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$\DeclareMathOperator\C{\mathbf{C}}$Motivation: this post discusses a simple criterion for a 1-parameter family of $n$-dimensional complex algebras to be a "trivial deformation", i.e., be generally isomorphic.

Example: the 3-dimensional Lie algebras with law $[e_1,e_2]=te_3$ (other bracket zero up to antisymmetry) are all isomorphic, except for $t=0$. While the 3-dimensional Lie algebra with law $[e_1,e_2]=e_2$, $[e_1,e_3]=te_3$ are all non-isomorphic, except changing $t\leftrightarrow 1/t$. Here verifications are easy but in larger dimension (e.g., dimension 50) checking isomorphism between two (Lie) algebras given as laws can be computationally hard, while checking whether a 1-parameter family is a trivial deformation is doable.


Let $B=B(t)$ be an algebra law in dimension $n$, with coefficients in $\mathbf{C}(t)$. That is, $B(t)$ is an element of $\mathrm{Hom}_{\C(t)\text{-mod}}(V\otimes V,V)$, with $V=\C(t)^n$. By evaluation, can view it as a 1-parameter family of complex algebra laws (for $z$ ranging over $U$, the complement of the set of poles of the $n^3$ coefficients defining the law).

Since a constructible equivalence relation on a cofinite subset of $\C$ (for the Zariski topology) either has a cofinite equivalence class or has finite classes of bounded cardinal, we have the alternative:

  • (a) there exists a cofinite subset $U'$ of $U$ such that the complex algebras $(\C^n,B(z))$ are all isomorphic for $z\in U'$
  • (b) the relation $z\simeq z'$ on $U$: $(\C^n,B(z))\simeq (\C^n,B(z'))$ has finite classes of bounded orbits.

(a)/(b) can be called trivial/nontrivial deformation.

Say that another law $c(t)$ is a 2-coboundary for $B(t)$ if there exists a $\C(t)$-linear map $f:V\to V$ such that $c(t)=B(t)(fx,y)+B(t)(x,fy)-fB(t)(x,y)$ for all $x,y$.

Then the claim is

The deformation is trivial iff $B'(t)$ is a 2-coboundary for $B(t)$.

Here $B'(t)$ is just the coordinate-wise derivative: $B'(t)(x,y)=\partial_t (B(t)(x,y))$.

At a computational level such a criterion is practical since it consists in solving some linear algebra problem (if $K$ is a computable field including coefficients, it consists in checking whether some column matrix of size $n^3$ belongs to the image of some $n^3\times n^2$ matrix with entries in $K(t)$). For Lie algebras this is the coboundary operator from degree 1 to 2 in adjoint cohomology.

I think I can check it using some basic differential calculus (constant rank theorem); at least I've written it down (edit: see linked pdf). The arguments are so basic that I believe it's well-known. Also the above setting may seem artificially restrictive (one should consider more general structures, more general families).

Question what is a reference for such a claim, or what's a known result having this as particular case?

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  • $\begingroup$ Hi! I was trying to understand the proof and got stuck right at the end. In lemma 6 at the end you write "taking a limit point z, we deduce by a limit argument that B'(z) is a 2-coboundary". Could you please elaborate a little bit on this limit argument? $\endgroup$ Jan 25, 2021 at 20:29
  • $\begingroup$ @Serge You're right the argument doesn't seem to work, I need a differentiable local section. I updated the above link, changing the proof of Lemma 6. $\endgroup$
    – YCor
    Jan 25, 2021 at 22:03
  • $\begingroup$ Thanks! I think there is still a little error though: M* switches from right to left in the proof of Lemma 6. I also rewrote the proof while reading it, maybe you will find this useful: drive.google.com/file/d/1S5X4-_NqFJYH8cgE9583M6tqun3HmGCz/… $\endgroup$ Jan 26, 2021 at 14:09

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