frac{1}{\log(t)}\big) ~dt$?

Mike Miller

Right

Pick points $x, y$ in the boundary and pick disjoint opens $x \in U, y \in V$

Then $U \cap I$ and $V \cap I$ are disjoint open subsets of $I$ which contain sequences with no subsequence converging in $I$

If $I$ is a half-open interval then this is a contradiction and you see there could only be one point in the boundary. If $I \cong (-1,1)$ then it follows WLOG that $(0,a) \subset U \cap I$ for some $a$ and $(b, 1) \subset V \cap I$ for some $b$ (otherwise, negate your homeomorphism; the point is they contain opposite outer intervals)

it follows that you can't have three points in the boundary, because $(0,1)$ only contains two outer intervals

Hausdorffness is necessary because you could take the line with two origins and the interval $(0,1)$

Sha Vuklia

6:11 PM

@MikeMiller Oh, apparently we can't even have a homeomorphism. By removing $x_n=0$ from both spaces, $\mathbb H$ yields a connected space, while $\mathbb R^n$ doesn't.

Mike Miller

And it's used here to show that a point in the boundary corresponds to an open end of $I$

@ShaVuklia I disagree for $n>1$! The plane without the origin is certainly connected.

Sha Vuklia

It's not just the origin, but all points with $x_n=0$, right?

Mike Miller

Homeomorphisms need not preserve that condition :)

user2103480

@EdwardEvans metal nerds are probably the worst in terms of categorising every millimeter of their musical spectrum, but it still speaks volumes that there is a need to have a subgenre defined by proximity to national socialism

Mike Miller

Call that say $P$ for (hyper)plane; then $\varphi: \Bbb H^n \to \Bbb R^n$ need not send $\varphi(P)$ to a hyperplane. It is sent to a subset which you know (at least) is a closed subset homeomorphic to the hyperplane. But who's to say why that should separate $\Bbb R^n$?

It is actually hard to show that $\Bbb H^n$ and $\Bbb R^n$ are not homeomorphic. You want to use either homology or homotopy groups.

Edward Evans

6:15 PM

@user2103480 your country's fault it exists tho

weeeyyy

user2103480

And norway

Thorgott

@MikeMiller Why can't both of them contain points arbitrarily close to both ends?

Edward Evans

@user2103480 fair

Mike Miller

@Thorgott Intermediate value theorem!

You contain a descending sequence $x_n \to 0$, so you contain $\bigcup_{n > 1} (x_n, x_1) = (0, x_1)$

Thorgott

@MikeMiller boundary points in H^n have contractible punctured neighborhoods, points in R^n don't, what am I overlooking?

Lukas Heger

6:21 PM

@ShaVuklia your approach with removing a hyperplane can be made to work, but it's not easy: you need the Jordan Brouwer separation theorem whose proof is not easy

Mike Miller

@Thorgott Proving that those punctured neighborhoods are not contractible

That's equivalent to showing $\pi_n S^n \neq 0$

Sha Vuklia

Right, well, I found the argument, but my homotopy theory is rusty enough for me to be content with just understanding it vaguely for now.

Mike Miller

@ShaVuklia The point is that to do it up to homeomorphism is actually very difficult. But you have just discovered a proof smoothly :)

Sha Vuklia

@MikeMiller Well, I still am not sure what you meant by tangent half-space, I'm afraid. Was it all vectors minus the outward pointing ones?

