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4:08 PM
> On the other hand, we can construct a division algebra without multiplicative inverses by taking the quaternions and modifying the product slightly, setting $i^2=-1+\epsilon j$ for some small nonzero real number $\epsilon$ while leaving the rest of the multiplication table unchanged. The element $i$ then has both right and left inverses, but they are not equal. (We thank David Rusin for this example.)
(This sounds nonassociative as well)
 
Is there a criterion of a semisimple algebraic group $G$ over $\mathbb{Q}$ for when its group of real points is compact?
 
What does semisimple mean?
 
decomposable as a direct sum of simple algebraic groups perhaps?
 
Every algebraic group over $\mathbb{Q}$ has a radical, the unique maximal connected solvable normal subgroup. It is semisimple if this radical is trivial.
The thing is that you don't get a symmetric space of noncompact type if the group of real points $G$ is compact, because then the maximal compact group is $K=G$ and $G/K$ is empty, which is not of noncompact type.
So I'm wondering if this can be seen algebraically by looking at $G$.
 
Hi all; I have a number theory question;
If anyone could help that would be great
P: Find all prime numbers p for which ${ p }^{ 2 }-1$ is not a multiple of 24.
So immediately, I broke it into cases
for all p^2<25, and all p^2>25
For all p^2<25, p = 2,3
Then, I started testing a few primes p such that p^2 (mod 24) =1
I tested 5^2,7^2,11^2, and 13^2. All had a residue of 1 modulo 24.
 
4:17 PM
0
A: What does it mean that a normal is inside a solid angle?

Ross MillikanYour solid angle here is essentially a rectangular pyramid with the tip at the origin. The base is $\sin \theta d\theta \times d\phi$ and you want the normal to be within that pyramid. It is no different from the cone case except for the shape of the region.

 
The problem is, though, I don't know how to prove (by induction maybe?) that this pattern holds for all primes
 
@DarkRunner Hint: Suppose $p$ is not $2$ or $3$. What are the possible options for $p\mod8$? What are the possible options for $p^2\mod8$?
Similarly, what are the possible options for $p\mod3$ and $p^2\mod3$?
 
@AkivaWeinberger, oh I should test out the factors of the modulo,
OK, Let me see
 
Remember that, if $p$ is prime and is not $2$, $p$ must be odd
 
So, p will always be 1 mod 2 (except for p=2)
Wait
In fact, it will be 1 mod any factor of 24!
p^2(mod 2,12,3,8,4,6,24) = 1, except for p=2
But I don't understand how that helps us
 
4:24 PM
@DarkRunner Well, p^2(mod 3) can be 0, if p is 3
 
True
But how can we use the factors of the mod to show that this residue holds for all primes? Am I missing something?
 
@DarkRunner Think of mod 8. What are the options for p(mod 8), when p is odd?
The final key is gonna be the Chinese Remainder Theorem
 
oh no
 
If we prove that p^2(mod 8)=1 and p^2(mod 3)=1 then by the CRT we'll know that p^2(mod 24)=1
 
So the options for p(mod 8), if I'm correct are 1,3,5,and 7
 
4:29 PM
Right
And then if you square all those you'll get the options for p^2(mod 8)
 
Wait, I just tried a few primes and divided them by 8; how should I know that the residues for all primes are 1,3,5, or 7?
Do I have to prove that; or should I just assume?
thanks for helping, btw @AkivaWeinberger
 
@DarkRunner Well, any number (prime or no) mod 8 is gonna be either 0,1,2,3,4,5,6, or 7
 
helo
 
However, if something is 0,2,4, or 6 mod 8, that means it's of the form 8k, 8k+2, 8k+4, or 8k+6
in other words, it's even
So if something is odd, it can't be 0,2,4, or 6 mod 8.
So our only options are 1,3,5, and 7.
 
