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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeSep 2nd 2024

    have added more references to classical monographs

    diff, v16, current

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeSep 10th 2024
    • (edited Sep 10th 2024)

    Here is a question:

    Given a Lie algebra 𝔤(T i) i=1 n\mathfrak{g} \simeq \big\langle (T_i)_{i=1}^n \big\rangle with bracket [T j,T k]=f jk iT i[T_j, T_k] = f^i_{j k} T_i, then alongside the BCH formula one can expand the Maurer-Cartan forms (left-invariant (e i) i=1 n(e^i)_{i = 1}^n with de i=12f jk ie je k\mathrm{d}e^i = -\tfrac{1}{2}f^i_{j k} \, e^j e^k) of the corresponding Lie group in these coordinates as

    e i=dx i12f jk ix jdx k+16f jk if kl kx jx kdx l+. e^i \;=\; \mathrm{d}x^i - \tfrac{1}{2}f^i_{j k} x^j \, \mathrm{d}x^k + \tfrac{1}{6} f^i_{j k'} f^{k'}_{k l} x^j x^k \, \mathrm{d}x^l + \cdots \,.

    Is this BCH-like series-expansion for the MC-forms citable from any source, for the first few terms?

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeSep 10th 2024
    • (edited Sep 10th 2024)

    Urs 2: it is an easy exercise (I had to do it 18 years ago) that the inverse of the MC matrix is the expression for the say left invariant vector field in terms of the same coordinate system on a neighborhood of unit element on a Lie group; in exponential coordinates – for example – one gets the linear part of the BCH formula, as it is clarified in our paper with Durov.

    • Nikolai Durov, S. Meljanac, A. Samsarov, Z. Škoda, A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra, Journal of Algebra 309:1, 318-359 (2007) math.RT/0604096, MPIM2006-62 doi published pdf

    Chapter 7 has a conceptual intro, chap 2-6 are brute force calculational approach.

    • CommentRowNumber4.
    • CommentAuthorzskoda
    • CommentTimeSep 10th 2024

    Here is the reasoning for the explicit formula in exponential coordinates, but a fresher version not mentioning MC is chapter 2 in LMP paper on Hopf algebroids,

    • S. Meljanac, Z. Škoda, M. Stojić, Lie algebra type noncommutative phase spaces are Hopf algebroids, Lett. Math. Phys. 107:3, 475–503 (2017) doi arXiv:1409.8188

    Given any basis e 1,,e ne_1,\ldots,e_n of the tangent Lie algebra T eGT_e G, extend it to a basis e 1 *,,e n *e_1^*, \ldots,e_n^* of the space of left invariant vector fields on GG, and consider the dual differential forms ω 1,,ω n\omega^1,\ldots,\omega^n, that is ω i(e j)=δ j i\omega^i(e_j) = \delta^i_j where δ j i\delta^i_j is the usual Kronecker delta times the unit constant function on group GG. Forms ω i\omega^i are sometimes called Maurer-Cartan forms, but more often nowdays the left-invariant 𝔤\mathfrak{g}-valued form i=1 nω ie i\sum_{i=1}^n \omega^i e_i, comprising all ω i\omega^i-s, bears the same name.

    exp:𝔤G\exp : \mathfrak{g} \to G is a C C^\infty-map, and it is a diffeomorphism when restricted to an open star-shaped neighborhood UU of 0𝔤0\in \mathfrak{g}. In particular, we may consider the pullbakcs exp *ω i\exp^* \omega^i and hence define functions A j i:X(exp *ω i) X( j)A^i_j : X \mapsto (\exp^* \omega^i)_X (\partial'_j) where X= kx ke k𝔤X = \sum_k x^k e_k \in \mathfrak{g} and j\partial'_j corresponds to e je_j under the natural isomorpshism between T eGT_e G and in T XT eGT_X T_e G. Denote also j:=d(exp X( j))T expX\partial_j := d (\exp_X(\partial'_j)) \in T_{\exp{X}}. Helgason shows~(Helgason, II 7, page 139), that for all 1jn1\leq j\leq n,

    1e adXadX(e j)= i=1 nA j i(X)e i, \frac{1 - e^{-\ad{X}}}{\ad{X}} (e_j) = \sum_{i=1}^n A^i_j(X)\, e_i,

    where adX(X j)= k( ix ic ij k)e k \ad{X}(X_j) = \sum_k (\sum_i x^i c_{ij}^k) e_k and the structure constants c ij kc^k_{ij} in symbolic are defined by [e i,e j]=c ij ke k[e_i, e_j] = c^k_{ij} e_k. Thus adX N(e j)=(m N) j ke k\ad{X}^N (e_j) = (m^N)^k_j e_k, where m j k:= ix ic ij km^k_j := \sum_i x_i c^k_{ij} are the components of some matrix mm, and m Nm^N is its NN-th matrix power.

