The h Bar

Slereah

5:35 AM

Superselection rules are a more general concept, though

ACuriousMind

6:56 AM

@Slereah sure, I'm just trying to understand what situation exactly you're thinking about :P

Slereah

Vague idea about how both are about reducing the tensor product of two Hilbert spaces to a subset of it :p

and symmetries are involved

Daniel Underwood

2:22 PM

I'm not sure if this makes sense, but are the morphisms in a category curried (in the programming sense)?
Like if you have a monoid $(\mathbb{N}, +, 0)$, your hom-set isn't $\{(a, b) \mapsto a + b\}$ but $\{ a \mapsto a + b | b \in \mathbb{N}\}$?

Or I guess in other words, if you want a morphism like $(a, b) \mapsto a + b$, you would need both $\mathbb{N}$ and $\mathbb{N} \times \mathbb{N}$ as objects in your category?

ACuriousMind

depends on what "your category" is

PM 2Ring

@ACuriousMind Here's some background info on physics.stackexchange.com/q/709508 The OP recently posted answers to astronomy.stackexchange.com/q/25369 which contain some weird ideas about spacetime curvature, especially in the 1st answer (now deleted). I tried linking physics.stackexchange.com/a/587025 in a comment, but it didn't help. I've given up trying to communicate with him. Maybe others will have more success...

ACuriousMind

@DanielUnderwood I mean, you're right that we interpret a monoid as an object in the category of monoids by mapping it to a single object that has one automorphism for each element $a$ of the monoid, and the concrete representation of these as maps is $b\mapsto a +b$, but currying would be that you have some function $f$ (the curried version of $+$) that takes an $a$ and produces the $b\mapsto a+b$ map from that - I don't see that here

@PM2Ring being rather "symbolic" myself, I don't have a history of success communicating with "visual thinkers", either :P

PM 2Ring

I really don't know how to respond to someone who says "our sun is part of our earth". ;)

ACuriousMind

@DanielUnderwood I think what you could say is the following: Currying is an operation that turns a map $f : X\times Y \to Z$ into a map $f_c : X\to \mathrm{hom}(Y,Z)$ (see nlab). Now, we take $X=Y=Z=\mathbb{N}$, and then what you're saying is that taking $f = +$, then $f_c$ is an isomorphism

Daniel Underwood

2:48 PM

ah sorry, my internal definition of currying was incorrect. I was thinking currying was "function with one argument" but neglected that it gets that way by being a higher-order function.
From what you say, I think proper currying would resolve part of my question and may actually answer a couple questions I didn't know I had

ACuriousMind

3:10 PM

@DanielUnderwood actually, I think what I just said doesn't quite work

it's correct that we can view a single monoid as a category with one object and its hom-set equal to the output of $f_c$.

but when we want to talk about the category of monoids - the "space" in which stuff like $\mathbb{N}\times \mathbb{N}$ exists - then we need a notion of what a morphism between monoids is and the obvious choice is to say that a morphism is one that sends the identity to the identity and preserves the $+$ as $f(a+b) = f(a) + f(b)$

but in this definition of morphism we have that the endomorphisms of $X$ are more than just its elements: For instance, there is the "trivial" endomorphism that just sends everything to the identity, which need not be represented by an group element

the problem is that there are various ways (at least) to view a monoid "in a category": As this standalone single-object category, as an object in the category of monoids, or even as a category whose objects are the elements of the monoid and there are no non-trivial morphisms but you have a tensor product (aka "monoidal") structure induced by the monoid operation

I mixed two of them above, and that doesn't really work

explicitly, in the setting of the category of monoids, $f_c$ for $+$ is not an isomorphism, but in the single-object category where I'd "like" it to be an isomorphism, I can't define what $f_c$ is.

