01:00 - 17:0017:00 - 00:00

1:08 AM
@TedShifrin A glide relfection would be orientation reversing, not preserving.
In the case of a sphere, a rotation is the analogue of a translation, so any glide reflection is a composition of a rotation with a reflection.

2 hours later…
2:45 AM
@anakhro thatâ€™s just any improper isometry. Is this a standard terminology? Glide reflections are usually improper isometries with no fixed point.

3:04 AM
I've read more things with "glide reflection" than "improper isometry".
But could just be my experience.

3:20 AM
For the sphere? Not the plane or 3-space?

5 hours later…
8:33 AM
@Balarka tf man

lol

8:52 AM
Any elegant way to project onto the Von Neumann neighborhood? With the Moore neighborhood I can just do np.sign(vec).

1 hour later…
10:14 AM
Anyone know any necessary and sufficient criteria for a ring to be cancellative?
(Or are they all cancellative?)

What does cancellative mean ?

If you have $ax=bx$, then you have that $a=b$. You can "cancel" terms.
I'm thinking they might all be cancellative, actually
But, then again, maybe not

I assume you impose $x\ne 0$
So that's the same as claiming $(a-b)x =0 \implies a-b = 0$

Yeah, of course (I always neglect to mention trivial terms)

So $cx=0$ implies $c=0$ or $x=0$

10:23 AM
Ah, so we want an integral domain
Perfect
Ah, they're equivalent, too

Yup, you can do the exact inverse reasoning of what I did to show integral domains are cancellative
(actually if you just want cancellative => integral, you can just take b=0, and you get ax = 0 => a=0)

I can get too bogged down in looking at the properties as just properties that I forget to do algebra with them, sometimes

11:10 AM
Problem: If $M$ is a module over a PID $R$, then $M$ is an injective $R$-module. Proof: Let $I$ be an ideal and $f : I \to M$ a homomorphism. Because $R$ is a PID, $I = (r_0)$ for some $r_0 \in R$. Define $\varphi : R \to M$ by $\varphi (r) = r f(r_0)$. Then $\varphi (\iota (rr_0)) = \varphi (rr_0) = rf(r_0) = f(rr_0)$, so $\varphi \iota = f$ on $I = (r_0)$.
Something about this seems off...I don't seem to use the fact that $M$ is divisible anywhere...How do I solve this problem?
BTW, I am trying to use Baer's lemma/criterion.

you defined $\varphi(r) = rf(r_0)$, so $\varphi(rr_0) = rr_0 f(r_0)$ not $rf(r_0)$
in any case it is instructive to look at examples
$R = \Bbb Z$, $I = 2\Bbb Z$, $M = 2\Bbb Z$, $f = \operatorname{id}$

11:47 AM
@LeakyNun Would $\varphi(n) = n$ if $n$ is even and $\varphi (n) = n+1$ if $n$ is odd work?
No...it wouldn't, because $\varphi (n+k) \neq \varphi (n) + \varphi (k)$ if $n$ and $k$ are odd integers.

12:08 PM
If f,g\in Z[x] has no common irreducible factor then f,g\in Q[x] has no common irreducible factor?
Someone said that that's because of Gauss lemma but I don't know why

12:18 PM
if $P \in \Bbb Z[X]\subset\Bbb Q[X]$ can be factored $P = FG$ with $F,G\in \Bbb Q[X]$. Then you can write $nP$ as the product of two $F', G' \in \Bbb Z[X]$
From then on using the gauss lemma you can eventually prove that $n=1$ works

By gauss lemma, if $p = fg$ over Q then $p=(af)(bg)$ for some a,b\in Q where $af,bg\in Z[x]$
If f and g has common irreducible factor in Q[x] then f = pf', g = pg' over Q. By gauss lemma, f = (a_1 p)(b_1 f') and g = (a_2 p)(b_2 g') over Z for some $a_i,b_i\in Q$.
It seems that we can get common irreducible factor in Z[x] related to p. But I don't know how

1:01 PM
I want to learn more about how algebras of functions on a space relate to the geometry of space. What field would this lie under? Operator theory? Algebraic geometry
/?

@Daniel Probably under the heading of operator algebras
Which tends to fall under the more general area of functional analysis

Ok thanks :)

@Daniel What kind of geometry do you have in mind? Some things to google might be Gelfand-Naimark duality or the book "rings of continuous functions"

1:17 PM
@love_sodam basically if p|n, you get that p|nP, thus p divides every coefficient of P. if you look at the highest coefficient of F' and G' that are not multiples of p (assuming they exist), you see that they produce a coefficient not divisible by p in nP, which is a contradiction
Thus either p|F' or p|G' and you can divide and start over with another prime divisor of n

If $\gamma\colon[0,1]\to M$ is an analytic path, is it obvious that $\mathrm{d}\gamma$ is nonzero except on a finite set?

