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12:00 AM
Hey @MatheinBoulomenos, how's it going?
 
Or how do I define a multiplication there?
 
part of the definition of an algebra is the operation
 
@quallenjäger depends on the context. People would definitely call $L^1(\Bbb R)$ and algebra of functions, but here multiplication is convolution
 
you need two things to define an algebra, a set and an operation on the set
 
12:00 AM
So there is no natural way to define such mutliplication?
 
@TedShifrin perhaps, is there semi-affordable colored chalk? (Or do people just leave that chalk laying around more elsewhere?)
 
I'd say there's more than one way, and you can't call any one way the "natural" way quall
 
@quallenjäger on the space of all functions, you can multiply pointwise. But if you have a subspace it might not be closed under that
 
there are several common ones
 
The good stuff isn't cheap, Demonark, but your department should buy it.
 
12:01 AM
composition, multiplication, addition, convolution, ...
 
@Daminark Pretty well, thanks. Just finished a post on rep theory
 
Thanks @GFauxPas @MatheinBoulomenos
 
sure
take care guys
 
That clarify things up.
 
Yeah we just have Crayola, also the white chalk, though it seems to not be as bad. Every now and then Hagoromo is laying around and we're all just like :O
@MatheinBoulomenos nice! Post meaning, MSE answer? Or something else?
 
12:03 AM
To extend a family of 2d functions into three dimensions can you just rotate them all
 
@Daminark blog post. Beware, shameless plug: wlou.blog
@Daminark How's it going yourself?
 
What is a square integrable function
 
Going pretty well. Tentatively I've decided on Galois cohomology because I'm mildly meh on doing elliptic curves with no AG, and there's probably not much of a point in doing it now if I intend to do it next year
 
Haven't eliminated modular forms from the question though
 
12:10 AM
both are really cool and important for NT
@Daminark you might enjoy the blog post, it's mostly examples for now. The story is that a friend of mine who doesn't know rep theory but is interested asked me if I could teach it to him by writing some blog posts about it, so now I'm doing a series on that
 
@Ted I forgot how quickly paced this chat room can be!! All for the good!
 
Suppose I know that $\psi(x_u)$ can be expressed as a linear functional of some function $S(x)_u$ and I know that $S(x)=\int_0^T S(x)_u dx_u$. How can I conclude that $\int_0^T \psi(x_u) dx_u$ is also a linear functional of S(x)
 
Demonark should read and comment. He'll be helpful.
 
Can I simply interchange the linear functional with the integration?
Is there a possible way to do that?
 
@quallenjäger: Your notation is confuzling. What are the $u$ subscripts?
 
12:18 AM
Sorry, $x$ is a bounded variation path and $x_u$ is $x(u)$, where u ranges from 0 to T
 
@TedShifrin I wonder if I have too much examples and motivation. I start with 4 paragraphs of philosophical musings and then I give 11 examples (some of them are really families of examples) before I prove anything
 
I'm OK with that, @Mathein. It's a blog. Not a thesis.
 
yeah, true
 
@quallenjäger: I'm still lost in your notation. So are the integrals really with respect to $u$? Yes. You need to fix all that and get it right.
 
My integrals are with respect to some bounded variation path $x_u$
This can be understood as kind of Riemann-Stieltjes integral.
 
12:23 AM
I don't understand, then. You can't put $\int_0^T$. You mean $x'(u)\,du$ then.
 
The problem is, $x'(u)$ must not be integrable for bounded variation path
So one need other notion of the integration of $\int_0^T dx_u$
This is then the Riemann-Stieltjes integral.
 
Right. That's fine.
Ah, OK, I guess your notation is consistent with R-S.
Apologies.
 
I have to apology. I am bit lazy and left a lot of things out.
 
This is quite good! If you are looking to get some results earlier, you could talk about how if you have a finite group, you get diagonalizability, and simultaneously if it's abelian. Probably not necessary but it's at least kinda "easy" somehow and gives a grip on things
 
The definition of $S(x)$ is given in the beginning of the Section 2.
 
