Thanks, you guys! It was quite late so I didn't finish your method, @LeakyN. That sum was also in a previous question so I guess that was the recommended method, @robjohn. :-)
Suppose I have a set $S=\{a,e\}$ which is closed and associative under the operation $\cdot$ (i.e. $S$ is a semigroup)
Given two axioms
1. $\forall x \in S, e\cdot x=x\cdot e=x$ 2. $e\underbrace{\cdots}_{\text{n times}} e =a$
*Collapse Theorem: $S$ is trival*
Begin with $$a\cdot a$$ Using axiom 1 twice $$a\cdot a = e\underbrace{\cdots}_{\text{n times}} e \cdot e\underbrace{\cdots}_{\text{n times}} e$$ Using axiom 2 $2n$ times $$a\cdot a = e$$
@Secret I don't know how your "collapse theorem" is interesting. By axiom $1$, $ex = x$. Now $eee\cdots e = e$ by repeatedly applying axiom $1$. So $a = e$ has to happen if axiom 2 is true.
@Kari Because when you expand a in terms of e, you can either get a.a=e or a.a=a depending on the sequence of axioms you applied to a.a. Therefore a.a=e=a as otherwise it will mean the order you apply the axioms to the proof matters (which I don't think that is a property of a semigroup
@BalarkaSen Division by zero is $\exists q, q0=1$. More generally $\exists r, r0\neq 0$. If we restrict ourselves to the reals, then $n1=1+\cdots+1$ n times. Therefore if we have the additive identity adding to something that is not the additive identity, it has the same property as introducing a division by zero element
But isn't that $\cdot$ defined above in the semigroup is not necessary the usual multiplication we understood in real numbers, thus the proof still holds even if we say $\cdot$ is the usual addition?
Definition of "division by zero" does not require anything because it's nonsense :) But yes, to say that, you first need to say what a "zero" is. An additive identity is not the same as a zero.
Ok, I guess that explains why I am not actually introducing a division by zero element, you are right I only proved a string of additive identity must add to the additive identity
@BalarkaSen I got a.a=e by a.a=eee.eee=e, and I got a.a=a by a.a=eeee.a=a. Because I got different results of a.a depending on how the axioms are applied in sequence shouldn't be that will mean a.a=a=e?
or do the sequence of axioms applied in the proof matters in general for an algebraic structure?
Sure, I just said $a^2 = a$ in itself does not imply $a = e$: you need some more things about $a$. But as I already said, your proof is overly complicated: $eeee \cdots e = e$ no matter how many times you multiply because of the axiom 1.
What in general I am trying to do in activities like above is trying to proof that no nontrival division by zero structure exists for the most general algebraic structure. I knew how the proof is done in the reals and complexes. But because any mathematician say it is nonsense, I want to prove it is by trying to build a proof that showed all conceivable structures that has the property of divison by zero is trivial
@BalarkaSen Actually on deeper thoughts, while in a semigroup an identity is not necessary an absorber, it always absorb itself because eeeeee...e=e, and as you have mentioned in that my proof is too complicated, this implies this property of identity is enough to prevent a sequence of identity to result in anything other than itself
in particular, I am trying to isolate the necessary condition that division by zero in an algebraic structure lead to trivality. Check various MSE plus myself doing some algebra in paper scraps, it seems both the distributive law and additive identity played a role.
@kari @BalarkaSen However I am not sure which one is the necessary condition, or that both are. While it is easy to show that knocking off additive identities will make it work, trying to work with non distributive structures are kinda intractable even for 2 elements because the number of possible elements grew out of control hence I cannot analyse it via a cayley table of x and +
Hey everyone! I'd like some feedback on my answer; I'm unsure of whether or not the generalization I make at the end holds (from "By induction.."). If it does, how can it be made less hand-wavy/intuition-based? Thanks!
@TedShifrin Apparently not. I didn't e-mail him for some time in fear of not getting a reply (perhaps out of a guilty feeling for being maybe partially responsible for his departure from the chat, and partially because I was progressing very slowly) - I did recently and got a couple replies. Still talking since then.