Another way to see that is to notice that the projection $\phi:\Bbb Z[X] \to \Bbb Z_p[X]$ sends $nP$ to $0$, So you have $\phi(F')\phi(G') = 0$, which implies $\phi(F')=0$ or $\phi(F')=0$ because $\Bbb Z_p$ is an integral domain

@Astyx I don't know what you are talking about. You mean if $p = fg$ in Q[x] then $p = fg$ in Z[x]?

$d\gamma$ is analytic too and if there are infinitely many zeros on $[0,1]$, they accumulate, but then identity theorem implies it's zero, so the path is constant or sth?

No

1:21 PM
Then what are you trying to show? I don't get it

@Thorgott yeah that's what I was thinking too

If P has integer coefficients, and is the product of two polynomials with rationnal coefficient, then a multiple of P (so nP with n an integer) is the product of polynomials with integer coefficients

Is analytic enough for the identity theorem? I've always seen it stated for holomorphic stuff

But as I said above, Gauss lemma shows that if p = fg over Q where p\in Z[x], then there is some a,b\in Q such that p = (af)(bg) and af,bg\in Z[x]

it's true for real-analytic functions
I've never thought about analytic functions between manifolds, but surely it should generalize from that

1:32 PM
It works locally by picking charts around the accumulation point and its image, so it should work

@love_sodamOh ok, so you already know that (I hadn't seen that message)

Yes, analytic dudes cannot have accumulating singularities
I'm doing my physics homework and writing $d\mathbf{F}$ for infinitisimal force
Let $X$ be a vector field. Consider $dX$
Kill me

@Astyx Yes.
If f and g has common irreducible factor in Q[x] then f = pf', g = pg' over Q. By gauss lemma, f = (a_1 p)(b_1 f') and g = (a_2 p)(b_2 g') over Z for some $a_i,b_i\in Q$.
Here is where I'm stuck

You can impose that a_1 and a_2 are both the smallest a such that ap\in Z[X]

What do you mean the smallest

1:40 PM
The min of $\{ a\in \Bbb N, ap\in \Bbb Z[X]\}$

smallest a\in N such that ap\in Z[x]?

Yes

Well, ap should be a factor of f
How do you guarantee?

Ah but you only have a_i, and b_i in Q. You can prove that the result is also true with a_i and b_i in Z

How can we know a_i, b_i in Z

1:49 PM
if you write $a_i = p_i/q_i$, then $a_1p = p_1q_2 \times p/q_1 q_2$ and $a_2p = p_2q_1 \times p/q_1 q_2$

What does it show?

So if you name $a'_1 = p_1q_2\in Z$ and $p' = p/q_1q_2$, you find that $a_1p = a'_1p'$, same result with 1 and 2 swapped

Ok, then what is the common irreducible factor?

ap', with a the smallest integer such that ap' in Z[X]

Same question. How can we know ap'|f in Z[x]?

2:01 PM
@Tobias it just dawned on me (remember my question with $V^\infty = \bigcup V^K$ with the $K$ compact open subgroups of a group $G$ and $(\pi, V)$ a representation of $G$?) I wanted to show that $V^\infty$ was $G$-stable and your advice was that $V^\infty$ should be $G$-stable and not invariant. I can pick $v^\infty \in V^K$ for some compact open $K$.
Shouldn't $K(g) := gKg^{-1}$ work? This is compact open because conjugation is a homeomorphism and for every $g \in G$, $\pi(g)v^\infty \in V^{K(g)}$..

Because f = kap' b_1f' for some $k\in \Bbb Z$

So you're saying a_1' = ka for some k\in Z where a is the smallest integer that ap'\in Z[x].
How do you know? a is just the smallest such thing

It's a subgroup of Z

I don't understand. Why a_1' should be a multiple of a

Because $\{a\in Z, ap\in Z[x]\}$ is a subgroup of $\Bbb Z$

2:15 PM
Oh, Now it makes sense to me. Is this generally true? I mean, if f,g has common irreducible factor in K[x] then f,g has common irreducible factor in R[x] where K is a quotient field of R

I'm not sure, You probably need R to be principal

Thanks. I didn't know that statement needs this much effort. TA said it like very simple consequence of gauss lemma

