Interpreting Intersection Form in H_1(K)

[Bacle]

Hi, Everyone:
The intersection form q(a,b) in dimension 1 (i.e., in H_1(K) , for any top. space K)
is symplectic/alternating , meaning that

q(a,b)=-q(b,a).

From this last, it follows that q(a,a)=0. How do we interpret this last.?. Does this
imply that any curve in any space K can be homotoped into a curve without
self-intersection.?.

But by self-intersection I don't mean in the literal sense, but in the sense of
intersecting the curve C with a parallel copy C' of itself ( i.e., we define a V.Field
X in C, and translate along the image of X.)
Does this lack of self-intersection then follow from the orientability of all curves
(seen as 1-manifolds.) , so that we can define a nowhere-zero V.Field X on C.?


Also: just curious: can we treat forms defined in homology in the same way we
treat differential forms.?


Thanks.

 

[mathwonk]

the word "form" has many different meanings and here it means "pairing" not at all what it means for differential forms. and your inetersection form on curves makes little sense unless your space is say a smooth oriented surface. In general on a manifold intersection is define between two gadgets whose dimensions add up to that of the manifold itself.

then yes, self intersection zero just means you can shove the curve a little ways parallel to itself until it no longer meets its previous self. for a non orientable surface like a mobius strip this may not be possible. try it on the center circle.

guillemin and pollack is a nice undergrad book on intersection theory, and pricey.

 

[Bacle]

Yes, I forgot the obvious, and all the curves "live" in the same surface, e.g.,
in R^2 itself. It seems harder to see, e.g., on a Mobius strip.

 

[zhentil]

The way you've formulated it, your setup only makes sense for orientable surfaces. In the setup, the q(a,a)=0 statement follows from the fact that the normal bundle of a 1-manifold embedded in a orientable 2-manifold is trivial. Intuitively, you can "push off" so that the curve doesn't intersect itself.

You can interpret this in terms of differential forms using Poincare duality. Then it translates to the fact that a^2 = 0 for any one-form.

 

[mathwonk]

to see why the center circle on a mobius strip does not push off itself, cut the strip to get a rectangle and use the intermediate value theorem.

 

[Bacle]

" The way you've formulated it, your setup only makes sense for orientable surfaces. In the setup, the q(a,a)=0 statement follows from the fact that the normal bundle of a 1-manifold embedded in a orientable 2-manifold is trivial. Intuitively, you can "push off" so that the curve doesn't intersect itself."

Yes, that is the geometric argument, but I don't know if we conclude the bundle is
trivial from the fact that q(a,a)=0, or the other way around. You can also use algebra
alone, using the fact that the intersection form on H_1 is alternating --this is a purely
algebraic fact -- which forces the self-intersection to be 0. And I agree, we can push
the curve along the normal vector field.



"You can interpret this in terms of differential forms using Poincare duality. Then it translates to the fact that a^2 = 0 for any one-form."

But this is not a differential form; it is a map from H_1(K)xH_1(K)-->Z, and not
a map defined on TM*, nor (TMxTM)*, and I don't know if you can wedge quadratic
forms that are not differential forms.

 

[Bacle]

One issue that confuses me here, tho, is why we call this a quadratic form;
AFAIK, a quadratic form is a homogeneous polynomial of degree two, and I don't see
any polynomials here, nor any ring F[X1,X2,..Xn] over which to define the intersection
form.

 

[zhentil]

Yes, that is the geometric argument, but I don't know if we conclude the bundle is
trivial from the fact that q(a,a)=0, or the other way around. You can also use algebra
alone, using the fact that the intersection form on H_1 is alternating --this is a purely
algebraic fact -- which forces the self-intersection to be 0. And I agree, we can push
the curve along the normal vector field.

How do you define q(a,a) if the bundle is not trivial?

But this is not a differential form; it is a map from H_1(K)xH_1(K)-->Z, and not
a map defined on TM*, nor (TMxTM)*, and I don't know if you can wedge quadratic
forms that are not differential forms.