    Let A 1A^{-1} be the inverse matrix of AA, that is, kA j i(A 1) k j=δ k i\sum_k A^i_j (A^{-1})^j_k = \delta^i_k etc. Let e l *= jχ l j je_l^* = \sum_j \chi_l^j \partial_j. Then (ω i) expX( jχ l j j)=δ j i(\omega^i)_{\exp{X}} ( \sum_j \chi_l^j \partial_j) = \delta^i_j, hence jχ l jω i( j)= jχ l jA j i=δ l i\sum_j \chi_l^j \omega^i (\partial_j) = \sum_j \chi_l^j A^i_j = \delta^i_l i.e. χ l r=(A 1) l r\chi^r_l = (A^{-1})^r_l. This yields our formula.

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeSep 10th 2024
    • (edited Sep 10th 2024)

    In general coordinates, one takes the MC matrix and calculates the equation for the inverse and then one compares it to the differential equation for function ϕ\phi in our paper with Durov (the equation is in the chapter 4). The equation was btw in the very first letter of Lie around 1871 in which he actually proposed the idea of Lie algebra by looking at the differential equations for infinitesimal action up to second order. In the modern terms, the differential equation and the letter have been explained in short in the historical appendix to Lie groups book of Bourbaki written by Dieudonne.

    • CommentRowNumber6.
    • CommentAuthorzskoda
    • CommentTimeSep 10th 2024

    Maybe you could look at Helgason and see that your formula is there implicitly.

    II 7, page 139 in

    • Sigurdur Helgason, Differential geometry, Lie groups and symmetric spaces, Am. Math. Soc. (2001). Acad. Press (1978)
    • CommentRowNumber7.
    • CommentAuthorzskoda
    • CommentTimeSep 10th 2024
    • (edited Sep 10th 2024)

    Take into account the inverse matrix (and make sure if you work with the same convention left vs right invariant diff forms) to have the agreement in signs and coefficients.

    BCH-like series-expansion for the MC-forms

    It is from only the linear terms of BCH in fact, that is those which are linear in one of the two variables. Or, in differential geometric terms, it is from the formula for the differential of the exponential map (hence similarity to the Todd genus etc. in our case).

    • CommentRowNumber8.
    • CommentAuthorzskoda
    • CommentTimeSep 10th 2024
    • (edited Sep 10th 2024)

    If one takes the Maurer-Cartan equation in ANY coordinates (not the normal coordinates induced by the exponential map) and uses the formula for the differential of the inverse matrix, and if the ϕ j i\phi^i_j are the entries of the inverse matrix you get the differential equation for the components of the vector field. That is if the left invariant vector fields are

    X j=ϕ j i(x) i X_j = \sum \phi^i_j(x) \partial_i

    then (A 1) j i(A^{-1})^i_j satisfies the equation for ϕ j i\phi^i_j iff AA satisfies MC. This is the exercise I was talking about.

    The equation for ϕ j i\phi^i_j is

    (δ ρϕ μ γ)ϕ ν ρ(δ ρϕ ν γ)ϕ μ ρ=C μν σϕ σ γ (\delta_\rho \phi^\gamma_\mu)\phi^\rho_\nu - (\delta_\rho \phi^\gamma_\nu)\phi^\rho_\mu = C^\sigma_{\mu\nu} \phi^\gamma_\sigma

    where CC are the structure constants and δ ρ\delta_\rho is the ρ\rho-th partial derivative.

    Our approach in the paper with Durov is having diff. operators acting from the right and then using the antiautomorphism of the Weyl algebra so some signs and orders are opposite to the commutative viewpoint. This is in order to look at X iX_i as coordinates in the noncommutative deformation.

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeSep 11th 2024
    • (edited Sep 11th 2024)

    The answer is: Helgason (2001), II Thm. 7.4.

    Thanks!

    Concretely:

    That theorem gives the MC forms at X=x iT iX = x^i T_i as

    e i=dx i(1exp(adX)adX( k))dx k e^i \;=\; \mathrm{d}x^i \left( \frac {1 - exp(- ad X)} {ad X} (\partial_k) \right) \mathrm{d}x^k

    where (by ibid, p. 36)

    1exp(A)A n=0 1(n+1)!(A) n, \frac{1 - exp(-A)}{ A } \;\; \coloneqq \;\; \sum_{n=0}^\infty \tfrac{1}{(n+1)!} (-A)^n \,,

    so that:

    e i=dx i( n=0 1(n+1)!(adX) n( k))dx k. e^i \;=\; \mathrm{d}x^i \left( \textstyle{ \sum_{n = 0}^\infty } \tfrac{1}{(n+1)!} ( - ad X )^n (\partial_k) \right) \mathrm{d}x^k \,.

    Unwinding this with

    (adX)=(x jf j ) (ad X) \;=\; \big( x^j f^\bullet_{j \bullet} \big)

    we get my little formula from #2

    e i=11!dx i12!f jk ix jdx k+13!f jk if kl kx jx kdx l+ e^i \;=\; \tfrac{1}{1!} \mathrm{d}x^i - \tfrac{1}{2!} f^i_{jk}x^j \mathrm{d}x^k + \tfrac{1}{3!} f^i_{j k'} f^{k'}_{k l} x^j x^k \mathrm{d}x^l + \cdots

    Will be adding this now to Maurer-Cartan form.