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Bob
12:04 AM
Good evening
@TedShifrin You might want to respond to my post on Math Meta
0
Q: What Counts as a duplicate Question?

BobImagine somebody who is studying differential equations. He comes to the section on first order linear differential equations. He attempts to do the first problem in the section, but he comes up with the wrong answer. He then posts the problem and his answer to Math Stack Exchange. Somebody point...

It deals with what counts as a duplicate question
 
 
2 hours later…
1:53 AM
thanks bob how about when a question is being voted for no good reason can you cover why that happens?
 
 
1 hour later…
3:21 AM
Convolution is odd.
Also, Wolframalpha does not like it, apparently.
 
3:45 AM
It's basically an integral of the form:
$$\int_{\Bbb{R}} (x-t)g(t) dt$$
which is in general hard to compute
 
vzn
4:21 AM
@Secret there are many other ways seems to be significant beyond those proposed/ "widely" acknowledged, not the least that maybe solving it would lead to new techniques that are very strong and applicable to other areas vzn1.wordpress.com/code/collatz-conjecture-experiments
 
f(x-t), there. The convolution with x is sort of ok.
 
lol, typos
but yup
 
Convolution has some nice properties but unless you're working with like Dirac deltas or related things... Not really worth trying to compute
 
I recall back in my honours year when analysing molecule spectra, I need to compute a rather scary convolution where $g(t)$ is some complicated exponential function
Both mathematica and mathlab just give up trying to find its closed form, and eventually it is computed very quickly by the good ol' trapezoidal rule
May 26 '17 at 14:24, by Secret
$$F(E,a,T,c,s)=\left(\int_{\mathbb{R}}a\frac{e^{ -\frac{(\epsilon-E_0)^2}{2\sigma^2} }}{1+(\epsilon+E)^2}d\epsilon\right) Ee^{-\frac{E}{k_BT}}$$
 
4:51 AM
@vzn I am not an expert in computational science, thus I don't initially saw that connection. But it is known that the collatz map exhibits some chaotic phenomenon, thus it will not be surprising to me that combined with the fundamental properties of primes, it has implications to PRNGs
 
@Danu The phrase I have always heard is this: Estimate how long it will take to finish. Multiply by 3. Then increase the unit. Then you will have successfully not lowballed the actual time. :)
So I look forward to seeing your paper in 18 years.
@Danu It does not. You just need to learn to talk about infinite dimensional manifolds.
The curvature approach is nice but I always had trouble understanding certain important things, eg relationships to other kinds of equivariant homology in a sufficiently natural way to use for applications. The Cartan model mostly confuses me.
But if we just want characteristic classes it's nice.
 
I'm just wondering: are the algebraic manipulations I do in my high school classes considered true algebra or are they considered arithmetic stuff? It seems different than the algebra you guys do.
 
5:06 AM
Really simple question. I am asked to determine if 2x - 10 is irreducible in Z_11[x]. It says that 2 is a unit in Z_11[x]. But 2x - 10 = 2(x - 5), doesn't x - 5 belong to Z_11 so it is reducible?
 
Hi, is there a command in GAP that can test whether a graph is planar?
 
5:20 AM
@CaptainAmerica16 arithmetic. you could call the study of systems of linear equations the beginning of algebra, maybe.
One often calls the first class in that subject "abstract algebra", and it will often cover groups and/or rings
 
polynomials are important!
the "algebraic manipulations" are all calculations within $\Bbb R[x,y,z,a,bc]$ right
many people don't even know the division algorithm
 
That's also fair
 
hi chat
 
5:46 AM
I hope 3B1B will make a video that visualises field extensions as they are hard to wrap the head around
 
So basically Maschke's theorem tells us that they're semisimple and then Wedderburn gives the sum of squares formula
 
6:05 AM
@MikeMiller @LeakyNun Thanks for the explanations guys! That makes a lot of sense based on what I was thinking.
 
