Lie Algebras 7 -- Engel's Theorem

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[@iabervon]

There are a couple of functions defined on quotients in this video that aren't obviously well-defined by theorems we've proven previously. In each case, the proof that it was well-defined was a single line, but involved looking at the domain and definition of the function to see why the difference between results had to be in the coset.

[@berry4862]

This one and Video Nr. 3 are missing on the Playlist.

[@justanotherman1114]

@ 30:37 you have to show that image of W under y is contained in W. proof Let w be an element of W. We have to show y(w) is in W. Equivalently,ay(w)=0 for all a in A. But ay(w)= [a,y](w)+ya(w)=0 Because [a,y] is in A due to ideal property and the definition of W. Thus, our claim follows.

[@Timwadsworth75]

Thank you! I did not understand this bit

[@AxelCabrol]

In the proof of Engle's theorem, I don't get why dim(V/U) = k as the induction hypothesis is on dim(L) not V. Using the quotient V/U makes sense as u doesn't depend on x, but then we need to show that L_bar is of dimension strictly below k+1. Which it is clearly, but can be lower than k imo.

[@SylComplexDimensional]

Lie-Bäcklund symmetry generator was a project paper of mine in undergrad (BA math) !!

[@bjornfeuerbacher5514]

42:05 Shouldn't that be gl(L) instead of gl(V)?

[@justanotherman1114]

@ 34:21 induction is over dimension of V?

[@willemesterhuyse2547]

Why do you need the "ad_x" if you can just specify [x, . ]?

[@zlodevil426]

Because it’s useful to consider the properties of adx as a map

[@7177YT]

There's no audio! Cheers!

[@lukewatson8848]

I can hear it!

[@pierfrancescopeperoni]

Usually it's a button on the upper right of your phone, or if you are using a computer it's a small key with a megaphone symbol and a plus sign next to it, on the upper part of the keyboard. That should work.

[@7177YT]

@@pierfrancescopeperoni I found it! It worked! Tysm!

[@pierfrancescopeperoni]

@@7177YT You're welcome.

[@ayzikdig1983]

5:00 why x(w)=0 ?

[@sicko_the_ew]

I also keep bumping my head against this one. x^1(of any vector) 0 (unless N = 2), so there has to be a "typo" (otherwise we have x(w)=0 AND x(w)0 -> upside-down-T or contradiction). With that in mind, I think what's meant is x^N(w) = 0 ? But then that also doesn't make sense WRT what follows. So I bump head again. (Doesn't help playing the video at .75X so it runs only slightly faster than I think, either.) Sorry not to offer a solution. I'm just letting you know you're not the only one.

[@divisix024]

Isn’t is because he set w=x^(m-1)(v) and v is in ker(x^m), noting that here x is a linear map and the power represents n-fold composition, for instance x^3(u)=x(x(x(u)))? Also regrading the other comment, since x is a nilpotent map on a nontrivial space, its kernel must be nontrivial. That is, x(w)=0 for some w≠0. In fact, it has been explicitly given above, yo can just take w=x^(m-1)(v) for any v in ker(x^m), where m

[@divisix024]

But I think one could’ve used x^m(v)=0 in a different way, as follows. Since x is nilpotent, the set W={v, x(v), … , x^(m-1)(v)} must be linearly independent, because any nontrivial relation among them would be of the form p(x)=0 on W where p is a polynomial of at most degree m, but the minimal polynomial for x restricted to W, which is the generator of the ideal of all polynomials q such that q(x)=0 on W, is exactly x^m by assumption. This means that p(x) must be a scalar multiple of x^m. Then the matrix representation of x in the subspace generated by W is (upper) triangular (by explicit computation of the matrix). Let W=V/ker(x), then x is surjective on W and all the above still applies. But then we may extend the matrix representation of x restricted to W to that of x on the whole space V with 0’s, as V=W+ker(x), where the + is vector space direct sum.

[@arnouth5260]

x(w) = x(x^(m-1)(v)) = x^m(v) =0.

[@AxelCabrol]

In the proof of Engle's theorem, I don't get why dim(V/U) = k as the induction hypothesis is on dim(L) not V. Using the quotient V/U makes sense as u doesn't depend on x, but then we need to show that L_bar is of dimension strictly below k+1. Which it is clearly, but can be lower than k imo.