incidentally, those intersection numbers do have a utility in computing contour integrals (in addition to being a way to compute winding numbers, i guess)
namely, it shows up in the Picard-Lefschetz formula for how such contour integrals transform under monodromy
On a complex manifold you have a splitting of $\Omega^i$ into $\oplus \Omega^{j,k}$, where $j+k=i$, into forms that have $j$ holomorphic parts and $k$ antiholomorphic parts. Then this is a bigraded chain complex with $\partial$ and $\bar \partial$. The Dolbeault cohomology $H^{i,j}(M)$ is the cohomology of this chain complex w/r/t $\bar \partial$.
@MikeMiller i've heard that a secret way to compute cup product is to Poincare dualize everything, where the cup product is the intersection, and then dualize back to cohomology. prof, as an example, said that the two generators of the homology of a torus intersect transversely at one point. the cup product of the two generators of the cohomology class of the torus also takes the value 1 at the whole 2-simplex.
while this was very surprising, I couldn't make head or tails of it
if one is just doing contour integrals of integrands which are holomorphic on a branched surface, though, i'd think that'd make the antiholomorphic part irrelevant?
I don't know anything about that. I didn't say it was relevant... just that it was the complex version of de Rham
whereas de Rham has an interpretation in terms of "dualizing the space of smooth simplices in $M$", I don't think Dolbeault does. That's all I was saying.
fair enough. i just mean that, in that restricted setting (which I think would just be $\Omega^{1,0}$ if I'm reading that right), that notion of dualizing might still make sense
fun fact, poincare duality shows up in statistical mechanics in a rather straightforward way
(sorry, @PVAL, i pinged you earlier seeing your avatar hanging in the user panel, but the question i was supposed to ask has already been answered. apologies for the unnecessary ping)
i got another one. when finding the order of $(\bar 5,\bar 4) $ in $\Bbb Z_{12}x\Bbb Z_{25}^*$ is there easier way without finding $4^a\equiv1 (mod 20)$?
@PVAL: OK, here's the question. Given a closed complex manifold $M$, $\text{Aut}(M)$ is finite-dimensional. Proof: it's a closed subgroup of $\text{Diff}(M)$, hence a Lie group. We just need to show that its Lie algebra is finite-dimensional. One calculates that its Lie algebra is the algebra of holomorphic vector fields, and the space of holomorphic sections of $TM$ is finite-dimensional by the standard elliptic theory.
If one wants to mimic this proof for a closed almost complex manifold $(M,J)$ one runs into trouble. What is the Lie algebra when you don't have a notion of holomorphic vector fields? I suppose you could pick a connection $A$ and work with $\bar \partial_A$. Then you get a sequence $\Omega^{0,i}(TM)$ - but can we always pick a connection such that this is a complex, and then apply the elliptic theory?
And is the space of "holomorphic sections" still the Lie algebra?
@MikeMiller Thanks very much for your time on that discussion. I understood what Hatcher does and probably can use the "secret method" in computations too. (although not knowing how it works).
@BalarkaSen: It will be harder to do for anything higher-dimensional (other than special things like $\Bbb{P}^n$), though you should be able to do the exact same thing for any compact surface. No problem.
I do not know off the top of my head. If $M$ is symplectic with $J$ a compatible acs, the obvious connection to try would be the LC-connection of $g(v,w)=\omega(v,Jw)$. This paper seems to answer the question in generality jstor.org/stable/1970219?seq=1#page_scan_tab_contents
It just implies that there is some kernel of truth about why something is a theorem; some principle reason that is abstracted away from the technical details
Which one is may not be abelian. Orders:4,31,55,39,121. Since 4 and 121 prime square both abelian. 31 is prime so abelian. 55=5*11 and 39=3*13 abelian. So all of them abelian. İs it correct?
@MikeMiller Can you give me the brief pointers on the computation of the cohomology ring $H^*(\Bbb RP^2, \Bbb Z/2)$? I just followed up until where they considered inclusion $\Bbb RP^i \hookrightarrow \Bbb RP^n$ and $\Bbb RP^j \hookrightarrow \Bbb RP^n$, $i + j = n$, such that the two copies intersect transversely. There's a bunch of algebraic computations afterwards, which I can try to decipher tomorrow morning, but just wondering if you can give the geometric idea playing behind.
@RobertCardona depending on the question, I might or might not help. I just did some basic chain complex stuff and Ext and Tor computations yesterday evening, and I'm content for now.
I'm trying to show $A \otimes \oplus_i B_i \cong \oplus (A \otimes B_i)$. I can define the $\rightarrow$ and prove it's well-defined in the standard way. The trick is then to define it's inverse.
surely an element of $\bigoplus (A \otimes B_i)$ will look like $(a_i \otimes b_i)$ where $a_i \in A$ and $b_i \in B_i$. that is, $a_i$'s needn't be the same.
@SohamChowdhury, tell me about it. I double majored in math and computer engineering. Biggest regret. I should have just spend all my time doing math. Would have enjoyed it way more.
@BalarkaSen Again, like before. Pick the generator of $H_1(\Bbb{RP}^2;\Bbb Z/2)$. A cocycle that corresponds to $\text{Hom}(H_1(\Bbb{RP}^2;\Bbb Z/2),\Bbb Z/2)$ is, well, intersection with a (slightly moved) copy of that curve.
since we're working in $\Bbb Z/2$-coefficients we're just counting intersection points, not oriented intersection points.
The obvious embedded copies of $\Bbb{RP}^{n-1}$ correspond to hyperplanes in $\Bbb R^{n+1}$. Take $n$ hyperplanes and intersect them. Generically, these will intersect in a line, i.e. a single point in $\Bbb{RP}^n$. So $\alpha^n$ - intersecting with copies of $\alpha$ $n$ times - is $1$
yeah, heard of transversality. I am not trying to formalize this geometric picture -- just want to keep it at the back of my head while looking at Hatcher's algebraic proof.
ok, gotta sleep. g'night, everyone.
dunno what will happen to me in the ergodic nt talk tomorrow. probably will fall asleep.
Which one is may not be abelian. Orders:4,31,55,39,121. Since 4 and 121 prime square both abelian. 31 is prime so abelian. 55=5*11 and 39=3*13 abelian. So all of them abelian. Is it correct?
@KevinDriscoll Besides, even if someone helps on there, I think it's better to have someone guide me if they are willing to, if not, the post couldn't hurt.
