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DIRAC1930

12:00 AM

In your comment 'all projective representations are linear representations of the universal cover, which in turn are linear representation of the algebra'

Nevermind

Sir Cumference

12:19 AM

what is up with the new profile layout

looks kinda weird

DIRAC1930

12:52 AM

Is a covering group $\tilde{G}$ of $G$ just a group such that $G \cong \tilde{G}/K$ where $K$ is a normal subgroup of $\tilde{G}$?

Physics Meta

3:13 AM

1

Q: My question got closed saying that it was opinion based

I cannot wrap my head around how can you not have a wrong answer inn physics it's not philosophy that we are talking about... Can anyone please suggest how can i improve my question to open it again? Link to the question : Can 0 acceleration be termed as constant acceleration?

Nihar Karve

3:49 AM

@DIRAC1930 1. It needs to be a topological group and 2. The kernel $K$ has to be discrete. But otherwise yes.

DIRAC1930

So in reference to a universal covering group, what is K?

Nihar Karve

$K$ will be the fundamental group of $G$, eg. $\mathrm{SU}(2)/\mathbb Z_2\cong\mathrm{SO}(3)$; $\pi_1(\mathrm{SO}(3))=\mathbb Z_2$

DIRAC1930

Is there any way to understand these things without topology?

Is $K$ the subgroup of $G$ that commutes with every element of $G$?

Nihar Karve

4:16 AM

Have you read the Wikipedia page?

DIRAC1930

There's just so many mathematical words that I have no idea where to even start with

Nihar Karve

ok

DIRAC1930

So I'm guessing I was wrong

Nihar Karve

do you know about ordinary group quotients (no topology involved)?

DIRAC1930

I think so

So $SU(2)/\mathbb{Z}_2$ would be $\{ \pm G_1,\; \pm G_2,\dots\}$

Nihar Karve

4:20 AM

right

and do you know why, in taking the quotient $G/H$, $H$ must be a normal subgroup of $G$?

DIRAC1930

I do not

DIRAC1930

4:53 AM

Okay, I think it is because otherwise the quotient group wouldn't satisfy the group axioms

It only does if the left cosets are equal to the right cosets

Nihar Karve

correct

so a covering group is a natural extension of this to topological groups: the projection homomorphism $G \to H$ should be continuous

DIRAC1930

What is a topological group in simple terms

Nihar Karve

1. A group that is also a topological space and 2. the functions $a,b\to a\cdot b$ and $a\to a^{-1}$ are continuous

Again, the Wiki page is pretty good

DIRAC1930

5:08 AM

I think I'm just going to give up

Nihar Karve

If this is a barrier to proceeding with something else, then you should just ignore the topology aspect of covering groups and think of quotient groups only

you can learn the rigorous definition later if you like

DIRAC1930

So in my definition $G = \tilde{G}/K$ where $\tilde{G}$ is the covering group and $K$ is a discrete normal subgroup, what would $K$ be for a universal covering group?

Nihar Karve

Like I said, for a universal covering group specifically,

1 hour ago, by Nihar Karve

$K$ will be the fundamental group of $G$, eg. $\mathrm{SU}(2)/\mathbb Z_2\cong\mathrm{SO}(3)$; $\pi_1(\mathrm{SO}(3))=\mathbb Z_2$

DIRAC1930

But how do I understand that without topology?

Nihar Karve

OK I recant my previous statement, you will likely need to learn some basic general and algebraic topology if you want to understand this specifically

DIRAC1930

5:29 AM

If $G$ is a (multiply) connected Lie group there
exist a simply connected group $\tilde{G}$ (unique up to isomorphism) such that $G$ is
isomorphic to the factor group $\tilde{G}/K$, where $K$ is a discrete central invariant
subgroup of $\tilde{G}$. The group $\tilde{G}$ is called the universal covering group of $G$.

