prepares to get slapped

Easier just to put you on ignore.

The differential is a functor from the category of smooth manifolds and the category of smooth vector bundles!

You're just trying to record the fibers without recording the consistency between them it seems

Doesn't strike me as useful

I have a question regarding the definition of topological manifold with boundary

ah, that is a very good point

that should actually make things easier

A boundary point $p\in \partial{M}$ is in the domain of a boundary chart, $(U,\phi)$ , in this case is $\phi$ a homeomorphism to $\mathbb{R}^n$ or $\mathbb{H}^n$?

the latter

is there a notion of the logarithm of a category?

Is that the definition or a consequence of the definition, @MikeMiller?

One of the two! I don't know your definitions.

@Thorgott So you seem to be just defining $C^D$ for categories in the more-or-less obvious way as the category of functors from $D$ to $C$.

@TobiasKildetoft Doesn't seem quite the case here --- this is like the disjoint union sqcup_{sets I} C^I

Since his objects are just raw collections of vector spaces

There's probably some name for this construction, though I don't know it

yeah, I'm not sure whether that's the same

How should I think of $(V_i)_{i\in I}$ as a functor from $\mathbf{Set}$ to $\mathbf{Vec}$?

@Thorgott Ahh, right, you seem to have forgotten the maps in Set

(in the construction I mean)

@MikeMiller It's a functor from a discrete category I guess

@AlessandroCodenotti That's a great classic. I finally worked it out when I started teaching point-set topology. Over the years, I believe I had two students who gave polished proofs.

Oh, howdy, demonic.

Nice timing! Hi Ted

@Alessandro you mean this is the functor category when we think of Set as discrete category?

@Thorgott Well, not the category Set actually

More like a discretisation of it (i.e. we remove all non-identity arrows)

No, a collection of vector spaces indexed by $I$ is a functor $I\to\mathsf{Vec}$ where you think about $I$ as being a discrete category

@AlessandroCodenotti Until he talks about morphisms, because we then have arrows between the different I's!

So it's functors from the category of discrete categories to the category of vector spaces

Yeah

I lost brain cells just by typing that out

wtf

I shall convince myself that that's true

It's clear!

A discrete category is a set (its set of objects) and a functor between discrete categories is just a map of sets

I guess everything can be phrased in terms of categories if you try hard enough, but I feel that in most cases (and in this case in particular) it's nothing but obfuscation

I think the obfuscating thing in this case is describing this construction to begin with :p

yeah, the object itself is just weird

smooth vector bundles are what I actually needed

:)

I convinced myself that the identifications $T_x\mathbb{R}^n\cong\mathbb{R}^n$ are in fact natural, in the sense that they give a natural isomorphism between the differential functor and a functor representing the classical derivative (by acting on the trivial bundle) as functors from the full subcategory of $\mathbf{Diff}$ generated by Euclidean spaces to the category of smooth vector bundles

lol

and that is clearly not obfuscatory, but the right way to think about this :p

Turns out there are ways to make differential geometry even worse!

no i'm pretty sure the correct phrasing here is that there's a canonical isomorphism $TV \cong V \times V$ given by translating the tangent space above zero

it does not seem that you are saying any more lol

(this is actually just saying that the differential of a map $\mathbb{R}^n\rightarrow\mathbb{R}^m$ in standard coordinates is precisely the Jacobian)

well, the entire point is to make the being canonical precise by giving a setup in which it is a natural iso

If $B_{H^n}(x,r)\subseteq \partial{H}^n$ is $B_{H^n}$ homeomorphic to $\partial{H}^n$?

This is not differential geometry. I keep complaining that you abuse the term. The notion of a tangent bundle is basic differentiable manifolds.

I have no idea what that string of symbols is supposed to mean, purple guy

What is B? What is H^n? Is it the same as $\Bbb H^n$?

Yes

Probably best not to change notation in the same line. What is B?

How can your hypothesis even hold?

what is it actually? differential topology? or just "differential stuff"?

Manifolds

$B_{H^n}$ is $B_{H^n}(x,r)$

is that classified as its own topic?

