« first day (2262 days earlier)      last day (2746 days later) » 

4:00 AM
@MikeMiller Now I know what Gaussian curvature is for surfaces. I have yet to understand what it means, however.
 
omg i solved the problem
 
Cheers!
 
and it is so nice
 
@Balarka It's curviness. Gaussian curvature is one of the most intuitive ideas in differential geometry.
 
I was so happy the day I finally learned what a Riemannian manifold was, so I could at least understand the entire line from Tom Lehrer's Lobechevsky song
 
4:03 AM
If it's positive, the surface curves in towards itself. Negative, it curves away. Magnitude measures how bad the curviness is.
 
lmfao it is so easy!!!
it cannot exist
 
I don't immediately see why it is the right measure of curviness. Maybe I should work out examples than think abstractly about it.
The sphere, the saddle and the plane (duh) should do it.
 
Try sphere of radius r and the pseudosphere too.
 
What's a pseudosphere?
 
Got it. I don't know the parameterization for the pseudosphere, but it's probably in one of the exercises in Ted's notes.
Yeah, it's there.
 
4:07 AM
@TedShifrin
lol
 
@Alan It's a silly name for some of the standard surfaces in R^3 with curvature -1. Spheres have curvature +1.
 
so your right any group where all elements have order 2 must be abelian
I just proved it
so this case can't happen
good thing you saved me from writting my initial argument
was very handwavy
thanks @TedShifrin
 
@Balarka Here's a very nice question which I'll let you keep at the back of your head. Let $\Sigma$ be a smooth closed surface. What smooth functions $f: \Sigma \to \Bbb R$ can be the Gaussian curvature of some metric?
 
Fun problem. I can't immediately come up with a nontrivial characterizing property of the Gaussian curvature, but I'll think about it once I familiarize myself with it.
 
Can someone tell me if a question like this is a better fit for Math.SE or Physics.SE?
2
Q: How do we prove that the angular velocity vector is equal to a limit?

Keshav SrinivasanThe angular velocity vector of a rigid body is defined as $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{|\vec{r}|^2}$. But that's not how most people intuitively think about angular velocity. Euler's theorem of rotation states that any rigid body motion with one point fixed is equivalent to a rotat...

 
4:13 AM
Thanks for this. However, I should focus on trying to not mess up my calculations.
 
Is there a way to view the whole question? Looks physiccs-y from what I see
 
once i asked a physics question on MSE and they caught on
3
Q: Where are the other solutions to this ODE disappearing in this analysis?

enthdegreeI am asked to solve the following ODE involving constants $\alpha, L, V_0 > 0$ and $E < 0$: $$-\psi'' - \alpha V_0\psi\cdot [\delta(x)+\delta(x-L)] = \alpha E\psi.$$ In particular, we want solutions $\psi:\mathbb{R}\to \mathbb{C}$ that are: Continuous $L^2$ We are given that the solutions l...

 
@Alan The only way to view the whole question is to click on the link: physics.stackexchange.com/questions/285843/…
 
smart people answered though
@keshav, this is a totally intuitionist answer, but my gut feel is that the reason a fixed point implies well-defined-ness of an axis of rotation is tautological
 
@enthdegree It's not tautological, Euler's theorem of rotations takes some work to prove:
In geometry, Euler's rotation theorem states that, in three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point. It also means that the composition of two rotations is also a rotation. Therefore the set of rotations has a group structure, known as a rotation group. The theorem is named after Leonhard Euler, who proved it in 1775 by an elementary geometric argument. The axis of rotation is known as an Euler axis, typically represented by a unit vector ...
 
4:21 AM
But it sounds like rotation is defined in terms of Euler's notation theorem: "When a sphere is moved around its centre it is always possible to find a diameter whose direction in the displaced position is the same as in the initial position."
*rotation
 
@Balarka You don't have the right tools to do anything with this yet. I'm just giving it to you.
 
Ah, alright. Got it.
 
wait, maybe not.
 
Oh wow, I did it for the unit sphere and it worked. K = +1.
Onto saddle.
@MikeMiller Without doing any calculations, I think nothing special should happen for radius r \neq 1. I and II are both multiplied by r, so they cancel off after taking det so you still get K = +1
 
@enthdegree If you rotate a sphere 90 degrees about the x-axis and then 90 degrees about the y-axis, it is not immediately obvious that there exists some rotation that will take you from the initial configuration of the sphere to the final configuration.
 
4:29 AM
@Balarka That is not correct.
 
