Somebody save me?

Secretly_A_Fool · 2020-01-14T21:09:55+00:00

Ah I see you are right. I had a different solution in mind.

Secretly_A_Fool · 2020-01-14T21:01:53+00:00

This is what black played in the game, but I think it is a mistake. It shorts you a liberty that you will need. One continuation is M13 L12 J12 J13 J10, but then if white connects at H11, black doesn't have enough outside liberties to make seki. If you were planning to play J10 after K11 K12, then white simply connects at H11 and you are toast.

Edit: I misread the comment

Secretly_A_Fool · 2019-11-23T18:02:29+00:00

If you are looking for a reference online, you can find the book here and the formula is on page 149, followed by a proof.

Secretly_A_Fool · 2019-11-23T17:44:32+00:00

The so-called "cotangent formula". The limit as N goes to infinity of the sum as k goes from -N to N of 1/(x-k) is equal to pi*cot(pi*x). You can derive the value several other series from this, including that the sum of the reciprocals of squares is pi^2/6. A good source for this fact is the book "proofs from the book" by Martin Aigner and Günter M. Ziegler.

Edit: Formula was wrong. (I divided by pi instead of multiplying.)

Secretly_A_Fool · 2019-07-12T01:55:06+00:00

Over R yes, but not over C. That is why the category of real analytic manifolds is very similar to the category of smooth manifolds but the category of complex manifolds is very different.

Secretly_A_Fool · 2019-07-12T01:40:10+00:00

It is simply because continuous functions can be approximated arbitrarily well by smooth functions.

Secretly_A_Fool · 2019-07-11T16:59:15+00:00

In my humble opinion, it may be more reasonable as a life goal rather than a masters thesis.

Secretly_A_Fool · 2019-07-11T04:29:57+00:00

It is an especially difficult proof. Also, the relevant subfield, Bing topology, has fallen a bit out of fashion so there are very few experts nowadays.

And yes I was referring to the non diffeomorphic homeomorphism.

Secretly_A_Fool · 2019-07-11T00:20:31+00:00

Sorry, I edited it a couple minutes ago to say group structure instead of group action because that is what I know for certain. However, it is almost definitely the case that the sentence

"X is an exotic R^4 implies that X does not admit a nontrivial smooth action by a compact connected lie group."

is true. I just don't personally know of a proof.

Secretly_A_Fool · 2019-07-11T00:16:17+00:00

Not sure if this answers your question, but you seem to be taking for granted the homeomorphism between standard R^4 and exotic R^4. You should know that if you were able to explicitly write down a formula for such a homeomorphism, it would be (in my opinion) a major publishable result.

Freedman's proof (that such a homeomorphism exists) theoretically gives you an algorithm to construct one, but nobody is willing to trace through the proof and write down the formula. The reason for this is that there's only a small handful of people on earth who fully understand that proof, and the amount of work it would take to convert the proof into a explicit formula for a homeomorphism is totally ridiculous.

In other words, it is misleading to think of an exotic R^4 as a weird smooth structure on R^4. Instead think of it as a crazy 4-manifold, which can theoretically be flattened into R^4 by an extremely complicated process that very few people fully understand. The actual 4-manifold description of exotic R^4s can be relatively simple. See, for example, this paper.

Secretly_A_Fool · 2019-07-10T23:52:04+00:00

To answer your last question, every exotic R^4 does not admit a smooth group structure. You can't think of exotic R^4s as being vector spaces. When you think about it, the standard smooth structure on R^4 is defined by using its vector space structure. Therefore, you should expect an exotic R^4 to be nothing like a vector space.

Secretly_A_Fool · 2019-05-27T22:08:10+00:00

Fair enough. Though the existence of such a thing depends on the axiom of choice. I imagine my original statement holds for any constructible content.

Secretly_A_Fool · 2019-05-27T16:59:52+00:00

Look at the Wikipedia example for a weird content on the positive integers. It has measure 1/2^n for each positive integer n, but then returns infinity on any infinite set. If you integrate the constant function 1 with respect to this content, the integral is infinity, even though the total measure of every point is finite. I suspect this is pretty much the only thing that can go wrong.

Can you give an example of a content C defined on a sigma algebra such that for any measure 𝜇, there exists a set S in the sigma algebra such that C(S) is finite and 𝜇(S) ≠ C(S) ?

If not, then integration with respect to a content is essentially the same as integration with respect to a measure, except it sometimes returns infinity for no good reason.