12:26 AM
Demonark: This is like the good stuff I used to get. Maybe UC can buy some.
 
I can't write it here, because the Latex gives always error. I don't know why.
And the part I don't understand is the Lemma 4.1
 
Matt was actually talking about that today since he was starting Fourier theory of finite abelian groups (first time he explained it that I actually understood). It built up quite nicely to the idea of characters of groups
(Specifically those representations which arise out of taking the free space given by a G-set
 
But I am working with the bounded variation case instead of p-finite variation. So you can think all the integration in R-S sense.
 
@Daminark ah, I intend to prove that stuff in some later posts
 
Oh okay
 
12:28 AM
I simply don't get the step that they can conclude that the integral is a linear function of $S(x)$.
 
I'm going to do a character-free approach first and prove a lot of stuff with some ring theory and then do a post after that on characters
 
Do you mean character in the sense of character theory? Or do you mean homomorphisms to S^1?
 
in the sense of character theory
 
Ah, I had in mind the latter
 
12:31 AM
That link doesn't help me at all, @quallenjäger.
 
when you have noncommutative (say finite) groups characters as homomorphis to S^1 can only give you information about the abelianization
 
Is there any possible way to see that $\int_0^T \psi(x_u) dx_u$ is a linear functional of $S(x)$?
 
so you can still do Fourier analysis on finite groups, but you do actually consider higher-degree characters in the sense of representation theory
@Daminark there's a nice book of Steinberg about rep theory of finite groups that takes a Fourier based approach
 
@quallenjäger: I'm trying to sort out unfamiliar notation. It seems to me it's just that the linear functional pulls out of the integral. That's just standard.
 
he also does the stuff for noncommutative groups and even random walks on groups
 
12:36 AM
How do you mean linear functional pulls out of the integral?
So I can interchange the linear functional with the integration?
 
Yes.
I'm sorry it took me so long. The notation scared me.
 
Yeah I've heard about Steinberg actually!
 
I have to admit that it is very confusing, because of the different notion of the multiplication.
How can I see that I can interchange the linear functional with the integration?
 
Demonark: You got my link on the chalk?
@quallenjäger: Think about writing down a finite sum. Does the linear functional interchange with the sum?
 
Yes#
But the limit?
 
12:40 AM
But the integral is the unique number between upper and lower sums.
(That should still work with R-S.)
 
I did see it
 
OK, Demonark.
 
That would suffice to justify that the limit can be pulled out of the linear functional?
 
Yup, @quallenjäger. Maybe there's something subtle with your stochastic setting, but I dunno.
 
There is no stochastic in there yet. It is all about integration against p finite variation paths.
 
12:43 AM
I think it's fine.
 
To be honest, I am still not really convinced that I can commute the limit with the linear functional. Could you be bit more specific?
 
Linear functional means continuous linear functional.
 
Ah I see, so the continuity allows me to exchange the limit
 
@MatheinBoulomenos so I'm not sure in general what kind of audience you're going for, though it does seem like modulo some categorical talk and some examples, it's the kinda thing that you can understand after having taken an algebra course. If that is what you're going for, it might be good to put a disclaimer that some of the examples present (e.g. algebraic groups and parallel transport) within are meant for readers with more background and aren't necessary for reading the rest
 
Are there any good books, which provides me a deeper insight into similar tools like this?
 
12:47 AM
That's a great suggestion, Demonark.
 
@Daminark yeah that disclaimer is a great idea, thanks
 
@quallenjäger: I'm not sure. This seems like standard analysis/introductory functional analysis. Maybe Riesz-Nagy. I dunno.
 
Ok, thanks Ted
 
Sorry I was so slow.
 
