Spinors for Beginners 16: Lie Groups and Lie Algebras

[@eigenchris]

At 32:15, I messed up the 2nd last commutation relation on the bottom left. It should be [Ktz, Ktx] = - Jzx.

[@kevinsellers7566]

You are the greatest teacher of applied mathematics and quantum mechanics ever.❤❤❤

[@aaronTNGDS9]

I can't express to you adequately how helpful your description of what I always considered a most cryptic statement "Lie Algebra is the tangent at the identity". Some have described this algebraically, but something seemed missing--- what has long eluded me as to what that looks like geometrically is now made clear and satisfactorily. THANKS A BUNCH!!

[@baruchspinoza4979]

Wow! A clear path through something that had appeared as a complex jungle. Thanks EigenChris.

[@rhke6789]

The only video I found that explains why Lie Algebra is important and has a use and how in the field of Quantum mechanics...that is also understandable. Thank you so much

[@DavidAspden]

Probably one of your best yet. The ground you covered was vast.

[@themightyquinn100]

Man, this was so insightful and helpful. I'm taking a Geometric Mechanics class and need to know this Lie group/algebra stuff. Great Job!

[@Impatient_Ape]

Bravo! Clap! Clap! Clap! This is a really really really good introduction to Lie algebras for motivated undergrad math majors. Lucid, accurate, and well visualized. For those in the comments... a typical math major in the USA might not see this material during their bachelor's degree; however, it is usually part of the graduate math curriculum.

[@kikivoorburg]

Very excited for this section! I’ve vaguely heard of Lie Algebras (and that they matter in physics) but never had quite a clear picture of what was going on. You have an incredible ability to make complex topics digestible and understandable, so I’m sure you’ll fill those gaps for me! Edit: I was not disappointed! It’ll take me a bit of time and a few rewatches to fully grasp it I’m sure, but wow have I already learnt so much! I finally get why “the Lie algebra is the tangent space of the Lie group” actually makes sense! Amazing work!!

[@TheJara123]

Block bustor man!! Your operational approach cuts through the topic...showing us way to learn!! So big thanks in order...!!

[@francoisfjag4070]

brilliant video. The best explanation I ever get on relationship between a Lie group and its corresponding algebra !!

[@Temp-s7l]

Always the best explanation anywhere anytime

[@fizikchy]

Amazing video! You knit things very naturally. Thanks.

[@hugoreytinas5170]

I am so happy : i could not wait for this new video ! This is like a Christmas gift ! Thank you so much Eigenchris ! You are welcome anytime in south of France ! 😂

[@richardneifeld7797]

Your videos keep getting better and better.

[@jimlbeaver]

I’ve been trying to understand this for a while. This was extremely helpful. Thank you.

[@richardneifeld7797]

Excellent! Also, I find that mathematician's normally pro ide more comprehensible video lectures than physicists. As of they better understand the underlying principles. I put you in the mathematician lecturer class. Clear and insightful.

[@enterprisesoftwarearchitect]

It’s a semester course in 30 minutes!!! ❤️❤️❤️

[@Anonymous-kj6cu]

I love your voice. Sounds like a robot.

[@alexleibovici4834]

Is it his voice?

[@AdrianBoyko]

Vocal fry

[@ryansamuel8835]

I like how he doesn’t get all emotional when he speaks. Like he has no emotions and is just presenting information

[@MiroslawHorbal]

It sounds like a robot who is also slightly annoyed that he has to explain things to you. But jokes aside. I love these videos and his teaching style.

[@AdrianBoyko]

@@MiroslawHorbal Marvin

[@pieterkok7486]

To find the generator from the group element, you apply the "standard" derivative rule to matrices. However, there are many types of derivatives one can define for matrices (for example Lie brackets, as in the Liouville/Von Neumann equation). Do we need additional justification (other than that it works in this case) for this choice of derivative? Great series, by the way! 😀

[@eigenchris]

I haven't thought about that too much. I'm just used to using the standard derivative. Maybe there are cases where you need another kind if derivative, but I'm not familiar with that.

[@gijsb4708]

Whenever a differentiable manifold (such as a Lie group) is represented as elements of a vector space (here the space of matrices) then the standard derivative gives the right answer when calculating tangent vectors (which are in this case the generators).

[@kikivoorburg]

Upon rewatching I've noticed how similar the structure coefficients f_ij^k are to the connection coefficients Γ_ij^k from differential geometry / GR! Both define the "multiplication" of basis vectors in terms of the basis vectors: [g_iᵢ, g_j] = Σ f_ij^k g_k ∂_i ∂_j = Σ Γ_ij^k ∂_k The placement of the indices on the coefficients is even identical (upper/Lower indices ftw, understanding co/contra-variance is so clarifying). Always more to learn in these videos!! Now that I think about it, doesn't the fact we can "multiply" basis vectors in differential geometry mean that it too is an algebra? Edit: in case it wasn't super clear, due to the definition of ∂ / ∂x^i = ∂_i = e_i, we have ∂_i ∂_j = ∂ e_i / ∂ x^j, which is the formula that Wikipedia uses with the Christoffel symbols.