Oops.
Normal is multiplied by r too
 
@enthdegree In any case, my question isn't about proving Euler's theorem of rotations, it's about relating Euler's theorem of rotations to the angular velocity vector.
 
@MikeMiller Alright, so it's 1/r^2.
 
A really small sphere is really curvy! A really big sphere is almost flat.
 
True!
The curvature of the saddle looks rather bad.
 
4:52 AM
@KeshavSrinivasan: The fastest way to see that visually seems to be to prove that a rotation can be decomposed into two reflections even if you can choose either the first or the second. This means that you can see that composing two rotations can be done by choosing the second and third reflections to cancel, leaving you only two reflections, which is a rotation.
 
Actually, it simplifies to $-1/(1 + u^2 + v^2)^2$.
 
@BalarkaSen Not what you're looking for, but it's the obstruction to the space being locally isometric to flat space
I'm sure what you're reading talks about quadratic approximations?
 
Nope
 
5:08 AM
Had to go away from keyboard. @0celo7 Can you elaborate what kind of thing you havei mind by quadratic approximation?
Also, $\kappa = -1/(1 + u^2 + v^2)^2$ makes sense for the saddle. As $u, v$ increases (so I go far away from the origin, on the surface of the saddle) $\kappa$ becomes $0$. At the origin $\kappa = -1$, so I suppose that's the local model for negatively curved surfaces.
Also the minus sign probably indicates that the Gauss map reverses orientation.
This is fun.
 
5:50 AM
anyone still here?
i've got a question for y'all.
bummer.
 
6:06 AM
I'm semi-here
 
6:23 AM
@Alan here?
 
Around, yeah
@dbliss @adeek What's up?
 
so I classified all non-abelian groups of order 8 it is generally $D_4$ and $Q_8$
I gave also presentation for them
but now I would like to give presentations
for
$Z_8$
$Z_2 \times Z_2 \times Z_2$ and $Z_2 \times Z_4$
what would be a presentation for those ?
I guess for $Z_8$ it is trivial
so $Z_8 = <a : a^8 = 1>$
$Z_2 \times Z_4 = <a,b : a^2 = 1, b^4 = 1>$ ?
 
@adeek you need ab=ba for the second
Without that, you'd get an infinite group
 
oh
 
Always remember to list how the generators interact, unless it's something like a free group where they don't
and by interact, I generally mean how do you change the order of them in a product
 
6:32 AM
for $Z_2 \times Z_2 \times Z_2 = <a,b,c : a^2 = b^2 = c^2 = 1,ab = ba, ac = ca,...> $?
yeah I see @Alan
 
yeah, and the third one, bc=cb :)
 
oh ok and then to show that the presentation given above for $Z_2 \times Z_4$ and $Z_2 \times Z_2 \times Z_2$ I have to show that they are isomorphic ?
to that group generated by those presentation
I mean for $Z_8$ it is just by definition
is a quick way to prove it for the other 2 ?
 
For isomorphisms when you have a presentation, they are generally easy to prove, you just need to prove that whatever elements you are assigning to each of the generators obey all of the relationships in the presentation
so for the first, you'd show that $(1,0)=a,(0,1)=b$ obeys the 3 relationships, and that's your isomorphism.
Through a function name on there to be technical :)
Throw, even
 
I guess I need to do it rigorously right ? I mean assign a function
$(1,0) \mapsto a$ and $(0,1) \mapsto b$
by the universal property of free groups this will induce a group homomorphism
into F(X) where $X = \{a,b\}$
then I show that the relations in the specified are satisfied
but then I need to prove that
hm it is gets complicated
Ok I will deal with it @Alan
I just have a quick question before you leave because I have been thinking about it all day
 
The relations are pretty trivially satisfied
(1,0) has order 2, (0,1) has order 4, and they commute.
Go ahead.
 
6:42 AM
@Alan can you find counter example of the following statement ? Suppose F is a free abelian group of finite rank n. I need counter examples for the following statements:
i)Every linearly independent subset consisting of n elements is a basis of F.
ii)Every linearly independent subset can be extended to a basis of F.
iii)Every generating set of F contains a basis of F.
 
Hmm, double checking my defintions...free abelian of rank n juist means it has n generators that have infinite order and all commute, right? And what do we mean by linearly independant subset in a group? (I'm familiar with that for vector spaces, not groups). Likewise, basis?
 
is that in the sense of regular vector space
It is same definition
just a sec
let me type it
0
Q: Questions on free abelian groups

SeoralIs the following true? $G$ is a torsion free abelian group of rank $>n$. Let $S$ be be a subgroup of $G$ generated by $s_1,s_2, \cdots ,s_n$. (1) If $m_1s_1+m_2s_2+ ...+m_ns_n$ is linearly independent modulo $p$ for every prime $p$, then $S$ is linearly independent. That is, if $m_1s_1+m_...