Mike Miller

Let me choose a notation. Write $C_x(M) = \{v \in T_x M \mid \exists \gamma: [0,\epsilon) \to M, \gamma'(0) = v\}$

Thorgott

6:25 PM

right, I guess that takes some work

user2103480

@EdwardEvans btw, since it may be "your country" as well some day (do you need to apply for visa now?), I feel compelled to show you another east-german music gem

Mike Miller

(This set makes sense in a very general context, much more general than manifolds, and can be called a tangent cone, whence my notation)

Thorgott

though you can just work it down to Homotopy invariance for integration and writing down a volume form on $S^n$

Sha Vuklia

Oii, a cone, is it categorical :D

Edward Evans

@user2103480 I don't think I need a visa for now, but I'm probably going to be applying for citizenship once I hit the 6 year mark

Mike Miller

6:26 PM

Then any diffeomorphism $\varphi: M \to M'$ has $d\varphi_x(C_x(M)) = C_{\varphi(x)}(M')$

Edward Evans

so go ahead

Astyx

I'm confused, are you talking about nazism or black metal?

Mike Miller

No, it means cone like the picture of cone you are used to

Generally all you can say about $C_x$ is that it is closed under scalar multiplication

For a figure-eight at the origin $C_x$ is two lines

So my point above is that because $C_0(\Bbb H^n)$ is a half-space, but $C_p(\Bbb R^n) = \Bbb R^n$ for all $n$, they can't be diffeomorphic

user2103480

hahaha the video is not safe for work enough for this chat

user 6232128

lol

Edward Evans

6:29 PM

@Astyx b o t h

Sha Vuklia

@MikeMiller Sry if I'm slow, but... aren't we again relying on the fact that half-space isn't diffeomorphic to $\mathbb R^n$?

user 6232128

@EdwardEvans :o

Mike Miller

Np, the point is that $C_0(\Bbb H^n) \subset T_0(\Bbb H^n)$ is a proper subspace

$d\varphi_x$ is an isomorphism from $T_0 \Bbb H^n$ to $T_{f(0)} \Bbb R^n$

user2103480

@EdwardEvans I hope you caught the link and that your german is good enough to appreciate this masterpiece

Mike Miller

So $d\varphi_x(C_0(\Bbb H^n))$ is still a proper subset

(In fact, still a half-space; a linear map sends half-spaces to half-spaces)

Edward Evans

6:32 PM

@user2103480 I didn't catch the link

Mike Miller

The trick is that we are reducing to linear maps by taking the derivative

user2103480

one more time

Sha Vuklia

Ahh, right, I think I understand it:0 I might have to reread some lines to be sure.

Mike Miller

I did not explain it as cleanly as I could have I think

Let me try one more way which I think might make everything much clearer

Edward Evans

fuck

I wasn't paying attention lmfaoooo

user2103480

6:33 PM

lmao

Edward Evans

juan more time

okay got it

user 6232128

lol

user2103480

without the thumbnail it's much friendlier

Mike Miller

Say a curve $\gamma: [0, \epsilon) \to M$ is strictly inward pointing if it doesn't extend to a smooth curve $\tilde \gamma: (-\epsilon, \epsilon) \to M$

Then $\Bbb R^n$ does not contain any strictly inward pointing curves

Edward Evans

@user2103480 loool loving it

Sha Vuklia

6:35 PM

By definition right? Because being smooth on $[0,\epsilon)$ means there is a smooth extension?

Mike Miller

However $\Bbb H^n$ does --- it's the curves with $\gamma_1(0) = 0$ and $\gamma_1'(0) > 0$

Right!

user2103480

@EdwardEvans they just rip apart that east german village nazi stereotype

Mike Miller

And "smooth map to $\Bbb H^n$" means "smooth map when considered as a map to $\Bbb R^n$"

So when we say $[0,\epsilon) \to \Bbb H^n$ is smooth, the point is that there is an extension $(-\epsilon, \epsilon) \to \Bbb R^n$

Not necessarily to $\Bbb H^n$ again

Edward Evans

@user2103480 ahhh I'm not so familiar with the stereotype, I don't know much about east Germany

Sha Vuklia

Ah, okay, didn't know that (/hadn't thought about it), but that's fair

Mike Miller

6:40 PM

Totally reasonable

Thorgott

@MikeMiller I don't follow this. Are you assuming the subsets are connected or which IVT are you applying here?

