hello @AkivaWeinberger

yeah it works @MatheinBoulomenos I figured it out even before knowing that Vakil put a hint in there :P

does any one knows what \sum_{k=1}^{n} k10^k equals to ?( i found one but messy)

@Adeek yes, that's correct

it is this number 1234........n

goood

@MatheinBoulomenos I should hopefully finish first 4 chapters + problems by July

@MatheinBoulomenos after Ravi Vakil I plan to move on to Shava both 2 volumes and finish Maybe Michael Atiyah and Eisenbudd and finally go back to hartshrone

and solve the problems in it

sounds good

@MatheinBoulomenos my alg top course explicitly follows Ch.0,1,2 of Hatcher

@edcharlie Are you familiar with geometric sums and derivatives?

Does anyone know of some resources that explain the basic manipulations of covariant and contravariant four vectors used in physics mostly? I'm trying to evaluate $\partial_\mu(x^2x_v)$ but I don't know exactly what to do. I see that I can do $\partial_\mu(x_v x^v x_v)$ but don't know how to continue

$$\require{AMScd} \begin{CD}
\operatorname{Hom}(C',A) @>{i_{C'}}>> \operatorname{Hom}(C',A')\\ @V{\operatorname{Hom}(f,A)}VV @V{\operatorname{Hom}(f,A')}VV\\
\operatorname{Hom}(C,A) @>>{i_C}> \operatorname{Hom}(C,A')
\end{CD}$$

@MatheinBoulomenos really, using library to draw category diagrams

why not?

nothing

The first time I had to make a commuting diagram was for a short paper on homology last quarter, but I found this which was rather nice

I draw commutative diagrams by whitespace manipulations

@TobiasKildetoft yes

Problem, Donald Knuth?

Hi

I just got a flu shot

I should have gotten it much earlier

Apologies to all the people I have endangered due to my reckless behavior

did you get it after catching a flu

No

then you're nowhere near as reckless as I

spitting noises

@edcharlie then note that your sum is 10 times the derivative of a geometric sum

(evaluated at 10)

hello, someone help me on derivative ?

@Arrow Doesn't "locally Euclidean" mean that it's a manifold?

@Akiva You have the charts but maybe you don't know they're compatible?

By locally Euclidean I mean the property of being locally homeomorphic to Euclidean space, not a differentiable structure.

@TobiasKildetoft Thanks a lot

I didn't read the question carefully but my first instinct would be to say something like the dense line in the torus, torus embedded in R^3.

That should work.

Well.

It depends on what "locally Euclidean" means

It sounds like it means it's topologically a manifold

so... a topological manifold :)

gah... ninja'd

(but not a smooth one)

That's not my point

Yo mama so smooth...

Does it mean a "topological submanifold of $\Bbb R^n$"?

sub is the important part

I.e., does it mean at each $p \in X$ there is a ball $B$ around $p$ in $\Bbb R^n$ such that $B \cap X \cong \Bbb R^n$?

@BalarkaSen Well, the Whitney embedding theorem could be applied, no?

oh... I see what you are saying

I mean that $X$, taken with the subspace topology from $\mathbb R^n$, is locally homeomorphic to some Euclidean space.

@Arrow OK, then you do mean what I wrote.

That means $X$ is a topological submanifold of $\Bbb R^n$ already

Wait... why does $X$ have a metric structure?

Forgive my ignorance, but what is the dense line in the torus?

It sounds like he wants it to be a smooth manifold

you just said it was locally homeomorphic to Euclidean space...

You are saying if a topological submanifold at each point admits a tangent space it is a C^1 submanifold

@XanderHenderson Subset of a metric space?

Not necessarily C^1, just differentiable.

Fair enough

does smash product in topology generalize categorically?

This should be true.

So the dense line in the torus (whatever it is) is not a counterexample?

@LeakyNun It is an integral part of the theory of spectra

@Arrow Nope.

@Daminark hello, please have an idea about derivative of a function defined on sobolev space ?

On any small nbhd the line hits arbitrarily many times

It's not a topological submanifold

Great, that's comforting (I also felt it should be true). Do you have any idea how to prove it?

@Arrow Is there an example that is one but not the other?

@AkivaWeinberger do you mean differentiable but not C^1?