@Alan this is the definition
 
check, and a basis would have everything having a unique representaiton as a sum of.? Hmm.....
 
yeah
 
Lets see.....take rank 2
lets see, playing around here, look at....$ab$ and $baa$.
 
6:51 AM
yes
 
Clearly linearly independant, you'd need the same number and opposite sign for the b's, and thhat wouldn't kill the as
now, is it a basis? Lets see if we can write a and b as a combination of those two
 
hm
 
pretty sure you can't, try assuming you can and see if you get a contradiction?
 
yeah I will try thanks yeah I think if I adjust this example around I get it for other ones too
 
nod I always try to look at the simple cases :)
 
6:55 AM
yeah I dunno why my mind like to think of complex cases
haha
 
though this does presume that n>1. For the n=1 case, you'd have to use $aa$
 
yeah
 
actually, there it is, even simpler.
use $aa, bb,cc....$ etc.
no way that's a basis, and it's linearly independant :)
 
oh those things are linearly indepedent
 
but not a basis.
 
6:57 AM
yeah
 
And can't be extended to a basis either, because you wouldn't have unique repesentation of $aa$ then
 
yeah
btw thinking about the earlier problem I would like to do it rigorously just want to see if I have the right steps so let us say we take $Z_8$ we have $\phi: F(X) \rightarrow Z_8$ where $X = \{a\}$ and the map $a \mapsto 1$
then it is easy too see that $R = \{a^8\} \subset ker\phi$
so $<R> \subset ker(\phi)$
it is also easy to see that the $\phi$ is surjective.
I would like now to prove that F(X) / <R> has at most 8 elements
why is that ?
 
In my class, we didn't do things from the standpoint of free groups, so...hrm.
Different kind of formalism :)
 
oh
 
Hi, can someone explain me log *(n) with an example?

like, if n = 65536
 
7:08 AM
What do you mean by the *?
and by log, do you mean common (base 10) log, or natural log?
 
@Alan This, to be specific:
In computer science, the iterated logarithm of n, written log* n (usually read "log star"), is the number of times the logarithm function must be iteratively applied before the result is less than or equal to 1. The simplest formal definition is the result of this recursive function: log ∗ ⁡ n := { 0 if ...
 
Ahh, it's a measure of order of magnitude that's way, way more condenseced than normal logs. log massively shrinks a number. This says how many times do we go through that process before we get below 1. So for your example....what base log do you want to use? 2, 10, e?
 
Anything works, let's just go with e :D
 
so, ln (65536)=11.090354889
That's 1 step of shrinking, we've shrunk it a bunch, but not below 1, so we go anotehr round
ln (11.090354889)=2.40607580166
ln (2.40607580166)=.87799712231
we've gotten below 1 in 3 steps of taking logs
so our answer is $3.87799712231$
it almost took 4 steps. It's just a way of shrinking large magnitutrdes even more than logs do :)
 
7:25 AM
hm
I think there is a mistake in one of my questions
would you like to check it @Alan
 
Sure, but I'll warn you, my group theory is rusty :)
Wow, just read that it's only slower than the inverse ackermann function. The latter function actually came up in one of my PDE class work problems in a bound.
 
Let $D_{2n} = <a,b : a^n = e,b^2 = e,baba = e>$ we would like like to prove that $Aut(D_{2n})$ ~$Z_n \rtimes_{\phi} Z_n^{*}$
I think the $Z_n^{*}$ should be $Z_2$
 
I'm really, really amused that I've gotten so much rep from this silly answer: math.stackexchange.com/questions/1964987/…
 
haha
 
Hrm.....ack, gotta run. RL just rang. Good luck!
 
7:49 AM
Hey guys, I'd appreciate some help with this question math.stackexchange.com/questions/1966340/…
Simple question on direction fields and differential equations but I've gotten a lot of...not so good replies so far.
 
8:13 AM
Awesome. Thanks a lot :)
One doubt: Why is it 3.87..... ? And not just 3 or 3+1?
 
8:29 AM
Ohh nvm. Got it :D
 
@BalarkaSen Sick, never heard of that definition
 
8:55 AM
Well, let's see if this computation manages to finish in time before I lose my office and the computer it is running on. The calculation for type $B_6$ took about 60 hours, and I have just over two weeks for it to finish type $B_7$.
 
