The h Bar

12:01 AM

I should have known better, of course, I'd met about twenty physicists-by-training with advanced degrees at various adult functions my parents dragged me to. Exactly two were still doing physics, and most were not even in obviously technical fields.

0celo7

@dmckee Thats a nice prospect for the physics students in here

dmckee

But I was such a snot nosed little twat that I figured that was a sign that they were defficient somehow.

@0celo7 The market for BS's is rather more traditional. You can more fairly smoothly into engineering or into tech.

Taking the grad school plunge is entering your future into a lottery for the pleasure of pure learning, and a big hit in accumulated compensation.

0celo7

I'm not a physics student

I wonder if the misery in this chat had anything to do with that

dmckee

@0celo7 You keep saying that.

0celo7

@dmckee I do, because it's true

I've done a good job of convincing myself of it

I really don't like physics

It's math I now have to rid myself of

DanielSank

12:10 AM

@dmckee Heh.

@dmckee You have to know people.

@dmckee Not your fault. Academia spreads that lie around and seems to be proud of it.

dmckee

@DanielSank And the people I knew had retired, so they had little swing.

DanielSank

@dmckee ah

WillO

@0celo7 @0celo7: I posted an answer to your question at MSE.

0celo7

12:26 AM

You missed the part in the OP where $p$ and $q$ are nonzero

WillO

@0celo7 You are absolutely right; I missed that.

You meant to say non-negative, not non-zero.

Anyway, yes, with that additional condition I think what you're claiming is likely to be true.

0celo7

You're right

Strange, I distinctly remember thinking "nonnegative" when I typed that

(I get those two mixed up a lot)

@WillO I hope it is!

WillO

Well, it's much less of a blunder than my missing the fact that there was any condition in the first place!

0celo7

I'd like an answer that doesn't use big machinery of homological algebra though

I'm afraid if I ask on MO they'll tell me to use a triple filtered Ext functor filter

or something stupid

WillO

:)

0celo7

12:30 AM

And my (geometric analysis) advisor will tell me to ask an algebraist

WillO

More likely they'll tell you it's off topic because the answer is obvious --- but it's not 100% obvious to me.

0celo7

This Ricci flow thing WILL kill me

WillO

Are there any other conditions on the $K^{\bullet,\bullet}$? E.g. are they projective?

0celo7

I don't know what that means

WillO

Or better yet, are they vector spaces?

0celo7

12:34 AM

Indeed they are

WillO

Oh!

Then it's just about got to be true.

0celo7

I eagerly await your proof :D

WillO

I have no proof, but here is how a proof will go:

Break up your double complex into a direct sum of much simpler double complexes, all concentrated in just two total degrees. If all the $K^{\bullet,bullet}$'s are vector spaces, you can do this.

0celo7

Concentrated in just two total degrees?

WillO

Do the same for $L$. Then you can break up your map $f$ into pieces that take one piece of $K$ to one piece of $L$ and analyze these....

Yes.... Total degree means p+q

0celo7

12:38 AM

I still don't know what you mean by that

WillO

So---- pick your favorite $p$.

0celo7

ok

WillO

Your life would be much easier if all the non-zero terms of K were one-dimensional vector spaces that live in K^{a,b}'s where a+b is either p or p+1

0celo7

I...don't know

WillO

Call those complexes "simple". (That is not standard terminology.)

You can write your complex as a direct sum of simple complexes.

I could be wrong, but I believe this will lead to a proof

0celo7

12:41 AM

I'm sorry, I don't see what that has to do with vector spaces

WillO

I am invoking a theorem that says that if the K^{pq} are all vector spaces, then your double complex is a direct sum of simple complexes

0celo7

uhhh

Can you describe this in detail please?

I don't know much homological algebra at all

WillO

I wish I had a way to draw a picture here. Let me go draw one elsewhere and I'll be right back with it.

0celo7

ok

WillO

okay---is there a canonical place to post a picture, or is it possible to post one into this chat window? or should i put it elsewhere for you?

0celo7

12:52 AM

imgur

see the upload button at the bottom

WillO

imgur wants me to create an account. try panix.com/~wotw/ocelo.gif

There you see four different double complexes (all non-pictured spaces are zero).

0celo7

what kind of imgur are you using

WillO

These are even simpler than they look, because all of these vector spaces are 1-dimensional.

0celo7

uhhh

WillO

Theorem: Any bounded double complex is a direct sum of double complexes of this sort (and their obvious longer generalizations).

I'm not sure what "uhhh" means :)

0celo7

12:56 AM

I don't know what this picture is telling me

WillO

You mean you don't understand what I'm saying, or you don't understand why it might be useful?

