24.05.24 (An infinite family of degree 0 Nil-Hecke idempotents deforming the trivial idempotent)
Here is some cool thing: Take polynomials in n variables A= k[x_1,\ldots, x_n]. Take symmetric polynomials A^(S_n) living inside all polynomials. Take End_{A^(S_n)}(A) and what you get is what people call the Nil-Hecke algebra denoted NH_n. This is some very cool algebra, containing A, containing the group algebra of S_n (since the action of S_n on A is by ring homomorphisms) and the so called Demazure operators \del_1, \ldots, \del_{n-1}. NH_n is generated by the \del's and the variables in A and putting the del's and the x's in degree -2, respectively 2 makes it graded. However it's degree 0 part is in general bigger then kS_n...
As a toy example consider NH_2 which is as algebra generated by x_1, x_2 (which commute) and \del (with the relations x_1\del=\del x_2 + 1, \del x_1 = x_2 \del +1 , \del^2=0). This algebra has a unique indecomposable projective module (up to grading shift) which Rouqier tells us can be obtained by the idempotent \del x_1. For general S_n one gets \del_{w_0}x_1^{n-1}x_2^{n-2}...x_{n-1}.
Now let's forget that I told you that and look at the degree 0 part of NH. It is 3-dimensional and spanned by 1, a=\del x_1, b=\del x_2. So a^2=a, b^2=b, ab=b, ba=-a. Note that this is a 3-dimensional non-commutative complex algebra, so it cannot be semisimple!
The question what I will answer is what are all the idempotents in this degree 0 part. First observe that the generator s of S_2 can be written as 1-(x_1-x_2)\del = s = \del(x_1-x_2) -1.
The trivial idempotent e_triv = (1+s)/2 then becomes (\del (x_1-x_2)) /2 = e_triv = 1 - 1/2 * (x_1-x_2)\del (Note the Jones--Wenzl expression) and the sign idempotent becomes (x_1-x_2) \del / 2 = e_sgn = 1 - 1/2 * \del (x_1-x_2).
There is some third interesting and evil element z which is \del (x_1+x_2) = (x_1 + x_2)\del. This element z satisfies z^2=0 since del^2=0. Morever note that e_triv * z = (1 - 1/2 * (x_1-x_2)\del )\del(x_1+x_2), however z * e_triv = (x_1+x_2)\del \del (x_1-x_2) = 0 . Summarizing the degree 0 part has basis 1, e=e_triv, z and relations ez=z, ze=0, z^2=0, e^2=e. (By the way if you take e'=e_sgn=1-e instead of e you get e'z=0 and ze'=z).
What does this mean? Take any a complex number \lambda. Then e + \lambda *z will again be an idempotent (since (e+ \lambda *z)^2=e^2 + \lambda ez + \lambda ze + \lambda^2 z^2 = e + \lambda z.
There is one particularly choice which is \lambda=1/2, which gives the idempotent \del x_1 (where x_1 is the product version of the half-sum of all positive roots for gl2). However the real idempotent in the back is \del (x_1-x_2)/2. If one generalizes this to other types the trivial idempotent will become (\del_(w_0)prod(positive roots)/|W|).
Finally note that there are even more idempotents (non-homogenous ones) in NH_2 for instance e +\del is an idempotent since by the same calculation (e\del= \del, \del e= 0, \del^2=0).
21.05.24 (Counter examples to injectivity of g into U(g))
Here some not complete, but funny overview of statements, regarding injectivity results of Lie algebras into their U(g)'s, in case you are interested:
First assume g is a Lie algebra over some commutative ring k.
If k is a field then g embedds into U(g)
If k is not necessarily a field, but contains the rationals, then g embedds into U(g)
There are counter examples if k contains F_p.
Now assume c is a Lie coalgebra over some field k, as in the diary entry from the 23.04.24 c doesn't want to embedd into U^c(c) but rather be it's quotient.
c is a quotient of U^c(c) if and only if every element of c is contained in a finite dimensional Lie subcoalgebra of c
Now assume (A,L) is a Lie-Rinehart algebra (concretely A is a commutative k-algebra, L is a Lie k-algebra and left A-module with a A-linear Lie algebra map L->Der(A) + action of A on L obays Leibniz rule)
If L is projective as left A-module then L embedds into U(A,L) (This holds for instance for the Weyl-algebra (=U(k[x], Der(k[x]) ), here the derivations of k[x] embedd into the Weyl algebra)
If the anchor morphism L->Der(A) is injective, then L embedds into U(A,L)
If the anchor morphism L->Der(A) is zero, then L is automatically an Lie algebra over A and we are in the first setting (for instance it is then true in characteristic zero, but wrong in char p)
Even if k is of characteristic zero, my advisor and me expect that it's not necessarily true that L embedds into U(A,L), however we don't know a counter example.