@CaptainAmerica16 you know the definition of topology right
 
@LeakyNun I guess if you mean topology as a subject.
 
the definition of a topological space
 
@LeakyNun Oh, then no I don't think so.
 
ok
 
6:13 AM
@LeakyNun Could you tell me?
 
how much time do you have :P
 
I don't have school in the morning...so as much as you're willing to spare :D
 
Given a set $X$, a topology on $X$ is $\tau$ a collection of subsets of $X$ such that:
1. $\varnothing \in \tau$
2. $X \in \tau$
3. If $E_i \in \tau$ for all $i$ then $\bigcup_{i \in I} E_i \in \tau$ (arbitrary union)
4. If $E_1, E_2 \in \tau$ then $E_1 \cap E_2 \in \tau$ (finite intersection)
We call the things in $\tau$ "open"
 
I'm pretty sure I understand #4, could you explain #3 a bit?
 
it allows you take the union of arbitrarily many elements in $\tau$ and still end up in $\tau$
the way we phrase this to have an "indexing set" $I$ to index the elements in $\tau$ that we want to take the union of
so for every $i \in I$, $E_i$ is an element of $\tau$
 
6:20 AM
Oh, now it makes sense.
 
great
 
So, what's the point of all these rules? Why is that these make something a topology?
 
yeah.... about that...
I still haven't figured that out :P
 
Lol, ok.
 
I mean, the literal answer is "because that's how it's defined."
 
6:23 AM
@ÉricoMeloSilva why are those the rules?
 
the real question is "why is topology a useful definition"
 
rules for what
 
for topology
 
the historical reason is that people took proofs from real analysis and extracted a core of what the basic concepts that kept getting repeated over and over again and then abstracted
 
@CaptainAmerica16 ok let's define the "usual" topology of $\Bbb R$
so $X = \Bbb R$
 
6:25 AM
If you are differentiating an equation like $x^2 + y^2 = 169$ with respect to t when does it become $2x\cfrac{dx}{dt} + 2y\cfrac{dx}{dt}$? i.e. when do you add the extra $dx/dt$ term because the derivative of $x^2$ is just like $2x$
 
@Startec the derivative of $x^2$ with respect to $x$ is $2x$
but now you differentiate it with respect to $t$
so the chain rule applies
 
@LeakyNun Ok
 
@CaptainAmerica16for In the "usual" topology, we define $A \in \tau$ to mean that for every $x \in A$, there is $r > 0$ such that $(x-r, x+r) \subseteq A$
try to draw a picture (in your head) to get a intuition of this lol
 
hmm yeah so I only have used the chain rule w/r/t U substitution, how does it relate to when there is some other term (like t)?
 
that's quite a lot of symbols @CaptainAmerica16
 
6:27 AM
the letter u is just that---a letter
 
lol
 
there's nothing special about it
the chain rule works just as well if you use t instead of u.
 
@CaptainAmerica16 so that means $A$ "locally" looks like $(-r, r)$ for small enough $r$
 
@LeakyNun I might have to draw an actual picture, lol
 
sure
 
6:28 AM
Okay so this whole "with respect to" is a little confusing to me. I thought the derivative of $x^2$ is $2x$ with respect to Y like an quation like $y=x^2$
 
oh, it would be read like "the derivative of Y w/r/t to x is 2x"
 
you're interested in the function $f(x)=x^2$
which we typically write as just $y=f(x)$ and therefore $dy/dx=2x$
 
so basically the extra $dt/dx$ terms come up when we are differentiating w/r/t extra variables and have to use the chain rule (in this case dt)
 
My internet completely went out for a second, but it kind of reminds me of the properties of $P$ from Spivak. $P$ meaning the set of all positive numbers. @LeakyNun
 
6:36 AM
indeed that is an open set
 
:D
I think I'm starting to get it a little bit - at least at a surface level. So, "locally" $A$ is defined as an interval?
 
So, the chain rule is always being used. Usually, when we're working with a particular variable, we don't see it, because $x=x$, and $\frac{dx}{dx}=1$, so the net benefit of deriving with respect to $x$ is just that you multiply the final equation by 1. However, if you have an equation like $x^2$ and you're deriving with respect to $t$, now you know that $x$ does not necessarily equal $t$, so you have to write $x$ in terms of $t$ and take the chain rule with respect to it.
 
@Rithaniel thank you
 
@CaptainAmerica16 right
 
Why do we have to use these bookmarks to render mathjax? Is it the client side performance issues?
 
6:46 AM
it's not built into the chat
dunno why not tbh, but probably there's some reason
 
@LeakyNun Well now that we've defined the usual topology of $\Bbb R$, what can we do with it?
 
@ÉricoMeloSilva amusingly, the first definition of topology included the Hausdorff condition iirc
 
@MikeMiller this makes sense to me honestly
i feel like the natural progression starts in metric town and then eventually ur like "oh shit maybe sometimes stuff isnt hausdorff wut"
 
idk the history well enough to guess when/why it was removed
Q: is there a (non locally compact, necessarily) hausdorff space for which continuous functions don't separate points?
 