0celo7

Either

WillO

Okay....the first is probably a prerequisite for the second....

0celo7

What is this picture supposed to be

@WillO Usually, yeah

WillO

It is not one picture. It is four completely separate pictures.

0celo7

12:58 AM

oh, that's a helpful hint

WillO

The first picture is a double complex in which all but four entries are zero, and the other four entries are one-dimensional.

The second (to the right) is a double complex in which all but three entries are zero and the other four are one-dimensional.

The third is another double complex with just four non-zero entries. The fourth has just five, and other "staircase-like" double complexes can have six, seven, eight, etc.

If you knew that your double complex was one of these, it would be very easy to calculate with. It's easy to compute the horizontal, vertical, and total homologies.

0celo7

Hmm

But in the third picture

don't you get things from three total degrees

WillO

You don't know that, but you do know something along those lines: Your double complex is a direct sum of complexes that are this simple.

0celo7

nevermind that

WillO

My total degrees are fixed along 45 degree lines that run NW to SE

0celo7

1:01 AM

@WillO now that I do not see

WillO

you might be in the habit of indexing differently

you do not see.....what?

0celo7

not all of the thingies in the complex are the same vector space

I do not see that $K^{\bullet,\bullet}$ is the sum of such complexes.

WillO

But they are all direct sums of one-dimensional vector spaces.

I am not claiming that it's obvious that your complex is a direct sum of such things. I am claiming only that it's true.

0celo7

Ok, let us suppose that this is true

WillO

I am also claiming that I've often found it a useful thing to know, and I suspect it will be useful here, though I make no absolute guarantees.

0celo7

1:03 AM

why is it useful?

WillO

Because it breaks big complicated problems up into much smaller uncomplicated problems (though at the cost of moving the complication over to how the solutions to the smaller problems fit together)

0celo7

ok

WillO

In the other direction, if you want to look for a counterexample....you can compute, from the structure of the double complex, maps $H^q(K^p,{\bullet} \rightarrow H^{q-1}(K^{p+2},\bullet}$$. You've got (by assumption) that the homologies are the same when you replace $K$ by $L$, but not necessarily that these maps between homologies are the same when you replace $K$ by $L$.

These are important invariants, and if they differ in some example, then that's a good candidate for a counterexample.

Okay, what you want is true. I have a proof, but it will take a little time to write it up.

But in brief (and recognizing that you might or might not know all these words):

Let C be the mapping cone of f. Then the first spectral sequence for the homology of C collapses at E^2. The boundedness condition says that the spectral sequence converges, and voila.

Secret

1:50 AM

6 hours ago, by dmckee

A lot of what goes on in these discussions is epistemology. Peolpe what to know what the quantum world "really is". Well, so do I, but I've had to come to terms with the lack of cog-wheels in there.

5 hours ago, by ACuriousMind

I get that it is somewhat dissatisfying that physics doesn't answer every "why" question you can think of, but that's, after all, not what it's supposed to do. It's supossed to predict nature, to give us a model we can work with.

One way to work out the why is allowing two "whys" of the same nature to interact with each other

As an example, before spectroscopy involving polarised light, we only knew something is an enantiomer because it interacts with another enantiomer

In the end at such level, we don't know what the "whys" are and why they act this way, but from these indirect experiments, we can identify what "why" we are dealing with

Another example is the stern gerlac experiment which told us an electron has a angular momentum like quantity that spans a 2 dimensional hilbert space. The maths then organise these findings into a systematic fashion so that prediction can be made from it. But what it "actually" is we don't know, other than it is some label of the representation of an more generalised notion of changing the orientation when looking at th…

So I argue that most of the time, because most things cannot be directly probed, most of the whys have to be inferred from the hows, just like how daniel mentioned about the electromagnetic vector potential is a nice bookeeping device to predict motions of charged objects under electromagnetism

@Bass You nailed what I am trying to say back there

> Those "harmfully wrong" answers can be utterly confusing to students if one does not recognize them as wrong / non-mainstream. You read the answer and you try to understand and learn the (wrong) stuff in it. Then some time later, somewhere else, you read contradicting (but correct) content.

DanielSank

3:41 AM

I'm annoyed that there's a featured question asking what a photon is.

It's good that the question "what is a photon" is getting attention, but it's less good that it's a duplicate which got away with being a duplicate because it 1) has a bounty, and 2) is (unnecessarily) long.

Also, @JohnDuffield, a question not having an answer is not an excuse to post a duplicate question. Instead, edit the original, put a bounty on the original stating that you want a better answer, campaign in chat for the original to get attention, or do nothing.

Don't post duplicates because you feel like it.

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