Some abstract algebra "findings"(?):
 
@Secret I have no idea how to read that
 
Caption: The convention of a cayley table is that the row element comes first followed by the column element. Therefore the table is basically the same as treating the row and columns as vectors and the entries of the table is given by the tensor product of Row and Column as $\textrm{Row}\otimes\textrm{Col}$
 
(Type $B_6$ did produce the nice looking dropbox.com/s/3vaxzpfwbn0jhy9/B6-twosided.pdf?dl=0 as well as the less nice looking dropbox.com/s/ovtl82q7lltow7b/B6-shape.pdf?dl=0 and the downright terrible dropbox.com/s/wa1rj8h9pvem0st/B6-full.pdf?dl=0)
@Secret why tensor product? It is just regular matrix product
(note that the page size on the last link is 100x230 inches)
 
(If I recall correctly) The matrix product of a nx1 row vector times 1 x m column vector is a special case of a tensor product because it gives a nxm matrix . The reason I use the tensor product instead is because the next step in constructing the nmp array cannot be written by a matrix product
Perhaps more accurately, the row vector times column vector matrix multiplication is a Kronecker product, and that is the tensor product wrt to a certain basis
Now, after this is done, all entries of the algebraic structure that corresponds to the equation (ab)c is then given by the tensor product of the cayley table to the column of elements in the algebraic structure
The resulting entries will then be evaluated based on the rules of that algebraic structure
Similarly all entries that corresponds to a(bc) is given by the tensor product of the column of entries to the transpose of the cayley table
If these two arrays (which formally should be tensors) are equal, then the algebraic structure is associative wrt the binary operator under consideration
Is this making sense to you, or still unclear?
 
9:11 AM
I am starting to get the idea
 
It's not very efficient as you are still computing all the terms unlike Light's associativity test, but it makes the organisation and perhaps automation easier in a computer as you effectively convert the abstract algebra problem into a matrix problem
 
 
2 hours later…
10:52 AM
Is $]0,1[$ standard notation for an interval?
I am reading it in a paper but I have never seen it before
 
@Lembik Yes, some places that is the standard notation for open intervals
 
@TobiasKildetoft oh that's interesting. Which subfields do you know?
 
It avoids the over-used $(,)$ at the expense of messing up when parantheses start and end
 
it is from physics?
 
@Lembik It is not so much about subfield as it is geographic as far as I know
 
10:54 AM
oh please tell me more
where do they use it?
 
As I recall, it is fairly common in Denmark at least, and I think I have seen one of the notations mentioned as belonging to some region, but I don't recall the details
 
aha.. scandanavia would make sense
I think it's a swedish author
 
user116211
11:18 AM
How can there exist a bijection between a subset and a superset?
 
easy, if the set is infinite
e.g. consider a subset of $\mathbb{Z}$ such as $\mathbb{Z}^+$
 
user116211
 
user116211
@Secret yes, the book is talking about $S\subset \mathbb N\,.$
 
user116211
But it still sounds quite odd ;/
 
@BalarkaSen I love this section of MS---it all looks so wonderfully playful and creative
 
11:22 AM
@MAFIA36790 It's because of infinity. An infinite subset of an infinite set may have the same cardinality
 
user116211
@Secret okay.
 
@Danu It's cool.
 
@BalarkaSen Hi
 
Hi @Tobias!
 
More than cool :)
 
11:24 AM
Are you onto Stiefel-Whitney numbers yet?
 
There now
I'm also typing everything up so I'm being slow
Gotta be able to present it all
The guy has given me an impossible task though
 
Sure, sure. Slow is good. Make sure you do some smallish computations after you have done this.
 
@MAFIA36790 Munkres, corollary 6.4 illustrates this, because any attempt to plug an infinite set into Theorem 6.2, it breaks down thus can no longer conclude that the subset is not bijective for that set in question
 
I'm supposed to present all of 4, plus 10 pages of appendix A to recall fundamnetal class and PD
In 1.5 hours
 
Oh no!
 
11:26 AM
no fucking way
Even just section 4 would already be very hard, I think
And I don't wanna omit any of it
I might ask for some more time... :\
 
user116211
@Secret Okay, I'm checking; Currently into Royden.
 
I can probably present section 4 in an hour.
There's really not much in there, just a set of nice ideas.
 
No way
You gotta write everything on the board
That takes insane amounts of time
(you also have to write out full sentences!)
 
Oh, you're doing it on the board, and not a slide-show?
Damn.
 