7:04 AM
ik this can happen but i dont know an example off the top of my
 
apparently integers w a dumb topology works
@MikeMiller kuratowski did this 1922 but idk y
 
Dieudonne would know. Rip
 
i guess it's called hausdorff condition bc hausdorff was the guy who gave the first defn
makes sense
 
 
1 hour later…
8:14 AM
@CaptainAmerica16 you can prove that it's a topology :P
(sorry, I rested because I wasn't feeling well)
@AlessandroCodenotti what's the interpretation for the "averaging" formula if we look at module over algebra over field?
 
So, here's a fact that we know, that if $H$ is a proper subgroup of $G$, then there's some conjugacy class $C$ which is disjoint from it
Serre says that this implies that you can find two distinct characters of $G$ which agree on $H$
And I'm becoming less and less sure of this as time goes on
You can find class functions, sure, but I don't see why you can find characters
 
@Daminark Please give some simple proof
and complete too
Don't just gossip
4
 
8:29 AM
I don't have time to help but I dunno if there are that many easy ways to prove it, simplicity of $A_5$ is a result that requires work
Like, I only know the standard proof
Also, as a general rule, please do not go around pinging people asking for help on your question
Link it once and that's it, if people answer they answer
 
@RockDock what the actual. is everyone here obliged to help you? they aren't even allowed to talk about any other thing?
12
who do you think you are?
@Daminark you weren't here when this happened
 
Oh no
 
@Daminark wow, I'm also reading Serre
 
So, in my mind, set theory and topology and other more abstract fields of math are the study of rule systems when you've relaxed almost all rules. Kind of the continuous question of "If you get rid of everything except a few specific qualities, what things are still assured to be true?" It makes me wonder how far you can continue to relax the rules and still get meaningful results.
 
This is a particular paper of Serre's, not his rep theory book
 
8:35 AM
@Rithaniel reverse mathematics
oh ok
 
It's called "On a Theorem of Jordan"
 
Reverse Mathematics? Should I google this term?
 
sure
@Daminark is this hard to prove? somehow I still can't prove it
 
In other news:
 
@Leaky so, do you know Burnside's Lemma?
 
8:41 AM
@Daminark btw I think it's only for finite groups
because I think I know a counter-example if you don't assume finiteness
 
Oh yeah finite groups
 
yes
 
But yeah so once you have Burnside's lemma, you know that if you have a transitive action on a set with at least 2 elements, some group element fixes nothing
 
Set theory and topology is hardly relaxing all the rules. There are things called domains which are an even broader generalisation to topology

Eventually, you end up with logic when you relaxes almost every rule
 
Then you let $G$ act on $G/H$ by left multiplication. There's some $g\in G$ which does not fix any coset. Well, assume $xgx^{-1} = h$. Then $xgx^{-1}H = H$, so $gx^{-1}H = x^{-1}H$, womp womp womp
 
8:47 AM
Fair enough. I guess it was overgeneralizing for me to say "almost all," but those fields do involve a degree of relaxation of the rules.

Where does category theory fall on the scale of "relaxedness?"
 
@Daminark I see
 
Category theory is like "the bag itself where everything is contained in it"
you have things like sheafs which can do a lot of very very weird maths
 
Sounds like fun.
 
It is not easy though, and typical recommendation is to have a rather solid grasp on abstract algebra before you can start to appreciate it
 
Well, I'm taking abstract algebra next semester, and I've been getting glimpses of it from meandering in this chat these last couple of months, so hopefully I'll soon have a grasp on it.
 
8:53 AM
@LeakyNun I was inspired by your doing scales videos and just did one too, lol.
 
This for spaceship sounds:
 
@WillHunting I think people still think I played it myself
 
@Leaky I don't know which formula you're talking about
 
9:10 AM
@AlessandroCodenotti so given a map $h : V \to W$ we make it $G$-linear by setting $h^0 = \sum_{g \in G} g_W^{-1} \circ h \circ g_V$ or something like that
 
What if, in Collatz, you were allowed to do $3x+1$ even if it weren't odd
 
@LeakyNun The group acts on the linear maps by defining $(gf)(v) = g(f(g^{-1}v))$
and you sum over $G$ with these to get a homomorphism, since you need a fixed point for the action.
 