It is pretty much Gauss lemma with a few technicalities
going from a rational to a integer is technical, but not hard
And we had to work from your consequence of Gauss' lemma, which can be worked out directly

I have one more question. If S is a polynomial ring with infinite variables i.e. S = R[x_1,x_2,...] where R is a ring, then S^{\times} = R^{\times}?

consider $(1+2x)^2=1$ in $\mathbb{Z}/4\mathbb{Z}[X]$

2:27 PM
Didn't consider torsion

the statement is true if $R$ is reduced

I will suppose R is a domain

domains are reduced, so yeah

What do you mean 'reduced'?

contains no non-trivial nilpotents

2:33 PM
So the degree arguement holds?

not in general, but it works for domains

Ok, thanks

2:47 PM
Random thought: Some person researching a new kind of property in math should name said property such that it looks like a misspelling of the name of another well-known property.

loool
evil

they should give generic names that are impossible to google
I'm surprised I get proper results for "very good cylinder object" (actual terminology that exists)

Oh jeez, like the "Going Up" or "Going Down" properties for overrings

Wiener sausages
Brownian sheets

3:04 PM
If $M$ is an $R$-module, where $R$ is a unital ring, and $f : R \to M$ is an $R$-module homomorphism, then $f$ is completely determined by $f(1_R)$, right?

@Thorgott Uh maybe I'm missing something now but what does it mean for $d\gamma$ to have a zero, bein a map into $TM$

it's a vector
being zero means being the zero vector
there is a canonical way to talk about zeroes in $TM$; the zero section

yeah, which is the same as to say that the differential is zero as a linear map
@user193319 ye

I don't understand. $d\gamma$ is a map $T([0,1])\to TM$, right? I see that $d\gamma_p$ can have zeroes, but $d\gamma$ itself? Sure sometimes it can be the zero map, but then I can't just blindly apply the identity theorem or am I missing something?

$d\gamma : [0, 1] \to TM$, $d\gamma(t) = \gamma'(t)$ is your map.
Derivative of a path is a path (of vectors)

3:10 PM
Hm fair enough

The zero set of $d\gamma$ is the intersection of the image with the zero section in $TM$
They are both analytic submanifolds of $TM$, so intersection locus cannot have accumulation points

is the image of an analytic map always a submanifold?

No I don't mean that but you can talk about intersection by pulling back the zero section
$(d\gamma)^{-1}(0_M)$

I'm not following what you are doing to get no accumulation points

@Thorgott And this is false, consider $t \mapsto (t^2, t^3)$

3:14 PM
right

@Alessandro The point is $d\gamma$ will not be transverse to $0_M \subset M$ but wherever they intersect they intersect with finite multiplicity (by analyticity)
So the intersection locus will still be a $0$-manifold
A finite discrete set

why finite

because $[0, 1]$ is compact

urgh I don't know anything about transversality
This was supposed to be a trivial observation lol

you can just write a proof using identity theorem, i'm not giving you a proof clearly, just saying why it should be obvious

3:18 PM
compact spaces can have countable discrete subspaces
oh, but it's closed by continuity

right :P
ToPoLoGy

t o p o l o g y

yes but I'm not seeing how to write it down properly with the identity theorem
Geometry was so nice without analysis, this is why general topology is much better

Make Geometry Great Again

@Alessandro It's really not complicated. Suppose you have a sequence $a_n \to a$ such that $\gamma'(a_n) = \gamma'(a) = 0$, OK?
Pick a chart near $a$ in $TM$ adapted to $0_M$, i.e., such that the zero section becomes $\Bbb R^n \times 0 \subset \Bbb R^n \times \Bbb R^n$

3:27 PM
Ok

Then you have a curve in $\Bbb R^n \times \Bbb R^n$ which basically wiggles up and down, hitting $\Bbb R^n \times 0 \subset \Bbb R^n \times \Bbb R^n$ at various times $a_1, a_2, \cdots$ as it goes, finally hitting it at $a$
This is the curve $\gamma'$ of course
Right?

There is also the curve $\gamma$ lying entirely inside $\Bbb R^n \times 0 \subset \Bbb R^n \times \Bbb R$, which completely agrees with $\gamma'$ at the points $a_1, a_2, \cdots, a$
Remember that $\gamma' = (\gamma, \gamma')$ as a curve in $TM$. It's an abuse of notation to say $\gamma'(b) = 0$, one really means $\gamma'(b) = (\gamma(b), 0)$
$TM$ keeps track of the base and the fiber, two components
Fine so far?