I found the above definition

Which doesn't seem to rely on topology

But I'm not sure if I'm interpreting it right

Nihar Karve

5:49 AM

Lie groups (and their isomorphisms) implicitly rely on topology :)

RewCie

7:24 AM

What do mathematicians do when they are heartbroken? and how is that different from what a physicist does? Discuss. 100 Marks

6

Slereah

The only physicist example I can think of is this : lettersofnote.com/2012/02/15/i-love-my-wife-my-wife-is-dead

7

RewCie

@Slereah That's very sad

Slereah

Heartbreaks will do that

RewCie

Wish never happened to him that way

Also god knows how people recover from it

I still get bad flashbacks and nightmares all time

the pain that's never leaving

f

Physics Meta

8:07 AM

0

Q: Should we offer more feedback when we vote against reopening a question?

Here on Physics.SE, it culturally seems to be much easier for us to close questions than to reopen them. Recently, I've noticed several cases where a closed question, which appears to be on the edge of being appropriate for the site, shows up in the reopen queue after being edited. Reviewers oft...

discussion closed-questions reopen

fqq

11:12 AM

@DIRAC1930 "simply connected" is a topological notion

as is the notion of (universal) covering space, and therefore universal covering group

RewCie

1:06 PM

@JohnRennie and we have got @EvilJohnRennie too!

what's the difference between the good and evil one?

satan 29

ones good

the other is evil

Physics Meta

3:58 PM

0

Q: A question (quoted below) was asked which I answered, and my answer was deleted. I would like to ask the logial reason for the deletion

The question: Was the universe's relative dark energy content close to zero right after the beginning? This seems to me to clearly be a question seeking a "Yes" or "No" answer. The question was preceded by reasoning clearly supporting the "Yes" answer, so I added to my answer, "Your reasoning i...

discussion deleted-answers

Nihar Karve

4:40 PM

The "angular momentum is not conserved" guy is back again...

DIRAC1930

Just out of interest, how easy did people find it to learn representation theory after a physics background?

Slereah

not very

DIRAC1930

I'm finding that maybe I should learning the basics of math from the first year of a mathematics degree first

should start*

ACuriousMind

yes, if you are interested in somewhat rigorous approaches to physics, learning the math separately is definitely worthwhile

I never understood what was going on with groups and representations until I learnt it from the mathematicians

similarly for relativity and differential geometry, but in both cases it seems many physicists have absolutely no problem with "the physics way", so ymmv

DIRAC1930

In what ways is topology used in condensed matter theory?

ACuriousMind

4:52 PM

not at all if you look at what you learn in math when you just ask for "topology" :P

however, differential and/or algebraic topology can play roles in physics because they relate the shape of some spaces (their "topology" in imprecise language) to physically relevant things - a common thing is counting zero modes of certain physically important operators, or having "topological" solutions that represent things like different bundles over the space

DIRAC1930

Are homomorphisms invertable?

ACuriousMind

sometimes?

"homomorphism" essentially just means a map that preserves whatever structure you currently care about

e.g. a homomorphism of vector spaces is one that plays nice with scalar multiplication and vector addition, but a homomorphism of groups is one that plays nice with the group multiplication

invertible homomorphisms whose inverse is also a homomorphism are called isomorphisms

DIRAC1930

So if two groups are homomorphic, you have to show explicitly that there is a map that is invertible, meanwhile, if there is an isomorphism, I can freely go back between each of the groups with a map?

And if you find the inverse in the former case, the groups are isomorphic?

Slereah

5:07 PM

I think any map from a group to the identity group is a homomorphism

Which is very much not invertible

So yeah you do need to define the invertibility

ACuriousMind

@DIRAC1930 "homomorphic" is not really a notion

be careful to not confuse this with homeomorphic, which is the same as "isomorphic" in the context of topological spaces

DIRAC1930

5:46 PM

Okay so the complexification of a Lie algebra is a monomorphism therefore if I complexify an algebra, I can always go back

Is this correct

My intuition is that it is because it's one to one

ACuriousMind

I don't really understand what you're trying to say

what does "go back" mean in this context?

DIRAC1930

I mean that you can invert it to get back to the lie algebra you had before complexifying

ACuriousMind

also, "the complexification" is itself a Lie algebra, it's not a morphism. There exists a morphism from the original algebra into its complexification, but the complexification is not that morphism

DIRAC1930

Okay thanks

ACuriousMind

if you view the complexification as a real algebra generated by $T_i$ and $\mathrm{i}T_i$ where the $T_i$ are the generators of the original algebra, there is a map $T_i\mapsto T_i, \mathrm{i}T_i\mapsto 0$ from the complexification to the original algebra, but I'm not sure why you'd look at that map

DIRAC1930

5:52 PM

I'm just trying to get a conceptual idea of what's going on

Here is where I'm at now

In what way are the lie algebra $\mathfrak{so}(1,3)$ and the lie algebra of $Spin(1,3)$ related?