Sha Vuklia

Ah ok, so I believe the argument then is; say we have a strictly inward pointing curve $\gamma\colon [0,\epsilon)\to M$, then we would be able to extend $\phi\circ\gamma$, and then we just apply $\phi^{-1}$ to get an extension of $\gamma$.

user2103480

@EdwardEvans the not so short tl;dr is that in the GDR, they rigorously replaced former nazis in the government (more than west germany!), but did not culturally reflect on the nazi past because "that wasn't us". Now combine that with greater poverty, a general feeling of "we don't matter to the government and culture" and a way less heterogeneous population, and you get the nationalism problem and GDR-nostalgia which is referred to

"Onkelz poster" refers to the band "Böhse Onkelz", which did far right rock in their starting times, but since then went mainstream and condemn their past

Mike Miller

To be honest this is strictly more irritating than just proving the classification theorem

Edward Evans

@user2103480 interesting cultural tidbit, thanks

Mike Miller

6:46 PM

But I'll start again

(And be more careful in stating assumptions etc; I agree what I said wasn't complete)

@Thorgott There probably is a concise argument but I'm finding it very irritating to construct. The point is that if you paste two intervals along each other in any way other than very carefully you get something non-Hausdorff.

Thorgott

yeah, I'm also finding this very irritating

Mike Miller

It follows from the Hausdorff hypothesis that $U \cap I$ is either exactly one or exactly two components, and both are outer in $(0,1)$

Thorgott

which is in itself irritating, this should be clear

Mike Miller

Look at Gale's article on classifying 1-manifolds

The first Lemma there is what you want

It's not phrased the way you want but it gives what you need

user16319

Suppose again that A is a domain, with fraction K, and let B be an A-algebra
such that K \otimes B is a domain. Must B be a domain ? I don't think so; but im having trouble coming up with a counter example.

anyone has any ideas

Lukas Heger

7:01 PM

@user16319 consider $A=\Bbb Z$, $K=\Bbb Q$ and $B=\Bbb Z \times \Bbb Z/2\Bbb Z$

$K \otimes B \cong \Bbb Q$, but $B$ is not a domain

Edward Evans

7:16 PM

@Lukas Vogel just gave us an exercise about two friends, Donald and Kim. Kim is playing with toy soldiers and Donald hates the idea that Kim might have more toy soldiers than him. But unfortunately, the surveillance media can't count Kim's soldiers exactly!

Lukas Heger

lol

I remember we had such an exercise in Algebra 1 by Vogel

it was probably easier than an ANT 2 question though

Edward Evans

It's a CRT exercise lol

Lukas Heger

lol

then it's the same I guess

you do CRT exercises in ANT 2?

Edward Evans

Apparently so rofl

nahhhhhhhhhhhh man

I just noticed, it was linked to the ANT2 page on Mampf hahaha, damn I was like "nice free points"

how silly of me

Lukas Heger

ah yeah Mampf has links across different lectures

Edward Evans

7:19 PM

I thought it was kinda suspicious

CRT exercise in ANT2

EM4

7:57 PM

quick question https://math.stackexchange.com/questions/731699/find-order-of-given-factor-group

how did they get cyclic subrgoup of H.

robjohn

@user6232128 Yes, I noticed your avatar earlier this morning.

Thorgott

I don't understand your question

user16319

@LukasHeger "$A=\Bbb Z$, $K=\Bbb Q$ and $B=\Bbb Z \times \Bbb Z/2\Bbb Z$

$K \otimes B \cong \Bbb Q$" Why is K \otimes B iso to Q ?