Now $\gamma', \gamma$ are both analytic curves in $\Bbb R^n \times \Bbb R^n$, agreeing on a set of times $\{a_1, a_2, \cdots, a\}$ with an accumulation point. So they must agree. This means "$\gamma' = \gamma$", which means $\gamma'$ has zero second component

3:33 PM
Problem: Show that if $f(z)$ is a nonconstant analytic function on a domain $D$, then $f$ is an open mapping. "Proof": Let $U \subseteq D$ is open, and take $w_0 f(U)$. Then there exists $z_0 \in U$ such that $f(z_0) = w_0$. If $f'(z_0) \neq 0$, then I can invoke the inverse function theorem for analytic functions to show that $w_0$ has an open nhbd contained in $f(U)$, so $f(U)$ is open....For the case of $f'(z_0) =0$, since $f'$ is analytic, it has isolated zeros, meaning I can find $r > 0$ such that $B(z_0,r) \subseteq U$ contains no zeros of $f'$ other than $z_0$. Is it possible to find

ie $\gamma' = 0$
so $\gamma$ is the constant curve
The last place is where I use the identity theorem of course

Ok that makes sense
Thanks

No biggie
@Thorgott pure geometry
analysis written in terms of pure geometry
stop being a mutant and convert to the other side

@user193319 I am sceptical that this works. You're only using properties which also hold for real-analytic functions (inverse function theorem and having isolated zeroes), but real-analytic functions don't necessarily map open sets to open sets, e.g. consider $\mathrm{arctan}:\Bbb R \to \Bbb R$ which has the non-open range $[-\pi/2,\pi/2]$

@LukasHeger This is the method my book suggested.

3:41 PM
I wish I was a mutant
just doing 2-cats is melting my brain already

That is a mutant wish

how do people do this unironically

People can't even agree on the correct definition of an $n$-category...

I'm currently at a point where I'm not able to figure out how to compose two things

Hello,
This is the link to the question:
math.stackexchange.com/q/â€¦

3:48 PM
@user193319 I'm still sceptical. You have to use some result specific to holomorphic functions which fails for real-anaytic functions on open subsets of $\Bbb R$. The approach you suggested doesn't do that

@Thorgott Why are you doing $n$-categories anyway
I don't see a $2$-category in the night sky
What's the point

I'm not doing any $n$-categories, just $2$-categories
I was doing some normal $1$-categorical stuff and it suddenly naturally required the language of $2$-categories to write down

That is an extremely motivating reason
I understand now

ok, let me give you a natural example
you know the Yoneda embedding

@user193319 the easiest proof of the open mapping theorem is using the inverse function theorem and Taylor's theorem to show that for every point one can always find a holomorphic chart in which the map looks like $z \mapsto z^k$ for some $k$, which is an open map by an elementary calculation (The last step fails for $\Bbb R$: for example the image of $(-1,1)$ under $x \mapsto x^2$ is $[0,1)$ which is not open)

3:53 PM
given a small category $\mathcal{C}$, it's a fully faithful functor $\mathcal{C}\rightarrow[\mathcal{C}^{op},\mathbf{Set}]$ sending $C$ to $\operatorname{Hom}(C,-)$

yeah ok

now, here's an observation: $[\mathcal{C}^{op},\mathbf{Set}]$ is a cocomplete category ($\mathbf{Set}$ is cocomplete and limits of functors are pointwise)

a motivating example for 2-categories is Cat, the category of 1-categories, with 2-morphisms natural transforms
or Top with 2-morphisms homotopies

yeah but what can i prove with 2-categories

@LukasHeger I have the following (which the book also suggest I use to handle the case $f"(z_0) = 0$): If the analytic function $f(z)$ has a zero of order $N$ at $z_0$, then $f(z) = g(z)^{N}$ for some function $g(z)$ analytic near $z_0$ and satisfying $g'(z_0) \neq 0$.

3:55 PM
say $\mathcal{D}$ is a cocomplete category and $F\colon\mathcal{C}\rightarrow\mathcal{D}$ is a functor, you can extend this through the Yoneda embedding to a cocontinuous functor $[\mathcal{C}^{op},\mathbf{Set}]\rightarrow\mathcal{D}$
this cocontinuous functor is not unique, but unique up to isomorphism (since colimits are only unique up to iso)

But even when $f'(z_0) = 0$, I don't necessarily have that $f(z_0) = 0$, so $z_0$ is not necessarily a zero of $f$...so I don't see how to apply the above when $f'(z_0) = 0$.