@TobiasKildetoft right, so I'm asking, what happens if we don't have the group anymore
but instead just any f.d. algebra over an alg.closed char-0 field
 
And you want to do what? Turn a linear map into a homomorphism?
 
right
I suppose you just pick a basis right
 
9:22 AM
You need some analogue of $g^{-1}$ to do this sort of thing
 
I see
 
(though note that algebraically closed and char $0$ are not relevant for the group version here)
 
char 0 is relevant if you want to divide by |G|
 
I am not sure if it is the most general version, but symmetric algebras precisely have this sort of thing
 
so let's say our algebra is semisimple then?
 
9:23 AM
Sure, but we didn't need to yet
 
what's a symmetric algebra?
what is so special with groups and what makes them always semisimple, lol
 
A f.d. algebra $A$ over $k$ with a trace function $\tau: A\to k$ (i.e. a $k$-linear map such that $\tau(ab) = \tau(ba)$) such that the associated symmetric bilinear form given by $(a,b) = \tau(ab)$ is non-degenerate.
 
hmm
why don't we just say a non-degenerate symmetric bilinear form lol
 
Given such an algebra and any basis $B = \{b_1,\dots, b_n\}$, we can define the basis $B^{\vee} = \{b_1^{\vee},\dots, b_n^{\vee}\}$ by requiring that $(b_i,b_j^{\vee}) = \delta_{i,j}$
Because not all such forms come from a trace function.
 
oh
 
9:27 AM
note for example that $(ab,c) = (a,bc)$ here
 
where?
 
when the form comes from a trace function
 
oh
 
Now, if you let the $b_i^{\vee}$ play the role of the inverse elements, then everything works out as in the group algebra case
 
nice
who cares about semisimple algebras over C, they're all C^n anyway
 
9:29 AM
Who said we assumed the algebra to be semisimple? Also, they are not, since there can be matrix algebras there as well
 
wait, is C[G] semisimple?
 
it doesn't look very C^n-ish to me
 
But a symmetric algebra need not be semisimple
@LeakyNun And it won't be like that unless the group is abelian
 
but every simple algebra is a division algebra right
and they can't exist over C
 
9:31 AM
no, simple does not imply division, since division is the same as left simple
whereas simple means twosided simple
 
:o
TIL two-sided simple doesn't mean left-simple and right-simple
lol
 
left
ok I still have much to learn
@TobiasKildetoft they look so nice
at the same time having much subtlety
it's like
linear algebra is nice because everything is simple (pun not intended)
and making them into algebras (pun not intended) add ever so slightly a touch of complexity
Is C[X]/(x^2) symmetric?
 
@LeakyNun Who played it? And did you compose those pieces?
 
@WillHunting it's just MIDI
yes
 
9:37 AM
@LeakyNun MIDI played your compositions too?
 
yes
 
I see. Interesting what machines can do.
 
@LeakyNun I don't think it is, but I don't recall many good ways to rule out an algebra being symmetric easily
 
ok
 
9:54 AM
@TobiasKildetoft so why is C[G] semisimple?
because it's simple?
lol
ok I just answered my own question
 
It is not in general simple
 
hmm?
 
In fact, it is only simple if $G$ is trivial
 
can't you get from anything to anything lol
I guess you can't get from g1+g2+...+gn to g1
the world is strange
 
0
Q: Why is $\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $?

mickLet $f_- (n) = \Pi_{i=0}^n ( \sin(i) - \frac{5}{4}) $ And let $ f_+(m) = \Pi_{i=0}^m ( \sin(i) + \frac{5}{4} ) $ It appears that $$\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $$ Why is that so ? Update This is probably not helpful at all , but it can be shown ( not easy ) that there exist a ...

 
9:55 AM
@TobiasKildetoft does it have as many components as conjugacy classes?
 
You can get to anything from a suitable place, but you need to choose that place nicely
it does indeed
 
this is so sad
 
In fact, it is quite easy to show that if $C$ is a conjugacy class then $\sum_{g\in C}g$ is a central element in the algebra, and that these elements form a basis for the center
So once we know it splits via Artin-Wedderburn, we immediately see that the number of components is the number of conjugacy classes
Since all simple $\mathbb{C}$-algebras have $1$-dimensional center.
So this is actually the "best" way to show that the number of irreps of a group is the number of conjugacy classes, but it requires Artin-Wedderburn (or similar), so I needed to do it in another way for the course I teach.
 