Don't be! It was great help as always.
You guide me to the right direction to think about
as always
 
12:49 AM
Hello:) I have an optimization question. When the exersice says 'write the kkt conditions for the problem' means that I should write 'the kkt conditions are $\min f(x) s.t. g_i(x)\le 0, h_i(x)=0$, x feasible, $I=\{i:u_ig_i(\overline x)\}$ and there exists constants $u_i\ge0 ,v_i$ not all zero such that $\nabla f(\overline x)+\sum_{i\in I} u_i\nabla g_i(\overline x)+\sum v_i\nabla h_i(\overline x)$' am I right?
becuase if we have those conditions, we can say that $\overline x$ is minimum
 
I am not really familiar in certain field like functional analysis and algebra. So sometimes I need tools there but I don't know where to look for.
But here, I always receive the right help.
 
For some things, I'm much better with specific references. This seems sort of generic and I'm not so sure. But I think you'll get it.
 
Sure. It helps already a lot if you tell me where to look for.
Because it is such a pain to look at wrong tools and at the end it doesn't work.
 
Riesz-Nagy is an old, but concrete text on functional analysis. Probably better than fancy new texts.
I gave away almost all my books, so I now longer have it. :(
 
@Daminark I added a disclaimer as you suggested
 
12:54 AM
Are you still keeping reading math books?@TedShifrin
 
No, @quallenjäger. Not really.
 
Nice
 
But if @EricSilva works on a paper of Bryant's, I've promised to try to help :)
 
I see.
Are you familiar with Hopf algebra?
 
Bryant is the Geo/forms guy?
 
12:55 AM
yes, Demonark.
No, @quallenjäger, but @mathein and others here are.
 
@quallenjäger I know a tiny bit about Hopf algebras
 
I see. Lol I'm finally starting to know more mathematicians outside of my school who are still around
 
haven't really studied them systematically, but they come up a lot
 
Good for you, Demonark. Sincerely.
 
What is actually a group-like element?
How can I see these things in a simple context
intuitive and simple context
My current topic involves so many areas in mathematics. I really have the feeling that I don't know enough
 
12:58 AM
Funny thing actually, combo final had a question (worth very few points but still) about naming 4 mathematicians that came up in class but whose names weren't written on the test already. In the review session, my prof stressed the importance of knowing who people are in the subject instead of just seeing some theorems
 
It's a challenge, @quallenjäger. Don't get discouraged.
 
But I don't want to give up. I try to learn as much as possible
@TedShifrin I won't!
 
Interesting, Demonark. I've never done that, but maybe I should have. I have some historical blurbs in two of my books.
 
@quallenjäger the motivation behind the definition of a group-like element is the group algebra. If you consider a group $G$ (say finite), then you have the group algebra $k[G]$, the free vector space generated by the elements of $G$ together with a convolution product (that's really just taking the multiplication of $G$ on basis elements and extending it linearly.) It turns out that this thing is a (cocommutative) Hopf algebra
and the comultiplication is given on basis vectors by $g \mapsto g \otimes g$
But this holds only for the basis elements, for example, for two different basis elements $g,h \in G$, we send $g+h$ to $g \otimes g + h \otimes h$ which is not the same as $(g+h) \otimes (g+h)$
 
@MatheinBoulomenos The "circle cross" is the tensor product between groups?
 
1:00 AM
Also in every class he teaches he asks people to spell the singular of the word "vertices", which amuses me for sure
 
where I don't even have a base?
 
it's the tensor product of the group algebra with itself
 
LOL, Demonark. That's one of my (many) pet peeves, too.
 
$k[G] \otimes_k k[G]$
For any Hopf algeba $A$ (or coalgebra, really), the comultiplication is a map $A \to A \otimes_k A$
This is the opposite direction of multiplication, which is a map $A \otimes_k A \to A$
 
Very stupid question, can one always find basis elements in the group $G$?
 
1:03 AM
no, every element of $G$ is a basis element
We take the vector space where we index a basis by elements of $G$
 
Ah I see, so you mean taking multiplication of G and then extending it linearly to the $k[G]$?
 
exactly!
I haven't given you the full Hopf algebra structure, though, we also have the counit $k[G] \to k$, that is given on basis elements by sending each $g$ to $1$ and then extended linearly
 
So I have two multiplications, one multiplication which is given by the Group structure. With that I will define the convolution product?
And the comultiplication is given by the tensor product.
 
no, the multiplication by the group structure is a kind of convolution product
 
So what does actually the comultiplication do?
 
