In the derivation discussion of D(ab)... what does a product of the vectors a and b mean? The only thing we added to make the vector space a Lie algebra was the bracket...
@@uptownfunk9446 So a general "algebra" has some kind of axioms on products of its elements? But then shouldn't a Lie algebra have those axioms, or is a Lie algebra not in fact an "algebra" in that sense?
Might be wrong but I think an F-algebra is just an F-vector space A equipped with an F-bilinear map A×A->A called the *product* (usually denoted by concatenation). And a Lie F-algebra is an F-algebra whose product is alternating and satisfies the Jacobi identity, and we usually denote the product by [.,.] and call it the *Lie bracket*
Yes, what @paulshin4649 wrote is correct. However it might be misleading, since "in the wild" many authors use the word algebra with implicit extra assumptions. The most common one is requiring the product to be associative, for instance, in which case you get an _associative algebra_, but which is sometimes also just called an algebra. Hence, a Lie algebra is an algebra in the most general sense as in the previous comment, but if someone requires an algebra to be associative then a Lie algebra will not anymore be necessarily an algebra in that sense. I hope that I wasn't too confusing!
I love that you do this video !! But I have a question, at 3:05, why does [x+y ; x+y] = [x;x] + [x;y] + [y;x] + [y ; y] ?? is this because of bilinearity ?
AAAAAAAAAAAAAAAAAAAAAAA IVE BEEN WAITING ON THIS SERIES FOR LITERAL YEARS DR PENN YOU HAVE NO IDEA HOW EXCITED I WAS TO SEE THE NOTIFICATION FOR LIE ALGEBRAS ON MATH MAJOR!!!!!
It seems to me that the first proposition shouldn't be an iff condition, rather it should simply determine the second implication, aka, "If char(F) =/= 2 then [y,x] = -[x,y] implies [x,x] = 0 (if char(F) =/= 2, anti-commutativity implies alternating property)". This is because bilinearity + alternating property already imply anti-commutativity in ANY characteristic (0 = [x+y,x+y] = [x,x] + [x,y] + [y,x] + [y,y] = [x,y] + [y,x], so [y,x] = -[x,y]). Let me know in case I'm not following the argument
Nice - i believe a derivation can have a different left and right action - i recall the Fox free differential - it (for the left action group ring) defined the right action as. The action after the trivialised of the group ring element to the ring only to D_x (vw) = v D_x(w) + D_x(v) tr(w).
Very excited to follow along in this series! For Exercise 3: Show [Eij,Ekl] = djk*Eil-dil*Ekj . I'm not sure how to proceed. I think I'm just struggling to parse the indices, what is this saying?
And finally Lie algebra, please give some time to the isomorphism between SO(3), SU(2) and 3 and 2 spheres. In addition, give some time to the concepts of connectedness, simple connectedness and compactness.
For the part on derivations, with the lie bracket defined by [D_1, D_2] := D_1D_2 - D_2D_1, can anyone elaborate a bit more on the underlying algebraic structure here? Unlike other purely algebraic vector spaces I've seen before, there appears to be another relationship between the vectors and the field by the fact that the field elements can be inputs to the vectors. It appears to me like derivations are special types of functions. So is this space a function space, or something?
Question: What does the trace of matrix (matrix of a linear transformation) mean i get that it is the sum of the entries of the diagonal but what does it mean
He used it for the reverse direction at 4:44. The definition of a Lie algebra uses the condition that is stronger if char(F)=2, so a Lie algebra always has both properties, but you could define it with either if you only use fields with characteristic 0.
@@iabervon I had to watch this 1st vid to familiarize myself with Lie algebra (I’m BA, MSc Math) he’s claiming characteristic 2 is not zero & showing [x1,x1] = -[x1,x1] => [x1,x1] + [x1,x1] = 0
@@SylComplexDimensional If char(F) is not 2, then [x,x]+[x,x]=0 => [x,x]=0. If char(F)=2, then [x,x]+[x,x]=0 doesn't tell you anything, because y+y=0 for all y in a field of characteristic 2.
Commutators = two paths. Abelian (commutes, symmetric, Bosons) is dual to non abelian (non commutes, anti-symmetric, Fermions). Enantiodromia is the unconscious opposite or opposame (duality) -- Carl Jung. Vectors are dual to co vectors (forms). Sine is dual to cosine or dual sine -- the word co means mutual and implies duality! Homology (syntropic) is dual to co homology (entropic) -- same is dual to different. Injective is dual to surjective synthesizes bijective or isomorphism. Subgroups are dual to subfields -- the Galois correspondence. "Always two there are" -- Yoda.
