Section 3 Constructions
Subsection 3.1 Enumeration of torsion-free gradings
Following the method suggested in Section 3.7 of [21], we now give a simple way to enumerate a complete (and finite) set of universal realizations of gradings of a Lie algebra using the maximal grading. For the rest of this section, letProposition 3.1.
The set
Proof.
Let \(\mathcal{V}\) be the universal realization of some torsion-free grading. Due to Lemma 2.9, the grading group of \(\mathcal V\) is some \(\mathbb{Z}^m\text{.}\) By Proposition 2.18, there exists a homomorphism \(f\colon \mathbb{Z}^k\to \mathbb{Z}^m \) and an automorphism \(\Phi \in \Aut(\mathfrak g)\) such that \(\mathcal V= f_*\Phi(\mathcal W) \text{.}\) Let
We are going to show that \(\mathcal V' = (\pi_I)_*(\mathcal W) \) is equivalent to \(\mathcal{V}\text{.}\) Then, a posteriori, \(\mathbb{Z}^k/\langle I\rangle\) is torsion-free and we have \(\mathcal V' \in \Gamma\text{,}\) proving the claim.
First, since \(\ker(\pi_I) = \langle I\rangle \subseteq \ker(f)\text{,}\) by the universal property of quotients there exists a unique homomorphism \(\phi \colon \mathbb{Z}^k/\langle I\rangle \to \mathbb{Z}^m \) such that \(f = \phi \circ \pi_I\text{.}\) In particular,
so \(\mathcal V\) is a push-forward grading of \(\mathcal V'\text{.}\)
Secondly, since also \(\ker(f)\cap (\Omega-\Omega) = I \subseteq \ker(\pi_I)\cap (\Omega-\Omega) \text{,}\) we deduce that \(\mathcal V \) and \(\Phi(\mathcal V')\) are realizations of the same grading. Since \(\mathcal V\) is a universal realization, it follows that \(\Phi(\mathcal V')\) is a push-forward grading of \(\mathcal V\text{.}\) Consequently, \(\mathcal V'\) is a push-forward grading of \(\mathcal V\text{.}\) Since the grading group of a universal realization is generated by the weights, we get that the gradings \(\mathcal V\) and \(\mathcal V'\) are equivalent by Lemma 2.6, as wanted.
Subsection 3.2 Stratifications
Definition 3.2.
A stratification (a.k.a. Carnot grading) is a
Lemma 3.3.
A nilpotent Lie algebra
Definition 3.4.
Recall that the lower central series of a Lie algebra
where
Proposition 3.5.
Let
such that, for each triple of indices
Proof.
If \(\delta\colon\mathfrak{g}\to\mathfrak{g}\) is a derivation that restricts to the identity on \(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]\text{,}\) then by Lemma 3.3 \(\mathfrak{g}\) admits a stratification
such that \(\restr{\delta}{V_i} = i\cdot \operatorname{id}\text{.}\) Since the terms of the lower central series are given in terms of the stratification as \(\mathfrak{g}^{(i)}=V_i\oplus\dots\oplus V_s \text{,}\) it follows that \(\delta(Y)\in i\cdot Y+\mathfrak{g}^{(i+1)}\) for any \(Y\in\mathfrak{g}^{(i)}\text{.}\) That is, a derivation \(\delta\) restricting to the identity on \(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]\) is of the form (3.2) for some coefficients \(a_{ij}\in F\text{.}\)
It is then enough to show that (3.3) is equivalent to the Leibniz rule
Indeed, this would prove that a linear map defined by (3.2) is a derivation if and only if the coefficients \(a_{ij}\) satisfy the linear system (3.3).
Since the basis \(X_i\) is adapted to the lower central series, only the structure coefficients with large enough degrees are non-zero, i.e., we have
By direct computation using (3.2) and (3.4) we get the expressions
Denoting \(\sum_k B_{ij}^kX_k = \delta([X_i,X_j])-[\delta(X_i),X_j]-[X_i,\delta(X_j)]\text{,}\) we find that the equation \(B_{ij}^k=0\) is up to reorganizing terms equivalent to (3.3).
