@JohnRennie ohh thanks for the good info @JohnRennie :) Most of it is above my head, but good reminder about assuming no acceleration

The fun thing about the current research is that I mostly have to search for the suffix -oid to find what I want

A question for the QFT experts:

Assuming there is a physical object that is described by the operator field, and assuming that this object has a vacuum state, is that vacuum state an energy eigenfunction?

This is related to the issue of whether vacuum fluctuations are real, since an energy eigenstate is time independent. If the field's vacuum is an energy eigenstate then by definition it can't fluctuate.

The vacuum is by definition an energy eigenstate

"vacuum fluctuation" just means that operators other than energy can have non-zero variance in the vacuum state (that's why "fluctuation" is a silly name - there isn't anything fluctuating in time here)

in other contexts it is not even about that, but only a story that people attach to the vacuum bubble Feynman diagrams

Thanks, I thought that had to be the case but I wanted to check.

The downside of learning aboug groupoids is that it differenciates $\mathcal{G}$ and $G$

Not easy to write down notes for it

The curious thing is the QCD lava lamp video is alleged to be the energy of the QCD vacuum as a function of time but it does fluctuate. I suspect the video has been misunderstood.

@JohnRennie That's a picture for a QCD on the lattice and at finite temperature, the "time" there is likely "lattice sweeps + cooling steps", largely unrelated to actual time

If it's for T > 0 presumably it isn't the vacuum anyway.

well, it's the ground state for that temperature - but yes, that's not the axiomatic Poincaré-invariant vacuum of zero-temperature QFT

When people talk about the "diffeomorphism gauge group", are they refering to the Haefliger groupoid?

big scare quotes because it's not even a Lie groupoid apparently

@ACuriousMind Aha, it's not the energy it's the action density.

that, too, yes

I guess it has been described as the "energy" in the popular science press since no-one would know what the action density was.

The frames in the animation are essentially different samplings of possible results for a measurement of that action density, not some sort of "time evolution"

"One can prove that the bisection group is a infinite-dimensional Lie group in the sense of Milnor"

Why can't anyone decide on what a Lie group is

@ACuriousMind Is the diffeomorphism group a Lie groupoid or not dagnabbit

It is a bisection of the pair groupoid, which is a Lie groupoid

@Slereah The problem is that there is no universal agreement what the correct infinite-dimensional notion of manifold is

and since both Lie groups and Lie groupoids are defined in terms of manifolds, we can't really answer the question whether any given infinite-dimensional object is "a Lie group(oid)"

I also vaguely remember that if the manifold is compact, the diffeomorphism group does have infinite dimensional manifold structure

Which I am guessing is because of those issues of mapping classes/diffeomorphisms that do not stem from flows

That may be another part of the issue

@Slereah but what does that mean? What notion of "infinite-dimensional manifold structure" are you using in that sentence?

@ACuriousMind Very good question!

I think it's in Schreiber's big introduction

let's see

ncatlab.org/nlab/show/manifold+structure+of+mapping+spaces

I suspect they're talking about Frechet manifolds

I'm guessing the full status of the diffeomorphism group is probably not that important

and I probably just need whatever the Lie groupoid(s) of the Lie algebroid of vector fields is

The rest is probably just weird diffeomorphisms that aren't that useful

Help me Mr. Lie

What's an algebroid

If we say a groupoid is just a group with a partial binary operation, then onwards we just use the usual Lie theory thinking, but every now and again things can break a little bit here and there, if the diffeomorphism group is a group then it's got a complete binary operation not a partial one (or the partial one is also complete, so the groupoid is a group in this case)

I think the "groupoid" language is usually used for locally defined diffeomorphisms

I think so yeah

the diffeomorphism groupoid is all diffeos $U\to V$ for some $U,V\subset M$ in your manifold, the group is the case $U=V=M$

There are hints that people usually only consider compactly supported vector fields

but it's hard to be sure because they tend to be a bit vague about it

But a lot of people feel pretty free to just write $$\mathrm{Lie}(\mathrm{Diff}(M)) = \mathfrak{X}(M)$$

and leave it at that

"He died the following year in 1899 at the age of 56, due to pernicious anemia, a disease caused by impaired absorption of vitamin B12."

Sophus Lie did not die for this

How did he get B12 deficiency, was he a vegan

Well the result is the same

Lie died in vain

Also apparently he was a friend of Friedrich Hegel

Two tragic tales of men whose ideas were improperly applied

Resulting in disaster

To be fair nlab have done it to every area of math and physics

@bolbteppa "impaired absorption" means that the problem wasn't that he wasn't getting enough B12, it was that his body wasn't processing it right - "pernicious anemia" is an autoimmune disease, not a result of lacking B12 in one's diet

@Slereah they're probably just saying "vector fields generate diffeomorphisms", I don't think they really want or need to claim anything about the manifold structure. However, see mathoverflow.net/q/153484/157071 for why even that statement isn't as simple as it might seem

I know damn well, that's why I am complaining

There's too many things "diffeomorphism group" and "Lie group" could mean in this context and it's rarely clear what they're saying

If we have two Lie algebras and we want to show that they are equal, e.g. $\mathfrak{gl}(V) = \mathfrak{sl}(V) + \mathfrak{s}(V)$, where the $+$ is a direct sum, is it sufficient to show that there is exists an isomorphism $\mathfrak{gl}(V) \mapsto \mathfrak{sl}(V) + \mathfrak{s}(V)$ ?

@ShikiRyougi Is this not exactly what it means for two structures in mathematics to be "equal"? The existence of an isomorphism?

@Charlie I do not find that convincing. The sets {a,b,c} and {1,2,3} are isomorphic but not equal.

Isomorphism seems like a weaker condition than equality

Sets alone are not interesting, you're talking about a Lie algebra which has lots of additional structure on top of it's set

One has to at some point draw a line and define "equal"

If the underlying vector spaces of the Lie algebras are not equal, how can the Lie algebras be equal? And if the underlying sets of the vectors spaces are not equal, how can the vectors spaces be equal?

My point is that they are not strictly equal, but "equal" in the sense of an isomorphism. That's just a different definition of "equal" in the context of, let's say, algebraic structures.

"Equal in the sense of an isomorphism" is what most people mean by equal in this context, what alternative definition of equal are you suggesting?

You would also need to be careful with what you mean by "equal" for vector spaces since all finite dimensional vector spaces of the same dimension are isomorphic

@Charlie Thank you! That clears things up :) I think that an appropriate definition of equal would be, e.g. for vectors spaces, that their underlying sets are equal. Might need to add a few things like impose the fields being the same as well etc.

Any sets of the same cardinality are equal up to a relabelling of their elements. Also you can certainly find objects who's sets have the same cardinality however there does not exist an isomorphism between their higher level structures so that doesn't seem like a good definition of equal

Equal after relabeling the elements does not seem actually equal to me :p

And yes, I haven't really thought of a good definition

Has anybody got any resources to help with this problem?

It seems like all notes and stuff I find are of the simplest 2 state case and don't really help with this