@BernardMeurer Ok, you have to be more careful.

@0celo7 yeah?

@BernardMeurer The thing you said is trivial, isn't. $2^a$ is not even if $a<0$.

And $3^b$ is not odd if $b<0$.

Why?

Because odd and even are for integers, and those are not integers.

How can I rigorously say that though?

Furthermore, you should give a proof that $3^b$ is odd (read: not even) for $b=1,2,3,\dotsc$.

@Danu I know. I wanted @BernardMeurer to tell me that.

Don't worry, I don't get what he said

@Danu Do you agree that if one is to give a very pedantic proof, there is more work?

$$\frac{\log 3}{\log 2} \notin \mathbb{Q} \\ \text{Let's say} \frac{a}{b}=\frac{\log 3}{\log 2}; a,b\in\mathbb{Z}; (a,b)=1 \\ 2^{\frac{a}{b}}=3 \\ 2^a = 3^b \\ \text{It is clear that }2^a\text{ will always be even, while }3^b\text{ will always be odd}$$

Not really, no---all the issues you raise are one-liners.

@BernardMeurer So everything works up until the last sentence

@Danu A pedantic proof includes the one-liners.

That's what I mean by pedantic here.

@0celo7 What? That 2 doesn't divide $3^b$ is really obvious, what more of a "proof" could one possibly give for that?

You should consider some separate cases for $a,b$

@0celo7 Sure, but they're not more work.

@ACuriousMind Uniqueness of prime factorization.

Lolwut

Just write $(2k+1)(2m+1)$

Much easier

Any product of odd things is odd

Dude I don't need to prove the commutativeness of addition every time I use it, I don't need to prove the uniqueness of prime factorization

You don't need prime factorization!

Unless my whole point is to prove the uniqueness of prime factorization of course

@BernardMeurer Uh, uniqueness of prime factorization is nontrivial.

@IceLord Depends---in Einstein convention, NO!

Maybe you don't need prime factorization.

100% you don't. I just gave you a proof.

@Danu He's making a very pedantic point that you don't know that $3\cdot 3$ is not divided by $2$ without appealing to the uniqueness of prime factorization.

But product of odds is odd? How does one prove that?

@0celo7 Write out $(2k+1)(2m+1)$

Notice that $1$ is odd

@Danu Is every odd of the form $2n+1$?

Seems like you need long division for that.

@0celo7 If you're not willing to accept that then I am not interested in your proof ideas

¬¬

@0celo7 wtf

@ACuriousMind Am I wrong? Does one not need either (a) long division or (b) prime factorization?

@IceLord I don't like to think about tensors in terms of matrix multiplication much.

IIRC, only things with one raised and one lowered index really work like matrices

@0celo7 You're not technically wrong, but you're taking a different (and unreasonably low) amount of prior knowledge as granted here.

You probably get away with all of that stuff in Minkowski space anyways

@ACuriousMind Good. That's all I wanted to hear.

All I know is that there are some tricky annoying things---like the fact that the determinant of a matrix is basis-independent, but the determinant of the metric is not (in GR).

@IceLord What seems to be the issue?

@0celo7 I would urge you to rethink your approach here though, since I don't think you said anything that actually helped @BernardMeurer.

@ACuriousMind I'm looking at this like I was taught to in my algebra class.

I still don't think one should say it's trivially true that $2^a\ne3^b$.

I would urge you to either say something constructive or just stop wasting Bernard's time

I'm willing to accept that $3^b$ is odd.

@0celo7 How would you prove that $3^b$ is odd?

@BernardMeurer Sorry, don't want to waste your time.

@0celo7 Which is different from the intro to analysis context that BM is in. In algebra one emphasizes things that are "obvious" in the reals because they fail in more general algebraic structures, while in the intro context this type of exercise is used to learn proof structures like proof by contradiction or induction while using arithmetic the students are already familiar with.

@BernardMeurer I told you twice, already

Prove that a product of two odd numbers is always odd

@IceLord Depends on where the higher $\mu$ is in the initial thing.