Slereah

10:28 PM

Does this seem like an appropriate coordinate-free definition of derivatives on product manifolds

Take the manifold $M = M_1 \times M_2$, with projection map $\mathrm{pr}_i : M \to M_i$

From a variety of proof, one can show that the tangent bundle splits as $TM \cong d\mathrm{pr}_1 TM_1 \oplus d\mathrm{pr}_2 TM_2$

Given a vector field $X \in \mathfrak{X}(M)$, you can use the canonical projection and canonical injection to define $\iota_i \circ \pi_i (X)$

Which would be roughly $(X_1, 0)$ or $(0, X_2)$

And then the derivative on the first or second coordinate would be $$d_i f(X) = df(\iota_i \circ \pi_i (X))$$

Or $$d_i = (\iota_i \circ \pi_i)^* d$$

ACuriousMind

I'm not quite sure what you're trying to do there

Slereah

Well coordinate-wise it would be something like $$\frac{\partial}{\partial x_1^a} f(x_1, x_2)$$

ACuriousMind

you're not defining a derivative, all you're saying is that the splitting $M = M_1 \times M_2$ induces a splitting $TM = \mathrm{pr}_1^\ast(TM_1)\oplus \mathrm{pr}_2^\ast(TM_2)$, which in turn induces a splitting of the co-tangent bundle.

Slereah

Basically just making it so that the gradient, when it is acted upon by a vector, is only acted upon by a vector that is in the kernel of $d\mathrm{pr}_i$

So that those derivatives go away

ACuriousMind

the derivative is defined all along, you just saying it's splits into the individual derivatives on the factors by composing the $T^\ast M$-valued $\mathrm{d}f$ with the projections $p_i : T^\ast M \to \mathrm{pr}_i^\ast(T^\ast M_i)$

Slereah

10:38 PM

Well yes, that was the point

Not a particularly amazing feat, but doing it entirely without coordinates isn't that fun

ACuriousMind

so yes, that works, but I wouldn't call it "defining derivatives"

Slereah

Well it's specifically to define derivatives on bitensors

They are usually defined entirely with coordinates

ACuriousMind

would've helped if you'd said that at the start :P

Slereah

Would it

It doesn't fundamentally change the question

ACuriousMind

sure, so you want to take $M_1 = M_2 = M$ to think about the bitensor as a function on $M^2$, and you wonder how you can say "take the derivative w.r.t. the first input"

Slereah

10:41 PM

Pretty much

ACuriousMind

that's perfectly reasonable, but to me, again, very different from "defining derivatives" in general :P

Slereah

Sorry it didn't amaze enough

ACuriousMind

if you'd said that I think my whole response would've been "yes, that works" ;)

Slereah

I tried many ways to do it

I thought I came up with a clever one by defining it as a product connection

But that involved lifts and it was not nice to deal with

And also it's fundamentally the same idea since it's the same short sequence thing

ACuriousMind

I mean, I guess you could've done this even "easier" if you first showed that $TM = TM_1\times TM_2$ (as manifolds, not as bundles over $M$)

Slereah

10:51 PM

I did show it, but idk if it works out well for defining sections that are specifically in one?

ACuriousMind

or, well, perhaps not, this not being a bundle isomorphism probably makes transforming a section of the l.h.s. into its "component sections" on the r.h.s. awkward

Slereah

a lot of proofs around theorems regarding products of manifolds involve the embedding function $\iota_i : M_i \to M_1 \times M_2$, but I'm not sure if it's entirely helpful?

It seems to only work point-wise

Colourful Spacetime

Hopefully in a few weeks\months, I will be able to follow what you are saying! :D

Slereah

I guess the canonical injection is basically the pushforward of that though

Colourful Spacetime

Have a good night everyone!

Mauro Giliberti

10:57 PM

Hello @ACuriousMind, I think that the comment section under my question is becoming a bit of a separate discussion, so I thought I'll continue it here.