Ah I see!
OK cool, nice
So trying all the options, the only possibility for p^2(mod 3) & p^2(mod 8) is 1
Unless p =3, but yeah
 
4:35 PM
@DarkRunner Incidentally: Another way to write "p is 1,3,5,or7 mod 8" is to write "p is 1, 3, -3, or -1 mod 8"
(because -3 and 5 are equivalent mod 8, and similarly for -1 and 7)
 
Right;
 
That way it's easier to square them all and see that they're all 1 mod 8 or 9 mod 8
but 9=1 mod 8 so we're left with that they're all 1 mod 8
 
if you have two surfaces how do you prove they are perpendicular without solving normal vectors ?
 
@AkivaWeinberger Yeah, that's better; so how can we use this to show this holds for any p^2(mod 24)?
 
which tag should I use for a question about Weingarten functions... as in en.wikipedia.org/wiki/Weingarten_function
?
 
4:37 PM
Or is CRT required?
 
if we think of situation with two planes $z_1$ and $z_1$ if their normal vectors $\hat{n_1}$ and $\hat{n_2}$ satisfy $$\hat{n_1}\cdot \hat{n_2}=0$$ then they are perpendicular. But does this translate to surfaces ?
 
@DarkRunner Well, I guess another way to finish, is to say that we just proved that $p^2-1$ is a multiple of $3$ and of $8$
And anything that's a multiple of $3$ and if $8$ is a multiple of $\operatorname{lcm}(3,8)=24$
 
could anyone help me with tag suggestions for math.stackexchange.com/questions/2616365/… please
 
@Tuki That's equivalent to saying their tangent planes are perpendicular
 
@AkivaWeinberger Hm ,ok I'll take it from here. Thanks a lot for the help!
 
4:43 PM
how do you even define that two surfaces are perpendicular ?
it's bit problematic
 
@Tuki You can measure the angle between two planes
 
@AkivaWeinberger and how would i do that ?
 
For surfaces, you look at their tangent planes at the point of intersection
You mean, geometrically?
A dihedral angle is the angle between two intersecting planes. In chemistry it is the angle between planes through two sets of three atoms, having two atoms in common. In solid geometry it is defined as the union of a line and two half-planes that have this line as a common edge. In higher dimension, a dihedral angle represents the angle between two hyperplanes. == Definitions == A dihedral angle is an angle between two intersecting planes on a third plane perpendicular to the line of intersection. A torsion angle is a particular example of a dihedral angle, used in stereochemistry to define the...
 
exactly
there is $$ \cos \rho = \frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_^2}\sqrt{a_2^2+b_2^2+c_2^2}}$$
can't find the error in that
anyway the angle is defined in that article
but this is for planes ?
if i have surfaces
$$ z=8-x^3-y^2 $$ $$z= \frac{23}{8}+\frac{y}{16}-\frac{\sin(\pi x)}{4 \pi } $$
well if i compute tangent planes of those functions at point $p(1,2,3)$ and then compute angle between these two
or possibly normal vectors of tangent planes ?
 
4:58 PM
Well the angle between two planes is the angle between their normal vectors
 
yes essentially same thing
 
Does anyone know where I could learn about energy/priori estimates online?
 
so i compute tangent planes for both functions at this point ?
that would work
 
You can also just compute the normal vectors, that might be easier
 
normal vector is hmm
if we have $z=8-x^3-y^2$
we can mark $$ w(x,y,z)=8-x^3-y^2-z $$
$ \nabla w(x,y,z) $ = normal vector ?
 
5:09 PM
Does anyone know about such sources? I couldn't find anything about energy/priori estimates on google at all.
 
@AkivaWeinberger I think the angle between the normals of the planes is precisely the minimum of angle between all pairs of lines lying in the two respective planes.
 
@BalarkaSen Should be, yeah
And that coincides with the dihedral angle
(The minimum would be when both lines are perpendicular to the intersection of the planes)
 
Right
 
is $\nabla w(x,y,z)$ normal vector for $z$ ?
 
5:21 PM
talking to me right now ?
@AkivaWeinberger
 
good good
 
He's actually replying to my questions which I pinged him a day ago
 
)))
 
@BalarkaSen So here's something you probably don't care about
Apparently, if an algebra satisfies $x(xy)=(xx)y$ and $y(xx)=(yx)x$, it satisfies $x(yx)=(xy)x$
This is not true for magmas, though. That is, you have to use $+$ in the proof
There's actually a pretty simple proof of it in the end
But you're essentially proving that any subalgebra generated by two elements is associative.
 