[@eigenchris]

Yeah, I'm not sure how deep or shallow the connection is, other than "two things come together to give a third thing". The Lie bracket can be thought of as a kind of derivative... the Jacobi Identity I show briefly can be thought of as a kind of product rule if you take one of the terms and put it on the other side of the = sign. Would be interesting if there was a deeper connection between the two, but I'm not aware of anything at this point in time.

[@kikivoorburg]

@@eigenchris Even shallow connections are interesting, though it would indeed be exciting if there was more going on. Thanks for the reply!

[@sair9850]

came from the joke videos, stayed for the spinors for Beginners 16: Lie Groups and Lie Algebras

[@kousarshaheen4801]

Thanks for your videos, it helps me alot in my thesis on particle physics, kindly make sure when we reach particle physics

[@thevikifalcon7670]

I have never seen a good explanation for convergences between geometric algebra and lie theory (lie algebras + lie groups). Oftentimes, you can use both to represent the same things- spinors in this case, but more generally rotation-like transformations on some manifold. In this series, you should consider explaining why this convergence exists. For instance, is there a way to directly convert the lie theory way of thinking about things into the geometric algebra way and vice versa? Thanks and keep up the great work!

[@eigenchris]

This might become more obvious in the next video. In the spin-1/2 representation, the generators are Clifford Bivectors, or equivalently, quaternions. And the SU(2) matrices are the equivalent of the Spin(3) group.

[@AkamiChannel]

Can't wait for the next video! Thanks!

[@HelloWorld-lv4we]

I need a physics expert! I am a layperson who enjoys physics, and I have a question about spinors. My understanding is that spinors are equivalent to space-time indices (indices on a 4D manifold). Do spinors ever exist on odd-dimensional manifolds? In other words, do spinors occur on even-dimensional manifolds only? I ask this because I’ve heard that bosons are “non-spinorial”, so I’m mainly curious to know whether the force fields are odd-dimensional. If I’m conceptualizing these things wrong, please tell me.

[@ericbischoff9444]

Again, great video! I have a question. At 27:27 you prove that the commutator of tangent vectors is a tangent vector. Am I correct to consider this does not prove the general case of a multiplication in a Lie algebra, using the "more abstract" definition of a Lie algebra based on the Jacobi identity? If I understood well, commutators are the most common example of a Lie bracket, but not necessarily the only ones. Or am I missing something?

[@sebastiandierks7919]

You are correct, it doesn't prove the general case, although if I remember my Lie Groups and Lie Algebras course correctly, the proof of the general case is very similar. You also take the conjugation of two paths through the manifold to show that for every Lie group an object called the Lie bracket can be constructed that satisfies alternating and antisymmetry properties as well as the Jacobi identity. (I'm a physicist, not a mathematician though, although the course at university was a pure maths course from a maths professor; it's just been 10 years ago so yeah...)

[@ericbischoff9444]

Thank you@@sebastiandierks7919 .

[@juleskurianmathew1983]

After finishing spinor series, kindly do a quantum mechanics/ quantum chromodynamics/ quantum electrodynamics/ quantum field theory series. I love your tensor series.

[@eigenchris]

Hi. While I will cover a little bit of QFT in the last part of this series, I won't be doing a full QFT series. It's just too big a topic and I don't understand it very well.

[@saber9468]

Hello Chris, thank you so much for your video. I was interested in knowing if you had a similar video covering lie algebra derivatives? I really like your teaching style and was hoping for a more visual representation of lie algebra derivatives i.e derivative of point w.r.t to rotation etc...

[@sebastiandierks7919]

Really great video! Can't wait for the next episode. At 9:28, the group element R_zx should have the minus sign in front of the other sin function if I am not mistaken.

[@eigenchris]

I think it's correct. If you put in theta=90 degrees, and plug in the z-vector [0,0,1]^T, you end up getting the x-vector [1,0,0]^T. So it rotates z into x, as expected.

[@sebastiandierks7919]

​@@eigenchris sorry, my bad, you're right of course. I always assumed that the signs in front of the sin functions were the same for all three rotation matrices so that if you deleted the column and row with ones and zeros you'd get the 2-dimensional rotation matrix. It's wrong though, turns out the rotation about the y-axis has different signs. I guess as a physicist you rarely rotate about the y-axis, you mostly use a coordinate system where you rotate about x and z, so my mistake never got caught.

