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01:08
Guys you are $\large{awesome}$! I got 2 upvotes yesterday and now I got an El Dorado hat for that question! W00T!
01:27
@Clarinetist not yet solved the ctrl+k issue for IE11, but you should not able to save and login using the menu on the right:
https://kasperpeulen.github.io/mathedit/
01:39
So the error margin of the function I'm working on turns out to be much smaller than I'd anticipated, which is great-- the downside is that multiple iterations are required, which can "grow" the error margin
Today I learned that it's actually extremely simple to calculate tenth-roots, though
 
1 hour later…
02:46
hello
Hey
@MikeMiller Would you happen to know the definition of a uniformly continuous function, with respect to the definition of a uniform space with covers?
I don't, sorry.
03:05
No problem.
Anybody happen to know what I asked Mike?
I have managed to avoid uniform spaces so far, and I am not too sad about it
Hah.
I remember they were in Kelley when I learned topology, and I skipped that part
03:14
For quite some time I successfully ignored simplicial homotopy theory. In retrospect this is unfortunate because I volunteered to give a talk about it in a week.
Sadly the definition there is different.. there's like 10 equivalent definitions, and neither is that clearly equivalent.
Sometimes that is the reason to give a talk, to force yourself to work out something in detail
And of course, my prof. used a definition that isn't in any of the books.
Is $\displaystyle \lim_{h \to 0} \frac{3^h-1}{h}$ the definition of some derivative evaluated at a point?
03:15
looks like the derivative of 3^x at x = 0
Thank you!
@CarlMummert: I was tricked (by myself, mostly) into thinking I'd be talking about topology until I started preparing...
I'm not at all in that area, but I would think it is somewhat related to topology. Too categorical?
For my taste, yeah. I'm not used to thinking this way. At some point, all this machinery we're building up in this seminar is going to be applied to do some topology I can comprehend; I was under the impression (that nobody offered but myself) that this starts being done in the section I'm covering.
03:31
Gotcha
 
7 hours later…
10:25
I will copy this here, in case someone has something reasonable to add this:
in Linear & Abstract algebra, yesterday, by johnny09
hey, i was wondering if it's possible to study rings and fields first and then group theory. does the order really matter?
10:37
@Jake1234 Have a look in the general topology chatroom.
11:22
Consider a surface patch $\sigma$, and the unit normal of the surface, $\textbf{N}$.

We have that $\sigma\cdot N=c$, where $c$ is a constant.

How do we conclude that $\sigma$ is an open subset of a plane?
12:21
@robjohn do you maybe have an idea?
12:32
$\LaTeX$ in chat? got to see þat
it's only 90 minutes to midnight here. That means that next year is 1200 in base 12. The dozenal folk are not dozing off at that one!
12:48
If you have an open patch of surface, then its surface is entirely determined by its boundary.
Suppose $\sigma$ is the normal vector of a surface (as it is in physics). Then for a closed surface $\sigma = 0$. You can prove this by noting that the enclosed volume of $\sigma$ is independent of coordinate, and $\sigma \cdot \vec{r} = V$ is independent of the position of the origion of $r$.
Morning.
Morning
We can replace divide the total surface into $\sigma$ and $\sigma'$, the sum of which is zero. By replacing $\sigma$ by any other surface bounded by the perimeter, the total sum of vectors is zero, and $\sigma = $\sigma_2$ etc.