Lukas Heger

tensor commutes with direct sums, so $\Bbb Q \otimes B \cong \Bbb Q \otimes \Bbb Z \times \Bbb Q \otimes\Bbb Z/2\Bbb Z =\Bbb Q$, because $\Bbb Q \otimes \Bbb Z/2\Bbb Z=0$

Thorgott

$\mathbb{Q}\otimes(\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z})\cong(\mathbb{Q}\otimes\mathbb{Z})\oplus(\mathbb{Q}\otimes\mathbb{Z}/2\mathbb{Z})\cong\mathbb{Q}\oplus\mathbb{Q}/2\mathbb{Q}\cong\mathbb{Q}$

user16319

8:04 PM

@LukasHeger ahh thanks. Sorry, i got confused over the direct product. But now everything is clear

Shaun

8:34 PM

Hi all, I just want to advertise a bounty. Here:

0

Q: Does there exist a topos with these $n+2$ truth values?

This question is based on the answers to this question. The Question: Let $n\in\Bbb N$. Let $N$ be a set with $n+2$ elements, labelled $0$ to $n$, and the $(n+2)$th element labelled $\infty$. Suppose we have a function $$\begin{align} t:N &\to N,\\ 0 &\mapsto 0,\\ m &\mapsto m-1,\quad\text{(for }...

logic dynamical-systems philosophy topos-theory intuitionistic-logic

schn

Suppose one has $m$ $n\times n$ matrices, then what kind of tensor is that?

Ted Shifrin

9:20 PM

@schn You need to impose more structure than just to say one has $m$ of something. What is a set of $m$ vectors?

love_sodam

Let $\gamma:[a,b]\to S$ is a unit speed curve in a regular surface and let the curve is injective. I already know the existence of parallel vector field along $\gamma$ satisfying some initial condition. Is there a parallel vector field that is linearly independent to $\gamma'(t)$ for all $t$?

schn

@TedShifrin If the vectors have $n$ elements, then it is an $n\times n$ matrix, no?

Ted Shifrin

You mean $m\times n$? No, you can associate it to one, but then I need an ordering, for one thing.

@love_sodam: What does that even mean? In principle, you can't compare tangent vectors at different points. That's the whole point of a connection and parallel translation.

schn

I have the matrices $A, ...,M$ all of which have $n$ rows and $n$ column.

love_sodam

@TedShifrin I mean for parallel vector field along $\gamma$, $w(t)$, $w(t)$ and $\gamma'(t)$ are independent for all $t$

Ted Shifrin

9:24 PM

So you want to stick them in a 3-dimensional array and call it a tensor. This is a popular game people play when they don't understand what tensors are. But you can play the game and say it's a tensor of rank $3$.

schn

Well what is it really then? :)

Thorgott

it's a collection of matrices

you can identify that with a tensor, but that's not meaningful in and of itself

Ted Shifrin

Ah, ok, @love_sodam. What if $\gamma$ is a geodesic to start with? Then you can choose $w$ to be a different vector at $t=t_0$ and you'll be OK. However, if $\gamma$ is not a geodesic, parallel translating $w$ will make it turn and it may well line up with $\gamma'(t)$ at certain values of $t$. You know examples.

schn

Alright.

Makes sense.

Tensors are more physics related then?

Maybe...

Ted Shifrin

No, they are mathematical in ways physicists don't think about them at all.

Boil your question down to the simplest case. Take $n$ real numbers. You can associate that list (ordered set) to a vector in $\Bbb R^n$. But vectors have lots more structure to them — like the vector space axioms. You're choosing to forget all that when you call your collection of numbers a vector.

robjohn

9:29 PM

@geocalc33 Is $x\le1$? I don't think the integral converges near $x=1^+$.

Ted Shifrin

@robjohn Have you had a response to that law of cosines post? I forgot to check today.

Thorgott

The question is ultimately why you have a collection of $m$ $n\times n$ matrices lying around and what you want to do with them. That should dictate how you think about the collection.

robjohn

@TedShifrin There are some upvotes. The OP had some concern that I replied to, but I don't know if they're satisfied yet.

Ted Shifrin

Ah, I should look.

robjohn

@TedShifrin I have updated the answer to be a bit better, I hope.

schn

9:32 PM

What is an example of what one chooses to forget when calling a collection of numbers a vector?

