I think you might have missed a step. When you take a z in L such that [X,z] =/= 0, you say such a z exists because L is non-abelian. But L being non-abelian just means there exists non-zero brackets, not that X has non-zero brackets with other elements. What you really need for such a z to exist is that X is not in the center of L. One way to show this is to show that any 2-dimensional non-abelian Lie algebra has a trivial center. Suppose x = av+bw is in the center of L. Then in particular [x,v] = 0 and [x,w] = 0. But [x,v] = b[w,v] = -b[v,w] = 0 and [x,w] = a[v,w] = 0. But as L is non-abelian, [v,w] =/= 0. Therefore we must have a = b = 0, so x = 0. So the only element in the center of L is the zero element.
@franzlyonheart4362
Жыл бұрын
Bingo. You're correct -- he didn't even use the 2-dimensionality condition properly. My similar comment then is a duplicate, I could have saved typing it up by reading other comments first! 😂
@cookieshade197
Жыл бұрын
Alternatively, since L is two-dimensional, there must exist some Z such that X,Z spans L. Since all Lie brackets are scalar multiplies of [X,Z], L being nonabelian directly implies [X,Z] nonzero.
@iabervon
Жыл бұрын
He'd shown that [av+bw,cv+dw]=(ad-bc)[v,w] and [v,w] is non-zero for this algebra. For any a and b, unless a=b=0, you can choose c and d so that (ad-bc) is non-zero. So it was at least a reasonably straightforward consequence of things on the board, even if he didn't actually say the step.
@hyperduality2838
Жыл бұрын
Commutators are used to calculate forces -- Riemann geometry. Abelian (commutation, symmetry) is dual to non abelian (non commutation, anti-symmetry). The covariant derivative:- rotating the vector whilst keeping the vector basis constant is equivalent or dual to keeping the vector constant and rotating the vector basis -- duality! Gravitational forces are measured with the commutator of covariant derivatives -- Einstein. Commutators = two dual paths. Forces are dual. Action is dual to reaction -- Sir Isaac Newton or the duality of force. Lie algebra measures forces in physics! "Always two there are" -- Yoda. Symmetry is dual to conservation -- the duality of Noether's theorem. Abelian implies symmetry or conservation (invariance) = zero forces. "May the force (duality) be with you) -- Jedi teaching. "The force (duality) is strong in this one" -- Jedi teaching. "The force" is duality -- Jedi teaching.
@franzlyonheart4362
Жыл бұрын
6:52, glossed over a few steps, because finding such a z is not immediately obvious, it is not trivial to see that L' = [L,L] is contained within Z(L), the centre of L. Don't you need to use the condition here that the dimension of L over F is not more than 2? 8:05, however that Claim is at this stage entirely trivial. Could skip that easily here, rather than the former claim of existence of z.
@franzlyonheart4362
Жыл бұрын
Grammar correction … it is not trivial to see WHETHER OR NOT* the L' is contained within Z(L) … needs to use the 2-dimensionality here. Otherwise the calculation wouldn't really have used the "size" argument of the vector space properly.
@franzlyonheart4362
Жыл бұрын
11:40, I really like how he underpins the classification by giving two concrete matrix algebras that essentially represent that classification comprehensively.
@CM63_France
Жыл бұрын
Hi, 13:50 : missing "stop".
@trumanburbank6899
Жыл бұрын
We know that what Michael says is true because he is an excellent Lie-er. 😁
@gp-ht7ug
Жыл бұрын
When is Lie algebra used?
@ericbischoff9444
Жыл бұрын
or even simpler: a 3 dimensional vector space with the cross product
@TheLethalDomain
Жыл бұрын
Just about any time you work with common mathematics, even if you don't immediately realize it. Many physicists, if anything, have downplayed the role of Lie theory in mathematics as if it only describes a select few special systems that we should be surprised if we come across, despite obeying Lie theory every step of the way there anyway.
@SuicideRedemption100
Жыл бұрын
The very categorization of elementary particles arises from the fact that 4D space can be mapped into two SU2 groups ( JF Cornwell, group theory in physics ), it's the description of spin. Not to mention the description of the polarization of light, of basically any 2 level system in physics
@MrFtriana
Жыл бұрын
The mathematicians also used the lie Algebras in the study of differential equations. A name that i recall is Olver, if i remember well.