Yeah

@Akiva Graph of a differentiable function which is not C^1?

idk something like that

^

@AkivaWeinberger C^1 means continuously differentiable though

@LeakyNun the tensor product over commutative rings is like the smash product, formally

@MatheinBoulomenos interesting

@LeakyNun Right

both are left adjoints to the internal hom functor (in topology you have to restrict yourself to locally compact Hausdorff or something like that)

Ah, so the squiggly thing

$x^2\sin(1/x)$

sure

> In fact [23], [72], the zero divisors of norm one in the sedenions form a subspace that is homeomorphic to the exceptional Lie group $G_2$.

@Arrow Consider a homeomorphism $f : B \to \Bbb R^N$ which takes $X \cap B \subset B$ to $\Bbb R^n \subset \Bbb R^N$. Your statement about curves should mean $f^{-1}|_{\Bbb R^n}$ is differentiable.

@AkivaWeinberger what the heck. I didn't know people seriously think about sedenions

Oh no. Oh-ho-ho. There are functions which have directional derivatives in all directions but does not have a derivative, right? What's a dumb example of such a continuous function?

I forget

I only remember the discontinuous ones

$f(x,y)=\frac{x^3y}{x^4+y^2}$ has all directional derivatives but is not differentiable.

> Moreno (1998) showed that the space of pairs of norm-one sedenions that multiply to zero is homeomorphic to the compact form of the exceptional Lie group G2. (Note that in his paper, a "zero divisor" means a pair of elements that multiply to zero.)

Okay, great. The graph is a topological submanifold of R^3 which has tangent space in the sense you said, I think, but should not be a differentiable submanifold

I think

Let $N$ and $P$ be submodules of $M$. Then, $(N+P)/N = 1 + P/N$

I don't think so: $f(x,x^2)$ is a tilted parabola in $\mathbb R^3$ and the tangent to this parabola at the origin lies in a non-horizontal plane. However there are also tangents to the graph at the origin in every single direction in the $xy$-plane. Particularly the tangent set cannot be a two-dimensional vector subspace.

@LeakyNun Stap

@MatheinBoulomenos lol

@Arrow Oh, I see. So the tangent directions wiggle too much

They don't form a subspace

I just graphed the function

:)

Good question, dammit

$\Huge\longrightarrow$ @Arrow

For once, hahaha. Thought about it for a day but I just don't see how to prove it. On the other hand I don't think there "should" be a counterexample.

@LeakyNun your argument is invalid.

Hm, if $z\in\Bbb C$ and $j$ is the unit quaternion then $\bar z=jzj^{-1}$

which is almost a pun on the double meaning of the word "conjugation"

So I think everything boils down to proving if $f : \Bbb R^2 \to \Bbb R$ is a function such that $Df_v(0)$ exists for all $v \in \Bbb R^2$ and $\{(v, Df_v(0)) \in \Bbb R^3 : v \in \Bbb R^2\}$ form a vector subspace of $\Bbb R^3$ then $f$ is differentiable at $0$.

$\overline z = jzjzjzjzjzjzjzjzjzjzjzjzjzj^{-1}$

This sounds like a question for Akiva or Leaky.

is alerted

@AkivaWeinberger are the two usages actually related?

I think it's definitely true.

@LeakyNun I guess they both form automorphisms, either of fields or of groups respectively

@BalarkaSen that's good to know :) I am not sure I understand your reduction yet - it may be that a graph is an embedded submanifold yet the function is not differentiable (if there's a vertical tangent).

Well it should just follow because that being a vector subspace means $Df_{-}(0) : \Bbb R^2 \to \Bbb R$ is a linear map

Take it's matrix. That's the derivative

Just go through the limit defn

Should pan out

@BalarkaSen I will try again. If you feel like posting an answer (which would be much appreciated), I asked this question here.

@Arrow I'm just throwing ideas. I thought about the vertical tangent, but I think one can reparameterize or something

I'll think about it perhaps

Drowned it way too many things right now

@BalarkaSen hopefully you find the time :)

Hi @MikeMiller, could you please help me out with the linked question? I'm trying to understand whether a locally Euclidean subset $X\subset \mathbb R^n$ with tangent spaces of the correct dimension is necessarily an embedded differentiable submanifold.

@Vrouvrou I know nothing of Sobolev spaces, sorry

mmm Sobolev spaces...

Lol, do you use them a bunch?

nope

But I saw one up close once

'Sobolev space' is one of those phrases I feel like separates math people from physics people

if you know what that phrase means, you're probably a math person.

heh

Dude, I hear you like derivatives, so I put some differentials in your $L^p$!

