0
$\begingroup$

The PBW Theorem

In the literature there are many sources discussing the PBW-basis for Lie superalgebras, see for example M. Scheunert - The Theory of Lie Superalgebras theorem 1 and corollary in section 2.3

In a more recent source (Ian Musson - Lie superalgebras and enveloping algebras) the theorem is stated in the following form (for finite dimensional Lie superalgebras):

Theorem 6.1.2 (Musson) Let $x_1, \dots, x_m$ be a vector space basis for $\mathfrak{g}$ consisting of homogeneous elements. Then the set of all monomials of the form $$x_1^{a_1}\cdots x_m^{a_m}$$ with $a_i \in \mathbb{N}$ if $x_i \in \mathfrak{g}_0$ (i.e. $\bar{x_i}=0$) and $a_i\in \{0,1\}$ if $x_i \in \mathfrak{g}_1$ (i.e. an odd element) is a basis for $U(\mathfrak{g})$ over $\mathbb{C}$.

Question/Counterexample (?)

My question is on the following situation, which seems to disagree with the result above. Let $\mathfrak{g} = \mathfrak{sl}_{2|1}$ having even generators $E_1,F_1,H_1,H_2$ and odd generators $E_2,F_2$ subject to the following relations \begin{align*} [H_i,H_j] &= 0 = H_i H_j - H_j H_i\\ [H_i,E_j] &= a_{ij} E_j = H_i E_j - E_j H_i \\ [H_i, F_j] &= -a_{ij} F_j = H_i F_j - F_j H_i\\ [E_i, F_j] &= \delta_{ij} H_i = E_i F_j -(-1)^{\bar{ij}} F_j E_i \\ [E_1,[E_1,E_2]]&= 0 = E_1^2E_2 - 2 E_1 E_2 E_1 + E_2 E_1^2\\ [E_2,E_2] &= 0 = 2E_2^2 \\ [F_1, [F_1,F_2]] &= 0 = F_1^2F_2 - 2 F_1 F_2 F_1 - F_2 F_1^2 \\ [F_2,F_2] &= 0 = 2F_2^2 \end{align*} with $a_{ij}$ some integer coefficients (see e.g. Scheunert - Serre-type relations for special linear Lie superalgebras).

N.B. I have included the relations in $U(\mathfrak{sl}_{2|1})$ on the right side.

The theorem states that a basis is given by monomials $$H_1^a H_2^b E_1^c F_1^d E_2^e F_2^f$$ with $a,b,c,d \in \mathbb{N}$ and $e,f\in \{0,1\}$.

The question will maybe seem rather silly after such a build-up, but here goes:

Question: How does one express the element $E_2 E_1$ in terms of such a basis? (Similarly $E_2 E_1 E_2$)

As far as I can tell that is not possible and the basis is actually of the form $$H_1^a H_2^b E_1^c F_1^d (E_2E_1)^e E_2^f (F_2 F_1)^g F_2^h$$ with $a,b,c,d \in \mathbb{N}$ and $e,f,g,h\in \{0,1\}$

This basis seems to contradict with the theorem above. Thus if the answer on the first question is, `this is not possible'. What then should I make of the theorem above? Comments are most welcome.

$\endgroup$

migrated from mathoverflow.net Jan 21 '15 at 19:34

This question came from our site for professional mathematicians.

  • 3
    $\begingroup$ I think you have misinterpreted the statement of PBW. Indeed, $E_\theta = [E_1,E_2]$ and $F_\theta = [F_1,F_2]$ are homogeneous basis elements of $\mathfrak{sl}(2|1)$, and hence allowed monomial generators in the PBW theorem. Compare $\mathfrak{sl}(3)$, which is generated by $H_i,E_i,F_i$ for $i=1,2$, but which again includes two extra basis elements. $\endgroup$ – Theo Johnson-Freyd Jan 21 '15 at 5:05
  • 3
    $\begingroup$ So the theorem in fact states that a basis for $U\mathfrak{sl}(2|1)$ consists of the monomials $H_1^aH_2^bE_1^cF_1^dE_2^eF_2^f[E_1,E_2]^g[F_1,F_2]^h$, or some such, with $e,f,g,h \in \{0,1\}$. $\endgroup$ – Theo Johnson-Freyd Jan 21 '15 at 5:08
  • $\begingroup$ Indeed, it is a common mistake confusing a basis for the space of generators with a vector space basis for the algebra. I would think that this is a question that should have been asked on MSE $\endgroup$ – Vladimir Dotsenko Jan 21 '15 at 8:08
  • $\begingroup$ @THeoJohnson-Freyd Thank you for your answer. I now see my mistake clearly wrt. the basis. $\endgroup$ – Tim Weelinck Jan 21 '15 at 18:50
  • $\begingroup$ @VladimirDotsenko If this was a MSE question I apologize for that. I am new to both websites, but will take more care from now on in evaluating which site is better fitted for the question. $\endgroup$ – Tim Weelinck Jan 21 '15 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.