1:07 AM
it's induced by the diagonal map $G \to G \times G$
So it allows us, if we have for example two actions on some things to define the "diagonal action" on the product of those two
that's how I've seen the comultiplication used
So the idea behind grouplike elements is that you want to recover $G$ from $k[G]$
Obviously $G$ is a subset of $k[G]$, but if we don't know what it is, how can we single it out by using just the Hopf algebra operations?
 
I see, good point!
And what is it relation to primitive element and exponential map?
 
The two conditions that turn out to give exacty the basis elements we started with are $\Delta(g)=g \otimes g$ ($\Delta$ is comultiplication) and $\varepsilon(g)=1$ (this is the counit)
 
I just want to say ... thanks @Mathein for helping out. It's great.
 
Definitely! It makes all the textbooks easier to read.
 
so that's what you take as definition for general Hopf algebras
@TedShifrin It's fun :)
 
1:12 AM
I was a physicist, so I always need the intuition to understand stuffs.
 
@quallenjäger hmm, I'm not sure about that
 
I'm all in favor of that, @quallenjäger.
 
I also need intuition for stuff, but I think I base my intuition on different things
 
Whoa. Where did the exponential map come from?
Yup @Mathein evil grin
 
I have read that
the exponential map maps primitve element to the group-like element
 
1:13 AM
Yikes. I don't know about this sort of exponential.
 
That is how they proved the Baker-Campbell- Hausdorff formula.
 
goes back to his corner
Whoa.
 
In mathematics, the Baker–Campbell–Hausdorff formula is the solution to the equation Z = log ⁡ ( e X e Y ) {\displaystyle Z=\log {\big (}e^{X}e^{Y}{\big )}} for possibly noncommutative X and Y in the Lie algebra of a Lie group. This formula tightly links Lie groups to Lie algebras by expressing the logarithm...
 
ah, this is about Lie groups, right?
 
The second bullet point.
 
1:14 AM
I know BCH. But ...
What does second bullet point mean?
Yup, @Mathein
Oh, this .... r = exp(s) is grouplike (this means Δ(r) = r ⊗ r) if and only if s is primitive (this means Δ(s) = s⊗1 + 1⊗s).
That's beyond me :P
 
Yes, does it tells us that exp maps primitive element to group like element?
 
If and only if.
 
@quallenjäger I heard of BCH for p-adic Lie groups, but I haven't seen this stuff with noncommutative power series (I love it, though!) But I can't help with the details of that right now, I have to go to sleep.
I don't know much about Lie theory
 
Schlaf gut, @Mathein.
 
Ok Thanks! I have learned a lot.
 
1:19 AM
Bis morgen.
 
ICh aber auch
Ist ja 3 Uhr schon
 
@TedShifrin Gute Nacht! Bis morgen
 
LOL
 
Gute Nacht!
 
Gute Nacht @quallenjäger
 
1:20 AM
I go sleep as well Ted
 
@heather !!!
 
Its 3 o'clock in germany
 
Das hab'ich verstanden, @quallenjäger.
 
haha:D ok see you.
 
See you Mathein!
 
1:21 AM
see you @Daminark
 
See you all
 
@TedShifrin hello
why the exclamation points?
 
I miss seeing people who used to hang around more frequently. Sorry if they upset you :D
 
?
i am rather confused right now.
 
LOL ... no matter.
I withdraw the exclamation points.
 
1:23 AM
i was just popping in to ask if anyone could help me with this question of mine on the simpson method.
 
I didn't pay attention to your Simpson question. Did you get it settled?
Oh, no. Darn.
 
@TedShifrin nope =P
 
I just love "cumbersome". GRR.
 
i'm debating placing a bounty, but the last time i did, i a. figured out the problem literally the next day, and b. got no real answers other than my own write-up.
@TedShifrin right?
 