Finally, we observe that when \(w_k\leq w_i+w_j\text{,}\) the condition \(B_{ij}^k=0\) is automatically satisfied: for \(w_k\lt w_i+w_j\) all of the sums are empty, and for \(w_k=w_i+w_j\text{,}\) the only remaining terms from the sums cancel out as
- Construct a basis
adapted to the lower central series. - Find a derivation
as in (3.2) by solving the linear system (3.3). If the system has no solutions, then is not stratifiable. - Return the stratification with the layers
Remark 3.6.
By Theorem 1.4 of [4], the existence of a stratification for a Lie algebra is invariant under base field extensions, so it suffices to work within any field
Subsection 3.3 Positive gradings
Definition 3.7.
An
- Can the Lie algebra be equipped with a positive grading?
- Can one find in some sense all positive gradings of the Lie algebra?
Algorithm 3.8. Positive realization.
Input: A torsion-free grading
- Compute the universal realization
of as in Subsection 2.3. Let be the weights of - Compute
- Solve the integer linear programming problem
and binary variables If no solution exists, then the grading does not have a positive realisation. - Let
be the homomorphism Return the push-forward grading
Proof of correctness.
If the grading \(\mathcal{V}\) has a positive realization, then it is a push-forward grading of the universal realization by some homomorphism \(f\colon \mathbb{Z}^k\to\mathbb{R}\) satisfying the inequalities \(f(\alpha_i)\gt 0\) and \(f(\alpha_i)\neq f(\alpha_j)\) for all \(i\neq j\text{.}\) Since the inequalities all have integer coefficients, the existence of such a homomorphism is equivalent to the existence of a homomorphism \(f\colon \mathbb{Z}^k\to\mathbb{Z}\) with the same properties. We may always write such a homomorphism in the form \(f(\cdot)=\braket{w,\cdot}\) for some \(w\in\mathbb{Z}^k\text{.}\) To prove the correctness of the algorithm, we need to show that the linear programming problem (3.6)–(3.9) has a solution if and only if there exists \(w\in\mathbb{Z}^k\) such that
and that this solution has the smallest possible \(\max_i \braket{w,\alpha_i}\text{.}\) Furthermore we claim, that if a suitable \(w\in\mathbb{Z}^k\) exists, then there also exists one with
where \(C\) is the constant defined in step 2. We prove this claim later.
The smallest maximal weight property is equivalent to (3.6) and the first half of (3.7), since a solution will necessarily satisfy \(z = \max_i \braket{w,\alpha_i}\text{.}\) The latter half of (3.7) is exactly the condition (3.10). The inequalities (3.11) and (3.12) are encoded in the inequalities (3.8) and (3.9) using the auxiliary binary variables \(b_{ij}\text{.}\) Indeed, if we have \(b_{ij}=0\text{,}\) then the inequalities reduce to
and if \(b_{ij}=1\) then the inequalities reduce to
Therefore it remains to prove the claim about the additional condition (3.12).
First we show that disregarding the other constraints, the system (3.10) has a solution if and only if there exists a solution with \(\abs{\braket{w,\alpha_i}} \leq 1+N2^N\text{.}\) The normal form of the system (3.10) is given by switching to the variables \(x_i = \braket{w,\alpha_i}-1\text{,}\) resulting in the system
where the matrix \(A\) is the matrix whose rows are \(e_i+e_j-e_k\in\mathbb{Z}^N\) for each linear relation \(\alpha_i+\alpha_j=\alpha_k\) (dropping linearly dependent conditions) and the right-hand-side vector is \(d=(-1,\ldots,-1)\text{.}\)
The non-zero components of the basic feasible solutions of the normal form system (3.13) are determined by \(B^{-1}d\) where \(B\) is some invertible square submatrix of \(A\text{.}\) Writing
where \(\operatorname{Adj}(B)\) is the adjugate matrix of \(B\text{,}\) we see that integer solutions are determined by the vectors \(\operatorname{Adj}(B)d\text{.}\) Since each row of \(A\) has the norm bound \(\norm{e_i+e_j-e_k}_\infty\leq 2\) every minor of \(A\) is bounded by \(2^N\text{.}\) Hence we can bound the norms of integer basic feasible solutions to (3.13) by
Consequently the original problem (3.10) has a solution \(w\in\mathbb{Z}^k\) if and only if there exists a solution \(w\) with
Finally, if \(w\in\mathbb{Z}^k\) is as above, we claim that \(\tilde{w} = M^{k}w + (1,M,\ldots,M^{k-1})\) is a solution to (3.10)–(3.12).