@hyperduality2838
Жыл бұрын
Commutators are used to calculate forces -- Riemann geometry. Abelian (commutation, symmetry) is dual to non abelian (non commutation, anti-symmetry). The covariant derivative:- rotating the vector whilst keeping the vector basis constant is equivalent or dual to keeping the vector constant and rotating the vector basis -- duality! Gravitational forces are measured with the commutator of covariant derivatives -- Einstein. Commutators = two dual paths. Forces are dual. Action is dual to reaction -- Sir Isaac Newton or the duality of force. Lie algebra measures forces in physics! "Always two there are" -- Yoda. Symmetry is dual to conservation -- the duality of Noether's theorem. Abelian implies symmetry or conservation (invariance) = zero forces. "May the force (duality) be with you) -- Jedi teaching. "The force (duality) is strong in this one" -- Jedi teaching. "The force" is duality -- Jedi teaching.
@kylecow1930
Жыл бұрын
could you not say, we know that [a,b]=AX for some X for all a,b. So take some vector y not in the span of X, then {X,y} is a basis and [X,y]=aX if a is non zero let Y=1/a y this is obviously still a basis and [X,Y]=X and if a is zero we're just in case 1 because we have a basis with bracket 0 which contradicts our assumption that the bracket is not 0
@francoisleyvraz3920
Жыл бұрын
Call a ``generalized Lie algebra'' a vector space with a bilinear alternating bracket. I think the generalized 2-dimensional Lie algebras were also classified in this video. Am I missing something'
@anthonyserafini
Жыл бұрын
One could say all 2D Lie algebra lies on this video
@christianaustin782
Жыл бұрын
Is the operation used in the "some representations" section always the commutator operator? Inclduding the last one with the differential operators?
@SuicideRedemption100
Жыл бұрын
I think it's an example, the only condition on the bracket is that it's a bilinear map An example of lie algebra that isn't with commutators is the cross product, the cross product itself is the bracket. Obviously z x z = 0 for any z in R3, and the Jacobi identity holds, but the representation of the bracket is actually a Matrix, which is the sum of the three generators of the symmetry transformations of SO3. To further understand why this is a lie algebra, there is a theorem ( JF Cornwell, group theory in physics ) that says that there is a 2 to 1 map from SO3 to SU2, and the bracket in SU2 is the commutator you see in the video. So although the commutator is not the only representation, it is a pretty fundamental one, since it seems to appear in any irreducible rep , of even non definite orthogonal groups, EXACTLY because of the mentioned theorem.
@cstockman3461
Жыл бұрын
Yes, the bracket in the representation that uses differential operators is the commutator. Intuitively, this makes since because differential operators are linear operators on the vector space of smooth functions, so their commutator is not defined
@samsonblack
Жыл бұрын
More generally, for any associative "multiplication" that distributes over addition, we can define a Lie bracket via the commutator: [v, w] = vw - wv In the first two examples, the operation is matrix multiplication, and in the third example, the operation is composition of differential operators.
@Apollorion
Жыл бұрын
Well, I didn't find it obvious that if [x,y] is linear and [x,x]=0 that then [x,y]=-[y,x] ... but by calculating [x-y,x-y] it becomes: [x-y,x-y]=[x,x]-[x,y]-[y,x]+[y,y] due to linearity [x-y,x-y]=[x,x]=[y,y]=0 due to the equality of their input vectors. Fill this in the above equation and it becomes: 0=-[x,y]-[y,x] Add [x,y] on both sides, and voila: [x,y]=-[y,x]
@samsonblack
Жыл бұрын
Nice: note that if you want to have fewer minus signs in this argument, you can instead expand [v+w, v+w] = 0 and reach the same conclusion.
@samsonblack
Жыл бұрын
Although this implication holds in any field F, the converse requires that 2 is invertible in the field. *Can you show it?*
@Apollorion
Жыл бұрын
@@samsonblack I agree with your first comment (the one with the minus was just the one I tried first & that was directly successful), but I don't understand your 2nd comment; what do you mean by "converse" and "invertible"?
@samsonblack
Жыл бұрын
@@Apollorion You've argued that if [x, x] = 0 for all x in L, then [x, y] = -[y, x] for all x, y in L. The converse is: if [x, y] = -[y, x] for all x, y in L, then [x, x] = 0 for all x in L.
@samsonblack
Жыл бұрын
The argument for the converse requires that if 2z = 0 then z = 0, which is obviously true in familiar fields like the real numbers or complex numbers because the number 2 is invertible there (the number ½ exists with the property that ½×2 = 1). But there are finite fields of characteristic 2 where 2z = 0 for all z, so just because 2z = 0, we cannot necessarily conclude that z = 0.
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