ACuriousMind

@MauroGiliberti sure, you mean

I'm not sure what you're saying - in my eyes, whether or not a definition including factors of $\mathrm{i}$ is a "matter of convention" or not is merely a historical artifact of whether or not we've been unfortunate enough to have different influential people make different choices. There's no good objective reason why you'd be free to choose your definition of $G$ but not that of a different quantity - ultimately all definitions are "matters of convention", it's just that in the majority of cases there is only one common convention. — ACuriousMind ♦ 10 mins ago

Mauro Giliberti

Yes That's what I meant

ACuriousMind

@ColourfulSpacetime don't worry too much if you don't, these ramblings are not really what you need to worry about when you learn geometry :P

Slereah

It's all about triangles

ACuriousMind

@MauroGiliberti I guess I don't just understand what you think sets this particular definition apart from any other. When I define some new function $f$, I could just as well have defined it as $f' = -f$ or $f'' = \mathrm{i}f$ - that will change any subsequent formulae it appears in, but there's nothing inherently wrong with doing so

there's nothing special about the propagator $G$ in that respect (except that you apparently had the misfortune of running into texts whose definitions differ and didn't realize it quickly)

Mauro Giliberti

11:02 PM

your claim that <all definitions are "matters of convention"> is, again, plausible, but not obvious to me. There are some things that are of course matters of convention: I can define the Fourier transform with e^ixp and the inverse with e^-ixp or vice versa, and nothing else will be affected.

Other things (like the sign of the delta used for the propagator, as I learned from the comments and answers to my question) are matters of convention, but aren't trivial: you have to keep track of it thoroughly.

Slereah

what you have to watch out for is the contour

Mauro Giliberti

@ACuriousMind Well sure, but what if you want your function $f$ to describe something physical (or to be used in a model for the description of something physical)?

@Slereah wait WHAT

Slereah

there's a bunch

Mauro Giliberti

I've never seen something like this. Do you care to explain?

Slereah

They are the various contours one can take to compute the propagator

The most obvious are $G^+$ and $G^-$, the advanced and retarded propagator

$G$ is the Feynman propagator

Mauro Giliberti

11:06 PM

@Slereah Now everything makes much much more sense!

Slereah

$G^{(+)}$ and $G^{(-)}$ are just the regular two point functions, $\langle 0 | \phi(x) \phi(y) |0 \rangle$ and $\langle 0 | \phi(y) \phi(x) |0 \rangle$

ACuriousMind

I think the contours are yet another thing

Mauro Giliberti

oh no.

ACuriousMind

the differential equation that defines the Green's function is underspecified - like all differential equations, you need to prescribe boundary conditions to make the solution unique

the different contours correspond to different boundary conditions

but you're not talking about changing the boundary conditions, you're talking about changing the defining equation

Slereah

They don't all satisfy the same equation

ACuriousMind

11:12 PM

@MauroGiliberti well, if $f$ is directly connected to something you can measure, it's easy - surely no one is tempted to redefine the sign of the electric field

but most objects in theoretical physics aren't quite that directly related to observable quantities

Slereah

The ones that don't involve time explicitely obey $(\Box + m^2) G = 0$

While the ones that do obey $(\Box + m^2) G = \pm \delta(x - x')$

Mauro Giliberti

@ACuriousMind That was precisely my question, but I probably didn't deliver it properly: is $G$ something close enough to "something you can measure" that "no one is tempted to redefine" its sign?

Thank you for helping me focus my question, it's usually the hardest part for me

ACuriousMind

perhaps for a simpler example because it doesn't involve Green's function math: There are two different sign conventions for the metric in special relativity, and it is extremely annoying to try to translate expressions from one text with one convention into the other because they never explicitly keep track of which signs are due to this convention

I'd argue that reasonable people that can all instantly communicate with each other would've settled on a single choice, but that's not what happened

3

Slereah

Isn't there like 3 different signs that you can change in GR?

Maybe we should have all equations with a "sign convention constant"

ACuriousMind

similarily, reasonable people that can all instantly communicate with each other would've settled on one meaning of "propagator" or "Green's function", but that's not what happened either

Slereah

11:17 PM

$g = \varepsilon \mathrm{diag}(-,+,+,+)$

Just input $\varepsilon$ for the appropriate sign

ACuriousMind

yeah, you're right, it's even worse than only two different conventions

Slereah

although even worse are the papers that use $\mathbb{R}^{p,q}$ all throughout

In case I want to solve it in 5 time dimensions

$\mathbb{R}^{p,q|N}$ also

ACuriousMind

@MauroGiliberti The Wikipedia article has a funny line about this: "For purposes of Feynman diagram calculations, it is usually convenient to write these with an additional overall factor of −i (conventions vary)."