10:33 AM
@TobiasKildetoft are you still here?
 
Let $E$ be a dense subset of a metric space $X$, and let $f:E\to Y$ be a continuous function, where $Y$ is complete metric space. Does $f$ has continuous extension from $E$ to $X$?
(I know that if continuous is replaced with uniformly continuous, then it does.)
 
@Silent no
 
oh!
 
take $X=Y=\Bbb R$ and $E = \Bbb R \setminus \{0\}$ and $f(x) = 1/x$
 
Wow!
thank you.
 
10:40 AM
I'm sure that something similar has been asked on main before, but how exactly do I use residues to evaluate the integral of sin(x)/(1+x^2) from 0 to infinity?
 
Write it as e^(iz)/(1+z^2) and use the upper semicircle contour
 
Will the integral over the semicircular part of the contour be zero?
Sorry, the limit.
 
This is guaranteed by the Riemann-Lebesgue lemma.
 
but sin(x) is odd?
 
If it indeed is the case, then I should get 2 pi i times the sum of residues over all poles in the region I'm considering
The imaginary part of which is zero
Guess I'm just gonna post this on the main site.
 
11:20 AM
ok so what formats can we add as far as CAS worksheets and data tables are concerned? Can I just go ahead and assume I can upload a maple worksheet?
ponders his future dictatorial powers once he has finally managed to gather up a band of comrades and seize means of production
 
11:35 AM
The Dictator will arrive at soonest in 2020
 
 
1 hour later…
12:40 PM
Ok I got an awesome Q for which he/her is morally obliged to drop everything he/her is doing and devote his/her time to my comprehensions
Consider the set of $N$x$N$ $(N \geq 3)$ square matrices with entries that are defined by $m_{i,j}=i\Biggl(\Bigl\lfloor \frac{1}{2}\bigl\lfloor \frac{i}{j}\bigr\rfloor\Bigr\rfloor+1\Biggr)$

Since: $$\operatorname{Det}:{\Bigl\{[m_{i,j}]_N}\Bigr\}\mapsto {\{0}\}$$

We say that the Determinant in this context is a [A] (INSERT IMPRESSIVE LONG ADJECTIVE) and that the set of such matrices is a subset o [B](SOME CRYPTIC NOTATION) Then finish off with another word declared by the SF to be common knowledge to anybody whos anybody
and solve for A and B also C
 
it's either he/she or him/her or his/her
there's no "he/her"
 
12:59 PM
Similarly, no "him/she" nor "his/she"
 
yeah but I'm referring to the authoritarian message above
 
I'm practicing combinatorics, and I came across this resource: web.eecs.utk.edu/~booth/311-04/notes/combinatorics.html
for question 4a, it claims that the answer is 24, but I got 32
my thought was that the 1 had 4 spots to go to, not just 3. Am I wrong, is the resource wrong, is it a matter of interpretation?
 
hi @CaptainAmerica16 !
5 hours ago, by Leaky Nun
@CaptainAmerica16 you can prove that it's a topology :P
5 hours ago, by Leaky Nun
(sorry, I rested because I wasn't feeling well)
so you only slept for 5 hours :(
 
I think your interpretation is correct and the resource is wrong. I can't see a way there would only be 3 places to put the 1 in a length 4 string.
Though, you could test it by writing the strings out.
 
hacked together with JS, this is what I got: Array(81).fill().map((_,i) => i.toString(3).padStart(4, '0')).filter(x => x.includes(1) && x.indexOf(1) === x.lastIndexOf(1))
which returns an array of length 32
 
1:10 PM
If $\Gamma \subset PSL(2,\mathbb{R})$ is a discrete subgroup of finite covolume, is the same true for its preimage under the projection $SL(2,\mathbb{R}) \to PSL(2,\mathbb{R})$?
 
1:57 PM
Having to ask questions on here has actually helped me learn LaTeX. I didn't plan on learning LaTeX.
 
user131753
Is there any way to define the intuitive notion of "hole" or "void" or "cavity" in arbitrary topological spaces?
 
user131753
I expect the following properties to be proved from such a definition. For example, the number of holes of any simply connected space should be $0$ and that of a torus should be $1$.
 
2:18 PM
@user170039 homology
 
user131753
@LeakyNun See the discussion in the Homotopy Theory room.
 
user131753
@LeakyNun, loosely speaking what I want is to formalize the following intuitive definition (quoted from WolframMathworld), "A hole in a mathematical object is a topological structure which prevents the object from being continuously shrunk to a point."
 