5:32 PM
@AkivaWeinberger really cool. I want to learn some nonassociative algebra at some point
 
That is kind of strange
 
on a related note, you don't have to require that addition is commutative in a ring, it follows from the other axioms
 
Assuming you have a unit, yeah
 
The way you prove it, is to define the associator $[x,y,z]=(xy)z-x(yz)$. The hypotheses are equivalent to saying that swapping adjacent inputs negates it
The conclusion is equivalent to saying that swapping the first and last inputs negates it.
And that follows easily (swap first two, swap last two, swap first two)
 
5:37 PM
makes sense
 
Cool
 
(Because I guess that, in general, for bilinear functions, saying $[x,x]=0$ is equivalent to saying $[x,y]=-[y,x]$?)
 
if your characteristic is not $2$, yes
 
Right
[x+y, x+y] = 0
 
@MatheinBoulomenos Not even, I think
@BalarkaSen Right, that works in characteristic 2
 
5:40 PM
you need char $\neq 2$
 
What's wrong with Balarka's argument
 
I don't see it either
 
It's always true that $[x,x]=0$ implies $[x,y] = - [y,x]$
but the other direction doesn't work in char $2$
 
Oh ok
 
5:41 PM
in char 2 one version is called antisymmetric and the other is called alternating, I think [x,x]=0 is alternating
 
In char 2, that's just "symmetric"
 
nonzero characteristic is a myth anyway
2
 
or even char, w/e
 
this distinction is also important if you do stuff like determinants or Lie algebras in char 2
 
I am stuck on a calculation. Maybe I'll have to get my hands dirty
 
5:49 PM
NO! DON'T DO IT!
You'll get Lie brackets under your finger nails!
 
ITS TOO LATE HAHA
I am using the Lie bracket right now, in the Frobenius theorem
 
Well, you better wash your hands before your father gets home!
 
Now I just need to trick you into doing a Poisson bracket.
 
@Xander Hey, hey, hey you. I am roleplaying an evil twisted scientist. You're role playing my mom. This ain't gonna fly
 
Young man, I will not have this disrespectful tone from you.
Now, go upstairs and clean your room!
 
5:56 PM
This sounds like an Earthbound game.
 
@Semiclassical I sat in on a class last year with a bunch of Poisson and Lie brackets... it hurt so much
 
lol
I only mean Poisson bracket in the physicist sense
 
well, it was a class on QM
 
Given a classical phase space with $n$ canonical coordinates $\{q_i\}$ and $n$ conjugate momenta $\{p_i\}$, the Poisson bracket of two functions $f,g$ is given by $$\{f,g\}=\sum_{i=1}^n \frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}$$
 
$\newcommand\poisson[2]{\left\{#1,#2\right\}}$
 
5:59 PM
With $\{q_i,p_j\}=\delta_{ij}$ by construction
 
$$ \begin{align*}
&\poisson{f}{\poisson{g}{h}} + \poisson{g}{\poisson{h}{f}} + \poisson{h}{\poisson{f}{g}} \\
&\qquad= f_q \left( g_q h_{pp} + g_{pq} h_p - g_p h_{pq} - g_{pp} h_q \right) && (\text{terms from $\poisson{f}{\poisson{g}{h}}$}) \\
&\qquad\qquad+ f_q \left( -g_q h_{pp} + g_p h_{qp} \right) && (\text{terms from $\poisson{g}{\poisson{h}{f}}$}) \\
&\qquad\qquad+ f_q \left( h_q g_{pp} - h_p g_{qp} \right) && (\text{terms from $\poisson{h}{\poisson{f}{g}}$}) \\
&\qquad\qquad+ (\text{other terms not containing $f_q$}) \\
 
In physics the commutator usually just means $[A,B]=AB-BA$
 
there is a thing that I had to do...
 
ahh, Jacobi
 
yarp
gotta show that the Poisson bracket satisfies the Jacobi identity, and is therefore a Lie bracket, or some such BS
 
6:01 PM
Right.
It's stuff like that which makes diagrammatic notation seem appealing to me
even though I can never really convince myself that it's worth it
 
The forms version is a bit nicer at first glance: $\{f,g\}=\omega(X_f,X_g)$ where $\omega=dq\wedge dp$ (in the 1D version you were just doing)
But then you have to explain what $X_f$ is
 
in general you could replace omega by any symplectic form right
 
Right.
Though by Darboux's theorem this is sufficiently generic?
 