so it's not only that people might disagree on which of the propagators they mean by "$G$" (which will usually be the Feynman propagator, that's the contour choice), but also people stick random factors of $\mathrm{i}$ in there because they think it makes the formulae look nicer

it's a mess

3

Slereah

Conventions may differ for the metric, Riemann tensor, propagator and gamma matrices

Not gonna be fun to do quantum gravity

Mauro Giliberti

@ACuriousMind I should've checked the least trustable source of them all

@ACuriousMind Yeah that's the impression that I'm getting

ACuriousMind

11:23 PM

and don't get me started on spinors

Slereah

Is it $\varepsilon_{123}$ or $\varepsilon^{123}$

Mauro Giliberti

@Slereah To be honest, I didn't have this much confusion for my QG exam: probably because I only used 3-4 books and they all had similar conventions?

ACuriousMind

I mean, it's not for nothing that "sign errors" are a bit of a meme among physicists :P

Slereah

also factors of 2, i and $\pi$

Mauro Giliberti

To be honest, it seems like QFT is particularly full of these "widely varying" convention, but that might just be my experience

ACuriousMind

11:26 PM

not only is it easy to make an actual error, you can also really shoot yourself in the foot by copying from different sources without carefully checking whether their conventions might differ

Mauro Giliberti

@ACuriousMind In Italian, "minus" and "less" are the same word. My professor said that a physicist, "regarding the minus, couldn't care less" which was indeed a prophetic pun.

@ACuriousMind and that's precisely what I do, all the time.

Slereah

Also occasionally you may have an extra factor of $\infty$

watch out for those

Mauro Giliberti

hahaha

ACuriousMind

@MauroGiliberti I think that's a consequence of QFT just being pretty complicated in the sense that a lot of the time you can't really "sanity check" the formulae

it's hard to make sign errors if you're at all times only like a few lines of maths away from computing an actual observable quantity where you can think about whether your expression for it makes physical sense

Mauro Giliberti

And maybe because a lot of different people have worked on it to craft it?

Slereah

11:31 PM

I mean that's true of all theories

Mauro Giliberti

That's always been my impression

That QFT has a lot of different names, while for example in GR almost all the "core equations" are Einstein-something. This might just be the impression I have from having only 1 GR course and 4 QFT ones.

Slereah

Obviously you've never seen the Newman-Penrose formalism

Or ADM

ACuriousMind

that's more a consequence of the personality cult around Einstein I think :P

a lot of the modern formalism is not at all what Einstein would've liked

Slereah

Newman–Penrose formalism

The Newman–Penrose (NP) formalism is a set of notation developed by Ezra T. Newman and Roger Penrose for general relativity (GR). Their notation is an effort to treat general relativity in terms of spinor notation, which introduces complex forms of the usual variables used in GR. The NP formalism is itself a special case of the tetrad formalism, where the tensors of the theory are projected onto a complete vector basis at each point in spacetime. Usually this vector basis is chosen to reflect some symmetry of the spacetime, leading to simplified expressions for physical observables. In the case...

Mauro Giliberti

I studied the tetrad formalism, but not with this name

Slereah

11:35 PM

it's what GR would look like if you spelt out every equation

Mauro Giliberti

Man, physics truly is messy. One really gotta love it.

ACuriousMind

I'd actually much rather let the mathematicians clean it up and come back in 50 years when they've figured out what's going on :P

Slereah

They do it all the time but physicists never listen

although really mathematicians aren't much better

I've looked up and down many places, almost no two books define the vertical and horizontal lift the same way

Mauro Giliberti

@ACuriousMind That's the impression I got from my mathematical physics course! It was terribly precise (and terribly boring)

It seemed like a field to approach when one retires

Doing the BRST with differentials and not with loops was very elegant

Anyhow, it's late here, and I can't afford to be tired tomorrow: I have a EFT matching to do (which I won't understand, so you'll hear from me)

good afternoon, and in case I don't see ya, good evening and good night!

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