Slide show math talks
not even once
Impossible for the audience to understand what you're doing
I could talk through the section in an hour, sure
 
11:28 AM
Yeah, I get your issue.
@iwriteonbananas It's pretty hip
 
11:42 AM
anyone know of a proof that is really easy to check but hard to come up with? i post challenges on facebook. i'm thinking something along the lines of prove that there is a irrational number that can be raised to an irrational power to get a rational number but that one is pretty well seen
does anyone know if there exist a proof about a irrational number raised to an irrational algebraically independent number to get a rational result?
 
@shaihorowitz What about $e^{ln(m)}$?
 
hmm good point
i forgot about the famous e^ipi
 
Is there a name for a magma with the following cayley table?
 
@Secret not sure it has a name (note that it is a semigroup though, not just a magma)
it is a good source of examples and counter examples to various things (along with the obvious generalizations with more elements)
Ohh, actually one might call it the set of constant functions on the finite set $\{a,b,c\}$.
or at least identify it with that set
 
One thing I noticed is that its associativity table is quite trivial: given that we knew an associativity table of 3 elements is a 27 entry array. If we partition that into blocks of 3x1 columns, the result is always a,b,c
 
@Secret Right, the multiplication can be described as "the left-most element" however many elements you multiply together
 
interesting
Btw, when reading wikipedia's article about semgroups, they said that absorbers are often interesting in semigroups. Why is that and how they are used in proving or investigating stuff in semigroups?
 
I have not really studied semigroups that much
 
me as well
 
12:02 PM
ok nvm then
 
i'll read the wiki though
If a magma has both a left zero z and a right zero z', then it has a zero, since z=z\times z=z'z=zz'=z'.
If a magma has a zero element, then the zero element is unique.
 
now, multisemigroups I have a bit more experience with :)
 
uniqueness is an interesting property
 
i see
 
it also seems like its minimun sufficient information for somethings, like if we have x*y and we know that x is an absorb-er we have x, for example what is the minimum between x=(-infinity to 0) y=? we don't need to know y to note the minimum is -infinity
 
12:09 PM
ah, didn't knew that absorbers can be used like that
I suck at any investigations involving extrema in algebraic structures and set theory
It often took a couple of reread to understand why something is the least element, why is that a supremum and so on
 
@shaihorowitz where did the 0 and the infinity come from?
 
the infinity is the min max absorber and i just pulled 0 out of the air
 
@shaihorowitz I thought we were talking about the general case of absorbers here
 
yeah but i gave an example to illustrate my point.
i should have used the words "for example" instead of "like"
it seems for operations there exist unique elements such that for operation , xy=y (identity) or x*y=x (absorb-er), the absorb-er then gains the interestingness of the identity in the groups where it exist @secret and thats why there looked for and are interesting
 
12:25 PM
i see
 
my favourite absorber so far is (NaN) not a number for simple arithmatic, x+NaN=NaN,x*NaN=NaN, GCD(1,NaN)=NaN ect...
the absorber is unique but, if you remove that element from the group and it still remains closed there may be another absorber
 
12:51 PM
@BalarkaSen There's a trick about writing a surface locally as a quadratic, then the coefficients of the quadratic expansion can be used to easily compute the curvature
 
I see. Maybe it's later in the notes.
 
@0celo7 why don't you calculate the second partial derivative instead?
 
@DHMO I think because the quadratic itself (it's the taylor expansion) has the curvature of the surface
it's a conceptual tool
 
alright
 
locally approximate the (possibly crazy) surface by a paraboloid/hyperboloid with the same curvature
 
12:53 PM
I don't yet conceptually understand why the Gaussian curvature measures what it should.
 
how is ted defining it?
 
Product of eigenvalues of the shape operator, or det I/det II if you prefer.
 
the eigenvalues are what's normally called the principal normal curvatures, right?
 
Yes. What do they "mean"?
 
They're the curvatures of lines along the surface that go through your point
geodesics, probably
 
12:55 PM
Which lines?
 
you probably learned about the curvature of lines from ted's other book
@BalarkaSen I think they're geodesics that go along your two orthonormal unit vectors.
 
Oh, the integral curves of the basis vectors.
 
right
 
Right, yes.
 
so the idea is that by looking at the curvature of lines (which is "easy" to understand) you can figure out the curvature of the surface
 
12:57 PM
hi
 
Makes sense.
 
I have a question regarding the definition of homology with local coefficients defined using a module over the group ring. Is there someone who knows about this construction?
 
@BalarkaSen has Ted talked about the circle interpretation of curvature yet?
 

« first day (2262 days earlier)      last day (2746 days later) »