A Lie algebra $\mathfrak{g}$ has a central extension $\mathfrak{𝔤}_{\mu}$ with central charge $\mu$. Is there a family of Lie algebras $\mathfrak{g}_{\alpha\mu}$, for which $\mathfrak{g}_{\alpha\mu} \cong \mathfrak{g}_{\alpha} \oplus \langle \mu\rangle$ for $\alpha \neq 0$ (where $\langle\mu\rangle$ is the one-dimensional Lie algebra generated by $\mu$) such that $\mathfrak{g}_{\alpha\mu} \to \mathfrak{g}_{μ}$ and $\mathfrak{g}_{\alpha} \to \mathfrak{g}$ as $\alpha \to 0$?


$\mathfrak{g}: (X, Y, Z : [Y,Z] = 0, [Z,X] = Z, [X,Y] = Y)$

$\mathfrak{g}_{\mu} : (X, Y, Z, \mu : [Y,Z] = \mu, [Z,X] = Z, [X,Y] = Y, [\mu,X]=[\mu,Y]=[\mu,Z] = 0)$

$\mathfrak{g}_{\alpha\mu} : (X, Y, Z, \mu : [Y,Z] = \mu + \alpha X, [Z,X] = Z, [X,Y] = Y, [\mu,X]=[\mu,Y]=[\mu,Z] = 0)$

$\mathfrak{g}_{\alpha} : (W, Y, Z: [Y,Z] = \alpha W, [Z,W] = Z, [W,Y] = Y)$

with the decomposition $\mathfrak{g}_{\alpha\mu} \cong \mathfrak{g}_{\alpha}\oplus \langle\mu\rangle$ for $\alpha \neq 0$ given by $W = X + \mu/\alpha$.


The relevant question you're asking is whether a central extension can be "trivialized" by a deformation. As you noted later in a comment, this is a wide-ranging process than (contrary to one of the replies posted) can also be done with Abelian Lie algebras.

As your subject line alluded to, it gets into deformation theory. The structure coefficients of a Lie algebra can be succinctly described as a Lie-valued order 2 differential form $f$ that satisfies the equation $[f,f] = 0$, where $[\_\ ,\ \_]$ is the Nijenhuis-Richardson bracket. A deformation $f+\Delta f$, by virtue of satisfying $[f + \Delta f, f + \Delta f] = 0$ then satisfies the identity $[f, \Delta f] + \frac{1}{2}[\Delta f, \Delta f] = 0$, which is the Maurer-Cartan equation, once you note that $[f, \Delta f]$ is just the Chevalley-Eilenberg differential (up to sign, which is $-$ for order 2 differential forms) on the complex of Lie-valued differential forms.

For a Lie algebra of dimension $n$, the order-1 differential forms act as $n\times n$ matrices on the complex. Each such differential form $g$, if invertible, produces a transform of $gf$ of $f$ that also satisfies the same equation: $[gf, gf] = 0$. The limit points of this are contractions.

I can't take it much further than that, but these observations may help to properly frame the question.

Worthy of note, by the way, is the example you posed in the follow-up comment. The Abelian Lie algebra given by $[X,Y] = 0$ has the Heisenberg algebra as its central extension, which is the foundation for Fourier analysis and spectrograms (where frequency is depicted on a linear scale).

The deformation you posed then possesses a trivialization involving Lie algebras that are all isomorphic to the one given by $[W,Y] = W$. That's the algebra associated with wavelet analysis and scalograms (where frequency is depicted on a logarithmic scale in octaves).

So, in effect, you're bridging between the two, the deformations producing a hybridization of the two.

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(not a full answer but too long for a comment)

Let me first address the problem for a fixed contraction. Since central extensions are equivalent to 2-cocycles valued in an abelian Lie algebra the question can be reformulated as follows:

  • given a Lie algebra contraction $\mathfrak g_\alpha\to \mathfrak g$ and an $\mathfrak a$-valued 2-cocycle $\Theta:\mathfrak g\times\mathfrak g\to \mathfrak a$ is it always possible to lift $\Theta$ to a $2$-cocycle of $\mathfrak g_\alpha$?

In this case, I would consider the answer to be no since in general The space of 2-cocycles of $\mathfrak g_\alpha$ can be smaller then the space of $2$-cocycles of the contracted algebra $\mathfrak g$ (but a ready example does not come to my mind).

More generally you are asking if for any given $\mathfrak a$-valued 2-cocycle on $\mathfrak g$ there exists at least one contraction $\mathfrak g_\alpha\to \mathfrak g$ to which the cocycle can be lifted. This sounds more likely to be possible.

  • $\begingroup$ The question rather acts whether there exists a "contraction" for which the cocycle can be lifted, and such that the lifted cocycle is a coboundary for $\alpha\neq 0$. (For instance taking $\mathfrak{g}_\alpha=\mathfrak{g}$ we always have a (trivial) contraction for which we have a trivial lift, but the requirement that the lifted cocycle is a coboundary for $\alpha\neq 0$ prevents this trivial choice unless the initial cocycle itself is a coboundary.) $\endgroup$ – YCor May 11 at 9:49
  • $\begingroup$ A ready example where the space of 2-cocycles of $\mathfrak{g}_\alpha$ is smaller than that of $\mathfrak{g}$ is simply when $\mathfrak{g}$ is abelian (and not $\mathfrak{g}_\alpha$). $\endgroup$ – YCor May 11 at 9:52
  • $\begingroup$ An Abelian example: 𝖌: [X,Y] = 0; 𝖌_μ: [X,Y] = μ, [X,μ] = 0 = [Y,μ]; 𝖌_α: [W,Y] = αW; 𝖌_{αμ}: [X,Y] = μ + αX, [X,μ] = 0 = [Y,μ]; where 𝖌_{αμ} ≌ 𝖌_α ⊕ <μ>; with W = X + μ/α if α ≠ 0. $\endgroup$ – Lydia Marie Williamson May 11 at 22:53

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