The intersection form is defined modulo torsion, so we can think of H_1(M)/Torsion as poincare dual to integer valued de Rham one-forms. We're not wedging quadratic forms, the "dual" of intersection product is cup product, which is equivalent (here) to integrating the wedge product of one-forms.

 

[WWGD]

I am sorry , I am kind of rusty here:
Why do you need a trivial bundle to define q(a,a).?. It would seem if the bundle is
trivial, then you have , as you said, a normal vector field to help you push your
curve along, avoiding self-intersection, so you get q(a,a)=0. But it seems like
a sort of strange discussion, in that , for curves, q(a,a) ==0 , as Bacle said,
because by definition, q(a,b)=(-1)^n *q(b,a), forcing q(a,a)=0. But I don't know if
q(a,a)==0 necessarily implies that the bundle is trivial.

I would think if the bundle is not trivial, you can still find a small tubular neighborhood
to push your curve 'a' along so that it does not intersect itself.

But its been a while since I have seen this.

 

[Bacle]

Sorry for my lack of clarity in my previous posts. And thanks to Wonk and to
Zhentil for your comments.

My points were these:

i) Could we extend the explanation for why Q(a,a)=0 for n=2 ,to any intersection

2n-form for n even.?. I don't know if even for 2n=6 we can also say that the

bundles over 3-manifolds are trivial, to guarantee that Q(m,m)=0 in this case.


ii) How some properties follow from algebra alone, tho there is a geometric/topological

parallel:

the existence of a regular surface ( or higher-dimensional analogues) , i.e., a

surface x such that Q(x,x)=Q(x,y)Mod2 for all other surfaces, follows from properties

of quadratic forms over Z/2 alone ( it would be great if someone can explain the

actual geometry behind this).



Zhentil: I am not sure I fully understood your argument using a^2 being 0. I

do understand --and sorry I missed -- the Poincare duality issue you brought

up. Still, wouldn't your argument imply that self-intersections are 0 for

higher dimension by wedging the dual of Q(x,x) ( x of dimension n, n odd)

itself.?. You would then get a 2n-form that would need to be a zero form.


Hope I am not too lost.

here

for all

 

[Bacle]

to see why the center circle on a mobius strip does not push off itself, cut the strip to get a rectangle and use the intermediate value theorem.

I think I got it:

We first set the circle as the zero section of the mobius bundle. Then we draw a
(strictly positive/negative) graph on the zero section/embedded circle , which is really
a (potential) section of the normal bundle of the circle in the strip. The problem will
happen at the right- and left- endpoints, where, because of the orientation, the
values at the right/left must flip , meaning that the graph/section cannot be fully positive.
By continuity , the section must be zero at some point. Then there is no nowhere-zero
section of the normal bundle, so at least at that one zero point, we will not be able to
lift the circle of itself.

Please Critique,
Thanks.

 

[Bacle]

The way you've formulated it, your setup only makes sense for orientable surfaces. In the setup, the q(a,a)=0 statement follows from the fact that the normal bundle of a 1-manifold embedded in a orientable 2-manifold is trivial. Intuitively, you can "push off" so that the curve doesn't intersect itself.

You can interpret this in terms of differential forms using Poincare duality. Then it translates to the fact that a^2 = 0 for any one-form.

Wow, what a ****tard, I have been. Of course this generalizes to any 2n-dimensional
manifold: a^2 will obviously never be a volume form. Sorry for my complete idiocy
here, Zhentil, and thanks for your patience. You to Wonk, thanks both.

 

[Bacle]

I also realized why torsion does not matter in the intersection form:

We mod out by the torsion submodule and we end up with a torsion-free

module over a PID ( i.e., the integers here.) . And every torsion-free module

is free as a module, so the quotient module does have a bsis, and then we

can define intersection. Wow, a good night.

 

[zhentil]

Hi Bacle,

Sorry it's taken a few days. The intersection q(a,a) is zero for an manifold of dimension 4k+2. This translates to a^2=0 for any differential form of odd rank. However, it's not necessarily true in dimension 4k. For instance, a symplectic structure on a four manifold is defined using a 2-form whose square is nowhere vanishing. Look at the intersection form on CP^2, for instance.