As you have it now, you should make it $\alpha_{\beta\nu}$

In general, though, you usually define tensors with their indices in a certain way

And then you just define the raised or lowered versions by an explicit equation involving the metric

So you just choose what you want to do

Typically the best thing is to make the convention so that you preserve horizontal order of indices

@IceLord Sure, small angle

@IceLord cc @Danu That's ambiguous phrasing once again. As you're saying it, it sounds as if you want to take $\lim_{\theta\to 0} R(\theta)$ where $R(\theta)$ is the matrix that rotates by $\theta$, which is wrong

Try it.

You'll see.

What did you get?

What, exactly, did you get for $\lim_{\theta\to 0} R(\theta)$?

:P

Yes, the identity, that's correct. How is that the "opposite of what it is supposed to be"?

That looks a bit similar, but isn't the "opposite" in any meaningful sense.

First take the general rotation matrix, and Taylor expand in the angle

@IceLord Well, how did you define "infinitesimal rotation"?

Limit of what?

Yeah, that is not right

@Danu I'd prefer not giving an answer until we've got a precise question, because not asking the right question has been a recurring problem here

@IceLord Yes, it's wrong. Where did you get that definition from?

@ACuriousMind I don't know IceLord at all

Aha.

Okay.

Well, but what's your definition of "infinitesimal rotation"?

The one you gave is non-sensical, because it'll always be the identity. And you know what it's "supposed" to be, but how can you know what something is supposed to be when you haven't even defined it?

See, now we're getting closer. So you're reading a book. There are three possibilities: 1. The book is really bad. 2. The book somewhere defines this but you skipped over it. 3. The book is way over your head and you should be reading something simpler.

Sooooo... what do you learn from that? :P

@IceLord So go back and read the book more closely!

I'm not going to tell you the answer because that would reinforce you relying on asking people here instead of reading carefully.

I don't know what "causing the trig functions to switch" means nor do I know why you "add it to the identity". How did you come up with that?

@ACuriousMind Ok, point taken.

@ACuriousMind lol my thoughts exactly

Yes, but that's unrelated to this.

I'm probably trolling.

@ACuriousMind sorry

I'm hoarding the glory

@BernardMeurer You should say "suppose $\frac{\log 2}{\log 3}=\frac{a}{b}$" instead of "let's say"

Already changed

Ok. Also that tower of stuff isn't really nice. If you're typing out homework, consider writing full sentences.

@IceLord What do you want...

@IceLord What do you want...

No need to bug @ACuriousMind

no need to bug @ACuriousMind

whoa lag

Nice, two "no need to bug ACM" pings :D

two?

My internet is freaking out.

@ACuriousMind sorry, didn't want to bug you

@ACuriousMind sorry

lol

what's happening

@ACuriousMind do you have ping noises on

@IceLord Which one?

Yes.

@ACuriousMind How do your ears exist? The noise is so loud!

How do you do them?

@IceLord New rule: if it's in Shankar and I can find it in 5 minutes, I will not help

@ACuriousMind did I used to be like this?

@IceLord Read chapter 11 in Shankar.

@IceLord What about Cahill though?

@0celo7 It's not that loud

@ACuriousMind Well you survived Wacken so that's understandable.

@ACuriousMind When we write a rotation operator in terms of Pauli matrices, what is that actually called?

I'm not sure what you mean

Is it a group representation?

Or rather, if I am, then I don't think there's a word for that

@0celo7 The act of choosing the Pauli matrices to be rotations is choosing a representation, yes

Writing an arbitrary rotation in those Pauli matrices is then, well, "writing the representation of that rotation".

Right. Just checking.

Oh, I though you were searching for a catchier term :P

@ACuriousMind No, just trying to make sense out of physics.

I want to learn about representation theory, but I feel like I'll learn the wrong things.

I asked my advisor for a rep theory book and he thought I wanted to classify the homogeneous spaces...

(I do, that's why I'm reading Helgason. But I want to learn physics rep theory too.)

I think the standard recommendation for more physically relevant rep theory is Fulton and Harris?

@ACuriousMind He said "that's too algebraic" when I asked him about it.

His copy is relegated to the unreachable, use-once-a-decade shelf.

@0celo7 Is that from a physics or from a geometer's standpoint?

@ACuriousMind From a geometer, of course.

But is that what the mathematical physicists use?

@ACuriousMind So, when we do the SHO stuff in string theory, are we assuming no degeneracy of the string number operator?

@ACuriousMind According to Shankar, one can show that there is no degeneracy in 1-dim bound states.