23.04.24 (k[SL_2] is universal enveloping coalgebra of (sl_2)^{star})
So this is some crazy stuff. I read 'Lie coalgebras' by Walter Michaelis and the paper contains the following statement: ''The finite dual of the universal enveloping algebra of a Lie algebra g is the universal coenveloping coalgebra of the finite dual of g''. This in particular means for our best friend sl_2(C):
The coordinate ring k[SL_2] has a universal property, namely it is the universal coenveloping coalgebra of the Lie coalgebra (sl_2)^{*}.
Let's make this precise. If we take the dual basis of sl2 call it e*, f*, h*, the Lie bracket on sl2 turns into the following Lie cobracket
[e*]=2(h*\otimes e*-e*\otimes h*)
[f*]=(-2)(h*\otimes f* - f* \otimes h*)
[h*]=e*\otimes f* - f* \otimes e*
Alternatively we might consider k[SL_2] with generators a,b,c,d as a coalgebra and turn that comultiplication into a cobracket by [x]=x_(1)\otimes x_(2)-x_(2)\otimes x_(1).
This gives the following cobracket on k[SL_2]:
[a]=b\otimes c - c \otimes a
[b]=(a-d)\otimes b - b \otimes (a-d)
[c]=-( (a-d)\otimes c - c \otimes (a-d))
[d]= c\otimes b - b\otimes c
This Lie cobracket becomes the Lie cobracket of (sl_2)^* if we regard it as a direct summand of span{a,b,c,d}
given by e*=b, f*=c, h*=(a-d)/2 with complement (a+d)/2 [Note the S_2 action on the Cartan of gl_2].
In other words this statement gives an alternative way to show that the finite dual of U(sl2) becomes the coordinate ring k[SL_2]. You just have to check that the quotient map k[SL_2]\to sl_2^* has the correct universal property (every Lie coalgebra map C -> sl_2^* factors through a coalgebra map C-> k[SL_2]). Conceptually just how one can recover in characteristic zero the Lie algebra from it's enveloping algebra U(g) as it's subset of primitives, one gets g* as the quotient of indecomposables (also called coprimitives) of k[G] (i.e. the kernel of the counit divided by it's square). Concretely if one considers the span of (a-1),b,c,(d-1) which all lie in the kernel of the counit one can see that (a+d)/2-1 get's killed in this module (Indeed we have 0=(a-1)(d-1)=ad-d-a+1=bc+1-d-a+1=0+2-d-a since bc is in the square of the coaugmentation coideal) and so (a+d)/2=1. This is absolutely amazing, because here the determinant = 1 relation from k[SL_2] gives us a trace condition.
12.04.24 (PBW follows in characteristic zero (not) from Ados Theorem)
I was reading the book ''Topics in Noncommutative Algebra - The theorem of Campbell, Baker, Hausdorff and Dynkin'' written by Andrea Bonfiglioli and Roberta Fulci. In chapter 6 of that book they try to explain how the PBW theorem and other theorems are related. The come to the conclusion that one can prove the following theorems circularily one from another in characteristic zero:
1) For every Lie algebra g the canonical map g \to U(g) is injective
2) Free Lie algebras on vector spaces/sets exist (rather the Lie subalgebra of T(V) generated by V is the free Lie algebra on V)
3) Free algebras in 3 variables exist and the Campbell-Baker-Hausdorff-Dynkin theorem holds (the expression e^x*e^y=e^z where z is a power series in Lie words in x,y: concretely z=x+y+[x,y]/2+...)
4) The PBW theorem holds
The proof is going 1->2->3->4->1.
I also saw the following proof of 1) for finite dimensional Lie-algebra in Cartiers and Patras Book ''Classical Hopf Algebras and their applications'': If g is finite dimensional, then by Ados theorem there exists an injective map \rho: g->gl(V) for some f.d. vsp V (In other words g has a faithful representation). This representation extends to an algebra map \rho hat: U(g)\to End_k(V)=gl(V) by the universal property of U(g) via the embedding i: g\to U(g). This implies i: g\to U(g) is injective since \rho hat \circ i = \rho and \rho is injective. In other words one could summarize that prove in the following way: A Lie algebra embeds as Lie algebra into some associative Lie algebra (viewed as Lie algebra with the commutator), if and only if it embeds into it's universal enveloping algebra.
If you ask me this proof is absolutely crazy. However the problem is that this is not enough to prove PBW upstairs since not every Lie algebra is finite dimensional, in particular free Lie algebras :'( .
Anyways this was a fun excursion.