Lie Algebras 4 -- Classifying Lie algebras of low dimension.
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- čas přidán 24. 09. 2023
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At 24:06 is that not supposed to be gamma * w?
damn I’m so early that the higher res versions aren’t even done processing
Not only is this a video about the low-dimensional Lie algebras - only the low-dimensional representations are currently available, too.
4h after submitting the video still only 360p is avaliable :/
@@NuclearCraftMod😂😂
And I’m finally confused …
this series is wonderful. I'm watching on another channel and just using this one to comment. almost finished with the complex analysis course. i wouldn't have the opportunity to learn with my hectic life were it not in part for these courses. i wish there was a way for me to compensate you. sorry that there isnt for the moment being.
He has a Patreon
You can always use the "Thanks" button to donate directly to the channel!
Physicist-speak: U(1) is the only 1-dimensional Lie group. But… commutators define only the local structure. Are there multiple possible global structures? E.g. periodic vs non-periodic?
do you mean the only *compact* (and connected) 1-dim Lie group? because RR is canonically a Lie group, and somewhat relatedly is also the universal cover of U(1)
Where did ep. 3 go?
It exists on the channel but it didn’t make it on the playlist
Say, does anybody know an "elementary way" to prove that E7 and E8 exist, without using the Lie algebra-Lie group correspondence?
I'm still working on understanding how Dynkin diagrams classify Lie algebras, but if you believe that correspondence then they're a few of the exceptional Dynkin diagrams as well
33:00 I don't get why the matrix of the adjoint has zero column? 🤔
Isn't the adjoint just another vector in L?
Applying the adjoint to another vector in L is a vector in L, yes. But the adjoint itself is an operator, namely ad_y is the operator that takes any vector w in L to [y,w]. It's a standard fact of linear algebra that any operator is representable as a matrix.
@@Uoper12 thanks for the information so far.
If [x, y] = ax then I don't see why the first column should be zero (?)
Transforming a vector to a multiple of itself would require a diagonal matrix, or more precise a multiple of the standard matrix a*I (?)
@@thomashoffmann8857 The way that you get a matrix from an operator is you see how it acts on the basis vectors. So in the example the basis of L' is {y,z} then applying ad_y to each of these gives ad_y(y)=[y,y]=0=0y+0z and ad_y(z)=[y,z]=ay+bz. Then the matrix for the operator ad_y is then given by the column vectors ad_y(y)=0 and ad_y(z)=ay+bz which is the matrix he has up on the board at 33:00.
Is there a numbering I'm not understanding[1], or is there some reason that "Lie Algebras 3" is missing from the playlist??
[1] Sorry, I haven't started watching yet, so don't know of any comments made within the videos that might answer this point.
Yeh it is a bit weird, just look through his yt page, not the playlists, ep 3 is there
Unless you already figured it out lol
@@fahim1943 I did do just that a few days later. Ty nonetheless.
NB (for others reading this):
Same applies to ep. 7 - it's not in the playlist but is in his videos. Also, videos only recently uploaded so the course (believe only 8 eps. exist to date) may not be complete yet.
Anyway...
I'm a ways of being well-versed enough in what I regard as prerequisites for this, so for now just putting all my chickens (despite skipping depths such as Peano arithmetic, and have only glanced at those pesky combinatorial pigeons) in a row, and then 'transposing', so to speak [Lin. Alg.], to an '[attack] vector' ;p for understanding higher mathematics in the form of this course, ..or something similarly weighty.
I'd love if this playlist got updated with those missing videos but also with the newer ones in the series