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20:01
@arctictern How would you describe the fiber bundle S^3 --> S^7 --> S^4?
Hi, I'm studying about the bisection method for the first time. If we approach a problem with this method, how do we select the tolerance? Any help would be appreciated, thanks.
@BalarkaSen disregard, I was mixing things up
no worries
S^7 is unit elements of H^2, and we use the map (x,y)->xy^-1 combined with sterographic projection
H being quaternions
much easier than octonions :P
20:04
This is the first one in one of those infinite series of Hopf-type bundles, right
There isn't an infinite series
There isn't? Let me check that
@user1952009 feel free to help with comments/answers in any of my questions. My purpose is improve my abilities in mathematics. Many thanks for your help.
well, depends if you want the base space to be spheres or not
there are only four with the base space a sphere. otherwise it's a projective space.
Yeah
20:06
Fiber bundles S^p --> S^q --> S^r occur only if (p, q, r) = (1, 3, 2), (3, 7, 4) and (7, 15, 18)
Oh, the base space being CP^n is booring
I mean the $S^3\to S^{4n+3}\to \Bbb HP^n$
Booring :P
$S^{n-1}\to S^{2n-1}\to S^n$ occurs with $n=1,2,4,8$ and these are all the ones with three spheres
Why do you say boring? Projective spaces are probably the best spaces ever :P
I mean it's trivially easy to describe that bundle
20:07
What do you mean? The one you are talking about is a special case, so then your is also boring :P
@arctic OK, so I buy your description. Fibers are lines x = ky, which are embedded S^3's
@Danu It's not interesting that there are infinitely many such things with bases projective spaces. It is interesting that there are finitely many bundles with base, total space and fiber all spheres.
not every quaternionic line is of the form x=ky, but yeah
Yes, I agree what you said is mine for n = 1
@arctictern True.
@BalarkaSen Meh, if they're in such a nice sequence with both total space and fiber a sphere... :P
I still like 'em
@BalarkaSen Now what about fibrations $S^p \to S^q \to M$?
20:12
i just had some insight guys
i think i found out how to solve another one of the problems on the qualifying quiz
Hi Mike
@MikeMiller Ah, ok, so it could potentially be an interesting question what M can be. Need they all be projective spaces?
I like that smiley
I see a creature singing loudly with its eyes closed.
20:14
It's someone shrugging
shrugging person
I know.
Ah
@Mike, Does something like sheaf cohomology of $SO(n)$ make sense to you?
I only know how to do sheaf cohomology for Abelian groups
How does "cohomology" even make sense in that context
20:16
If $\mathcal S$ is a sheaf of groups, then you can of course define $H^0(\mathcal S)$, but also $H^1(\mathcal S)$. This is a set - no group structure whatsoever. You cannot define higher cohomology.
No $H^2$?
How come?
Try to define H^1 without using commutativity
You'll see that you get a raw set
Then try to define H^2 and you'll see where things go bad
OK. So what does a long exact sequence mean in this context?
20:17
It doesn't
Hmm... So you can't say something like $S^1\to \rm{spin}^c(n)\to SO(n)$ induces a long exact sequence?
hmm... That's sort of what our lecturer said though :P
Note that if $\mathcal S$ is the sheaf of maps $X \to G$, $H^1(\mathcal S)$ is the (pointed) set of $G$-bundles over $X$
This is no more than unraveling definitions
Reframing existence and uniqueness of $\rm spin^c$-structures in terms of at least maps between $H^1$'s
Yeah, I know that
20:20
Then of course you get maps S^1 bundles -> spin^c bundles -> SO(n) bundles
can i ask for homework help in this chatroom?>
Since BS^1 -> Bspin^c -> BSO(n) is a fibration, you should get an LES -> [Sigma X, BSO(n)] -> [X, BS^1] -> ...
@MikeMiller How do those maps work?
Oh
but [Sigma X, BG] = [X, G]
it's just the induced bundle
Homotopy stuff---I dunno
20:21
The first thing is not homotopy stuff. You take the induced bundle of the S^1 bundle given the homomorphism S^1 -> spin^c
anyone?
So in most problems, I find that the tricky part is figuring out whether there are any theorems that are applicable, by looking at things in the right way.
i need help with logarithms
Why don't we cut to the chase and you tell me why you want this?
@MikeMiller So just by saying "maps into $S^1$ are a special kind of maps into $\rm spin^c$ so an $S^1$ cocycle is a $\rm spin^c$ cocycle"?
20:23
Sure
hello?
@MikeMiller Oh, I'm just going through the first bits of the notes on gauge theory.
can someone help me?
In this problem, the tricky part is that we know way too much, so we'd be toying with things back and forth until we find which theorem actually takes us to the solution
20:23
please
@Macro Yeah?
Existence and uniqueness of $\rm spin^c$ structures and all that
it's condensing a logarithm
Just ask; don't ask to ask (see side bar)
how can i type a logarithmic expression here?
20:24
I really dislike working with definitions of spin and spin^c as groups... I much prefer defining a spin^c structure in terms of Clifford multiplication
\log(x) (inside dollar signs)
@Danu You should be comfortable with both of them, and understand why they're the same.
how do I make $\lfloor \frac{\sum a} {\prod b} \rfloor$ look right? The floor things are too small
@MikeMiller Yeah, it's all in the notes. Just so many definitions...
Before the brackets, type \big
20:28
thanks
No problem!
You should use \left and \right instead of \big
If you want it to be even bigger, type \bigg
It is dynamic, and read the situation, and figures out how big things need to get for you
\big\lfloor \frac {\sum a} {\prod b} \big\rfloor: $\big\lfloor \frac {\sum a} {\prod b} \big\rfloor$
$\left\lfloor \frac {\sum a} {\prod b} \right\rfloor$
^ \left \right one
Huh, I was not aware of that
20:36
What Paul said
So the task is to prove that for $1 < p < \infty$, if you consider some sequence $\underline{x}_n\in \ell^p$ which converges to $\underline{x}$ weakly, and also that $\|\underline{x}_n\| \rightharpoonup \|\underline{x}\|$, that the sequence converges strongly.
I've got two ways in mind for going about this
One would be to use that $(\ell^p)^* \simeq \ell^q$ and play with it from there
The other would be that weak convergence in $\ell^p$ is equivalent to convergence in each coordinate
Would you guys agree with me that the latter is probably a more fruitful way to go about this problem?
Hello, I think I misunderstand what is being asked of me here:
http://i.imgur.com/kIooVcW.png
could someone elaborate ?
although you've removed your response, assuming it was valid, does being asked for continuity on the left not mean i find which value f(x) is approaching from the left? I'm thinking I may have misunderstood the question completely
20:54
Is your answer 1, -3, 0, 2 ?
1, -3, 0, 0, 2
Why does 0 appear twice ?
And why are some of the values not values where the function is discontinuous ?
because of discontinuity @ x=2, 3
Oh I see what you did
That's indeed not what's asked
They do not ask the values of the left-sided limits of f at the discontinuity points
They want to know where the left sided limit is equal to the value of the function on the discontinuity points
ohhhh i see
20:57
So the answer is ..?
Right
Hey guys :)
hi @Drag
Hi/bye @Dragneel
And bye everyone else
Gotta go now
20:59
@Astyx thanks got it. Confusing wording there *sighhh
@Will Glad to help !
Ever since i linked that fergie clip I've been listening to fergie while working on the questions
So, I came across this statistics question, and I believe it's a Negative Binomial Distribution. Can anyone confirm? Here is the question: Accidents recur in a factory at the rate of $3$ per week. Assume that accidents happen randomly and independently of each other. What is the probability that there is at most $1$ accident in a week?
21:21
0
Q: Prove that the characteristic of the direct sum of $n$ rings is equal to the lcm of the characteristics