6:04 PM
mhm
 
mmkay
 
locally at least
 
right
 
my office-mate does something with forms
it looks like the rantings of an insane chimpanzee to me
 
Main point of comparison with QM is that the basic commutator identity is $[\hat{q},\hat{p}]=i\hbar$
 
6:06 PM
i somehow get dot product of $\neq 0$ for the problem :/
I posted the computations here
0
Q: Two intersecting surfaces

TukiProblem If we have surfaces $$ z=8-x^3-y^2 $$ $$z= \frac{23}{8}+\frac{y}{16}-\frac{\sin(\pi x)}{4\pi} $$ Show that these surfaces intersect in point $(1,2,3)$ and that they are perpendicular at this point. Attempt to solve I know that if two planes are perpendicular their normal vector's dot...

cannot find the error myself
 
$\hbar$!
huzzah!
 
Ugh... I have to go get setup for office hours soon :(
 
I'm waiting for the front office to tell me what rooms are available for my office hours
 
all office hours are going to be online this quarter
since there are no rooms available
 
6:10 PM
wtf
 
on the bright side, I have a 27" WACOM tablet to work with
 
now i am doing this completely wrong or ?
 
Mostly I'm annoyed that I emailed them at 8:30 am and it's 12 am now
does it really take that long to check the schedule?
 
you from united states or something ?
 
6:12 PM
@Tuki After a brief skimming of your question, your approach appears to be correct. There is probably an arithmetic error in there somewhere
track it down and murder its family, then all will be good
 
@XanderHenderson sure
 
that being said, I don't think that I would get hung up on unit normal vectors
if all you want to do is check orthogonality, then you don't need to normalize, no?
 
i don't think i need to ?
 
also, use \hat or \widehat; it makes things easier to read; for example $\hat n_1$ is rendered at $\hat n_1$
 
if two vectors are perpendicular it's dot product is zero independent from their length ?
 
6:14 PM
So why normalize?
@Tuki Yes.
it is an extra computation, meaning an extra place to make a mistake
 
Does anyone know where I could learn about prori estimates online? Any notes?
 
hmm where did i normalize it ?
 
at the very beginning, you make a big deal about $\hat{n}$ being a unit normal vector
and putting a hat over a vector generally indicates that it has been normalized
 
oh
sec
unit normal would be easier to compute in this case
if we were dealing with planes
 
I could not find anything online about priori estimates?
Could anyone help me out? Notes are helpful.
 
6:19 PM
priori estimates we are talking about bayesian statistics now ?
 
Yeah, I just wanted advice. Sorry to intrude.
 
no no not intruding at all. I was just curious
priori was distribution you define before you start updating your posterior distribution ?
so you would like to know that what makes good priori distribution ?
what is good estimate for priori distribution
 
@Tuki $w_1(x,y,z)=8-x^3-y^2-z\implies \nabla w_1=?$
 
@BalarkaSen REEEEEEEEEEEE
 
Check your work.
(Your work for $\nabla w_2$ seems right.)
 
