I am reading Humphrey's "Introduction to Lie Algebras and Representation Theory". However, when goes to the proof of PBW theorem, I met some problems in the following lemmas:

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Where $T_m$ is the filtration of tensor algebra of $L$

and $J$ is the ideal generated by $<x\otimes y-y\otimes x-[x,y]>$

and $I$ is the ideal generated by $<x\otimes y-y\otimes x>$

$\mathfrak{T}$ is tensor algebra of $L$ and $\mathfrak{S}$ is symmetric algebra. and $S_m$ is the filtration of $\mathfrak{S}$ (if I understand correctly)

My question is mainly on the proof of Lemma B and C

  • First of all. In lemma A, we defined $f_m: L\otimes S_m \to \mathfrak{S}$. But then in the proof of lemma B, it says Lemma A allows us to define $f: L\otimes\mathfrak{S} \to \mathfrak{S}$ I believe the reason why we can define such an $f$ is that since $S_m$ is a subset (or subspace) of $\mathfrak{S}$, and so that by the uniqueness part of Lemma A, if we restrict $\mathfrak{S}$ to some $S_m$, we can always get a unique $f_m$. So we can then extend all of the $f_m$ to an $f$ defined on $ L\otimes\mathfrak{S} $?

  • Secondly, in proof of Lemma C, to consider $\rho:\mathfrak{T}\to End\ \mathfrak{S}$, it is OK to say that $\rho(t)=0$ since $t\in J$ and $J\subseteq Ker\ \rho$. But then by lemma B, we can deduce that $\rho(t).1$ is a polynomial whose highest degree term is a combination of $z_{\Sigma(i)}$. Then the highest degree term is $0$ indeed. But the highest degree term now is NOT $t_m$, (since t_m is a linear combination of $x_{\Sigma(i}$). Am I right ? Then how can I deduce that $t_m$ now is in I?

The last 2 sentences of the proof dose not make any sense to me... Could anyone please help me with that?


  • $\begingroup$ There is a Peanut Butter Sandwich theorem ? $\endgroup$ – Rene Schipperus Feb 8 '17 at 16:00
  • $\begingroup$ sorry. I should make what PBW means clear. Already edited the title :) $\endgroup$ – Lalala Feb 8 '17 at 16:04
  • $\begingroup$ Hey its ok, I'm just being stupid. $\endgroup$ – Rene Schipperus Feb 8 '17 at 16:07
  • $\begingroup$ I should be as precise as possible anyway :-) Do you have any clues for my question? $\endgroup$ – Lalala Feb 8 '17 at 16:10

The answer to your first question is that $L\otimes\mathfrak{S}$ is the direct limit of the directed system $\{L\otimes S_m\}$ and the map $f$ is just the direct limit of the $f_m$.

In regard to your second question, I'm not sure why you are saying that $t_m$ is not the highest degree term. We have that $t\in T_m$, so the homogeneous component $t_m$ IS the highest degree term AND is a linear combination of the $x_{\Sigma(i)}$. What exactly is the problem?

  • $\begingroup$ Sorry, what I said might be confusing. I can understand that $t_m$ the highest term, which is a linear combination of $x_\Sigma$ What I want to say is that I do not think that $t_m$ is the highest term of $\rho(t).1$...and how can we get $t_m$ is in I? thanks! $\endgroup$ – Lalala Feb 10 '17 at 21:12
  • $\begingroup$ Again, I don't understand your problem. Why are you skeptical that $t_m$ is the highest term? What do you mean? $\endgroup$ – David Hill Feb 11 '17 at 3:19
  • $\begingroup$ Do you understand that the relation $x\otimes y - y\otimes x =[x,y]$ is degree lowering in the tensor algebra? When you take the associated graded of a filtered algebra (which is what is happening) you pick out the top term. $\endgroup$ – David Hill Feb 11 '17 at 3:28
  • $\begingroup$ hi, sorry for the late reply...I was being stupid...Now I can understand those lemmas fully. $\endgroup$ – Lalala Feb 13 '17 at 19:32

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