You say a torus should have one hole, but if you fill that in can it be shrunk to a point?
Also what about a sphere? It's simply connected so you suggest it should have no holes, however it cannot be shrunk to a point
 
user131753
@AlessandroCodenotti Yep. (Provided we can give a precise meaning to that intuitive thing.)
 
user131753
@AlessandroCodenotti Maybe I am only thinking about $1$-dimensional holes only?
 
user131753
2:30 PM
Let's try to be more localized. Does there exist a concept of annulus for arbitrary topological space?
 
I aggressively object to saying that homology measures holes
 
user131753
If there exists then my intuition says that "generalized annulus" should have exactly one "hole".
 
It measures something like incompressible surfaces / submanifolds
If you draw a loop around a point you deleted, sometimes that loop can be seen as detectimg a hole
But homology mainly detected that loop
Anyway I doubt there is a satisfying answer to your question.
 
user131753
Yeah. I know my question is very vague at this point.
 
user131753
Actually this this question was asked to me by an undergraduate who started reading complex analysis recently.
 
2:40 PM
I saw in the other room. Doesn't much change my answer.
They're thinking about H_1 of domains in R^2.
In that case rank of homology very precisely DOES measure the number of "holes". In a sense, this is the Alexander duality theorem.
This is basically all one gets, though the discussion of H_(n-1) M where M is a domain in R^n in the other room falls prey to the same theorem.
 
user131753
But $\mathbb{R}^n$ is an incredibly restrictive space.
 
user131753
That's why I formulated the question for arbitrary spaces because I have found that putting a question in its most general form possible make it simpler.
 
Domains in it are less so.
I know what you asked. I answered what you'll actually be able to get.
 
user131753
Any idea where I may begin searching for the answer of such a question?
 
user131753
I was actually hoping to construct a functor from the category of topological spaces to the category of graphs using this idea.
 
2:47 PM
You will not find an answer to such a question.
Ok, do whatever you find fun.
 
user131753
The basic intuition was this: All the generalized annuluses will be mapped to a loop, all the generalized doughnut to $K_2$,...
 
user131753
@MikeMiller That's bad.
 
0
Q: Lattices in PSL(n,R) vs Lattices in SL(n,R)

abenthyBy definition, a lattice in a Lie group $G$ is a discrete subgroup $\Gamma \subset G$ such that $\Gamma \backslash G$ has finite volume. Consider now the natural projection $$ \pi : \text{SL}(n,\mathbb{R}) \to \text{PSL}(n,\mathbb{R}), $$ which is surjective with finite kernel. It is know that i...

 
@abenthy Yes, it goes both ways. Should be true more generally for any $\pi: G \to H$ a homomorphism of Lie groups with finite kernel.
 
@MikeMiller Nice, I searched for it for quite some time, but couldn't find anything.
 
3:01 PM
The covolume of your lattice downstairs and your preimage lattice upstairs are the same (the quotients are the same!)
 
And why is the preimage discrete?
 
And the discreteness, if you like, is given by the fact that your projection is a covering map
But covering spaces over discrete spaces are trivial
 
And you apply this to the covering $\pi^{-1}(\Gamma') \to \Gamma'$ induced by $\pi$, right?
 
Yup
 
And the quotients are the same by the bijection $\pi^{-1}(\Gamma') \backslash SL(2,R) \to \Gamma' \backslash PSL(2,R)$?
I see. You feel like posting this as an answer or should I delete it?
 
3:05 PM
You should write this as an answer to your question. :)
I don't have the energy to write it myself write now, but maybe someone will find it useful later.
So I trust you!
 
Okay, thank you :)
Maybe I'll delete it for now, write up the answer clearly at first.
 
Up to you.
 
3:34 PM
@MikeMiller do you do representations?
 
3:48 PM
@LeakyNun What does that mean?
Both "do you do" and "representations".
 
 
1 hour later…
4:54 PM
@LeakyNun I'm back. I saw your last message, I ended up falling asleep by accident because it was 2 am where I am. I had some weird dreams.
Woah, did I sign on here again at some point? Why did you say hi a few hours ago?
 
So, the square of a quaternion with no real part is always a purely negative, real number? That's neat.
 
That's how you define "imaginary" in general in an algebra
 

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