Intersection Form of Connected Sum of CP^2

[Bacle]

Hi, Everyone:

Sorry if this is too simple: I guess the intersection for for CP^2

(complex projective 2-space) is (-1), right?. Since H_2(CP^2,Z)=Z,

which is represented by CP^1, which has self-intersection=-1.

Then, if we had a connected sum of CP^2's, the intersection form

would be : (-1)(+)(-1)(+)...(+)(-1) ?

And, if we reversed the orientation of CP^2, then we would

substitute a (-1) for a (+1) , right, or do the orientation

changes of the self-intersecting copies of -CP^2 cancel out

to give us a negative self-intersection?

Thanks.

 

[quasar987]

mmh. Why -1?

The cohomology algebra of CP^n is generated by an element a in H²(CP^n;Z) with <a^n,[CP^n]>=1, where [CP^n] is the fundamental class for the usual orientation of CP^n.

So as a bilinear form Z x Z --> Z, the intersection form of CP^2 is just multiplication: (p,q)-->pq

 

[Bacle]

Well, Quasar, but if CP^1 generates H_2, then the intersection form Q(CP^2) would
be defined by/as:

CP^1 . CP^1 =-1 (CP^1 has negative self-intersection in CP^2)

(Formally, given a basis {e_1,..,e_n} for H_2, Q is the matrix Q:={Q(e_i,e_j);

i,j=1,..,n).

But ,then, I am not 100% on whether the form is additive under connected sum,

nor on whether the sign of the intersection changes when we reverse orientation.

I guess I am assumming that when we do a connected sum M#M', the basis

elements for H_k( M ) can be made to avoid those of H_k(M'), so that the

intersection form would be additive . I am not 100% sure this is always true,

tho it does work for cases like S_g, the genus-g surface, whose homology is

Z^2g , and in which we can have a choice of symplectic basis

{x₁,y₁,...,x_n,y_n} , with

Q(x_i,y_j)= Del_i, ^j

 

[zhentil]

Why does CP^1 have negative self-intersection? That would seem to contradict a lot of things (positivity of holomorphic intersections, the fact that CP^2 is symplectic,...).