@user193319 the composition of open maps is open. $z \mapsto z^k$ is open, as I mentioned
@user193319 you can always use a translation to ensure that $f(z_0)=0$

@Thorgott gotchu

in different terms, you get an equivalence of categories $[\mathcal{C},\mathcal{D}]\cong[[\mathcal{C}^{op},\mathbf{Set}],\mathcal{D}]_{cocont}$

coconut lol

3:57 PM
if this were an isomorphism, natural in both variables, you would have an adjunction

You mean define a function define $h(z) = f(z)-w_0$?

this is not quite an adjunction, but it is a so-called biadjunction

@user193319 yeah

Does that avoid the need to distinguish cases?

this biadjunction describes the Yoneda embedding as the "free cocompletion" of the category $\mathcal{C}$

3:58 PM
you can avoid the need to distinguish cases yeah

im getting lost haha
@Thorgott what the f is this sentence
i have to start reading it from the first word after i finish 1/3 of it every time

He's speaking the language of the gods

@LukasHeger Okay, without loss of generality $f(z_0) = 0$. Then apply the theorem to write $f(z) = g(z)^N$ with $g'(z_0) \neq 0$, and then apply the inverse function theorem to $g(z)$, and use the fact that $g(z)$ and $z \mapsto z^k$ are open to conclude that $f$ is open?

in general, if you have two functors $F\colon\mathcal{A}\rightarrow\mathcal{B}$, $G\colon\mathcal{B}\rightarrow\mathcal{A}$ and an isomorphism $\operatorname{Hom}(GB,A)\cong\operatorname{Hom}(B,FA)$, which is natural in both variables, this is an adjunction

@Astyx Les mutants

4:01 PM
@user193319 yes this works

You know plenty examples of those

@BalarkaSen You can call them however you like

clearly a very good thing

@Astyx I was referencing Grothendieck's usage of that term, he wrote down some 18 people or something which he called "Les Mutants" I think

Is it difficult to show that $z \mapsto z^k$ is open?

4:02 PM
@Thorgott ya agree

wassup

Or the adjunctions (the "a") in fiat 2-categories

Oh I didn't know that, cheers @BalarkaSen

So, a 2-category (strict 2-category to be precise, but they're nicer to talk about) is a category enriched over the category of small categories. This means that each $\operatorname{Hom}$-set is itself given the structure of a category in such a way that composition is functorial. The classical example is the 2-category of small categories, where each Hom-set is of course given the structure of a category as the functor category.

ergh

4:04 PM
@user193319 the image of an open disk of radius $r$ around $0$ is the open disk of radius $r^{1/k}$ around $0$
use polar coordinates for that

Now, there's one natural way to define what an adjunction of 2-categories should be: it should be the same thing as what I said above, except these isomorphisms of sets should now be isomorphisms of categories

yeah

The sad reality, however, is that this definition is oftentimes too strict

time for lax bifunctors

E.g. in the example with the Yoneda embedding, a functor C->D correspond to an isomorphism class of cocontinuous functors [C^op,Set]->D
This says the hom-categories are equivalent
And even when just doing 1-categories, you may perhaps notice that equivalence is sometimes a better notion of sameness for categories than isomorphism

4:07 PM

@Thorgott agree

But then, I am special(ized) :)

Equivalence exhibits a pattern that is very pervasive through lots of stuff in 2-category theory, namely that the classical notion of a commutative diagram in a 1-category theoretical setting (think of two functors being inverse as example), we instead just require that diagram to be commutative up to a 2-morphisms (a natural transformation in the case of functors, thus giving equivalence)
stuff like lax functors are relaxations of functors in the same way
instead of requiring your functor preserves composition and identities, you require it preserves them up to a choice of 2-morphism (if you require those to be invertible, you get a pseudo-functor, I think) and that these collections of 2-morphisms themselves adhere to a couple necessary coherence conditions
another relevant notion is that of a pseudonatural transformation between 2-categories, which is the same as a natural transformation, except you require that all the squares only commute only up to a choice of invertible 2-morphism satisfying some coherence conditions
the reason I mention this is cause the equivalence of categories for "adjunction" given by the Yoneda embedding turns out to be pseudo-natural in both variables

it's an example of what's called a biadjunction
(except those work even more generally between non-strict 2-categories and non-strict 2-functors, but that gets very messy)
so, the point I'm trying to make is that the Yoneda embedding is a natural thing and it embeds a small category into a cocomplete category
and this embedding has a universal property, except it turns out that this property is not about existence of unique morphisms making a diagram commute, but only about the existence of unique up to 2-morphism morphisms making a diagram commute
and when we try to express that in the language of adjunctions, we naturally arrive at the 2-categorical weakening, the biadjunction
it's called the "free cocompletion", because it's an embedding into a cocomplete category that's universal in this precise sense