Spoiler Alert................. I think the exercise number 2 is just trivial. As any 1-dimensional vector space is spanned by a single vector, say v, one has [x, y]=[av, bv]=ab[v, v]=0
There should be a way to measure how far down the rabbit hole you are when watching a given math video. Not being a mathematician, this one feels a good length down, but how can I tell...
the "on the other hand" proof at around 4:14 you proved "[x,x] = 0" when you claimed and structured the proof as if you were proving "-[x,y] = [y,x]". This was definitely a mistake, and it causes confusion from mixed/befuddled messaging. edit: in the proof around 16:09 when moving from line 3 (ie: "=D1( aD2(b) + D2(a)b ) - D2( aD1(b) + D1(a)b )" to line 4, you repeatedly say "because D1/2 is a derivation" when justifying the distributive property, but you actually mean "because D1/2 is a derivation and derivations are linear maps and therefore have the property of linearity". It is confusing for new players, but not incorrect the way you said it.
Yeah, it reminds me of how my quantum mechanics professor teaches. No exposition, no explanation or motivation, just pure, hard results. Like, the only definition we get of Lie algebra is the minimum required to start messing with bracket algebra. Probably we'd need to watch and completely understand the whole abstract algebra series first, which appears to go farther than my modern algebra class did. As a physicist, i don't understand how the pure math people can make sense of anything with how arbitrary and unmotivated these definitions seem. Maybe it'll come together later
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
someone trying to make you do representation theory? just say no, they cannot legally make you learn representation theory
@20:50 the first entry in the cross product should say a2b3-a3b2
yess! I've been waiting for this one, lovely to see you upload again. Thanks for the high quality material and thanks for providing it for free!
Reeally hoping we get some exponential map stuff, although it sounds like no.
And in style!
That Witt algebra video was really something. Very excited to see this series about math's greatest lie.
aaaaa no way! I just started taking a class on Lie algebras and now you're doing a series on them!
How'd your class go?
So excited for this! I’ve been looking forward to it since you announced it months ago. Thanks Michael!
Omg I’ve been wanting to learn more about Lie groups/algebras for a longgg while now, so this is amazingg
So happy too see new uploads on the channel, and a great topic too !
In the formula for the cross-product, the second term of the first component should be a_3*b_2, not a_3*b_1.
Great timing! I recently picked up a book called Lie Algebras in Particle Physics, so this series will be super helpful for me
can you pass methat book ?
I have waited so long for this series! Thanks Dr Penn!
Very excited the math major videos are continuing!
I’m excited for this series
In the derivation discussion of D(ab)... what does a product of the vectors a and b mean? The only thing we added to make the vector space a Lie algebra was the bracket...
In that discussion, a and b are elements of A, which is not a Lie algebra but just an algebra, so you can multiply elements of A
@@uptownfunk9446 So a general "algebra" has some kind of axioms on products of its elements? But then shouldn't a Lie algebra have those axioms, or is a Lie algebra not in fact an "algebra" in that sense?
Might be wrong but I think an F-algebra is just an F-vector space A equipped with an F-bilinear map A×A->A called the *product* (usually denoted by concatenation). And a Lie F-algebra is an F-algebra whose product is alternating and satisfies the Jacobi identity, and we usually denote the product by [.,.] and call it the *Lie bracket*
Yes, what @paulshin4649 wrote is correct. However it might be misleading, since "in the wild" many authors use the word algebra with implicit extra assumptions. The most common one is requiring the product to be associative, for instance, in which case you get an _associative algebra_, but which is sometimes also just called an algebra. Hence, a Lie algebra is an algebra in the most general sense as in the previous comment, but if someone requires an algebra to be associative then a Lie algebra will not anymore be necessarily an algebra in that sense.
I hope that I wasn't too confusing!
I’ve been needing to learn some of this for years so thank you for putting this up and forcing me 😌
I love that you do this video !!
But I have a question, at 3:05, why does [x+y ; x+y] = [x;x] + [x;y] + [y;x] + [y ; y] ??
is this because of bilinearity ?
Yes, [x+y, x+y]= [x+y, x] + [x+y, y] and you do the same with first coordinate and get exactly that.
what a bout de char(F)/=2 c
ondition?@@szymonkauzny2931
AAAAAAAAAAAAAAAAAAAAAAA IVE BEEN WAITING ON THIS SERIES FOR LITERAL YEARS DR PENN YOU HAVE NO IDEA HOW EXCITED I WAS TO SEE THE NOTIFICATION FOR LIE ALGEBRAS ON MATH MAJOR!!!!!
Awesome. Would have loved to hear a bit more about Lie groups, though.
So glad to see you again!!!
Years we have waited
sir keep up the good work. I have learnt a lot from you from both channels.
Really awesome content! I can't wait for the next video about Lie algebras, please upload it soon🙏
I love lie groups and lie algebras and I love this series!
It seems to me that the first proposition shouldn't be an iff condition, rather it should simply determine the second implication, aka, "If char(F) =/= 2 then [y,x] = -[x,y] implies [x,x] = 0 (if char(F) =/= 2, anti-commutativity implies alternating property)". This is because bilinearity + alternating property already imply anti-commutativity in ANY characteristic (0 = [x+y,x+y] = [x,x] + [x,y] + [y,x] + [y,y] = [x,y] + [y,x], so [y,x] = -[x,y]). Let me know in case I'm not following the argument
Much easier than I thought. Random question, is gluing computational? Thanks.