6:24 PM
hmm why i keep messing up these kinds of things
should be straight forward
 
@tuki Oh, well you don't have to explain them to me. I just want to know where I can learn about them online.
If there is any such resource you know about.
 
you probably know bayes formula ?
/ theorem
$$ P(A | B) = \frac{P(B|A)P(A)}{P(B)} $$
$P(A|B) \text{&} P(B|A)$ are conditional probabilities
but if you dont know anything mayby these are good for starters
In Bayesian statistical inference, a prior probability distribution, often simply called the prior, of an uncertain quantity is the probability distribution that would express one's beliefs about this quantity before some evidence is taken into account. For example, the prior could be the probability distribution representing the relative proportions of voters who will vote for a particular politician in a future election. The unknown quantity may be a parameter of the model or a latent variable rather than an observable variable. Bayes' theorem calculates the renormalized pointwise product of...
In Bayesian statistics, the posterior probability of a random event or an uncertain proposition is the conditional probability that is assigned after the relevant evidence or background is taken into account. Similarly, the posterior probability distribution is the probability distribution of an unknown quantity, treated as a random variable, conditional on the evidence obtained from an experiment or survey. "Posterior", in this context, means after taking into account the relevant evidence related to the particular case being examined. == Definition == The posterior probability is the probability...
@majormaki are these what your looking for ?
then i also read something about this
In the theory of partial differential equations, an a priori estimate (also called an apriori estimate or a priori bound) is an estimate for the size of a solution or its derivatives of a partial differential equation. A priori is Latin for "from before" and refers to the fact that the estimate for the solution is derived before the solution is known to exist. One reason for their importance is that if one can prove an a priori estimate for solutions of a differential equation, then it is often possible to prove that solutions exist using the continuity method or a fixed point theorem. A priori...
has something to do with partial differential equations
 
Claim: If $X$ is a connected normal space with more than one point, then $X$ is uncountable. Proof: Suppose that $x,y \in X$ are distinct. Then $\{x\}$ and $\{y\}$ are closed, disjoint sets. By Urysohn's theorem, there exists a continuous function $f: X \to [0,1]$ such that $f(x)=0$ and $f(y) = 1$. If $f$ isn't surjective, then there exists an $r \in [0,1]$ such that $f(x) \neq r$ for any $x \in X$.
Hence, since $f : X \to [0,r) \cup (r,1]$ is continuous, and $[0,r)$ and $(r,1]$ are open in $[0,1]$, we see that $X = f^{-1}([0,r)) \cup f^{-1}((r,1])$ is the disjoint union of two open, contradicting connectedness. Hence, $f$ is surjective and therefore $X$ is uncountable.
How does this sound?
 
6:50 PM
sounds good
 
@MatheinBoulomenos Thanks!
 
@MatheinBoulomenos are there two non-isomorphic groups with the same subgroup lattice (including index)?
 
@LeakyNun interesting question. I'd say probably not, but I have no counterexample
 
yo @MatheinBoulomenos
just want to verify something with you
I verified with Daminark but I am not sure of something
 
hey @Adeek
 
7:00 PM
 
what about Q_8 and D_8
 
I am not sure of the English here
what does it mean for the maps to commute what diagram exactly?
 
@BalarkaSen D_8 has more subgroups than Q_8
 
are we assuming that there is maps $f : B \rightarrow C^{\prime}$ as well ?
 
s'pose thats true
i guess the lattice of normal subgroups of Q_8 and D_8 are isomorphic
 
7:02 PM
@BalarkaSen yeah, that's true
 
sure @BalarkaSen
 
Hello there nerds
 
hi @Daminark
 
i am stuck on a calculation
help
 
@Adeek it means that we have a natural transformation between the Hom functors $\operatorname{Hom}(-,A)$ and $\operatorname{Hom}(-,A')$
 
7:07 PM
ohh
okayyy
thanks @MatheinBoulomenos
 
so for any morphism $f:C \to C'$ we have a commutative diagramm $$\require{AMScd} \begin{CD}
\operatorname{Hom}(C',A) @>{i_{C'}}>> \operatorname{Hom}(C',A')\\ @V{\operatorname{Hom}(f,A)}VV @V{\operatorname{Hom}(f,A')}VV\\
\operatorname{Hom}(C,A) @>>{i_C}> \operatorname{Hom}(C,A')
\end{CD}$$
 
@BalarkaSen just ask don't ask to ask scrub
 
i wasnt actually calling for help lol
 
it seems like a dull calculation i need to do somehow
 
7:10 PM
yeah, I know what you mean
 
What kinda calculation? Does it involve calculus?
 
it involves differential forms
and indices
lots of them
 
indices are the worst
 
i am using the einstein index notation actually
 
>when you can't reduce your suffering so you increase it instead
 
7:11 PM
I see @MatheinBoulomenos
 
@Daminark l m a o
 
How many indices?
 
just three so far lol
 
But yeah now I feel validated in my choice to not do differential geometry
 
that's not sooo bad
 
7:11 PM
i am suffering but i am still not a physicist
 
@BalarkaSen
 
Details?
 