4:19 PM
@user193319 just for reference, here's the lemma that I would use to prove the open mapping theorem (which is also intersting in it's own right). Suppose that $f:X \to Y$ is a holomorphic map between Riemann surfaces. Then for all points $x \in X$, there exist charts of $X$ and $Y$ in which $f$ looks like $z \mapsto z^k$ for a unique $k \in \Bbb N_0$.
Proof: Using Taylor's theorem, we can expand $f$ in a chart around $0$ as a power series $f=\sum a_n z^n$. Now let $k$ be the smallest $n$ such that $a_n \neq 0$. Then $f=a_k \cdot z^k (1+\sum_{n \geq 1} b_n z^n)$ with $a_k\neq 1$. Rescale the

you can also freely cocomplete large categories, but then you're only allowed to take small presheaves on the category or something along those lines

4:31 PM
I feel like B is probably left unsatisfied

B has left. Full stop

It's not Thorgott's fault if Balarka cannot appreciate modern art
(I can't either apparently)
I can tell you about some more sane maths if you're not busy though

I am doing physics homework
@MikeMiller Let $X$ be a vector field
Consider $dX$

I mean, I can't answer the question of what you can actually prove with 2-categories; I don't know enough for that

@BalarkaSen That's probably worse than higher cats

4:34 PM
I'm just illustrating how you can naturally arrive at them from the point of language

@Alessandro yeah but "what's the use of physics" has an easy answer: i get passing grades

@BalarkaSen Ah but then it's easy, you should go to Bonn and take the higher cats course there to find some motivation

looking for motivation isnt the same as finding inspiration
i have found inspiration in physics because i have to pass
that's a "have to", not "could, maybe"

Fair enough

@LukasHeger Why not the disc of radius $r^{k}$?

4:41 PM
@BalarkaSen Wat

@user193319 oops, you're right

I'm just doing this to train myself to keep track of 3 layers of diagrams at once and am utterly failing

@Lukas so what are you doing mathematically these days?

@Alesssandro still working on modular forms, automorphic reps and Galois reps

4:44 PM
@MikeMiller physics
I am writing $dF$ for infinitisimal force lol

yeah

Cool
How much more time do you have before it has to be ready?

I want to be finished by the end of January

What does that mean though
Divergence?
1-form valued in vector fields?

@LukasHeger I see, do you already have plans for what to do afterward?

4:48 PM
@Alessandro I have multiple options, I haven't decided yet

@MikeMiller yeah 1-form valued in vector fields
stupid notation

Makes sense
Come to Muenster :P

either I stay in Heidelberg to do my masters asap or I apply to ALGANT
the latter would come with a scholarship and I could work one year in Italy
but if I stay in Heidelberg, I can likely finish my masters sooner, so I can start with a PhD earlier
Does Münster do NT?

I would probably write like $d_\theta F$ to put the connection in my notation somehow

I think there are some arithmetic geometry people here but I'm not sure
Peter Schneider and some other people I think
It's very far from the stuff I'm interested in though so I have very vague ideas of what they do

4:51 PM
ah, I heard of Peter Schneider, iirc he works on representations of p-adic groups

@MikeMiller yeah thats good
i am also doing a terrible sin

[Awful joke I can't make work] what, you're defining it with degrees instead?

I even have a book from Peter Schneider on p-adic Lie groups I think

i dont get the joke
oh
LOL

@Lukas we are working from a book of Peter Schneider in p-adic Hodge theory
I believe

4:55 PM
i am writing $F = \int dF$, then observing by symmetry that while doing the integral (which is integral of a vector valued function so integrate componentwise), the x and y coordinates vanish by some "symmetry argument" (think of integrating normal vector of a hemispherical bowl)
so I am writing $F = \int dF = \int dF \cdot e_3$
LOL
i feel like the grader deserves this for being a physicist

oh lord
you're making me reconsider taking physics this semeste

I thought you wanted to pass

I'm gonna take physics next semester

HAHAH rekt all
@Astyx i will
this is standard physics notation

I got in touch with someone though and they told me to skip to the senior-level mechanics course so I don't get bored, which makes sense

4:58 PM
rather a mutant than a degen