Ted Shifrin

Start with scalar multiplication and addition.

schn

If I put $n$ numbers in a vector, then these operations are respected, or?

Ted Shifrin

There's no meaning to those operations.

schn

Alright, but there was no meaning BEFORE one put the numbers in the vector, or?

Ted Shifrin

Right.

schn

9:35 PM

So what's forgotten?

Ted Shifrin

This is the point that Thor and I am making about how people abuse meaning/notation.

I'm saying that once you encode this information in a vector, it acquires algebraic/geometric properties that the original list can't make heads or tails of.

I mean, if all you want to do is store numbers in an array, call it an array. Don't give it an algebraic/geometric form.

schn

I see.

love_sodam

@TedShifrin Actually the original question (in this strange book) is for that unit speed $\gamma$, if $\gamma$ is injective, then there exists a single surface patch that covers the trace of $\gamma$. This problem exercise didn't mention parallel transport or parallel vector field actually. They just talk about some basic properties of geodesic in that section

Ted Shifrin

Oh, this is the idea I used in constructing that counterexample for you.

Just go reread that. (By the way, the author finally responded to my email. He sheepishly says he will get around to starting an errata sheet for his book this spring or summer. LOL. Maybe he doesn't appreciate us.)

schn

Out of curiosity, is an array something defined in mathematics? If one has a list of numbers, what is it one...has?

Ted Shifrin

9:40 PM

It's just a list.

schn

A set?

Ted Shifrin

Well, what if my list is 1,2,2. Is that $\{1,2,2\}$?

Lukas Heger

formally, a list is an element of a cartesian product

schn

No.

Ted Shifrin

glad to pass the mantle to Lukas

Karim Mansour

9:47 PM

any small category can be embedded in some larger category right maybe category of functors?

Lukas Heger

you can always take the Yoneda embedding

which goes from $C$ to $[C^{op},\mathbf{Set}]$

Karim Mansour

I don't know Yoneda embedding but I know the following

how does it go it goes from $\mathcal{C}$ into set by taking Hom sets?

ok just searched google yeah

Lukas Heger

The Yoneda embedding takes $c \in C$ to the contravariant Hom functor $\mathrm{Hom}_{C}(-,c):C^{op} \to \mathbf{Set}$

the Yoneda lemma says that this is fully faithful

Karim Mansour

cool

Thorgott

well, if you just want to embed a small category into some larger category, just take disjoint union with the trivial category

Lukas Heger

9:51 PM

that's true of course

Karim Mansour

I see.

love_sodam

@TedShifrin I read it but still I can't see

Ted Shifrin

We actually constructed a chart whose coordinate curves were geodesics.

What can't you see?

In this case, the original $\gamma$ curve is not, but you still use geodesics perpendicular to it, say.

Mike Miller

@Thorgott Did my reference satisfy you

Ted Shifrin

Howdy, @MikeM.

Mike Miller

9:59 PM

Hello

love_sodam

@TedShifrin Geodesic perpendicular to $\gamma$ and geodesic parallel to $\gamma$ you mean?

Ted Shifrin

I don't know if you know Hirsch better than I do. I was shocked to discover (perhaps wrongly?) that he never proves the transversality extension theorem (if you have a map that's already transverse to $Z$ on a closed subset $C$, then you can homotop slightly to a map that's everywhere transverse to $Z$). Someone asked about essentially this on main. @MikeM

Well, I don't know what parallel to $\gamma$ means, @love_sodam. But fix one geodesic perpendicular to $\gamma$ and follow geodesics perpendicular to it for a while.

love_sodam

You mean like tubular neighborhood?

Ted Shifrin

Remember: I'm doing this by using the ODE result that you always have coordinates (locally) if you have two everywhere linearly-independent vector fields and the coordinate curves are tangent to them.