JessyunBourneGiven that $R_{i}$ are rings, I need to prove that $char(R) = char(R_{1} \bigoplus R_{2} \bigoplus \cdots \bigoplus R_{n}) = lcm(char(R_{1}), char(R_{2}), \cdots, char(R_{n}))$ Recall that the characteristic of a ring $S$, denoted $char(S)$, is the smallest positive integer $n$ such that $...

21:46
@Akiva well i'm pretty discouraged
given how stubborn my mom can be at times, i don't think she'll let me go even if i get in
:(
Have you even asked her yet
not yet
but i'm pretty sure she won't
she'll be home soon though
youtube.com/watch?v=NYFhWBCfoX0 i just found this gem of a song (nsfw)
22:03
@Dragneel What level of stats? Also, why do you say negative binomial?
@akiva I'll report back when i talk to her about it
22:16
how do i solve an exponential equation if both sides are raised to the power of x
example: 5000(1.035)^t = 4000(1+1.05)^t
can someone help me please?
anyone on?
$P$ is a multivariate polynomial. If it's irreducible, is the set of solutions of $P=0$ connected?
In $\Bbb C$. I guess elliptic curves give counterexamples in $\Bbb R$.
22:31
@MacroGuy You can take the logarithm of both sides. In this example, I'd first divide both sides by one of the exponentials so there's only one $t$ to worry about.
I havn't thought about it much, but what topology @AkivaWeinberger ?
@PaulPlummer The usual topology of $\Bbb C^n$
Homeomorphic to $\Bbb R^{2n}$
23:02
@AkivaWeinberger Yah, not really sure. It should be connected in Zariski though
(I don't really know AG, so I may be mistaken)
@AkivaWeinberger Maybe you can try to break it into a bunch of 1 complex dimensional pieces, and use some complex analysis to see if you can get them to fit together. Anyway I have been up way to long I should leave

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