@heather: You should put more of the final quote. So what are the $a,b,c$?
Plus, what is Fig. 1?
 
1:25 AM
editing to update right now =)
 
We don't want to all go read the original paper!
 
yeah =)
 
Ugh, that paper looks very 1960s.
 
it is
 
sad face
 
1:28 AM
okay, added fig 1 and the definitions of a, b, c
i have to pop out for now
see you around
 
Thanks. So let's focus on your question. You're trying to find a parabola that fits three non-equally spaced points.
You can't pop in and pop out! :D
 
Rip
 
Why do I bother, Demonark?
 
Sometimes, amidst all the popping out, you find one who stays and you solve a problem together. And that makes it worth it
:P
 
Thanks for the platitude :P
 
1:40 AM
So what've you been up to?
 
Hi @TedShifrin :)
can I ask a simple question on here about dimensional / dimensionless units?
say we had a length scale of 0 to 5 cm, and we divide this length scale by say 1/2 cm, this gives a new set of numbers that's "dimensionless". what's the difference? does dimension here mean something like in linear algebra?
and does dimensionless also mean something like in linear algebra - like, scalars?
I have little physics background, so just need to pick up some basic lingo
 
2:06 AM
There's really not a linkage between the physics meaning of 'dimension' and the linear algebra one.
With physics, the lingo is that a given quantity has dimensions of 'mass' or 'length/time'
Whereas with linear algebra a dimension is always a positive integer.
 
2:24 AM
@Semiclassical so 'dimension' in physics just means 'units'?
yeah, seems to be, based on a google search
thanks for clarifying that there's no connection to linear algebra
could I ask you why the units "cancel" to give us dimensionless numbers?
is it because dividing one length scale, in centimeters by some length, also in cm, the result is a set of ratios, which are dimensionless?
 
Well, more broadly I'd say that units act like 'non-numerical' bases, e.g. cm^a*cm^b = cm^(a+b)
 
the difference between 'units' and 'dimensions' is that the same dimension can have multiple units for it
e.g. cm/inches/miles are all units of length, and it's length which is a dimension
the point, though, is that a ratio in one set of units has to be the same in any other set of units
if you say that a rectangle is five times wider than it is tall, then that's true whether it's 5cm-by-1cm or 200 miles-by-40 miles
the proportions are the same
 
I see
is that a "similarity" argument that's ... also not connected to linear algebra at all?
 
2:38 AM
I see
 
there is a linear algebra connection, but it only emerges once you have more than one dimension to worry about e.g. energy has dimensions of mass^1 * length^2 * time^-2
 
so it won't matter at all for when you're doing just 'length/length = dimensionless'
 
I see
if I consider the set [0, 5] with units = cm, then is the "length scale" centimeters or is the "length scale" the entire interval [0, 5]?
 
the former.
 
2:40 AM
ok, so call it a centimeter scale ...
 
an element in that case would be some length between 0 and 5 cm
 
I see
if I have several dimensional or dimensionless parameters, why is one of them referred to as the "control parameter" - I saw that in a paper today. Aren't all of them control parameters in a physical model?
I think they are fixing all other parameters but letting one vary ...
do you think I have that right?
so that if we fix all parameters but let 2 of them vary, then we have 2 "control parameters" ...
 
not quite. the speed of light is c=3.0x10^8 m/s is a dimensionful quantity, but it's not a quantity that the experimenter can control
 
so there can certainly be parameters that an experimenter can't control, or at least can't control in the given experiment
but often times you'll just pick one of the controllable parameters and vary only that one
for instance, you could look at the amplitude of oscillation of a pendulum of length L with a mass M at the end
you could vary either the length L or the mass M, but one may be more practical than the other
 
2:48 AM
I see
for graphs, say the interval for length is originally [0, L/3], from some paper, where L is length, using a centimeter scale. I've heard that it is typical to divide by L/3, in order to use the interval [0,1] instead. Is this called "normalization" like in math / linear algebra?
the division also throws me off, because the lengths have now all changed ...
shouldn't we be analyzing the true lengths of the rigid body?
 
