Best regards

Vincenzo

*[This particular article was reprinted as Section 2.3 of this book: https://terrytao.wordpress.com/books/hilberts-fifth-problem-and-related-topics/ . In general, the web URL should be fairly stable. -T.]*

It’s important to restrict to the characteristic zero case (Levi’s theorem fails for positive characteristic, and there may be other places in the above post where characteristic zero is implicitly used). But I think most of the discussion on the complex case transfers to the real case without difficulty (Lie’s theorem isn’t true over the reals, but that theorem isn’t really used in the above post except as motivation). And as you say, the complex case of Ado’s theorem implies the real case as a corollary.

]]>Thanks for writing this illuminating article. Regarding the statement

“One can of course define Lie algebras over other fields than the complex numbers {{\bf C}}, but in order to avoid some technical issues we shall work solely with the complex case in this post.”

made at the beginning of the post, could you please point out which part of the argument depends on this restriction to complex Lie algebras? I’ve not been able to figure out why the argument won’t go through for real Lie algebras, although Fulton and Harris deduce the real case from the complex case.

Thanks ever so much!

]]>This as a very enjoyable post. Thanks for posting it.

Regarding remark 3, I don’t know if this qualifies as a more “direct” proof but corollary 8 can be proved via Lie’s theorem and vice-versa. In one direction, corollary 8 can be proved using Lie’s theorem by first showing that any derivation maps the radical of into itself and second by using Lie’s theorem on the solvable ideal , where is the image of in under the ad map, to show that is nilpotent.

Conversely, corollary 8 can be used to prove Lie’s theorem inductively by taking a solvable Lie algebra , a codimension 1 ideal , and then showing that the space generated by 1-dimensional ideals of is both a non-trivial subspace (by induction) and an ideal in Showing that is an ideal requires cor. 8 and this produces a common eigenvector of the adjoint representation, showing that every solvable ideal has a complete flag consisting of ideals. From there, we can work out Lie’s theorem for any representation by considering the direct sum , where is considered as an abelian ideal.

As a bonus, theorem 7 also follows from cor. 8 and Lie’s theorem.

Best Regards,

Alejandro Ginory

*[Corrected, thanks – T.]*

If is not of codimension 1, it is not obvious how to locate a complementary space which is itself a Lie algebra (i.e. closed under Lie bracket). In the codimension one case, this is not a difficulty as any one-dimensional subspace is automatically a Lie algebra. (It is automatically also abelian, but this is a bonus which, as you say, is not particularly essential to the rest of the argument.)

]]>I was wondering why one need to take the ideal A to be of codimension 1? Already in the nilpotent case for example, i don’t see in which step we need A to satisfy that (also do we need H to be abelian?). I get the feeling that it is enough to be able to decompose the lie algebra into A \oplus H with A an ideal of the lie algebra.

Thanks for your answer!

Best regards

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