oh my goodness. I think I almost cried two times. Once out of frustration, once our of joy :P Thanks Ted. I'm slightly thrilled and traumatised by modular arithmetic now XD
Perfect, @ShaV. So you have a complete proof ... and you understand it! :)
I apologize for taking you out of your original line of reasoning, but it looked really confusing to me. And I think this way — once you understand it — shows the main ideas very clearly with very little technicality.
I can barely find time to start Algebraic Topology either, time management, as I've come to realize, is a very valuable skill to learn in second year :(
@Ted That reminds me of something I was wondering about a while back. Do you know a smooth way of getting Z/2 cohomology? I can do it with duals of submanifolds but that's not really what I want.
I was out for a bit because things are picking up fast. Laci did a rapid review of group theory in class today so that was nice, though there are quite a number of problems :P
So it turns out we're using a bit more Milnor than GP for manifolds
In the bazillion number of times I taught diff top, @Balarka, I did that precisely one time (at MIT). All the other times, I did a bit more on deRham cohomology ...
@Ted Here's what it was: Fold one corner of a rectangle to the other, then fold the resulting figure along the perpendicular bisector of the side such that we get a resulting quadrilateral. Prove that all 4 vertices lie on one unique circle
A little bit on cyclic quadrilaterals. You didn't need that though, if you just notices that two of the angles are right, and therefore one of the diagonals must be a diameter for both circles
It's beautiful, I love those proofs, but yeah I was trying to write out an element the commutator subgroup to find out what happens. Turns out it's just $g^{-1}hg = h^{-1}[h,g]$
@Daminark Take a rectangular slip of paper, fold one vertex to the other, then, fold the resulting figure over the perpendicular bisector of the fold you just made
@TedShifrin i have something else i wasn't sure to solve: $f(x,y) = \dfrac{x-y}{(x+y) \ ^ 3}$ , i need to show that fubini's theorem does not apply here, and to explain what is the problem with using it here , at $[0,1] \ ^ 2$
Zach, you still here? So I get an isosceles triangle there. AB and AD' have the same length. I don't see why the circle through those passes through E.
@BalarkaSen The beauty of it is that it's really easy to get back into. It's just a few rules and then basically manipulating polynomials. The background theory, you probably don't really forget.
