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Anonymous Wed 12 Aug 12:51:17 2020 No.12000740 View Reply Original Report

Quoted By: >>12000745 >>12000773 >>12000789 >>12000802 >>12001328 >>12001354 >>12003851

someone explain to me what the fuck is a spinor.
I understand tensors, Clifford algebras and Lie groups but I can't understand what Wikipedia is trying to define. I don't really know much topology so I don't understand the double cover of the orthogonal group definition, trying to understand the algebraic way.

inb4 "array of numbers"

Anonymous

Anonymous Wed 12 Aug 2020 12:53:01 No.12000745 Report

Quoted By: >>12000753

>>12000740
just use ricci flow with surgery

Anonymous

Anonymous Wed 12 Aug 2020 12:57:29 No.12000753 Report

Quoted By:

>>12000745
thats more understandable to me right now than these wierd fuckers

Anonymous

Anonymous Wed 12 Aug 2020 13:07:30 No.12000773 Report

Quoted By: >>12000820

>>12000740
spinor is an element of a vector space which is a (fundamental) representation of the spin group

now tell us what exactly are you not getting. do you know what a representation is? do you know what a spin group is?

Anonymous

Anonymous Wed 12 Aug 2020 13:11:57 No.12000789 Report

Quoted By:

>>12000740
It either spins ORRRRRRRRR it doesn't.

duhhh.

Anonymous

Anonymous Wed 12 Aug 2020 13:17:04 No.12000802 Report

Quoted By:

>>12000740
if you understand lie groups (and thus i assume lie algebras) you have a much better chance of appreciating this shit than i do. the space of rotations is hard to wrap your head around.

Anonymous

Anonymous Wed 12 Aug 2020 13:25:52 No.12000820 Report

Quoted By: >>12000852

>>12000773
The notes im looking at define the "twisted Clifford group" as the tensors in the Clifford algebra that act on vectors by a left and right multiplication of a(X)vX^-1, where v is a vector (also a grade 1 element of the Clifford algebra) and a is the automorphism of the Clifford group that swaps the sign of the odd graded elements.

then the spinor norm is defined as XX* where * flips the order of the tensors and multiplies by (-1)^n where n is the grading of X.

then they claim the spinor norm maps the twisted Clifford group into the multiplicative group of units of the field F and the Pin group is the kernel of the spinor norm

The elements of the twisted Clifford groups are also representable orthogonal matrices,so the define the Spin group as the restriction of the determinant of the respresentation to elements of Pin

this is all just a random construction to me, what does this have to do with rotations that change sign?

Anonymous

Anonymous Wed 12 Aug 2020 13:46:10 No.12000852 Report

Quoted By: >>12000877 >>12000892

>>12000820
because it IS a random construction. the definition of Spin(n) as a double cover of SO(n) is the "correct" definition because it captures the essence of the object. it's a theorem in the theory of Lie groups that every group has a universal cover which is also a Lie group. so we know that thusly defined Spin(n) exists, but you cannot really touch it, you just know it exists. the Clifford construction is a specific realization of this abstract object.

I can try to explain something, do you know anything about covering spaces? or at least fundamental group?

Anonymous

Anonymous Wed 12 Aug 2020 13:55:01 No.12000877 Report

Quoted By: >>12003704

>>12000852
I know the fundamental group, I don't really know what a covering space is but the definition as "a fibre bundle with discrete fibres" makes sense in terms of what the words mean but I have no intuition of what they do or what they are good for

Anonymous

Anonymous Wed 12 Aug 2020 13:58:13 No.12000892 Report

Quoted By: >>12000992 >>12003704

>>12000852
also this is very helpful anon thanks, I might learn some topology anyway to crack the more "intuitive" definition

Anonymous

Anonymous Wed 12 Aug 2020 14:29:18 No.12000992 Report

Quoted By: >>12003704

>>12000892
I don't have time right now, maybe I'll write something later. good luck in your study