[@eigenchris]

@@sebastiandierks7919 You can imagine taking the xy rotation matrix and moving the cos/sin 2x2 block down 1 row and to the right 1 row. If you do this twice, while "wrapping around" from bottom to top and from right to left, you get the zx rotation matrix. And doing it a 3rd time returns you to the xy rotation matrix.

[@n810h]

This was really good. Great job!

[@dariushimani8956]

Banger video

[@kevinsellers7566]

Can you please make these lectures available for purchase?❤❤❤ Thank you for sharing your hard work.

[@eigenchris]

The slights are on my github account.

[@Valine-e6c]

Is there going to be a QFT series after this, similar to hoe you had a relativoty series aftef tensor calculus? Love your content btw.❤

[@eigenchris]

I'm afraid not. QFT is too big and confusinf. I don't understand it, and don't have the energy to make a series long enough to cover it. I'll talk a bit about QFT at the end of this series.

[@homosapien5684]

Can you please release a video series on group theory? It would be so incredibly helpful

[@eigenchris]

I don't plan on doing a series on that. If you're interested in representation theory of groups, later videos in this playlist cover that.

[@pavgogo]

Excellent. If aba(inverse) leads to commutator ab-ba, what about bab(inverse)? also is it also ab-ba or ba-ab?

[@lazzatv]

I'd love to see a well made derivation of rank equal spin quantum number relations hinted at 1:24, since I only saw some details about it during a lecture. If I'm not mistaken, from commutation relations you get |spin angular momentum vector|^2=hbar^2*(s+1)*s, the |spin angular momentum vector| is then expressed with the sum of the single components. Single components are expressed with independent spin-projection operators eigenvalues and communtation relationship when not independent, so I think the tensor field dimension can be expressed as 2*rank+1. Spin-projection operators apparently have a rank dependentent representations as shown later in the video which determine their rank dependent eigenvalues. I always thought it was a cool explanation of the fact that rank of the particle tensor field implies a certain value for the spin quantum number s and then the non-relativistic spin angular momentum can actually be derived, since it just becomes a first order term for the wavefunction angular momentum derived when the SO(3) operator for small rotation is applied to the tensor field.

[@himanshuchaudhary4197]

Dear Eigenchris, Could you please create a video or provide information on deriving the Einstein field equation from the Einstein-Hilbert Action? Alternatively, if you could share some notes or relevant links, that would be greatly appreciated.

[@eigenchris]

Hi. Sorry but I'm done with making GR videos. I'm focusing on the spinor videos now.

[@tomgraupner171]

I understand that the "physics part" is coming along next, but please allow one question beforehand: Does the [Ktz, Ktx] = - Jzx means that a boost in the x-y plane automatically includes a rotation around the z axes?

[@eigenchris]

I'm not sure what the physical meaning of 2 boosts commuting to a rotation is. In the Lie Group, the basic rotations and boosts are independent from each other. You can always do a boost with no rotation and vice-versa.

[@tomgraupner171]

@@eigenchris I assume my question was silly. I found this: en.wikipedia.org/wiki/Wigner_rotation So, it looks like a second boost in a different direction (e.g. -> y), while performing the first boost (e.g. -> x). That leads to the rotation, as the product/sequence of these 2 boosts must no longer be a "single boost". Does that make any sense? Thanks a lot for all your work!

[@sgurdmeal662]

This made my day

[@eduardoGentile720]

One question, why should a Det=0 Matrix to the zeroth power be the identity if it's non invertible?

[@eigenchris]

Why does the invertibility matter?

[@eduardoGentile720]

@@eigenchris We say that a number to the zeroth power is 1 because by the proprieties of exponents by multiplying the same base with different exponents the exponents should add up, and since the negative power is equal to the inverse of a of a number, a number to the zeroth power is the number divided by itself, witch is 1 (when it's not 0). If you define it the other way around, so a number to the zeroth power is 1 and then you apply the rule of exponents to obtain that a number to the -1 is its inverse then you should be allowed to exponentiate 0^0 or even a 0 Matrix to the 0, I'm asking this just because this doesn't make much sense to me

[@bantix9902]

the goat back at it again!!!!

[@Cat-yz1tk]

in the part where you prove that the lie algebra is closed under commutators you multiplied two vectors in the tangent space as is required for the commutator a*b-b*a this makes sense where the tangent vectors are matricies but how does that work in a general lie group? how do tangent vectors multiply with each other?

[@AMADEOSAM]

Great explanation …

[@Daniel-nm4kt]

Can you make a series on the basics of linear algebra?

[@eigenchris]

I don't plan on doing any more long series after this. Have yoh watched 3blue1brown's "Essensce of Linear Algebra"?