It's used in physics as vector area, in $\vec{m} = I.da$, where $I$ is the current in a loop, and $da = the bound area
It's down to 50 minutes to midnight here. Still, we shall be the second major area into next year (the kiwis are already there!).
13:38
@MaryStar what do you mean by $\sigma\cdot N$? A surface patch is a map from $\mathbb{R}^{n-1}$ to the surface. Do you mean that $\sigma$ is the image of some small shape (a rectangle?) in $\mathbb{R}^{n-1}$? under $\sigma$?
@rob
@robjohn, if you suppose that $\sigma$ is a portion of a bounding surface, of N-1 dimensions, then it has a definite vector-area (of N-1 dimensions), defined entirely by the boundary.
It is true in every dimension.
The boundary does not have to 'co-planar' or even 'flat'. It works for any closed N-2 ring.
Huy
Huy
@MikeMiller do you know what it might mean for a representation to admit an invariant subspace? in my notes, a rep of a Lie algebra $A$ over $k$ is a vector space $V$ together with a Lie algebra homomorphism $\rho: A \to End_k(V)$, and $A$ acts on $V$ via $av := \rho(a)v$.
maybe existence of some subspace $W \subset V$ such that $\rho(a) W \subset W$ for all $a \in A$?
14:01
@Huy: That's correct, yeah.
@wendy.krieger What you seem to be talking about is the vector $\int_\Omega n\,\mathrm{d}\sigma$, which is dependent only on the boundary. If this is what Mary Star meant by $\sigma\cdot N$ being constant, that was definitely not clear.
@robjohn - that's how i read mary star's question.
@wendy.krieger Is ChatJax working for you?
@wendy.krieger Did you go to the installation page?
14:09
@robjohn maybe it needs a newbie page to get it to work. Kind of like in LaTeX in chat.
@wendy.krieger Did you drag the "start ChatJax" link to your bookmark bar?
@robjohn Yes i got it working now.
@wendy.krieger $\ddot\smile$
Huy
Huy
$$\Huge\ddot\smile$$
I use the $\int \vec n.dA$ to prove interesting things in higher dimensions, but most of my high-dimension stuff is done mentally. :)
happy new year!
14:16
HAPPY NEW YEAR guys.
2
Someone should pin the 'chat-guidelines / Latex in Chat, if it isn't done. It's really handy to have that stuff there.
Huy
Huy
it is pinned
That's a good thing.
@Balarka: What's the correct statement of implicit for Banach spaces?
14:31
Night to everyone. It's nice to meet you all.
15:09
Happy New Year, everyone!
2
15:21
@Huy: Minor note: For Lie group actions this is the same as saying that $gW = W$, rather than being a subset, because the action of $g$ is invertible. When we're working with actions of algebras this isn't true anymore. (Sometimes you'll see people say that an invariant subspace is one with $gW = W$ for all $g$; they're talking specifically about group actions.)
I am looking at this exercise: http://math.stackexchange.com/questions/1586043/open-subset-of-a-plane

At the edit part I added what I have done so far...
Huy
Huy
15:38
@MikeMiller ok, thx. I have a different dumb question: I'm looking at a proof that for a connected, simply-connected Lie group $G$ with nilpotent Lie algebra $\mathfrak{g}$, we can equip $\mathfrak{g}$ with some product such that $G \cong \mathfrak{g}$. for the last step, they make use of a proposition that if two conn., simply-conn. Lie groups have isomorphic Lie algebras, they are isomorphic.

in this proposition, I assume I need a Lie algebra homomorphism and not just a vector space isomorphism, correct? (the latter would be easy to obtain since the exponential map is a diffeo in a nbhd
Really? $G \cong \mathfrak g$?
Huy
Huy
yes
Huy
Huy
product as in "group operation"
Yeah I got you
Didn't know all s.c. nilpotent lie groups were R^n
Huy
Huy
15:40
I can outline the rest of the argument if you didn't know it
That'd be cool. Anyway, yes, you need an isomorphism of Lie algebras, meaning a vector space iso that preserves the bracket
$f([g,h]) = [f(g),f(h)]$
Huy
Huy
I figured, otherwise it would have been too easy
Well otherwise it would say that every connected simply connected Lie group of dimension $n$ is isomorphic
which is trouble when you have say $SU(2)$ and $\Bbb R^3$
Huy
Huy
@MikeMiller: on nilpotent Lie algebras, BCH formula only is a finite sum of homogeneous Lie polynomials, say up to degree $N$, in a nbhd of the origin. then define
$$X \star Y := \sum_{n =1}^N Z_n(X,Y)$$
on the whole Lie algebra. using the inductive formula for the Lie polynomials, you can show that it is indeed a group operation.
($Z_n$ are of course the Lie polynomials)
the derivative at identity of the exponential map gives the isomorphism, and I think since in a nbhd of the origin we have $\exp (X \star Y) = \exp(X) \exp(Y)$, you get a Lie algebra homomorphism
I was here when the star board wasn't filled with wishes for a 'Happy New Year'.
15:47
@Huy: Oh, I see. Very cool.
Huy
Huy
I agree, very surprising
@Khallil: Back when it was filled with Huy screaming at bystanders.
It was a great time to be here, @MikeMiller. ^_^
@MikeMiller Was working on some exercises to send to Ted, so hadn't thought about it. I'll answer it, I promise. I need to look at the proof of ImpFT more carefully.
Just write out the whole word. :P
15:53
Too much work :P
Ok, @Bal
Oh no.
Point taken.
Implicit function theorem it is, then.
Hahahahaha!
Whoever subbed this show is amazing. The context is that he's watching somebody climb down an extremely tall tower without footholds and windows.
Oops, gotta go do schoolwork.
16:44
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plz dont flag this as extra-noise
17:04
Could you take a look at my question math.stackexchange.com/questions/1589378/… and especially at the edit part? @DanielFischer @robjohn
17:45
Morning @AndrewT.
Early evenin' @MikeMiller
How's New Years wherever you are?
S'alright. Trying to get some work done. Pretty bleh.
Sounds kinda bleh indeed.
yours?
Yeah, its been good. Ate too much, going to the math departments bar with some friends.
Huy
Huy
17:59
wow ur maths dept has a bar?

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