My impression was that the acceptance was what really mattered... everyone can say they submitted their paper to the Annals, but once it's been accepted, that's that
@Danu: Following stars in the physics room I saw your discussion of what you want to do. Maybe I'll make it my project to push you into the corner of $G_2$-manifolds. :)
@TedShifrin I meant tools... there's not a lot of tools one can study G_2 manifolds with. There's some stuff in the way of constructions, but again, not much. What's worse is the complete inability to tell G_2-manifolds apart.
I think that's what's really stopped it from making a lot of progress: nothing much to use.
On the constructive side: only finitely many smooth 7-folds are known to have a G_2-struct. (Somewhere in the thousands, I think.) On the other hand, I think it's very hard (if not impossible to tell currently) whether or not two G_2-folds are different.
That should change, moderately, if people can successfully get various enumerative invariants off the ground
@TedShifrin: I guess I checked out a book on special holonomy by Joyce a while back (for unrelated reasons). Maybe I should make that my new bus reading.
If $a,b,c,d>0$ satisfy the condition ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1$, find the maximum value of $ab+ac+ad+bc+bd+3cd$.
I'm not progress in this inequality problem. Please help.
Thank you.
@Clarinetist Thanks! Here is a very simple question whose answer I think would help clear this all up. Let $M$ be square $n$ by $n$ and non-singular. We know that all such $M$ give us entropy exactly $n$ and yet their determinants can be hugely different. Why doesn't our analysis in terms of multivariate CLT apply?
I dont know... I didn't do anything....actually I was from a rural area, there we dont have all those teaching things available , so mostly what I did, I tried to solve all those previous year's question papaers
@BalarkaSen: I'm trying to do an exercise that $Isom(S,d)$ is finite for any hyperbolic metric on $S$, where $S$ is a closed, connected, orientable surface of genus $\geq 2$. I'm given a hint to first show that if it was finite, then $\forall \epsilon > 0$ there is an isometry that is not the identity such that $\sup_{x \in S} d(x, \varphi(x)) < \epsilon$. How does this property follow from infiniteness of the group of isometries?
@BalarkaSen: the problem is that on $H^2$ the Mobius transformations are the isometries, and with translation you obviously get the statement, even though the starting condition isn't met.
@Danu Here's a fun space: consider the subspace $A$ of $\Bbb R^2$ which iss union of circles with center at $(0, 1/n)$ or radius $1/n$. Fact: $A$ is not homotopy equivalent to the infinite wedge of circles.
$A$ is knows as the "shrinking wedge of circles" or the "Hawaiian earring".
If you're interested, try proving this. Use $\pi_1$.
And $\pi_1(S^1 \vee S^1)$ is not abelian, as you mentioned.
@Danu Similar logic would show that $\pi_1(A)$ is not abelian. Here's a hint on how one could show that A is not homotopy equivalent to wedge : do not think about $\pi_1$, but of abelianization $\pi_1/[\pi_1, \pi_1]$ of $\pi_1$.
@BalarkaSen: I'm looking at a textbook on mapping class groups now, and they define it to be the orientation-preserving homeomorphisms that restrict to the identity on the boundary modulo connected component of the identity. I also see in my personal notes later a definition of the MCG with that condition on the boundary. are they equivalent?
in my personal notes, we first work only with closed manifolds and there define the MCG without the boundary condition. a few weeks later, we work with compact manifolds and then include the boundary condition. I should have just kept on reading.
let me parse that. so the homeomorphism $\Sigma \to \Sigma$ is obtained from doing a Dehn twist on tubular nbhds of the longitudinal curves in $\Sigma$ and leaving everything outside the nbhd alone? you're claiming these generate the MCG?
Oh, I think I see what you mean. It takes the curve around the hole in $\Bbb T^2$ to the curve which goes around the whole, makes one twist - tight around the donut - and then follows the same track, yep?
@BalarkaSen: use the fact that for all non-seperating oriented scc $\gamma_1, \gamma_2$ on $S$, there exists a product of Dehn twists (or of some representatives of them) $\varphi$ such that $\varphi(\gamma_1) = \gamma_2$. for the theorem, then use induction on the genus.
I was just wondering about a trigs question. So, this is a normal smoothing funciton, right? Both input and output go from 0 to 1 and in a smooth function
this is usually used instead of the normal x function, althought even in that case both input and output go from 0 to 1, the problem being that it starts and stops suddenly
I use them for animations, not really important, just saying that how did I get to my question
at one point I was wondering whether the distance in the y of the lines would make another curve, so all I had to do was subtract one from the other
and I got this yellow line
it's a bit small, but by finding the local minimum/maximum and scaling it by 1/that amount I could make it go from 1 to -1 like a normal wave
I then compared it to a negative sin wave with double the frequency
they didn't quite match, but it was close
I then wondered if the blue line could actually be a wave (never mind the fact that google graphs doesn't let you chose x mod 1 instead of just x, then it would have been periodical), and all I needed for that now would be to check that the derivatives at x=0,1 would be the same
so the line would be continuously derivable, and continuous, therefore it could have been a valid wave
according to wolfram alpha the derivative of the yellow line is pi * sin( pi * x ) - 1, which is the same at 0 and 1, making the blue line an actual wave, although I can't figure a good transformation to make it look like any other wave I know of. Does it have a name? Wouldn't even know what to google
Some sort of bootstrapping argument. Pick a sequence f_n of isometries; find a sequence of points so that f_n(x_n) converges after passing to a subsequence; do the same for df_n at x_n; should be able to bootstrap this up to a neighborhood; do this for finitely many points.
There are proofs in probably any Riemannian geometry book. I don't remember them. This is my best approximation of what they probably look like.
So if you believe these two things, if Isom(S) was infinite, it would be a positive-dimensional Lie group, and hence have a 1-parameter subgroup, and hence have elements arbitrarily close to the identity, and hence your question follows.
Lee's Riemannian book is not so good, dunno if it's in there.
I know it's in Kobayashi-Nomizu but it would be mean to send you there.
I wonder if one can prove by hand that there are no killing fields, because of the curvature constraint somehow? (I don't know how, just thinking aloud.)
If $a,b,c,d>0$ satisfy the condition ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1$, find the maximum value of $ab+ac+ad+bc+bd+3cd$.
I'm not progress in this inequality problem. Please help.
Thank you.