I understand that $Spin(1,3)$ is the universal covering group of $SO(1,3)$

And that $Spin(1,3) \cong SL(2,\mathbb{C})$

Okay so 'Isomorphic Lie groups have isomorphic Lie algebras but the converse is not necessarily true'

ACuriousMind

6:11 PM

that's a very strange statement

DIRAC1930

I found it on here

Lie group–Lie algebra correspondence

In mathematics, Lie group–Lie algebra correspondence allows one to correspond a Lie group to a Lie algebra or vice versa, and study the conditions for such a relationship. Isomorphic Lie groups have isomorphic Lie algebras but the converse is not necessarily true. One obvious counter example is R n {\displaystyle \mathbb {R} ^{n}} and T n {\displaystyle \mathbb...

ACuriousMind

more carefully one should say that to each Lie groups there is only a single Lie algebra you can associate to it, but to a Lie algebra there can be many different Lie groups that have this algebra as their Lie algebra

what is true is that there is a unique simply-connected Lie group for any given Lie algebra, and all other Lie groups with the same Lie algebra are covered by this group

DIRAC1930

So can I say that the Lie algebra of $Spin(1,3)$ is isomorphic to $\mathfrak{sl}(2,\mathbb{C})$?

ACuriousMind

yes

DIRAC1930

Okay, so there exists a monomorphism between $\mathfrak{sl}(2,\mathbb{C})$ and $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}$ and in turn, we also have $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C} \cong \mathfrak{so}((1,3)_\mathbb{C}$

and $\mathfrak{so}(1,3)_\mathbb{C} \cong \mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C})$ which we know how to deal with

My issue is that since the monomorphism between the Lie algebras $\mathfrak{sl}(2,\mathbb{C})$ and $\mathfrak{so}(1,3)_\mathbb{C}$ is injective, don't we have to cut some of $\mathfrak{so}(1,3)_\mathbb{C}$ out?

ACuriousMind

6:38 PM

cut out to do what?

remember, we're doing this for representation theory, and we want to use that the complex representations of $\mathfrak{g}$ and $\mathfrak{g}_\mathbb{C}$ are the same for any algebra $\mathfrak{g}$

are you trying to figure out why this is the case?

It's a very simple argument: A complex representation of $\mathfrak{g}$ is given in terms of the generators as some $T_i\mapsto \rho(T_i)$. Since $\mathfrak{g}_\mathbb{C}$ also has $T_i$ as a basis (just as a complex vector space instead of a real one), this also defines a representation of $\mathfrak{g}_\mathbb{C}$ by sending the $T_i$ to the same $\rho(T_i)$

Conversely, any representation of $\mathfrak{g}_\mathbb{C}$ is a representation of $\mathfrak{g}$ - just restrict the map to "one half" of the complexification, or more precisely the image of the monomorphism $\mathfrak{g}\to\mathfrak{g}_\mathbb{C}$ you already talked about.

DIRAC1930

Sorry, I was trying to understand why we can use $\mathfrak{so}(1,3)_\mathbb{C}$ in the first place so I was showing that there was a map from the Lie algebra of the universal covering group $\mathfrak{spin}(1,3)$, to $\mathfrak{so}(1,3)_\mathbb{C}$

ACuriousMind

well, $\mathfrak{spin}(1,3)\cong \mathfrak{so}(1,3)$ by our earlier discussion of covering groups

DIRAC1930

Okay well that simplifies things a lot lol

So now I'm on the issue of complexification

I'm just going to read your thing

one sec

What is the difference between a complex lie algebra and the complexification of a real lie algebra?

ACuriousMind

7:00 PM

"a complex Lie algebra" is a Lie algebra where you allow linear combinations with complex coefficients

the complexification of a real Lie algebra is a complex Lie algebra you have gotten by starting from a real algebra

DIRAC1930

Why aren't they the same thing?