Anonymous

Anonymous Wed 12 Aug 2020 16:47:44 No.12001328 Report

Quoted By: >>12001918

>>12000740

Particle Physics Lecture 8: Spinors I
https://www.youtube.com/watch?v=-BFBIryWLMg

Particle Physics Lecture 8: Spinors II
https://www.youtube.com/watch?v=gctstk_IVhs

Particle Physics Lecture 9: Spinors III, Action Principles and Free Lagrangians
https://www.youtube.com/watch?v=DL5jym-usNY

Anonymous

Anonymous Wed 12 Aug 2020 16:55:37 No.12001354 Report

Quoted By:

>>12000740
The diagram is actually a good explanation for this.
If you want a really concise summary of them I think Sattinger and Weaver explains in like a few pages on the Möbius group what the spinor construction is. I think Roger Penrose also has a book or article explaining them.
But I visualise it like this. You can recall that for each continuous group of transformations we can associate a geometric invariant. For the rotation group SO(3) this is obviously just the sphere, and you would to think of the natural coordinates being the Euler angles (most often).
However you can show that SO(3) is isomorphic to SU(2), and this is done by a fairly simple transformation from Euler angles to the Cayley-Klein parameters. As the Euler angles go around a circle, they return to their starting position. However, the Cayley-Klein parameters only complete half a circuit, so only return to their starting positions after two revolutions, not one, like you might think of a normal angle doing. The Möbius strip is used to show you this visually and for the technical details of what I just described, I again recommend Sattinger and Weaver.

Anonymous

Anonymous Wed 12 Aug 2020 19:16:00 No.12001716 Report

Quoted By: >>12001918

>understands Clifford algebras and Lie groups
>doesn't understand/accept the definition of spinouts as a particular representation of a Lie group
Why?

Anonymous

Anonymous Wed 12 Aug 2020 20:30:00 No.12001918 Report

Quoted By:

>>12001716
I don't really know enough representation theory/algebra to properly understand it I think. My background is weird, mainly a lot of analysis and differential geometry, and I'm using the differential geometry as a crutch to try and understand some fun related areas of pure math like ricci flow, chern-weil theory, spin geometry and cohomology, but these are heavy fields it seems and I'm getting fucked by my topology weaknesses

>>12001328
Looks cool, I'll check it out thanks anon

Anonymous

Anonymous Wed 12 Aug 2020 20:59:51 No.12001996 Report

Quoted By: >>12003541

If anyone was interested, I found this physics overflow post extremely informative, particularly in explaining why they are the "square root of geometry" and made the clifford algebra construction "click" for me

https://www.physicsoverflow.org/41625

Anonymous

Anonymous Thu 13 Aug 2020 08:01:50 No.12003541 Report

Quoted By:

>>12001996
not bad for a physishit

Anonymous

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Anonymous Thu 13 Aug 2020 09:35:25 No.12003704 Report

Quoted By: >>12003705 >>12003740 >>12004420

>>12000877
>>12000892
>>12000992
the thing about covering spaces is the path lifting property. let $p \colon \tilde{X} \to X$ be a covering and let $\gamma \colon I \to X$ be a path with $\gamma(0) = x$ . then there exists a lift $\tilde{\gamma} \colon I \to \tilde{X}$ and this lift is unique up to the choice of basepoint $\tilde{\gamma}(0) \in p^{-1}(x)$ . see pic related. the important fact is that if $\gamma$ is a loop, the lift is not a loop in general, it can connect different points in the same fibre. this correspondence between {loops in $X$ } and {paths connecting points in the same fibre in $\tilde{X}$ } depends only on homotopy classes and it makes covering spaces essential in the study of the fundamental group. there are a lot of things to be said, I'll just say how it goes in the particular case. however I suggest you study this at some point, it's really a very nice part of algebraic topology.

let's now look at $Spin(n)$ and $SO(n)$ . the fibre of the identity element $1 \in SO(n)$ contains the identity element $\tilde{1}$ of $Spin(n)$ and also one another element, let's call it $-\tilde{1}$ . consider now a loop $\gamma$ in $SO(n)$ based at $1$ , and lift it to a path $\tilde{\gamma}$ starting at $\tilde{1}$ . there are two possibilities: the end point of $\tilde{\gamma}$ is either $\tilde{1}$ , or $-\tilde{1}$ . this depends exactly on whether $\gamma$ represents the trivial or the non-trivial class in $\pi_1(SO(n)) = \mathbb{Z}_2$ . thus the lifting procedure gives a bijection $\pi_1( SO(n) ) \cong \{ \tilde{1},-\tilde{1}\}$ .