So they not show up in physics? I heard they're quite useful for PDE

They are useful on the theoretical side of PDE

if you are a physicist, you just assume that all of your functions are "nice"

Ah

But I am not a physicist, so anything I say about physicist is suspect, and likely governed by motivated reasoning. ;)

Another classic to distinguish physicists from mathematicians

If $f(x,y)=x^2+y^2$, what is $f(r,\theta)$?

r^2 + theta^2?

Yes

Wait what's the other answer?

and you are not a physicist

$r^2$

heh

:thonk:

Polar/Cartesian

Oh

Lol it's been years since I've ever touched polar coordinates

Dude, you gotta make sure you wear gloves when you touch polar coordinates.

Otherwise, you risk frostbite.

Eh, I don't even wear gloves in Chicago, surely the poles can't be that much worse

@Arrow (cc @Akiva) I was trying to write an answer and immediately realized I am using local flatness. So that makes me wonder how the tangent sets of the Alexander horned sphere at the "bad points" look like.

I mean hell, Germany isn't even that bad

Poland surely can't be different from its neighbor

@BalarkaSen unfortunately all I know about that thing is fear.

lolol

I feel like they have vector subspaces as tangent set

Think about this animation

Even if that's true they may not have the right dimension. For instance the points over the axes (except the origin) in the graph of $f(x,y)=|xy|$ have tangent lines but not tangent planes.

Or maybe the curves which lie on the horned sphere crossing the bad points wouldn't really be differentiable

It's too hard for me to think about it. We need Akiva

@Arrow I agree probably

@BalarkaSen hmm... what about looking for a natural differentiable structure on $X$? That is, a sheaf of commutative rings making $(X,\mathcal O_X)$ locally isomorphic to an open subset of Euclidean space locally ringed with its differentiable functions?

@Arrow Ah, you imposed "set of derivatives of differentiable curves". So you're not really assuming the curves are always differentiable.

Definitely not - that would be unreasonable. We can always take a "sharp turn", even in vector spaces.

Differentiable structure on $X$ is not the problem; it's being a differentiable submanifold that's the problem

The Alexander horned sphere has a natural differentiable structure

It's just not naturally a submanifold

Wouldn't know anything about a differentiable structure on the locus of evil.

hey guys, how do I find the number of subsequences (x1,x2,x3) of {1,...,12} such that $x1 \leq x2 \leq x3$?

My thoughts so far are the following: I fix x3=k (which is at least 3). Then I choose pairs smaller than k, there are k-1 choose 2 of them.

@John11 I don't know how to count so I may well be wrong, but this sounds like a stars and bars problem.

hmmm @Arrow I'll try to work with that, thank you!

@Daminark

So I read that every topological space can be realized as a quotient of some Hausdroff space

And I was like "dafuq"

I think I rediscovered the Riemann curvature tensor...

lol

It pops out to be $d\Omega + \Omega \wedge \Omega$

Well, the curvature 2-form, rather

Where $\Omega$ is the "moving frames apparatus"

@BalarkaSen the curvature form is a generalization of the Riemann tensor I guess

Right

Hm, the curvature operator in the bundle valued differential forms theory looks exactly like that

I suppose that's not a coincidence

Can someone help me with finding the number of subsequences (x1,x2,x3) of {1,...,12} such that $x1 \leq x2 \leq x3$? :(

@Eric Oh, by the way. Suppose $\{e_1, \cdots, e_n\}$ is a frame on a chart $U$ on a manifold. This is integrable, i.e., there's local coordinates where $e_i = \partial/\partial \mathbf{x}_i$ iff $[e_i, e_j] = 0$ by Frobenius, right?

Suppose $\omega_1, \cdots, \omega_n$ are dual to the frame. $d\omega_k(e_i, e_j) = e_i \omega_k(e_j) - e_j \omega_k(e_i) - \omega_k([e_i, e_j])$. If $k \neq i, j$ everything is zero. If $k = i$ the first and last thing vanishes, and the middle vanishes because $\omega_i(e_i) = 1$ is constant function; similar if $k = j$.

So $d\omega_i = 0$...

We chose the frame such that our symplectic 2-form $\omega = \omega_1 \wedge \omega_2 + \cdots + \omega_{n-1} \wedge \omega_n$ (ie omega is standard on that frame). Since $d\omega_i = 0$, $d\omega = 0$. Just to explicitly establish it as a integrability criterion

What we'd need to show is that local sections of the subbundle of $F(TM)$ consisting of frames $\{e_1, \cdots, e_n\}$ such that $\omega$ is standard on that frame are integrable. It's not obvious to me how one goes about doing that.