To see that \(\tilde{w}\) satisfies (3.10)–(3.12), we consider base-\(M\) expansions of the integers \(\braket{\tilde{w},\alpha_i}\) and \(\braket{\tilde{w},\alpha_i-\alpha_j}\text{.}\) Since \(\norm{\alpha_i}_\infty\lt M\text{,}\) we have
A similar computation using \(\norm{\alpha_i-\alpha_j}_\infty\lt M\) gives the bound
showing (3.12) so it remains to verify (3.11). Expanding in terms of powers of \(M\text{,}\) we have
Since \(\abs{\braket{e_h,\alpha_i-\alpha_j}}\leq \norm{\alpha_i-\alpha_j}_\infty\lt M\) it follows that \(\braket{\tilde{w},\alpha_i-\alpha_j}\neq 0\) as soon as at least one \(\braket{e_h,\alpha_i-\alpha_j}\neq 0\text{.}\) Since \(\alpha_i\neq \alpha_j\text{,}\) this latter condition is always satisfied for some \(h\text{.}\)
Remark 3.9.
To obtain any positive realization, there is a much simpler polynomial time algorithm: Solve the linear programming problem
in the rational variables
as in the proof of correctness to guarantee distinct weights. Then the push-forward grading by
Subsection 3.4 Maximal gradings
In this section we provide an algorithm to construct a maximal grading for a Lie algebraAlgorithm 3.10. Maximal grading.
Input: A Lie algebra
- Compute the derivation algebra
Set - Compute a basis
for the centralizer - Repeat for each basis element
compute the Jordan decomposition If the semisimple part is not in the linear span of extend by and go back to step 2. - Determine the
-grading induced by the torus - Compute and return the universal realization of the grading
Remark 3.11.
If the Lie algebra
Lemma 3.12.
Let
Proof.
First we claim that if \(S_i\in\mathfrak{t}\) for all \(i=1,\ldots,n\text{,}\) then the centralizer \(C(\mathfrak{t})\) is a nilpotent Lie algebra. By Engel's theorem the centralizer is nilpotent if and only if each map \(\ad(A_i)\colon C(\mathfrak{t})\to C(\mathfrak{t})\) is nilpotent. By definition \(\mathfrak{t}\) is central in \(C(\mathfrak{t})\text{,}\) so we have
Since each \(N_i\in\der(\mathfrak{g})\) is nilpotent, so is \(\ad(N_i)\) and the claim follows.
Next, we claim that the Jordan decomposition of a sum of basis elements is
By assumption \(S_i,S_j\in\mathfrak{t}\text{,}\) so also \(S_i+S_j\in\mathfrak{t}\) and hence the sum \(S_i+S_j\) is semisimple. Moreover since \(\mathfrak{t}\) is central, \([S_i+S_j,N_i+N_j]=0\text{,}\) so all that remains is to show that \(N_i+N_j\) is nilpotent.
Since the centralizer \(C(\mathfrak{t})\) is nilpotent, it is also solvable. Since the field \(F\) is an algebraically closed field of characteristic zero, Lie's theorem implies that there exists a basis of \(\mathfrak{g}\) such that all the derivations \(A_i\) are represented by upper triangular matrices. Then \(N_i\) and \(N_j\) are both strictly upper triangular matrices, so also the sum \(N_i+N_j\) is strictly upper triangular, and hence nilpotent.
The Jordan decompositions (3.14) and the assumption that \(S_i\in\mathfrak{t}\) for all \(i=1,\ldots,n\) imply that the semisimple part of every linear combination of the elements \(A_i\) is also contained in \(\mathfrak{t}\text{.}\) Hence there cannot exist any semisimple elements in \(C(\mathfrak{t})\setminus\mathfrak{t}\text{.}\)
Remark 3.13.
The Jordan decompositions required in Step 3 of Algorithm 3.10 can be efficiently computed using the algorithm given in Appendix A.2 of [9].
Remark 3.14.
The relevance of the assumption that the field
When the Lie algebra
For the Lie algebra