One of the most important things to do while studying linear algebra, @AsMa, is to learn all the definitions. You can't do anything if you don't know definitions.
BTW, you gave an example of a disconnected half dimensional submanifold which has trivial self-intersection but nontrivial normal bundle (blow up CP^2 at a point not on the line at infinity: look at union of the exceptional divisor and the line at infinity).
@TedShifrin In your example, the normal bundle of one is O(-1) and the other is O(1). If you homologue it by tubing, you get trivial bundle as a normal bundle.
@TedShifrin I mean if you homologue the disjoint union to a connected sum inside $\Bbb{CP}^2 \#\overline{\Bbb{CP}^2}$, the resulting manifold has trivial normal bundle, so not a valid example. I don't know how else to make your example connected.
@TedShifrin What's the proper transform again? (I think the geometric picture should be making "X" to ") (" - that's what I meant by tubing - in which case I am not convinced that'd have a nontrivial normal bundle).
@AsMa: Next, consider things in a 3-dimensional space. Suppose U is 1-dimensional and V is 2-dimensional. What are the possible dimensions of U+V? Same if U is 2-dimensional and V is 2-dimensional?
For your original question, the only way you can get 3 as the answer is if U=W.
@Balarka: Sorry, forgot you asked something. If you blow up a point $P$ and take a line $\ell$ passing through that point, the proper transform is the closure of $\ell-P$ in the blowup.
Again, @AsMa, use 2 and 3 dimensions to get intuition. If you take 2 subspaces of dimension 2 in 3-space, what are the possible dimensions of the intersection?
@TedShifrin OK, yes, that's what I had in mind. I doubt that has nontrivial normal bundle. It's normal bundle should be O(-1) tensor O(1), which is the trivial bundle, no?
Oh. My claim was simply that a connected submanifold which self-intersects itself with 0 intersection number has to have trivial normal bundle inside a 4-manifold.
I guess I didn't understand the definition of "proper transform": I was implicitly thinking of tubing because there's no other way I can come up with to make it a connected submanifold. In the case of tubing, the normal bundle becomes the tensor product.
I don't know what you know how to prove and what you don't know how to prove, @AsMa. Do you know things about systems of linear equations? Or do you know about linear independence and basis? I don't know what you know.
@Balarka: This tubing stuff is out of the realm of complex geometry, of course.
The idea of proof of there not being any connected example is the following: if you take a section of the normal bundle, then the zero set has to have total sum of intersection no. = 0. So it suffices to prove that a 2-plane bundle on a surface which admits such a section is trivial. Pass to complex line bundles - there it's true by realizing it as pullback of the tautological bundle on $\Bbb{CP}^1$ (a global section of which always has zero set a point).
Pulling back that global section would give me a section with nontrivial intersection with the zero section, which is impossible, so that map to $\Bbb{CP}^1$ has to have degree $0$. Aka, the bundle is trivial.
@TedShifrin Let me look for it.
(There's a few details I need to work out in my proof above though.)
(It'd have oriented intersection $-1$ with the zero section, which could never happen if it was holomorphic - or so I think is the logic behind there not being any section)
Well, that's part of the point of that exercise you're looking for. But can you actually give me a continuous (or smooth) section with one zero and compute that the intersection # is -1?
Actually, it had better be smooth ($C^1$?) if you're talking intersection numbers.
@TedShifrin I can't tell you an explicit section, but: compactifying the tautological line bundle fiberwise gives $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ (I can tell you a proof of this if you want - I found it out a few months ago). Then computing the cohomology ring seems to say that the circle you're taking connected sum along has self-intersection no. $-1$).
Maybe I should try to build such a section by hand though. It shouldn't be hard and would be a good exercise.
@BalarkaSen Any suggestions on how to approach a proof that any general algebraic structure involving division by zero implies trivality. I knew the case is easy to show for fields, and some rings, but what about all algebraic structures that involve some notion of a multiplication and addition operator?
$X$ a space with infiinitely many elements. In the cofinite top, $\Omega$ is open iff $\Omega = \emptyset$ or $\Omega^c$ contains finitely many elements.