To me "differentiable manifolds" is the basic subject ... differential topology and differential geometry are more advanced things having that underlying structure.

You still haven't told me what that means!

This is incredibly frustrating

$\mathbb{H}^n$ is the half space

ugh, $H^n=\mathbb{H}^n$

Well, I haven't been told, either, @MikeM :D

ah, that makes sense

This statement is preposterous unless $B$ doesn't mean what I think it means

(as Ted said)

LOL

Well, Ted didn't say preposterous, although he thought it.

Let me write it clearly

Tell us what $B_{\Bbb H^n}(x,r)$ is.

If $B_{\mathbb{H}^n}(x,r)\subseteq \partial{\mathbb{H}^n}$, is the ball then homeomorphic to the boundary of the half space?

smack

@TedShifrin $B(x,r)\cap \mathbb{H}^n$

I said EXPLICITLY for you to define $B$ ....

where $B(x,r)$ is a ball in $\mathbb{R}^n$

Now, how can that hypothesis hold? That's what I asked 5 minutes ago.

If you say "if P, then Q" and P is false, then ....

@TedShifrin hey ted!

howdy @Stan

Well, for instance, if you have a boundary chart $(U,\phi)$ then such a ball exists, no?

No.

Draw a picture.

@TedShifrin So I was invited to join a reading group this summer and they are reading a book called "An Introduction to Manifolds" by Loring W. Tu

apparently these are useful robotics?

it's part of some group called Robotics and Optimization for the Analysis of Human Motion

Yes, manifolds appear in robotics. I personally would opt for Guillemin and Pollack, but people love Tu's book. I know Tu (from years ago) but have never looked at his book.

But you need multivariable analysis. Derivative as a linear map. Inverse and implicit function theorems. Essential background for this stuff.

yeah i'm sure

You can get a good start with my videos, of course (a proper subset thereof). That plus some exercises on the stuff would prepare you for the beginnings of a manifolds course.

then what's wrong with the following reasoning: Let $p$ be a point in the manifold boundary $\partial{M}$. So, there exists a chart $(U,\phi)$ where $\phi : U\rightarrow \mathbb{H}^n$ is a homeomorphism and $\phi(U)\cap \partial{H}^n$ is non-empty. Since $\phi(U)$ is open in $\mathbb{H}^n$, there exists a radius $r>0$ and a ball $B_{\mathbb{H}^n}(\phi(p),r)\subseteq \phi(U)$. Hence, $B_{\mathbb{H}^n}(\phi(p),r)\subseteq \phi(U)\cap \partial{H}^n$

@TedShifrin fabulous, which lecture should i start with?

Where did the "hence" come from?

why would a half-space ball lie in the boundary of the half-space?

Did you draw a picture ...?

Differentiation in the first semester, @Stan (through chain rule). Then you want to look at the intro to manifolds and then stuff on inverse/implicit function theorem in the second semester, plus a few more on manifolds there.

LOL, @MikeM, I could swear I said that too :D

You'd think we were the same mathematician until we try to talk math

yes, I see it geometrically, but i'm trying to fix the technical mistake

@TedShifrin gonna be a math filled summer :')

You sort of just asserted that the ball lies in the boundary at the end

LOL, @MikeM, we overlap a little :D

It's mysterious where that claim comes from

but wait, what is the difference between the category of discrete categories and the category of sets if functors between discrete categories are just maps between the objects?

oh, I guess the former has class-sized objects

No you're right and Alessandro's phrasing is wrong

but I don't want those either way

@topologicalorientablesurface: Why don't you start with $n=1$?

hmm, a discrete category is only specified by it's objects up to iso, if I were to be precise

No, that's false

so the category of small discrete categories is equivalent to the category of sets via forgetful functor

The category of small discrete categories and the category of sets are isomorphic by the forgetful functor

yeah, I assumed $\phi(U)$ is contained n the boundary.. ugh

Its inverse just remembers that Mor(x,y) = *

but you can technically choose different singletons as hom-sets

I'm honestly not going to talk to you again

LOL

@topologicalorientablesurface I think there may be some misunderstanding of what a boundary chart looks like. We're not talking about making the interior and the boundary into manifolds, separately. A boundary chart should incorporate both the boundary and the interior. For $H^2$, a boundary chart about $0$ looks like the top hemisphere of a disc

Glad you joined my team, @MikeM :D

Someone's going to teach me G&P now, so time for me to bounce

when does the topological boundary of a manifold and the manifold boundary coincide?