Thorgott

@Mike I haven't looked at it yet, still trying to figure this stuff out myself (though with as of now little success)

Mike Miller

10:12 PM

@Thorgott It's not a proof, it's a guided exercise sheet

Amin Idelhaj

10:23 PM

Hey guys

Astyx

hi nerd

Ted Shifrin

Hi Demonark and Astyx.

Astyx

hi Ted

Amin Idelhaj

The pot calling the kettle nerd I see

But yeah what's up with you guys?

Astyx

Feel like I'm behind schedule with regards to my maths courses, and lockdown isn't helping

Ted Shifrin

10:37 PM

Why behind schedule?

Alessandro Codenotti

Hi Dami

Ted Shifrin

Hi, demonic Alessandro.

Astyx

I was lacking in algebra, and by the time I "caught up" I've been struggling to keep up with my courses.

Hi Alessandro

I've once again been too optimistic with my choice of courses as well

Ted Shifrin

Well, I guess this is because you dabbled in physics/engineering at the beginning.

Astyx

Yup, that's probably not helping

The good news is I'll have an engineering degree at the end of this year

Ted Shifrin

10:41 PM

Oh, you're still doing it. I didn't remember that.

I don't think you should fret.

Astyx

Well, I'm doing pure maths, but as a concluding applied (lol) year in a general engineering curiculum

Amin Idelhaj

Astyx that's a mood

It's a bit rough not doing math with others

I think I get focus from being around people

Alessandro Codenotti

I always went to the library to study because I'm way more productive than at home

Astyx

It probably is. I'm worried because I'm to start my PhD next september, in continuation to my masters thesis which starts in January, and I'm feel like I'm absolutely not up to level

Amin Idelhaj

Things can change quickly. If you're really not sure do you think you can take a gap year or smth and try to get up to speed?

Astyx

10:50 PM

I'm wondering that as well. I'll ask around.

Ted Shifrin

Wow, all the kidlets are growing up :D

Sha Vuklia

11:14 PM

@AminIdelhaj I mean, there's still ways to do maths together?

I often work with people on Discord

It's pretty chill

@TedShifrin Lol, I do feel like I partly grew up here mathematically speaking

@Astyx PhD in what:0

Amin Idelhaj

In being a nerd probably

Sha Vuklia

x''D

Amin Idelhaj

@ShaVuklia tru but I guess what I have in mind is more like, idk my pset sessions with friends when we would colonize a classroom or the lounge and just kinda make dinosaur noises until we had some ideas for pset problems and try them out

Occasionally someone makes a joke or something, if we go late we order Subway

That sorta dynamic can't be recreated over Zoom you know?

Sha Vuklia

Ye, that's true, but honestly, I sometimes get close with Discord sessions. I have a group of friends/fellow-students who like me usually start working on their homework the day before the deadline, so sometimes we'll be working together on Discord for like 12+ hours straight, which does yield fertile soil for memes. But maybe I'm more appreciative of a simpler social life than others, so I do understand some struggle more than I do probably

geocalc33

can't you still make dinosaur noises over zoom

but you can't really capture the experience I see what you're saying

Ted Shifrin

11:42 PM

I would miss the mobs of students working together in my office hours ...

geocalc33

did your office have the fungshwei?

*feng shui

what makes a great office?

Amin Idelhaj

Hagoromo

geocalc33

In my opinion what makes an office "great" is this: (1) desk (2) chair (3) space (4) some decoration. All of the highest caliber

Alessandro Codenotti

What makes my office great is proximity to free coffee. Also the view is nice being on the 8th (and last) floor, but coffee

geocalc33

black board (fairly large)

chalk (the good kind)

yes definitely access to good coffee

schn

11:56 PM

Suppose I have a Gaussian function defined over the unit square. If it is rotated, should one expect the integral of the rotated version versus the non-rotated version, integrated over the unit square, to change?

geocalc33

this is my cousin's office lol

Thorgott

the Gaussians I know are rotationally symmetric

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$Mathematics$ Mathematics