ACuriousMind

it's rarely relevant in physics, but in fact there are complex Lie algebras that are not the complexification of a real Lie algebra, see e.g. math.stackexchange.com/q/1106548/143136

essentially, in order to have a corresponding real algebra, you need to have at least one set of generators where the commutation relations have no non-real coefficients in them (because these don't make sense in a real algebra)

but it can happen that there is no such set, and then your complex Lie algebra is not the complexification of any real Lie algebra

DIRAC1930

Okay thanks

so what is the definition of a complex representation?

ACuriousMind

it's just a representation on a complex vector space

i.e. we're looking at representations on $\mathbb{C}^n$

DIRAC1930

So the real representations of so(1,3) are only relevant in classical physics where the vector space is real?

ACuriousMind

7:09 PM

yes (but of course every real representation is also a complex representation by just replacing your $\mathbb{R}^n$ with $\mathbb{C}^n$, it's just that there can be complex representations that aren't real ones, just like with the complexification of algebras)

Slereah

Have some mercy @ACuriousMind :p

ACuriousMind

I was asked :P

DIRAC1930

I'm so confused

ACuriousMind

well, I didn't learn all these things in a day either, as with most math, it takes a few encounters with it to not feel daunting anymore :P

DIRAC1930

So in most circumstances real representations have real components?

because $R^n \rightarrow R^n$ which is only possible if the elements of the representation are real

But the real representation with real components can still act on $C^n$

ACuriousMind

7:18 PM

yes, that's the idea

DIRAC1930

but there are particular complex representations where the elements are complex therefore $C^n \rightarrow C^n$

Or even $R^n \in C^n \rightarrow C^n$

ACuriousMind

note that it's really more like "there is a basis in which the matrices have real components" - if you choose a random basis in $\mathbb{C}^n$ even a real representation might have complex entries

DIRAC1930

What if I have a real representation but I just multiply everything by $i$?

Okay so in my example, there is a basis that has real components so it's still a real representation?

No wait I think that's wrong

Okay I'm going to read a book on this first

ACuriousMind

just be careful not to fall too deep into the math rabbit hole :P

plenty of promising physicists have been lost to it

it starts with group theory and soon you're wondering whether you need the axiom of choice

Slereah

7:36 PM

I've seen people use Zorn's lemma for some proofs

ACuriousMind

Zorn is wonderful - proving the existence of things without having to show any work in actually constructing them feels strangely satisfying. "Look ma, I constructed this maximal ideal with no hands"

Slereah

ACuriousMind

man, I miss the goose

Slereah

It was fun when he wasn't complaining about girls not wanting a nice guy

ACuriousMind

right, the non math/physics comics could be pretty hit-and-miss

fqq

8:29 PM

@ACuriousMind the Chern-Simons stuff is actually topological no?

I might be stretching my definition of condensed matter, because I like condensed matter theory and maths. But stretching is allowed in topology :)

ACuriousMind

@fqq read my next message ;) "topology" without qualifier to mathematicians usually means stuff like separation axioms (e.g. Hausdorffness), while what physicists mean by "topology" is more properly differential and algebraic topology

fqq

yes I agree

ACuriousMind

I just wanted to caution against asking mathematicians about "topology" and expecting to find anything that physicists mean by "topology" :P

fqq

Ironically in my (physics) undergrad we studied quite well that general topology stuff, and almost nothing of the "useful" bits

ACuriousMind

well, it's much like set theory - it forms the basis for all the rest, but you don't really need to understand much of it in detail to appreciate the things that build on it

fqq

8:37 PM

yes, we did a lot of that too. I actually had to use Zorn's lemma to prove something in an exam

DIRAC1930

10:28 PM

If I have a vector space over the field $R^n$, why do the vectors have to be real?

There could easily be a vector $\imath V_1$ in that vector space so it doesn't really matter if the vector space is over the field $R^n$.