i'd say it's similar, in that you're choosing to express quantities in a way that's more convenient
 
the normalization in that case would be something like s=x/(L/3) = 3x/L
and you can recover the original quantity as x=(L/3)s
 
one point is really that the analysis should be expressed in a form that doesn't care about whether you're working in units of centimeters or feet or whatever
 
2:56 AM
May sound dumb, but quick question: suppose a function $\vec{v}(t)$ parameterizes a Jordan arc for all $t \in I$. Then can we express its tangent vector as a function of $\vec{v}$ alone, which is defined over $I$?
 
@Semiclassical wouldn't that be bad, though? because then it would lack contextual background ...
say, in describing meteorites ...
 
depends on the problem
 
however, in that scenario, the issue is that there's some extra length scale in addition to the one that you're using
for instance, if you're talking meteorites, then probably the radius of the earth is relevant in addition to the size of the meteorite itself
 
3:00 AM
that radius of the earth will matter regardless if you work in miles or kilometers
another way to look at the [0,L/3] -> [0,1] is that you're choosing to describe the lengths in terms of proportions, e.g. a length is one sixth of L -> s=(L/6)/(L/3) = 0.5
 
ah, right ...
should I call [0, L/3] a length interval?
I had mistakenly been calling it a length scale
 
sure, or more colloquially it's a range of lengths
 
I see
the paper uses the term "control parameter" while a professor I talk to often talks about the paper and says "control variable" instead. Is it ok to use both terms synonymously?
 
yeah, I'd say so.
 
3:08 AM
One can draw distinctions between them but
people tend to not always be precise about that
 
I see
thank you so much for your time and help, @Semiclassical - I really appreciate it
 
np. you may want to talk with your prof to make sure you're using the same terminology, mind
 
ok, will do
thanks so much
have a great night
(or great day)
 
night
 
 
2 hours later…
4:52 AM
Hello one and all!
Does any one have any examples of differentiating under the integral sign for 2 dimensional integrals?
 
@Rumplestillskin What do you mean, "differentiating under the integral sign"?
 
Excuse me. You know the Leibniz integral rule?
I would like to see this in action for higher dimensional integrals
 
Oh, not sure about it for higher dimensions, sorry :/
 
Zee
5:41 AM
@Daminark you like Schopenhauer?
 
5:52 AM
6
Q: Differentiation under the integral sign for volumes in higher dimensions

Pete CaradonnaConsider a smooth convex/compact domain $D\subset \mathbb{R}^n$ and a smooth, concave function $F:D\to \mathbb{R}$. Then we can define the function that simply takes the volume of the upper contour sets determined by the argument: $$G(t) = \int_{\{x\in D \; : \; F(x) \ge t\}} d\lambda$$ where ...

 
6:33 AM
Anyone keen to go on wild tangents regarding primes?
 
depends, how wild?
 
I'm not sure if you saw the post regarding primes and how they curiously form fractal patterns when distributed in triangular structures?
 
nope
 
Zee
When people say submanifold , do they usually mean immersed or embedded submanifold ?
 
they should make that clear in context, no?
 
6:39 AM
@student novaspivack.com/science/… - it's not a paper, by any stretch of the imagination, but it is compelling
2
so I got thinking about triangular numbering systems, and factoradics came to mind immediately
 
thanks for sharing the link :-)
 
there is a pattern there too. gist.github.com/jcdickinson/47883fa3acc8df6070312d235b6f9fa9 (note the repetition of 210 at semi-regular intervals)
the entirety of 4! never contains a digit of more than 4 (same with 3 and 2).
4! also always ends with either 210 or 010
as does 5! (so far as I have taken the experiment)
 
@Zee^
::still reading::
 
Zee
If they did , I would not ask this ;)
 
6:56 AM
another pattern: the last N-1 digits of each factoradic group appear always as the last digits in the next group
i.e. to find primes, you'd need to find the factoradics of previous primes. take the last N-1 digits and ensure that your current prime ends with those digits.
(there would be false positives, but not false negatives?)
 