@Daminark geometry is amazing
in all of its manifestation
 
@Semiclassical the details of the calculations are boring. i am trying to show that a certain frame is integrable
 
>just three
 
7:12 PM
ah
 
Standards are crashing
 
I couldn't get behind differential geometry, so yeah your choice was wise @Daminark
 
why @MatheinBoulomenos
 
too many differential equations
 
a lot of motivation of algebraic geometry comes from differential geometry and algebraic topology
 
7:13 PM
too ugly calculations
 
yeah true
 
I think there are different approaches to algebraic geometry, you can take your motivation from differential geometry and algebraic topology sure. but you can also motivate things by commutative algebra and algebraic number theory. (inb4 Balarka calls me mutant again)
 
mutant
i absolutely agree with what Mathein said btw
 
How horrible of an expression do you actually have right now?
 
it looks bad but i think i might have misunderstood
it shouldnt look that bad
 
7:19 PM
I see @MatheinBoulomenos
 
Also I think when it comes to motivating algebraic geometry, complex geometry is more worthwhile than real differential geometry, because the ties with algebraic geometry are much closer
 
Is complex differential geometry not as bad?
 
I never seriously thought about it, but I did a few things on Riemann surfaces and I find it nicer than real differential geometry so far
 
sure
but reading complex geometry is easier if you do differential geometry first
 
7:21 PM
i motivate algebraic geometry by topology :3
 
because most books skips most details
 
I mean, it's still a good thing I took the differential geometry course probably, but it's definitely the course that gave me the least motivation for things we were doing
 
I didn't like commutative algebra until began learning algebraic geometry. I think pictorially, but other people think formally.
 
I had a bit of curves/surfaces in my analysis camp last summer and I wasn't too fond. It felt like half annoying calculus computations that I could do but didn't really enjoy doing, and half problems that actually required geometric reasoning that were completely impenetrable
I don't even like to think about topology too pictorially since at this stage it feels vague-ish, but I can kinda manage, if only with guidance
 
maybe part of the reason I developed geometric intuition is reading Hatcher very carefully and struggling with it for hours
 
7:27 PM
But with geometry I'm just deadlocked
 
I don't have a problem with pictures per se. My intro alg top course was close to the style of Hatcher and I liked it, but the diff geo course was a different beast
 
Hm yes it seems I have been careless with the fundamentals which my calculation was based on
Shame
i can't blame it on indices
It seems fixable. Back to blaming indices
 
Hello
any idea for this question:
1
Q: How to calculate the partial derivatives Sobolev space $W^{1,p}$

Poline SandraI have the following function $f: \mathbb{R}\times W^{1,p}(\mathbb{R}^N)\to \mathbb{R}$ given by $$f(t,v)=J(t(u-v)),~ u,v\in W^{1,p}(\mathbb{R}^N)$$ where $J: W^{1,p}(\mathbb{R}^N)\to\mathbb{R}$ which is a $C^1$ function. How to find that $$D_1 f(t,v)=J'(t(u-v))(u-v)$$ and $$D_2 f(t,v) w=-J'(t(u...

 
7:56 PM
LOL
@MatheinBoulomenos just want to check with you my answer is correct
so in part A we can just choose C to be A
and in particular we have $id \in Mor(A,A)$ which must gets mapped to $\psi : A \rightarrow A^{\prime}$ and such $\psi$ satisfies the commutativity of the diagram
so we are done
we have constructed such $\psi : A \rightarrow A^{\prime}$ it is also unique by the properties of mapping
right ?
 

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