Anonymous

Anonymous Thu 13 Aug 2020 09:36:28 No.12003705 Report

Quoted By: >>12004420

>>12003704
consider now an action of $Spin(n)$ on some vector space $V$ . of course $\tilde{1}$ acts as identity on $V$ , but suppose that $-\tilde{1}$ is a non-identity transformation. let's "pretend" that we actually have a $SO(n)$ representation and take a loop $\gamma$ in $SO(n)$ based at the identity. intuitively this loop is a continuous rotation which ends in the same state as it started (because it's a loop, it starts and ends at the identity). your representation is not $SO(n)$ really so you cannot rotate $V$ along $\gamma$ . you can however lift $\gamma$ to $\tilde{\gamma}$ and "rotate" along this curve. if $\gamma$ is non-trivial in $\pi_1(SO(n))$ , then the lift is not a loop and the transformation of $V$ along this path doesn't end at the identity, it ends in the state $-\tilde{1}$ . going around $\tilde{\gamma}$ twice ends in identity for sure, it's because the concatenation $\gamma * \gamma$ is 1 in the fundamental group (it's $\mathbb{Z}_2$ ), therefore it lifts to a closed loop. this is what it means that "if you rotate a spinor by 360 degrees, it ends in a different state". you're not really rotating it by 360 degrees. you're lifting the rotation, i.e. a loop in $SO(n)$ , to something which is not a loop in $Spin(n)$ .

Anonymous

Anonymous Thu 13 Aug 2020 10:00:32 No.12003740 Report

Quoted By:

>>12003704
>duodenum

Anonymous

Anonymous Thu 13 Aug 2020 11:01:03 No.12003851 Report

Quoted By:

>>12000740
>I don't really know much topology so I don't understand the double cover
Then go and learn it you dumbfuck. No one can tldr that here with you ending up understanding it without you understanding the basics before.

Anonymous

Anonymous Thu 13 Aug 2020 11:08:42 No.12003870 Report

Quoted By: >>12004420 >>12004430 >>12005617 >>12008023 >>12008030

Can you give an explicit representation of Spin(n) as a matrix Lie group?

Anonymous

Anonymous Thu 13 Aug 2020 15:18:19 No.12004420 Report

Quoted By: >>12004683

>>12003704
>>12003705
amazing, did you get this from a book? pretty much explained it perfectly for me, thank you.

>>12003870
I don't have them on me right now but the notes I was looking at had a big list of some examples, most of the time it's a bunch of matrix groups over the quaternions

Anonymous

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Anonymous Thu 13 Aug 2020 15:25:34 No.12004430 Report

Quoted By: >>12005617

>>12003870
$Spin(4) \cong SU(2) \times SU(2)$

Anonymous

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Anonymous Thu 13 Aug 2020 15:58:06 No.12004523 Report

Quoted By:

It's a list of numbers.

t. Python expert

Anonymous

Anonymous Thu 13 Aug 2020 16:43:54 No.12004648 Report

Quoted By: >>12005681

This is an example of a good sci thread.

