Happy haloween people

Happy Halloween

Happy Halloween

Or a morbid one maybe

They are one and the same on Halloween

Can someone explain geocalc's joke?

Gromov invented J-holomorphic curves. Gromov is God.

Oh ok

I didn't know J-holomorphic curves were invented by God

fourth ICM proceedings, I think, that one

Maintaining the spooky trend, how does one keep track of what is happening in Dimension theory of rings. It all seems unmotivated random stuff.

You should learn dimension theory of nonmetrizable spaces, and then dimension theory of rings will look much better

why, dimension is fine

Krull dimension?

It might seem unmotivated, but the main examples are fairly obvious

@BalarkaSen which?

rings

Hausdorff dimension is also fairly intuitive I think

Krull dimension is quite intuitive in the definition, but then you get all kind of weird issues like dim+codim is not dim of ambient space etc.

but thats because of singularities

I do not understand the plethora of definitions it has. There's Krull dimension, there's the transcendence degree one for affine algebraic sets. Apparently they are the same in this case and I do not understand why.

oh that you just work out

for affine algebraic sets everything is clear

@AlessandroCodenotti how's codim defined?

@Astyx height of a prime ideal

I think dim + codim = dim of ambient is exactly the Cohen-Macauley condition

So you get dim of ambient >= dim + codim (?)

I don't know actually if the dim+codim thing can fail for rings now that I think about it

Is the equality for Noetherian?

It can fail for schemes

So the question is does it fail for affine schemes? Probably yes I guess

Everything can go wrong in schemes lol

Probably, you can have two independant maximal sequences of ideals of length a and b. If a<b and you take an ideal in the sequence a, its dim+codim will be a, yet the dimension is >b

Noetherian just means the dim is finite I think

@SayanChattopadhyay transcendence degree = Krull dimension follows from Noether normalization if you know that $k[x_1, \dots, x_n]$ has Krull dimension $n$

for f.g. $k$-algebras

@LukasHeger Yeah I do know the proof using Noether Normalisation but why should I expect these two definitions to coincide?

just think $\Bbb C$

Krull dimension is obviously the manifold dimension of the nonsingular locus

Right

The point of the trdeg definition is the space of meromorphic functions on your variety is a finite extension of $\Bbb C(x_1, \cdots, x_n)$ ($n$ being dimension)

so the variety is a branched cover of $\Bbb C^n$ in some way

which is exactly the statement of the Noether normalization anyway

the thing has a finite map to $\Bbb C^n$ -- a branched cover

@robjohn "You Go, gourd!" Nice work! And such a good exemplar wearing a mask!

somehow my question on philosophy was viewed 27,000 times. I guess this is in part due to people sharing the question elsewhere

@AlessandroCodenotti Yeah this is fine for affine guys

you can always choose a regular point of the affine subscheme or whatever

will someone take a look at this post: math.stackexchange.com/questions/3889262/…

do you agree that there are still nonsensical claims in the proof?

Anyway just forget commutative algebra

its useless

Just google stacks project whenever you need a result

lol

its the same gain with less effort

Oh don't worry, I forgot all of that stuff

And I never need it, so I even save the googling time

good

i learnt a lot of cube complexes last class

What are cube complexes good for?

Curiosity not accusation

But all meromorphic functions do not have to be rational functions right? Or am I missing something. The trdeg definition as far as I know, is the trdeg(k(V)/V), where k(V) is the field of fractions of the coordinate ring which in our case is C[x_1,x_2,..,x_n]

yeah I meant rational functions

@MikeMiller Hm let me think

I never really learned about those properly. I think they come up in ggt sometimes?

A postdoc was telling me yesterday about some interesting applications of asymptotic dimension to descriptive set theory, cool stuff

@bfff the proof looks ok, bar the issue littleO remarks

did I effectively fix the little issue you though?

like i edited the post to fix the thing i thought he was talking about... now he is saying there is a different problem with the proof

@MikeMiller You mean something other than the fact that every hyperbolic $3$-manifold group is RAAG, maybe?

That's one concrete application I can think of

What does RAAG buy me

I just use them as an alternative method of computing the homology of complex topological spaces

RAAGs are linear groups but that's not very fascinating, you already have that from hyperbolicity.

Hmmm

Yeah ok you can phrase it as follows. Every hyperbolic group which acts geometrically on a $\text{CAT}(0)$ cube complex virtually embeds in a RAAG. But this is a small generalization

Let me think a bit more

What was the Haken conjecture again?

Sorry, I should be able to tell you an appealing application. I am having trouble sifting through the details

@robjohn For such a great costume for Halloween, I've "treats" to offer you:

In a Poisson process, is it accurate to say that lambda is the probability that a unit length interval contains exactly $1$ arrival, or the probability that a unit length interval contains at least $1$ arrival?

@user10478 Where are you getting that from? N_t, the number of arrivals in [0, t], follows Poi(lambda t). So P(N_1 = 1) = P(Poi(lambda) = 1) = lambda e^-lambda, and P(N_1 >= 1) = 1 - P(N_1 = 0) = 1 - e^-lambda.

So the answer is "neither"

Expected number of arrivals in an interval of unit length is lambda.

lambda is not a probability in any way

@BalarkaSen Hmmm, was getting it from this chart: youtube.com/…

Arrival rate, correct, that's the expected number of arrivals in an interval of unit length. Not a probability.

Actually I guess I was inferring that it was symmetric with $p$ in the Bernoulli process, which can be thought of as the probability that any given time contains an arrival.

No, you should not think of it as probability, that is meaningless. It is a coincidence that expectation of Ber(p) is p, and the probability Ber(p) takes 1 is also p.

Okie, thanks for the clarification