Mathematics

user76284

5:33 AM

What's the right notation to convey $A \vdash B$ but under intuitionistic predicate logic, not classical logic?

Leaky Nun

$A \vdash_{\text{intuitionistic}} B$

Mathphile

8:03 AM

just a test please ignore

$\tiny{\LaTeX}$
$\scriptsize{\LaTeX}$
$\small{\LaTeX}$
$\normalsize{\LaTeX}$
$\large{\LaTeX}$
$\Large{\LaTeX}$
$\LARGE{\LaTeX}$
$\huge{\LaTeX}$
$\Huge{\LaTeX}$

6

Rudi_Birnbaum

11:04 AM

Hi all!

I have a question. I am dealing with "divergenceless" vector fields $v$: $\nabla\cdot v = 0$

The question is how to call that property in English (an German).

We had started with "divergence-free" and "divergence free" then went to "divergence-less" and "divergenceless"

and at some stage I have used "fullfilling the continuity equation" (its time-independen cases).

And in the haste of writing I ended up with "continuous" which is now obviously plain wrong (right?).

Happy for suggestions and comments!

Thorgott

I would go with "divergence free" ("divergenzfrei") and I think in the case of fluids they're called "incompressible" ("inkompressibel"), but I'm no expert

Rithaniel

1:17 PM

Is it possible to have a field that has both elements of finite order and elements of infinite order?

I feel like I might know that the answer is "no," but I'm unsure

Alessandro Codenotti

Hint: characteristic

Rithaniel

Yeah, I seem to recall that all elements of a field have to have the same order, and if that order is finite, then it has to be prime, but again, a proof doesn't immediately leap to mind

Wait, use the order of the identity. Got it.

Alessandro Codenotti

Right. If you want a somewhat more abstract argument for every ring $R$ there is a unique ring hom $\Bbb Z\to R$ ("$\Bbb Z$ is the initial object of $\mathsf{Ring}$") and it has kernel $n\Bbb Z$ for some $n$. That $n$ is the characteristic of $R$, from which you quickly get what you want by properties of homomorphisms

Thorgott

1:38 PM

this is true for order as elements of the additive group, the multiplicative group can have both elements of finite and of infinite order

Rithaniel

Like $i$ versus $2$

Alessandro Codenotti

Sure, the multiplicative group of $\Bbb C$ contains elements of any prime order

Thorgott

of any order was correct

Rithaniel

All of which, if they are of finite order, sit on a single circle

Thorgott

for each natural $n$, $\mathbb{C}^{\times}$ contains exactly one subgroup of order $n$ and it is cyclic

Alessandro Codenotti

1:43 PM

Every finite subgroup of the multiplicative group of a field is cyclic

Rithaniel

Fields are almost disappointing in that sense. You have all these interesting things in more abstract constructs. Once you get to fields, everything becomes roughly uniform

Alessandro Codenotti

Algebraically closed ones are even more boring, they're completely classified by characteristic and cardinality if they are uncountable

Thorgott

I mean, this is true for integral domains too

Rithaniel

Yeah, integral domains are still fun for other things, though. There are a lot of weird things with Krull dimension and rings of formal power series

Thorgott

You can still say some really interesting (and arguably surprising) things about algebraically closed fields, like Artin-Schreier

What's true though is that fields are a pretty badly behaved category

Alessandro Codenotti

1:58 PM

I'm a little confused. Let $\Omega\subseteq \Bbb R^n$ be as nice as you wish (open, bounded, smooth boundary, whatever). Of the inclusions $C^\infty_c(\Omega)\subseteq W^{1,p}_0(\Omega)\subseteq W^{1,p}(\Omega)\subseteq L^p(\Omega)$ which are dense? I'm interested especially in the case $p=2$

Thorgott

Isn't $C_c^{\infty}(\Omega)\subseteq L^p(\Omega)$ dense already?

Alessandro Codenotti

Hmm maybe you're right (and that would make sense, from a result I'm reading it does seem like they should all be dense)

Thorgott

I think $C_c^{\infty}$ is dense in $C_c$ by mollifying and $C_c$ is dense in $L^p$ by some tedious approximation arguments, but take this with multiple grains of salt

Alessandro Codenotti

I agree with the first part and I think I read about the second one a long time ago

Thorgott

2:14 PM

I can sketch an argument for the second part in a couple minutes if needed. Not sure how much niceness is needed on $\Omega$ for both parts to go through, but I feel like openness should work.

geocalc33

2:33 PM

Thorgott what's up

What's stronger than a diffeomorphism?