Anonymous

Anonymous Thu 13 Aug 2020 16:55:00 No.12004683 Report

Quoted By: >>12005617

>>12004420
>but the notes I was looking at had a big list of some examples
I would like to see this too

Anonymous

Anonymous Thu 13 Aug 2020 18:30:38 No.12004962 Report

Quoted By:

Basically physicists found that when trying to come up with a theory of rotations in Hilbert spaces there was some ambiguity due to the fact that physical states aren't actually defined in the unit sphere of the complex Hilbert space but in the projective Hilbert space as multiplying your vector by a unit modulus complex number doesn't provide or change the physical relevant information. So if you rotate your lab frame of reference there is an ambiguity as to how you can express mathematically this physical rotation in your Hilbert space for objects like spin so it seams like if you rotate your vector by 2pi, it doesn't come back to the original one. This is obviously a retarded take as you already know your relevant states are on the projective hilbert space not on the sphere and what really is happening is that you are just using a set of parameters where your "angle" has been halved.
https://en.wikipedia.org/wiki/Projective_representation
I may be wrong but I don't understand why there is so much focus into wanting to think of spinors as weird vectors under rotation while it just how to do with how you define "rotation" on your hilbert space. The "spinor" property is on how you define rotations, not on the vector itself.

Anonymous

Anonymous Thu 13 Aug 2020 18:54:21 No.12005031 Report

Quoted By: >>12005064

so you retards are trying to go down the weinstein rabbit hole, lol.
the guy hasn't even released a paper so you are wasting your time.

Anonymous

Anonymous Thu 13 Aug 2020 19:02:48 No.12005064 Report

Quoted By:

>>12005031
Absolute cringe
Imagine being this fucking uneducated

Anonymous

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Anonymous Thu 13 Aug 2020 21:59:36 No.12005617 Report

Quoted By: >>12007107

>>12003870
>>12004683
here's the page excerpt that calculates up to n=4, I can't quite tell if it agrees with >>12004430 , would be interesting if someone knows

$a$ and $*$ are the involutions I mentioned above:

$\left(\mathbf{x}_{1} \wedge \mathbf{x}_{2} \wedge \ldots \wedge \mathbf{x}_{k}\right)^{*}=(-1)^{k} \mathbf{x}_{k} \wedge \mathbf{x}_{k-1} \wedge \ldots \wedge \mathbf{x}_{1}$

$a\left(\mathbf{x}_{0}+\mathbf{x}_{1}\right)=\mathbf{x}_{0}-\mathbf{x}_{1}$

I think $\mathbf{x}_1$ is the part of the tensor with an odd length, but if someone knows if I'm understanding the $\mathbb{Z}_2$ grading wrong tell me

$\text{CL}_n$ with capital L is the twisted Clifford group I mentioned, Wikipedia also calls it the Lipschitz group, while $\text{Cl}_n$ with lowercase l is the Clifford algebra.

Anonymous

Anonymous Thu 13 Aug 2020 22:18:25 No.12005681 Report

Quoted By:

>>12004648
No, it's someone wanting for a concept to be mysterious and interesting and telling himself the thing is more than the plain definition of it.

Anonymous

Anonymous Fri 14 Aug 2020 08:38:53 No.12006951 Report

Quoted By:

bumping thread so i can think about the link to >>12004203 a little more

Anonymous

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Anonymous Fri 14 Aug 2020 10:12:08 No.12007107 Report

Quoted By:

>>12005617
$SU(2) \cong U( \mathbb{H} )$

Anonymous

Anonymous Fri 14 Aug 2020 17:44:41 No.12008023 Report

Quoted By: >>12008030 >>12008180

>>12003870
sure. Spin(1) = {1,-1}. Spin(2) = SO(2).

Anonymous

Anonymous Fri 14 Aug 2020 17:48:47 No.12008030 Report

Quoted By:

>>12003870
>>12008023
also Spin(3) = SU(2)

Anonymous

Anonymous Fri 14 Aug 2020 18:46:42 No.12008180 Report

Quoted By: >>12008195

>>12008023
>Spin(2) = SO(2)
How can SO(2) bei its own universal covering space? SO(2) is isomorphic to S^1 so its fundamental group is Z

Anonymous

Anonymous Fri 14 Aug 2020 18:49:16 No.12008195 Report

Quoted By:

>>12008180
Nevermind, I just read that this only holds for n >= 3

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