My argument goes two ways until that last step. $d\omega = 0$ does not necessarily imply $d\omega_k$'s are zero

@BalarkaSen an idea for the submanifold problem - perhaps we can construct an exponential map giving a diffeomorphism berween a neighborhood about $p$ of the tangent plane and an open neighborhood of $p$ in $X$? Would this be enough?

@Arrow This is certainly impossible without more rigor.

$X$ can be quite bad

I agree it needs more rigor, but I don't see how $X$ can be bad enough to preclude such a construction with the strong conditions I imposed.

I do not see a construction. I just see a vaguish idea

What is "the exponential map" in this context? Why is it a diffeomorphism?

I thought about somehow laying down straight lines in the plane onto geodesics. This is described in 4-6 of do Carmo's Differential Geometry of Curves and Surfaces but not quite in this generality.

I know the Riemannian geometry construction, but this does not sound possible here.

I don't buy this idea unless you actually give me a construction

I have nothing more to offer. Was just wondering if you think there is any substance to this idea.

I am 90% sure the ideas I gave solves your problem. Pick a ball $B$ such that $B \cap X$ is homeo to $\Bbb R^k$ by $f : \Bbb R^k \to B \cap X$. If $\gamma$ is any smooth curve through origin in $\Bbb R^k$, $f(\gamma)$ is a smooth curve in $\Bbb R^n$ and it's derivative at $p$ is well-defined. Then $f$, as a map to $\Bbb R^n$, has directional derivatives well-defined: $D_{\gamma'}f(0) = (f \circ \gamma)'(0)$

Since it's differentiable in every direction, and the directional derivative is a linear map in the vector component, it should be differentiable at $0$ as a multilinear map

Well, the problem is $f(\gamma)$ might not be a smooth curve in $\Bbb R^n$. Somehow the dimension condition says for any $v \in \Bbb R^k$, there's a curve $\gamma$ through origin such that $\gamma'(0) = v$ and $f(\gamma)$ is a smooth curve

Unless I misunderstood something it also doesn't seem that having a directional derivative linear in the vector component implies differentiability (see here).

That's not what I mean by linearity. Think again.

I mean $F(v) = D_v f(0)$ satisfies $F(v + cw) = F(v) + cF(w)$.

That seems to be exactly the question I linked to.

As in it's an actual linear map, and comes from a matrix.

@Arrow The answer by John is the example you gave me. How is it's dir. derv. a linear function of $v$???

The tangent set is not a 2-plane.

No, the tangent set is not a 2-plane, which is why the function isn't differentiable, but all directional derivatives are zero, so taking a direction to the directional derivative along it is the zero map which is linear. (Sorry if I'm being stupid, it's not on purpose)

The question itself says $\partial_vf(x)=L(v)\,\forall v\in\mathbb{R}^n$ where $L$ is linear.

The tangent set is not a 2-plane because directional derivatives cannot detect the derivative of the composite of $f$ with a "curved" curve i.e not a straight line. (If I understand correctly)

You're thinking of the wrong map. My $f$ parametrizes the surface; their $f$ graphs the surface.

What I am saying is literally equivalent to saying "tangent directions form a vector subspace"

It's not a very deep statement.

(I have not proved it does imply differentiability, of course, but it should be true)

Anyhow I have lost the enthusiasm to think about this question.

Apologies... Thanks for all the help though.

@Arrow I like your question but I don't know the answer

@MikeMiller thank you. Balarka Sen suggested an approach above if you're interested :)

I am, but I think I don't have the energy to catch up right now - I have writing to do before upcoming deadines

have you run into robjohn at all on campus?

Good night and good luck!

@Arrow I didn't suggest an approach per se, just that it's an easier and a more doable question that if a function's graph has a tangent space at a point then the function is differentiable at that point.

And perhaps the key to the answer

Understood

If you take the matrix the tangent space is a graph of, and fiddle around with the definition of a Jacobian, it might be worthwhile

One would want to show $\lim_{h \to 0} \|f(h) - f(0) - Ah|/\|h\|$ after all

For $h = tv$ where $v$ is a fixed direction, as $t \to 0$, this does go to $0$.

Given a differential 2-form w on a mfld M, then on a nbhd of a point m in M I can write it in local coords. I might have more than one local coord chart covering the same point, and I read that the compatibility condition we want is $\phi_{ab}^*w_b=w_a$ where \phi_ab is the transition function between local chart a and b. The question is: point-wise is this process just applying a change of variable to the matrix of coefficients of w?

Who wants to put a bounty on this