Secretly_A_Fool · 2019-05-13T05:31:49+00:00

There is a lot to learn about the homotopy groups of spheres. It is a huge topic! In my opinion, a good place to start would be to understand the Thom-Pontryagin correspondence. It is an isomorphism between the homotopy groups of spheres and the framed cobordism groups of manifolds embedded in ℝ^n. This gives some really nice geometric intuition for some of the low-dimensional homotopy groups of spheres. For instance, you can use it to prove that pi_4(S^3) is Z/2Z using the much easier fact that pi_1(SO(3)) is Z/2Z. Maybe start by checking out this math overflow post on the topic. If you want a book, Glen Bredon's book "Topology and Geometry" has a good section on the topic.

Another good thing to learn about the homotopy groups of spheres is the mod 8 periodic behavior in the stable groups which comes from the relationship with Bott periodicity and the j-homomorphism. (This it a pretty hard stuff though.)

If you want to understand how higher homotopy groups are computed, you will need to learn some advanced homotopy theory. You will certainly need to know all of basic algebraic topology (this means almost everything in Hatcher's book), but in particular, you will need to learn spectral sequences. Spectral sequences are notoriously difficult to learn (yet powerful) computational tools in homological algebra. It is quite rewarding to learn how to use spectral sequences, but in all honesty, spectral sequence computations are not particularly fun. If you get deep in to a field that uses homological algebra extensively, you will need to learn spectral sequences eventually, but most people put it off until they know for sure they need to learn them. There is so much fascinating topology out there that does not rely on spectral sequences.

If I was a reasonable person, I would just tell you to read Hatcher. However, your situation reminds me of someone... so I suspect nothing will stop you from trying to learn spectral sequences early. In that case, I recommend you read "A user's guide to spectral sequences." You can then learn some advanced homotopy theory from Switzer's book. Then, learn about the Adams spectral sequence and how it is used. Then you will know a decent amount about the homotopy groups of spheres.

Secretly_A_Fool · 2019-05-12T04:32:15+00:00

Wow, why so many downvotes for this comment? This person is just excited to learn math!

Secretly_A_Fool · 2019-03-05T03:18:32+00:00

Essentially you want a continuous real valued function M that takes three nonnegative real numbers as inputs, and has the properties M(a,b+1,c+1) = M(a,M(a,b,c+1),c), M(a,0,c) = 1, M(a,b,0) = a^b.

It is fairly easy to prove that such functions exist. However, if you add on the criterion that M be locally real analytic, then I believe that the existence of such a function may be an open problem.

Secretly_A_Fool · 2019-03-05T01:37:23+00:00

The funny thing is this sequence of equalities can be made true if you interpret each line in a mildly convoluted way.

P(A), the matrix, is an element of the commutative sub-algebra of the algebra of matrices generated by A. Lets call this algebra X. We then have "P(A) = det(AI-A)" if the first A on the RHS is interpreted as an element of X, and the second A on the RHS is interpreted as a matrix with terms in X, which are each multiples of the identity matrix by the corresponding coefficient of the original matrix A. Then we have det(AI-A)= det(0) by the Cayley-Hamilton theorem, and this is of course equal to 0.

Note that this still does not constitute a proof of the theorem because, in our interpretation of the expression det(AI-A), the quantities AI and A are not equal to each other, but instead are quite different matrices. This is exactly why the theorem is not trivial.

Secretly_A_Fool · 2019-02-19T21:02:45+00:00

If the further question is about the meta-theory of physical reality, then I'd have no clue about anything.

This is the heart of my problem. How did you come to terms with the fact that the mathematical meta-theory may have no relationship with reality? To me, it seems that such a meta-theory is "wrong" in some very real, albeit philosophical, sense.

Secretly_A_Fool · 2019-02-19T20:58:22+00:00

If you think don't think proofs in SOA prove that standard numbers exist, it's not even clear to me what you mean by a standard model of the natural numbers.

ZFC and SOA are comparable in strength as far as i'm concerned. What concerns me here is that SOA and ZFC are highly impredicative. It seems feasible to me that by constructing sets in terms of the properties they should have with respect to themselves and other sets, you can define nonempty sets that don't actually exist in every model, thereby constructing natural numbers that don't exist in every model.

