Lie groups: Lie groups and Lie algebras

[@drakoz254]

Thanks professor! I'm amazed at how fast these are coming out given their quality. Leaves me with no excuse but to finally get around to learning this stuff :)

[@sachalucienmoserferreira2233]

Its a pleasure find your channel in mathematics a hugs from Brazil!

[@GiovannaIwishyou]

Thank you professor Borcherds, this really means a lot.

[@anthonymurphy5689]

At 22:20, in switching slides, did he mix up G and H? On prior slide, H was the original group and G was meant to be the [simply connected] universal cover. He seems to use G as the original group from here on.

[@richarde.borcherds7998]

Yes, I accidentally switched G and H.

[@anthonymurphy5689]

@@richarde.borcherds7998 Thank you. You really are to be commended for the excellent series of lectures you have been posting over the past year. My wife of over 30 years is an algebraic geometer, while I was a mere physicist - you’re helping me understand her at last!

[@えいき-f3c]

Should N be unipotent and not nilpotent? 26:40

[@faisalal-faisal1470]

It's both

[@AsvinGothandaraman]

The group of unipotent matrices is a nilpotent group (it's lower central series terminates at 0) .

[@えいき-f3c]

@@AsvinGothandaraman Thanks Asvin for the clarification.

[@anthonymurphy5689]

I think the confusion arises in the terminology between the Lie group and the Lie algebra. The Lie algebra is nilpotent and I think the corresponding term is sometimes carried across to the associated Lie group. In matrix terms, the exp function will take a nilpotent matrix (ie in the Lie algebra) to a unipotent matrix (ie in the Lie group). Borcherds will cover exp, relating Lie algebra and Lie group, in the next lecture of the series. See also the later lecture on Engel’s Theorem where nilpotency is discussed in detail.

[@faisalal-faisal1470]

​@@anthonymurphy5689 It's also worth noting that any unipotent group (=linear algebraic group consisting of unipotent elements (e.g. N in the video)) is automatically a nilpotent group. Sketch of proof: As Asvin noted, it's easy to check this directly for the subgroup N of GL_n consisting of upper triangular matrices with 1s on the diagonal. The interesting part is that *every* unipotent subgroup of GL_n is conjugate to a subgroup of N.

[@grog-i9m]

Nice video as always! Could you provide a reference for the facts you did not prove?

[@richarde.borcherds7998]

Bourbaki "Lie groups and Lie algebras" has proofs of nearly everything.

[@domc3743]

Thank you so much for this

[@klmnps]

What you need in the questions about the correspondence between Lie subalgebras and Lie subgroups is the Malzev closure of the Lie subalgebra

[@hannesstark5024]

Thanks for the video

[@migarsormrapophis2755]

YEEEEEEEE

[@criskity]

Today I learned it's pronounced "lee", not "lye".