Alessandro Codenotti

@Thorgott Nah I can believe it

I guess it's in Brezis if I really want to see the ugly details

geocalc33

I think that the problem with your question is that "diffeomorphism" is too weak a notion to be relevant to general relativity (GR). In GR a "spacetime" is a Lorentzian manifold, that is, a differentiable manifold equipped with some additional structure. You are pondering what transformations you can apply to a spacetime without destroying all this structure. Now, a diffeomorphism is not enough; it does preserve the differentiable structure, but it needs not preserve the Lorentzian one.

The "right" notion you are after is that of a conformal mapping of Lorentzian manifolds.

Lukas Heger

3:31 PM

@Alessandro mollifying $L^p$ functions is not more tedious than mollifying $C_c$ functions actually. If you take the convolution of a locally integrable function and a $C_c^\infty$ function, then the result is $C^\infty$, so if you take an approximate identity you get that $C^\infty \cap L^p$ is dense in $L^p$ without more difficulty than showing that $C_c^\infty$ is dense in $C_c$.
If you actually want to $C_c^\infty$, you need to multiply by suitable cut-off functions, but that's not really that tedious: Write $\Omega = \bigcup U_i$ where the union is directed and all $U_i$ are open and b…

Leaky Nun

@LukasHeger did you see my message?

Lukas Heger

no

which one?

Leaky Nun

@LukasHeger show that Q has no Z-basis, using the most high-powered way possible

Lukas Heger

hmm

geocalc33

haha

most high powered way :)

Lukas Heger

3:41 PM

upon taking Pontryagin duals wrt discrete topologies, this is equivalent to showing that $\Bbb A_{\Bbb Q}/\Bbb Q$ is not isomorphic to some product of $S^1$s

Leaky Nun

ah, I forgot about Pontryagin dual

geocalc33

Pontryagin duality?

Thorgott

I see, that's a good point.

geocalc33

Can you convolute the differential structure of

a real analytic man-fold

Leaky Nun

and we know that $\Bbb A_\Bbb Q = \Bbb Q \times \widehat{\Bbb Z} \times \Bbb R$? wait that isn't right

geocalc33

3:44 PM

depends on how fat the man is!!

Leaky Nun

what's the decomposition? @LukasHeger

Lukas Heger

uhm, I don't think there's a decomposition like that for the additive adele group

Leaky Nun

ok

Lukas Heger

but I might be wrong

Leaky Nun

but we can embed $\widehat{Z}$ inside right

geocalc33

3:47 PM

decomposition in what adele class?

Lukas Heger

@LeakyNun yes

Alessandro Codenotti

@LeakyNun Can you compute Ext^1(Q,Z) without knowing that Q isn't free Abelian?

(this is essentially the same thing as Lukas product of circles thing but on the other side of the duality to be fair)

Leaky Nun

0 -> Hom(Q,Z) -> Hom(Q,Q) -> Hom(Q,Q/Z) -> Ext(Q,Z) -> Ext(Q,Q) -> Ext(Q,Q/Z) -> 0

how does Ext work again

geocalc33

ah Hom(Q,Z)....

Alessandro Codenotti

@LeakyNun I never remember lol

geocalc33

3:52 PM

hey alessandro

Alessandro Codenotti

Hi

Leaky Nun

looks like Ext(Q,Q) = 0

Alessandro Codenotti

You only need that Ext to be nonzero, you don't care what exactly it is here

geocalc33

Off topic w.r.t. the conversation atm but, can one convolute the conformal structure of a smooth and bounded lorentzian submanifold

Alessandro Codenotti

The point is showing that Q is not projective as a Z-module

Leaky Nun

3:54 PM

so we want to show that Hom(Q,Q) -> Hom(Q,Q/Z) is not surjective

Hom(Q,Q/Z) = Hom(injlim (1/n)Z,Q/Z) = projlim Hom((1/n)Z,Q/Z) = projlim nQ/Z = ???

Mike Miller

4:29 PM

@LeakyNun take any diagram of abelian groups, so that for any arrow f in this diagram, there is another arrow g with gf = 0, and furthermore every object has a morphism-out. then the colimit of the diagram is zero, because the map from each object of the diagram to the colimit is forced to be 0

there seems to be an issue with this argument lol

oh I see, none of the composites are actually 0 as I demanded

geocalc33

@MikeMiller do you have a minute?

geocalc33

4:48 PM

well, if you feel like editing or answering feel free :) math.stackexchange.com/questions/3562412/…