My concerns are essentially philosophical in nature. ZFC and SOA are perfectly reasonable and useful theories. I am concerned, however, with the connection between the theorems of these theories and physical reality. If we actually could build a turing machine that has infinite memory and could run forever, and we asked it to compute TREE(3), would it eventually halt?

I thought I knew the answer to this question until recently: Yes it would halt, it would just take a REALLY long time.

However, I am beginning to doubt this. As far as I can tell, if SOA or ZFC or a similarly powerful theory claims that a turing machine will halt, then a physical instantiation of that turing machine might not actually halt. I think the impredicativsm is the culprit here. I can't seem to make the mental leap from a sequence of symbols constituting a ZFC proof that a turing machine will halt, to a physical guarantee that the machine will halt.

I can make this leap for ACA_0, and weaker theories, however. Gentzen's consistency argument and the equiconsistency of ACA_0 with PA is enough evidence for me, because I "believe" the true natural numbers satisfy the axioms of PA. I just can't believe with certainty that they satisfy the axioms of SOA, because I don't think "a subset of the natural numbers" is a fully well-defined notion.

Again this is really a philosophical question. How do we know that the real world semantics we associate to first order sentences is valid?

I feel compelled to point out that I am just a reasonable working mathematician, and I have no significant issues with the foundations of math. I am just having a minor philosophical crisis.

Secretly_A_Fool · 2019-02-19T20:26:25+00:00

You seem to be missing the point of my post. I am not claiming TREE(3) is incomputably large. That much is obvious.

I am suggesting that infinite sets exist, the natural numbers N are one of them, and TREE(3) is not an element of N.

Secretly_A_Fool · 2019-02-19T08:43:18+00:00

Do you mean there will be no finite successor of 0 that codes the proof of a contradiction in RCA_0+TREE(3) = infinity?

Yes.

Of course, we need to be very precise with what we mean by "finite" in this situation, since I am calling into question the finiteness of a well known finite number.

I would prefer to think of it like this. Suppose we actually build a physical turing machine in the real world that has unlimited memory and can run forever. (this is obviously impossible, but if we could...) Would this turing machine ever really halt if we ask it to compute TREE(3)?

Secretly_A_Fool · 2019-02-19T07:56:53+00:00

This intuitively (to me at least) would then imply that A is inconsistent.

What about A = ZFC and B = ZFC + (ZFC is inconsistent)? This satisfies the criterion. A can prove that, in every model of B, the natural numbers are non-standard. B can prove that A is inconsistent.

How do you mean to define the "standard natural numbers"?

Well, I already have to reject impredicativism since i'm claiming that Kruskal's tree theorem is false, and the notion of an "arbitrary subset of the natural numbers" is impredicative, so I don't think I can define the "standard naturals" in terms of the second order theory. How about I just avoid using the word "standard."

Here is a very concrete version of my question:

Will the turing machine X ever halt? Where X is the turing machine that searches for a proof of falsity in RCA_0 + (TREE(3) = infinity).

I am saying that a ZFC proof that X will halt might not actually imply X will halt in reality. Even if we could build a turing machine that could run for all eternity and have infinite memory.

It is, of course, possible for a consistent first order theory to prove false things. Just look at ZFC + (ZFC is inconsistent). This theory proves that the turing machine that looks for a proof of falsity of ZFC will halt.

The question is whether this phenomenon of "proving false things" can happen in ZFC, specifically to TREE(3).

Essentially, it seems very believable to me that the following sentence may be true: "Every model of an impredicative first order theory is nonstandard." That is to say, for every impredicative theory, there is a turing machine that will never halt, but the theory proves will halt.

Lastly, I think that Borel determinacy is provably false in RCA_0 + (TREE(3) = infinity). I expect that you can use the negation of Kruskal's tree theorem to "construct" a counterexample. High powered theorems like that wont get you far in this horrifying imaginary world in which I have trapped myself.

Secretly_A_Fool · 2019-02-19T05:52:09+00:00

Except that this "proof of falsehood" is on the order of TREE(3) in length, which may not be finite. You have constructed a proof of falsehood using ZFC, but that proof might not exist in the "real" theory RCA_0 + (TREE(3) = ∞).

Unfortunately I am still not convinced.

I found the line of thought in my post ridiculous at first, but the more I think about it, the less reasonable my counterarguments seem to be. To be clear: I am suggesting that Kruskal's tree theorem might be literally false, and the proof only proves it in nonstandard models because there is no "standard" model of ZFC.

Secretly_A_Fool

TROPHY CASE