ACuriousMind

@DIRAC1930 what does $\mathrm{i}V_1$ mean in a vector space over the reals?

you can't multiply by $\mathrm{i}$ in a real vector space

DIRAC1930

But there could already be a vector in that vector space that is $iV_1$

ACuriousMind

again, what does multiplying by $\mathrm{i}$ mean there?

you just can't do that over the reals

fqq

also $\mathbb{R}^n$ is not a field, you cannot have a vector space over it

ACuriousMind

if you view the complex numbers $\mathbb{C}$ as a real vector space, it's just $\mathbb{R}^2$. You can call one of the basis vectors 1 and the other $\mathrm{i}$, but you can't talk about what $\mathrm{i}$ times some vector is- scalar multiplication on a vector space is only defined for scalars from the field it is over

DIRAC1930

10:37 PM

If I have a vector space over the field $0<\mathbb{R}<1$ , it doesn't stop a vector $(-1,0,0,0,0)$ living in the vector space though

ACuriousMind

I think you're confused about what a field is :P

fqq

that's also not a field

however the statement is "a vector space over $\mathbb{R}$ must consist of tuples of real numbers", that's not true, as ACM's example shows

but it's true that any $n$-dimensional real vector field is isomorphic to $\mathbb{R}^N$

DIRAC1930

are negative reals a field?

ACuriousMind

no, the reals are a field

a field is a bunch of numbers you can add, subtract, multiply and divide, essentially

if you multiply two negative numbers you get a positive number, so just the negative numbers make no sense as a field

unless you're doing number theory there are only three fields you should care about - the rationals, the reals, and the complex numbers :P

DIRAC1930

Say in a vector space there are three vector $X_1 = (1,0), X_2=(2,0), X_3 = (\imath,0)$

ACuriousMind

10:43 PM

I don't know what that means

DIRAC1930

amongst other vectors that satisfy the vector space axioms

It doesn't matter if I define the vector space to be over a field $R$ because $\imath X_1 = X_3$

ACuriousMind

in order to write vectors as components $(a_1,a_2,\dots)$, you have to choose a basis $e_1,\dots, e_n\in V$. Any abstract vector $v\in V$ can be written as $v = \sum_i v_i e_i$ and we use $(v_1,\dots, v_n)$ as its component representation

if you have a real vector space, $(\mathrm{i}, 0)$ doesn't make any sense

$\mathrm{i}$ is simply not an allowed value for one of the $v_i$

DIRAC1930

Okay so elements of a vector space are basis vectors?

ACuriousMind

not really

you probably need to take a class/read a book on linear algebra

DIRAC1930

How do I know $V \times V \rightarrow V$ and $F\times V \rightarrow V$ if I havent even defined what constitues being in the vector space

ACuriousMind

10:50 PM

ah, that's a common issue with abstract math definitions

you will rarely be given a vector space in terms of a set $V$ and these maps

bolbteppa

@DIRAC1930 You really should try to focus on the basics like linear algebra first before talking about things like spinors, Lie algebra isomorphisms, complexification of Lie algebras etc...

DIRAC1930

I feel like those parts are easy compared to finding out what mathematicians mean by a real vector space

ACuriousMind

yeah, I said most of what I said assuming you were familiar with more elementary algebra - if you don't even understand the mathy definition of a vector space, you can't understand the mathy definition of a Lie algebra

simply because a Lie algebra is a vector space + a Lie bracket

DIRAC1930

That's true

bolbteppa

A 'real vector space' is a short-hand for saying a vector space over the real numbers i.e. the underlying field of the vector space is the real numbers

ACuriousMind

10:54 PM

one difficult thing to wrap one's head around is that math definitions aren't made to be intuitive - they are designed to be "elegant" in a sense one can only care about after having seen a lot of them :P

bolbteppa

There are plenty of subtleties when you go from vector spaces to Lie algebras and in physics you can basically skip these subtleties to a large extent at least up through QFT though when you get to QFT it really encourages you to face those subtleties

ACuriousMind

they capture the "essence" of a thing in a way that's not concerned with how you actually use it

DIRAC1930

If I have a vector space over the field $C$ that, in and of itself is a vector space over the field $R$, it doen'st matter because it will still be a vector space over the field $C$.

That was my point

ACuriousMind

it depends on how you look at things

$\mathbb{C}$ can be viewed as a 2d real vector space

if you do that, then there is no meaning to multiplying by $\mathrm{i}$ - because you can't do that in real vector spaces

you can also view it as a one-dimensional complex vector space, and then the multiplication by $\mathrm{i}$ has meaning

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