[@BakedAlaska187]

All of the tangent vectors pictured are actually matrices, right?

[@eigenchris]

Yes.

[@breezy1118]

Great video! However, the summation indices at 8:38 should start at n=1 after taking the derivative.

[@siguc]

Thank you for this amazing series! Speaking of the proof at 14:21, isn't it obvious that dR^T/dTheta=(dR/dTheta)^T? After all, it doesn't matter whether we first differentiate every cell of a matrix and then transpose it, or, alternatively, transpose the matrix first and then differentiate every of its cells. If we accept the above, then we can differentiate dR/dTheta->MR (as you explained earlier) and then substitute R(0)=I, which completes the proof.

[@eigenchris]

What's obvious for one person might not be obvious for someone else. I didn't want to skip too many steps and risk confusing people.

[@siguc]

@@eigenchris Agree. Most of your logic seems very clear. It’s also very helpful that you keep reiterating the same conclusions across the course, especially when you approach those conclusions from different angles.

[@ritahall8148]

Typo at 32:19?, on penultimate line, should have bracket of tz and tx boosts equal to the negative zx rotation

[@eigenchris]

Yup. My bad...

[@Dc4nt]

Similar typo at 10:58 ? Last line, right hand side should have g_zx

[@Dc4nt]

Just spotted you use the same slide again at 16:44 , same typo.

[@tobiaspower381]

I cannot say thank you enough. :)

[@cmilkau]

Actually, we only proved the properties of an algebra, but not the Jacobi identity to make it a Lie algebra (antisymmetry is obvious). You can prove that this identity always holds when the multiplication (Lie bracket) [a,b] looks like a commutator a·b - b·a and "·" is associative and distributes over +.

[@fjaresj]

Long round of applause! ❤

[@Temp-s7l]

I think you could run this at the local cinema and get a great reception

[@cmilkau]

All hermitian or antisymmetric matrices have an EVD; once you have that exponentiation is super simple

[@phitsf5475]

No offence, but I think your script combined with Taylor Swift doing the reading would make for a badass combination.

[@김익명-n9v]

24:36 one slide summary of Lie group and Lie algebra!

[@tedsheridan8725]

Great video! It really helped demystify Lie groups/algebras for me. (BTW, I love your content, but as a constructive criticism from someone who also makes KZbin videos, I'd encourage you to work on your narration. It's clear, but very monotone and over-pronounced. I'd really try to adopt a more conversational style as opposed to reading off a powerpoint presentation. I think you'd find it much more engaging and pleasant to listen to.)

[@ericbischoff9444]

I'm of the contrary opinion. I find Chris' voice appeasing and it lets me concentrate on the contents. I would certainly not be able to do the same with a more "conversational" or even "journalistic" tone, like there is in so many youtube videos.

[@eigenchris]

My voice seems a bit divisive among some people. Some like it, others really don't. I'm probably just going to keep doing it in the current style since that's why I'm used to.

[@onebronx]

@tedsheridan8725 Just increase video playback speed

[@tedsheridan8725]

​@@ericbischoff9444 I agree with you on the 'journalistic' tone - it's so overused. It's quite distracting to listen to a narrator pretend to be an TV news anchor.

[@curtishorn1267]

This is the way.

[@omegapirat8623]

Can you extend this session to (quantized) spinor fields, spinor structures, spinor bundles, etc on a manifold? I'd like to understand modern physics in terms of a proper mathematical model of spacetime following this amazing lecture series kzbin.info/www/bejne/jGWcmpKCgp11Zqc Unfortunately, in this lecture, there is not so much said about spinor matter since it mainly focuses on differential geometry (and tensor matter). Spinor matter is the last import ingredient that I don't fully understand to get a complete picture of modern standard physics (standard model and general relativity).

[@eigenchris]

I'm not sure I'm going to get that far in this series. What exactly do you want to see covered? By "spinor field", do you just mean the Dirac Equation in QFT?

[@omegapirat8623]

@@eigenchris I am talking about spinors in curved spacetime (maybe as a smooth section of a spinor bundle?). Or to be more precise how to define it on a lorentzian manifold. In contrast to tensor fields, you need a metric for spinor fields. Therefore it won't work for general manifolds I guess. That's something that I didn't master so far and I don't find good literature related to that topic either. Tensor fields and spinor fields are key aspects in understanding matter and as they are constructed on a 4D lorentzian manifold in physics one should understand this to get a complete picture of modern physics.

[@funkfusiontale]

Δ++ (uwu)

[@GroovyLynx]

If I have found a Genie, I would ask for your videos minus all the proves

[@eccotom1]

God his voice is so hot

[@miguelaphan58]

.. a very dicult subject

[@stevelamprou]

Why differentiate to get the Lie group from the algebra though?