The h Bar

5:00 AM

Listen science, like every human endeavor, is always to some degree a social thing.

Obviously science has evolved to try to eliminate personal opinions. That was the insight of empiricism.

which now is just called the Scientific Method.

Yet they things we choose to focus on and the results that we wish to obtain, are very much subjective. ANd as such , they are vulnerable to the same pathological obsesions as any religious zealot.

But this wasn't reallyi the point. I'm really try to get science to address just how inbred it has become and anti-social.

It's very much like a wizard in a high castle. Once in awhile it accomplishes something unique or stunning and makes a contribution to the population at large.

I guess I've annoyed everyone

@secret: I suspect you are trying to enter the conversation, yet I've been very absorbed in the current dialog.

Secret

@theDoctor Not really, as many h barers can tell you, I am pretty weird in that I can maintain separate conversation streams without actually participating in a conversation

theDoctor

interesting.

Secret

Because of how chat messages highlight your own message in your window, I have no problem tracking different conversations

theDoctor

nice.

did you actually have a question or comment. I like those.

sorry missing question mark.

Secret

I am also weird in that most of the time I am actually talking to no one.

Not yet, I have not reach that part of the chat log yet

theDoctor

5:14 AM

I understand. I know the feeling. You are putting out "feelers" for lack of a better word, to see what connects....

Shoot me a line.

I take bait from any well-meaning questioner.

Balarka Sen

long transcript

theDoctor

I'm very serious, I don't mock (except to the most freshman arguments).

(meaning arguments of unearned boisterousness)

@BalarkaSen: hi

Balarka Sen

h

theDoctor

$pi$

*2, would have been funnier, I guess.

/2$pi$...

Do Kalabi-Yau manifolds have a meaningful "dimensionality"?

I ask, because when I read mathematicians talk about such manifolds, they use the term dimensionality in way that sounds rigorous.

Balarka Sen

dimension is well-defined for a manifold, yes. and that is irrelevant of them being Calabi-Yau or whatever

theDoctor

5:21 AM

...and because I have done considerable work to work out a metaphysics or philosophy of math in which to untangle the mess.

So, I hand you this complex surface made with my knitting yarn.... You can state it's dimensionality?

its

Balarka Sen

if it's a complex surface it's 4 dimensional by construction

theDoctor

hmm, interesting, and you can say that because the object presumably exists in the 3rd dimension?

I mean the actual yarn object.

What dimension is the Mobius strip?

Balarka Sen

I don't understand your yarn thing. C is 2 dim, C^2 is 4 dim. Complex surfaces are modeled on C^2.

Mobius strip is 2

it's an actual surface.

theDoctor

hmm, interesting. So this, to me, is a topolocy-specific use of the term "dimension".

And C refers to the surface?

Balarka Sen

C refers to the complex plane.

theDoctor

5:26 AM

topology models complex surfaces in the complex plane, or just the Calabi-Yau manifold?

I mean a torus is not in the complex plane....

Balarka Sen

Complex surfaces are modeled on C^2, not C. Torus is a complex line, it's modeled on C.

You're not "in" the complex plane... you just locally look like it

theDoctor

wierd, please excuse my ignorance. I'm working out what I call domain theory inorder to sort out some of these metaphysical confusions.

a complex line, how bizarre. I've never heard it described as such.

Balarka Sen

Sorry, I meant a complex curve.

"line" is not the right word.

It's literally a 1 dimensional curve when modeled on C. But it's 2 dimensional when modeled on R

theDoctor

ah, thank you, though in my geometry a line and a curve are equivalent except when you look at it from a higher dimension.

Interesting, I had imagined a torus as a 3d solid, but mathematically speaking, it could be seen as a surface that connects on one edge to itself.

and then conform to your statement about being 2d when modeled in R.

Balarka Sen

Torus doesn't have stuff filled in it. What you are thinking of is the solid torus, which is a 3D solid with boundary.

All of these would be considerably cleared up if you start by learning the definition of a manifold.

theDoctor

5:32 AM

Yes. I've avoided it because to me topology is just messy geometry.

can you give me an example of a 3dimensional manifold?

and is there such a thing as a 1 dimensional manifold?

scratch last question, i see it's answered on wikipedia

Balarka Sen

You have one already: the solid torus. Or ball bounded by the 2-sphere. Or.... it should be noted that these are all manifolds-with-boundary. On the other direction there's the 3-sphere $S^3$ defined by $x^2 + y^2+ z^2 + w^2 = 1$ in the $xyz$ space ($\Bbb R^4$.)

1-manifolds are all either intervals (open/closed/clopen) or circles.

theDoctor

Yes. I've avoided it because to me topology is just messy geometry.

Balarka Sen

*$xyzw$

theDoctor

But why call it a manifold, why not simply a morphic solid (for the case of topology)?

(re: torus)

Balarka Sen

Because it satisfies the definition of a manifold.

theDoctor

5:46 AM

wow, I see that you are a different kind of mathemetican than I.

Like different in the same way that the domain C is different than the domain R.

You make assertions, for example, that are entirely different than the way I make them. You appear to make them as a test of truth, while I make them as a result of it.

yuggib

7:14 AM

@0celo7 the strong operator topology is, in purely topological terms, the topology arising from the uniform structure of simple convergence. The uniform or norm operator topology is the topology arising from the uniform structure of convergence on bounded sets

The weak* topology is the weak topology induced by the predual (i.e. the coarsest topology on the continuous dual $X '$ of a top vec space $X $ that makes all the $x\in X $ (seen as linear functions from $X'$ to the field $\mathbb {K} $ of the vec space) continuous.

Therefore you see why yosida says that. Finally, in Banach spaces the continuous dual is naturally a Banach space, and its norm emerges from the so-called strong or $\beta (X',X) $ topology, i.e. the topology given by the uniform structure of convergence on bounded sets.

user228700

@JohnR: Morning :-)

yuggib

In Bourbaki's TVS book you can find all of that explained in good detail

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