Bubye.

I think i'm confusing these two concepts

What do you mean by topological boundary?

are your manifolds embedded in Euclidean space?

@Thorgott no

then what does boundary mean here?

One ordinarily talks about boundary for a subset ...

Hence my question. What do you mean by topological boundary?

@TedShifrin You take the closure and throw away the interior

that's the empty set

cause every topological space is clopen in itself

What's the topology? Isn't the thing closed in itself to start with?

Work with examples. I think you don't do that enough. Take a closed interval in $\Bbb R$ or a closed disk in $\Bbb R^2$.

and, for contrast, think of the closed disk in $\mathbb{R}^2$ as subset of $\mathbb{R}^3$ (lying in the $xy$-plane, say)

(Yeah, but top said his manifolds weren't embedded anywhere :P But yes, this is important.)

yeah, this highlights why the notion of topological boundary does not always capture what the boundary of a manifold intuitively ought to be (which isn't that surprising, the topological boundary is, in a way, extrinsic as it can depend on the ambient space, whereas the manifold boundary should be and is intrinsic)

this makes me wonder: is there a result stating that lower-dimensional subsets of Euclidean space have empty interior?

for a more general definition of dimension, like Hausdorff maybe

@Thorgott: Sort of like von Neumann's proof that a ball or rectangle (in $\Bbb R^n$) cannot have measure $0$.

hmm, I don't think I'm familiar with that

I learned it in Guillemin & Pollack (the appendix on measure 0 and Sard).

oh oh, @A is here.

I'm here to make sure you stay on your PT routine, professor :-)

Don't let the mass hysteria get in the way of your routine.

Gee thanks, A.

Clever way of getting me to capitalize your a.

np

By the way, this post turned out a bit more interesting than I'd expected.

How many non-negative integer solutions are there to the equation 2a + b + c + d = 12?

I tried using stars and bars, but there's a coefficient of 2, so I'm not sure how to approach that

I also tried setting another variable e = 2a, and solving # positive solutions to e + b + c + d = 12

That gives me a total of (12+4-1) choose (4-1) = 35 solutions

But I'm not quite sure if that's right

Good question. Why not use the $2$ to limit your choices for the location of the first bar?

I don't know how you did the $e$ approach. How did you force $e$ to be even?

Hi guys, I'm so excited I've just discovered a fundamental theorem. I need someone who can endorse me before I can publish

"fundamental"?

what makes you think it's publishable?

Welcome @Nicco

:-)

Hold down the fort, A. See you anon.

cya

Yes, of vector fields.

Have to leave now, but I'll be back in a bit.

@Balarka To answer your question on whether $Q$ has some kind of connectedness the answer is no, it is totally disconnected. For $T_1$ spaces $X$ $\dim X=0\implies X$ has a basis of clopen sets $\implies X$ is totally disconnected, but you need something like compact Hausdorff to reverse the implications

If anyone could take a look at my question, that would be great

AskAway

May i ask a question too? It's probably very simple for u guys

How many non-negative integer solutions are there to the equation 2a + b + c + d = 12?

If α,β,γ,δ, are the solution of the equation tan(θx+ Pi/4)= 3tanx . After simplifying previous expression i am getting a biquadtric equation in Tan(x) .My question is α,β,γ,δ, Or tan(α),tan(β) ...roots of that equation i got

Let's see what you are getting?

(aka show your work please :-)

I am getting tan^4x-2tan^2x+8/3tanx-1/3=0

Do u want me to explain the procedure too?

Sure.