@Zee I dunno the guy
 
Zee
7:12 AM
@Daminark and you call yourself a philosopher :p
 
Do I? I have a mild interest in it but that's all
 
7:30 AM
for the experiment to hold, 1 has to be a prime
factoradic(73) = 30010, 0010 doesn't appear unless factoradic(1)=10 appears first
 
8:30 AM
Hi @Daminark
 
Hey @MatheinBoulomenos!
 
Currently writing on the follow-up post. It's not going to be as elementary. I basically need one theorem from ring theory and then a bunch of stuff for representations of finite groups (in the case where Maschke applies, so e.g. characteristic 0) follows
 
8:46 AM
Okay so the proof you gave of Maschke seems to work, if we're being optimal, whenever the characteristic doesn't divide the order of G, right?
 
yeah
One can show that it's an iff for finite groups, but I planned that for a later post where I plan to study exactly the case when the characteristic divides the order
 
Okay yeah, nice. Before reading your thing I only knew about the inner product business
 
yeah, the inner product stuff is really nice, but a bit specialized wrt the ground field
 
Yeah
 
I want to go into that, too, but for now, I'll take a different approach
 
8:52 AM
That's fair for sure
I should consider getting a blog honestly
 
yeah, it's quite fun imo
Forces you to write some things down properly that you wouldn't otherwise, too
 
Earlier today I talked to my professor asking about topic choice and one of his recommendations was essentially about having Galois actions as representations and trying to do some computations of that sort, the logic being that when people introduce either, they often don't include the perspective of the other in examples, so it'd be the kind of thing that's worth having on a website for people to look up
In the midst of that he said "Do you have a website? In grad school that's a thing you should have for sure"
 
I'm actually going to do a project on computations with Galois representations with a prof this summer
 
Ah nice!
 
it's about a magma algortihm. My prof has seen some regularity in certain Galois reps, but so far he could only compute very few examples, because the stuff is computionally expensive, that's why he hesitates to call it a "conjecture" so far
 
9:04 AM
Oh you meant computations with a computer
 
9:16 AM
yeah
 
When exactly is "summer" for you?
 
starting august, I meant the summer break we have
it's true that we already have summer
 
Oh you start summer break in August, holy crap
(I know you meant summer break, I just didn't know when that was for you given that every school in the states is probably done now)
 
10:05 AM
My boys, $k$ be field, and let $A,B$ be $k$-algebras, say finite-dimensional. Does it ever happen that the inclusion $\operatorname{Alg}_k(A,B)\hookrightarrow \operatorname{Hom}_k(A,B)$ splits?
It probably does, but what are some equivalent conditions?
 
Seems pretty unlikely to me except for trivial examples. I don't see a way to associate to each $k$-linear homomorphism an algebra homomorphism.
 
Morning @Mathein @Daminark et al.
 
Hello, I have been struggling with a question if someone could help that would be appreciated. If I have two points lets say x and y, those points are over edges of an image. I can calculate the tangent lines l_x and l_y. Now I have a homography H, and I want to compare the tangent lines l_x and line l_y so I want to know how similar are they, using the dual homography l_x' = H^{-T}l_x. and I want to compare how 'similar' are l_x' and l_y this is all in P^2, I don´t know what to do
 
Morning @ÍgjøgnumMeg
 
@Mathein how's it going? :) Read your blog post on representations, was cool
 
10:21 AM
@ÍgjøgnumMeg pretty well, thanks, writing on the follow-up
@ÍgjøgnumMeg how are you doing yourself?
 
@Mathein slight hangover, taking a look at p-adic stuff in neukirch
lol
 
feel free to ask me questions
 
Thanks :D
 
10:42 AM
@ÍgjøgnumMeg I think I'll actually do algebraic topology instead of ANT during the summer in the end :/
 
@Alessandro that's cool man, no problem
 

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