Anubhav Mukherjee

But is that proof clear to you?

I can do that tomorrow

I may email you... Prof Dan Margalit send me that paper once. I don't have my laptop with me now. But if you believe that result, is my proof make sense to you now?

But you can ask me questions, I can clear your doubts.

@subhankar I'm sorry..I don't know any link. But I've read that paper sometimes back. And as I said the paper used more technical terms like Johnson homomorphism and all.

Torelli group generated by bounding pair of maps, but what Morita proved that, for constructing homology sphere all you need is torelli element genberated by sperating curves.

That is, given somne homology sphere there exists some heegard g- splitting of S^3 and some torelli elemet from that g-genus surface such that , regluing that will give you the resultant manifold.

@SubhankarD. Here I am not restricting genus... genus one case is the trivial... the theorem says that you can constructed homology sphere from torelli element when you allowed to vary your genus.

It's a trivial action on the homology. That doesn't imply it fixed curves.

@SubhankarD. Torelli group action doesn't fix any curve.

@SubhankarD. In that paper he mainly proved the statement which I mentioned in the 2 nd paragraph. I think the actual statement is a bit more complicated in terms of Johnson's homomorphism. If you want I can send you the paper. Are you convinced with the proof?

@BalarkaSen I forgot that...let me go through it

@BalarkaSen math.stackexchange.com/questions/83181/…

@BalarkaSen I never thought in this way. One way I like to think is by considering interesction number

@BalarkaSen can you explain or suggest me some reference

@BalarkaSen I don't know this

@BalarkaSen happy new year :)

they used some spectral sequence idea.

@BalarkaSen i can search for that paper and send you. you might find some interesting ideas there

I didn't check mike's answer...I'll probably go and check it

yes....I read that paper recently...it is very interesting

Its by milnor ...

the answer is false,

@BalarkaSen I forgot to tell you, but I got a nice answer of a question you asked (one year back) , ie. two vector bundle homeomorphic implies vector bundle isomorphic or not.

finite is good for me

@BalarkaSen even I forgot the exact obstruction argument...but thanks :)

btw it's still a open question, what are the homotopy group of $S^2$. As of my knowledge, a recent paper proved that there are infinitely many non-zero homotopy group of $S^2$. That might help for $S^2\vee S^2$.

I can't think of a proof right now

assume orientable as well.

@BalarkaSen if a fibration has contractible fibers, is that enough for existence of a section?

@MikeMiller hey, how are you?

@BalarkaSen thanks. I think I can construct such things.

?

But how do you ensure existence of such a curve

In that case it is true

do you assume all $\gamma_i$ intersect at a single points to the leaf?

@BalarkaSen how do you know that there are no other point on the leaf where those curves can intersect on the surface?

are you in bangalore now?

good night

bbye

Wait..I just lost you..let me read your whole construction once more

but now how do you manage the intersection number here (in the original problem)?

yes, that I can see

can you make it a simple closed curve?

yes, I was trying to make a simple closed curve

and omit other parts

and re join them by a diagonal

YES, then you can take them very close

NO PIRACY :p

I'm scared of using that :P