If anyone has any ideas for my problem that would be great

It's pretty simple use tan(x+pi/4) and tan(3x )formula and simply cross multiply and re arrange the terms

How about let y = tan x

Are roots α,β,γ,δ, Or tan(α),tan(β)..?

y^4 -2y^2 +8/3y -1/3 = 0

As it's given is α,β,γ,δ, satisfy the initial relation

Substitute them in and find out.

Substitute what?

can someone explain in really basic terms what a Lie algebra is?

@StanShunpike it's the set of $n$-dimensional (real) vectors for some $n$ with a choice of cross product that satisfies the Jacobi identity

I don't need the solution, i want to clarify whether is α,β,γ,δ, Or tan(α),tan(β) ...roots of that equation

root means solution

@LeakyNun ok thanks! that's a starting point at least

Read the initial question

So according to u as it's given α,β,γ,δ satisfy the relation thus they are the roots,right?

anyone have any recommendations for finding the roots of a system like: f(x,y) - g(x,y) = 0, f(x,y) + g(x,y) = 4?

Hi all, I've seen in many texts sometimes the assumption of "right directional derivative is finite"

why do we assume that?

what does it imply?

@APerspicaciouslyCuriousMind Thanks bro

@TedShifrin because after doing some research, I got convinced that the said theorem can solve a number of problems in number theory

Hey everyone.

Suppose I write: "Consider the algebra generated by the following: a sort $S$; elements $a, b : S$; an operation $\circ : S \times S \to S$; and the identity $x \circ y = y \circ x$."

To an algebraist, is it pretty obvious what I mean by that, or should I explain myself a bit more?

@DarkRunner I responded to you. Did you miss that?

@CharlieShuffler This leads you to $f(x,y) = 2 = g(x,y)$. Now you're on your own to solve that!! Depends a lot on $f$ and $g$, of course.

@TedShifrin Yes, I'm not sure how to proceed on the problem

I think you should use bars & stripes still. How many choices do you have for where the stripe after $a$ should go?

You'll then have to do cases after that. In other words, take the cases $a=0,1,2,\dots,6$ and then count for $b+c+d=12-2a$.

Ah yes! I will do casework

OK let me see what I get

The expert counters may know something slicker, but that's what I can suggest.

@TedShifrin what music have you been listening to lately?

Not much, @Stan. Mostly listen when I drive, and haven't been doing that much. I have zillions of CD's, one third of which are uploaded to iTunes and on my phone.

@TedShifrin :') well that's gonna definitely make it hard to listen if driving is your favorite time. i honestly haven't really traveled at all recently tho either. hard to go anywhere at the moment

Hello, what's doing here?

Hey everyone! In particular @Ted

How've you been?

hi Demonark

Hi Dami, long time no see

I have 8 identical pineapples and 5 identical bananas to distribute to four different students. In how many ways can I do this if two students only receive pineapples (each must receive at least one) and the other two students only receive bananas (each must receive at least one)?

As suggested, I followed the Stars and Bars principle and got 28, but I'm not sure if that's plausible

Because I tried with a smaller case of the problem, and it didn't seem to work

If anyone could check or verify if I got the right answer (or just let me know I'm on the right path) that would be great

ok, no one probably cares, but it turns out that the category discussed earlier is the standard example of a fibred category

So, if anyone's got some ideas, please let me know

I got 28 by stars and bars and multiplying

Hmm stuck

how many possible ways of splitting the four students into pairs of two are there? how many ways of distributing 8 pineapples on the first pair such that each gets at least one are there? how many ways of distributing 5 bananas on the second pair such that each gets at least one banana are there?

x^8-x^5+x^2-x+1 = or < 0, what is x?

plot the function

@ArnoldFernández No solution in real number x, because its global minimum is around +0.675

Whitout computer?

Is a problem for secundary

@Thorgott so are vector bundles tho

with morphisms fiberwise linear maps; the functor to Diff is (restriction to 0 section)

why not send the bundle to its base?

though I guess that's effectively the same

Hey, I've done some research about derivatives, but I just don't understand what they are! All I know is they have something to do with constant. Thanks!