Nice - i believe a derivation can have a different left and right action - i recall the Fox free differential - it (for the left action group ring) defined the right action as. The action after the trivialised of the group ring element to the ring only to D_x (vw) = v D_x(w) + D_x(v) tr(w).
I think there's a plus instead of a minus in the very last formula you say.
Lots of videos on Lie Algebras would be really interesting !
Thanks, Professor
Very excited to follow along in this series! For Exercise 3: Show [Eij,Ekl] = djk*Eil-dil*Ekj . I'm not sure how to proceed. I think I'm just struggling to parse the indices, what is this saying?
I would love to see calculus on Lie Groups/Lie Algebras
And finally Lie algebra, please give some time to the isomorphism between SO(3), SU(2) and 3 and 2 spheres. In addition, give some time to the concepts of connectedness, simple connectedness and compactness.
This is great! Do you have a recommendation of a book (preferably a Dover book) to go along with this? Thanks!!
For the part on derivations, with the lie bracket defined by [D_1, D_2] := D_1D_2 - D_2D_1, can anyone elaborate a bit more on the underlying algebraic structure here? Unlike other purely algebraic vector spaces I've seen before, there appears to be another relationship between the vectors and the field by the fact that the field elements can be inputs to the vectors. It appears to me like derivations are special types of functions. So is this space a function space, or something?
Having trouble with 4... I computed it to be Jx+(x^T)J
in the definition itself why didn't you mention bi linear property of the lie bracket?
Question: What does the trace of matrix (matrix of a linear transformation) mean i get that it is the sum of the entries of the diagonal but what does it mean
it is also the sum of the eigenvalues
Wait a min............ that L sub n......... that looks a lot like Ln (natural log). I would use some different letters.
LOVE YOU
cool!
But how did you use the characteristic of not 2?
@@freedomfilms974 well I suppose, doing nothing with it lies within that category
He used it for the reverse direction at 4:44. The definition of a Lie algebra uses the condition that is stronger if char(F)=2, so a Lie algebra always has both properties, but you could define it with either if you only use fields with characteristic 0.
@@iabervon I had to watch this 1st vid to familiarize myself with Lie algebra (I’m BA, MSc Math) he’s claiming characteristic 2 is not zero & showing [x1,x1] = -[x1,x1] => [x1,x1] + [x1,x1] = 0
for all vectors x
@@SylComplexDimensional If char(F) is not 2, then [x,x]+[x,x]=0 => [x,x]=0. If char(F)=2, then [x,x]+[x,x]=0 doesn't tell you anything, because y+y=0 for all y in a field of characteristic 2.
Love it!!!!!!!!!!!!!!!
3:10
yemin ederim ki sabah akşam sadece ve sadece bunlara çalışacağım
yeeeee Lie algebras finally!
If I said this wasn't an amazing lecture, I'd be Lie-ing!
Commutators = two paths.
Abelian (commutes, symmetric, Bosons) is dual to non abelian (non commutes, anti-symmetric, Fermions).
Enantiodromia is the unconscious opposite or opposame (duality) -- Carl Jung.
Vectors are dual to co vectors (forms).
Sine is dual to cosine or dual sine -- the word co means mutual and implies duality!
Homology (syntropic) is dual to co homology (entropic) -- same is dual to different.
Injective is dual to surjective synthesizes bijective or isomorphism.
Subgroups are dual to subfields -- the Galois correspondence.
"Always two there are" -- Yoda.
Spoiler Alert................. I think the exercise number 2 is just trivial. As any 1-dimensional vector space is spanned by a single vector, say v, one has [x, y]=[av, bv]=ab[v, v]=0
There should be a way to measure how far down the rabbit hole you are when watching a given math video. Not being a mathematician, this one feels a good length down, but how can I tell...
the "on the other hand" proof at around 4:14 you proved "[x,x] = 0" when you claimed and structured the proof as if you were proving "-[x,y] = [y,x]".
This was definitely a mistake, and it causes confusion from mixed/befuddled messaging.
edit: in the proof around 16:09 when moving from line 3 (ie:
"=D1( aD2(b) + D2(a)b ) - D2( aD1(b) + D1(a)b )" to line 4, you repeatedly say "because D1/2 is a derivation" when justifying the distributive property, but you actually mean "because D1/2 is a derivation and derivations are linear maps and therefore have the property of linearity". It is confusing for new players, but not incorrect the way you said it.
is this inspired by the mathemaniac series?
if he works hard enough teaching may be possible
2 min 20 sec into this and I’m already lost. It seems to be removing understanding from my mind.
Yeah, it reminds me of how my quantum mechanics professor teaches. No exposition, no explanation or motivation, just pure, hard results. Like, the only definition we get of Lie algebra is the minimum required to start messing with bracket algebra.
Probably we'd need to watch and completely understand the whole abstract algebra series first, which appears to go farther than my modern algebra class did. As a physicist, i don't understand how the pure math people can make sense of anything with how arbitrary and unmotivated these definitions seem. Maybe it'll come together later
I am a hack.
Nah bro you lieing! 😂