A rigid Leibniz algebra with non-trivial \(HL^2\) (Q2178529)

The paper under review presents a finite dimensional Leibniz algebra that is geometrically rigid, but not algebraically rigid, and hence represents a non-reduced point in the variety of Leibniz algebra laws. Let $k$ be a characteristic zero, algebraically closed field, such as the complex numbers. Let $V$ be a finite dimensional vector space over $k$. A left Leibniz algebra structure on $V$ is given by an element $\mu \in \Hom_k ( V^{\otimes 2}, V)$ that is a derivation from the left, meaning that \[ \mu (x, \mu (y, z)) = \mu (\mu (x, y), z) + \mu( y, \mu (x, z)) \] for all $x$, $y$, $z \in V$. Often $\mu (x, y)$ is written $[x, y]$, although the bracket in a Leibniz algebra is not necessarily skew-symmetric. All such elements $\mu \in \Hom_k ( V^{\otimes 2}, V)$ form an algebraic variety, $\mathcal{M}$. Now, $\mathrm{GL}(V)$ acts on $\mathcal{M}$ by \[ (g \cdot \mu)(x, y) = g \mu (g^{-1}x, g^{-1} y), \] where $g \in \mathrm{GL}(V)$ and $\mu \in \mathcal{M}$. The quotient ${\mathcal{M}}/\mathrm{GL}(V)$ is the space of Leibniz algebra laws, and $\mu$ is geometrically rigid if its $\mathrm{GL}V)$ orbit is open in the Zariski topology. A Leibniz algebra $\mathfrak{h}$ with structure map $\mu$ is called algebraically rigid if $HL^2( \mathfrak{h}; \mathfrak{h}) = 0$, where $HL^* (\mathfrak{h}; \mathfrak{h})$ denotes Leibniz cohomology with adjoint coefficients. Recall [\textit{D. Balavoine}, Commun. Algebra 24, No. 3, 1017--1034 (1996; Zbl 0855.17021)] that a point $\mathfrak{h}$ in the space of Leibniz algebra laws is reduced and geometrically rigid if and ony if $HL^2 ( \mathfrak{h}; \mathfrak{h}) = 0$. To produce the non-reduced point in the Leibniz algebra case, the authors use ideas from [\textit{R. W. Richardson jun.}, Pac. J. Math. 22, 339--344 (1967; Zbl 0166.30301)], where examples of non-reduced points $\mathfrak{g}$ in the variety of Lie algebra laws that are geometrically rigid but satisfy $H^2_{\mathrm{Lie}}( \mathfrak{g}; \mathfrak{g}) \neq 0$ are given. Here $H^*_{\mathrm{Lie}}( \mathfrak{g}; \mathfrak{g})$ denotes Lie-algebra cohomology with adjoint coefficients. For a Lie algebra $\mathfrak{g}$ and a $\mathfrak{g}$-module $M$, the semidirect product $M \rtimes \mathfrak{g}$ is the Lie algebra with underlying vector space $M \oplus \mathfrak{g}$ and Lie bracket given by \[ [(m_1, x_1), (m_2, x_2)] = (x_1 \cdot m_2 - x_2 \cdot m_1, [x_1, x_2]), \] where $x_1$, $x_2 \in \mathfrak{g}$, $m_1$, $m_2 \in M$. The hemi-semidirect product $M \overset{\bullet}{+} \mathfrak{g}$ is the Leibniz algebra with underlying vector space $M \oplus \mathfrak{g}$ and Leibniz bracket given by \[ [(m_1, x_1), (m_2, x_2)] = (x_1 \cdot m_2, [x_1, x_2]). \] For the Lie algebra ${\mathfrak{sl}}_2 (\mathbb{C})$, let $M_k$ denote the irreducible module of highest weight $k$, $k \geq 0$. Recall that $M_k$ has dimension $k + 1$. Let $\widehat{\mathfrak{g}} = M_k \rtimes {\mathfrak{sl}}_2 (\mathbb{C})$. Richardson [loc. cit.] shows that for $k > 10$ an odd integer, then $\widehat{\mathfrak{g}}$ is a non-reduced point in the space of Lie algebra laws, i.e., $\widehat{\mathfrak{g}}$ is geometrically rigid, but $H^2_{\mathrm{Lie}}( \widehat{\mathfrak{g}}; \widehat{\mathfrak{g}}) \neq 0$. For the Leibniz case consider another standard irreducible $\mathfrak{sl}_2 (\mathbb{C})$ module $I_{\ell}$ of highest weight $\ell$, $\ell$ odd, $\ell > 2$. Let $\widehat{\mathfrak{h}}$ be the hemi-semidirect product \[ \widehat{\mathfrak{h}} := I_{\ell} \overset{\bullet}{+} \widehat{\mathfrak{g}} = I_{\ell} \overset{\bullet}{+} \big( M_k \rtimes {\mathfrak{sl}}_2 (\mathbf{C}) \big). \] Proven is that $\widehat{\mathfrak{h}}$ is geometrically rigid and $HL^2 ( \widehat{\mathfrak{h}}; \widehat{\mathfrak{h}}) \neq 0$. A key step in the proof is to show that the map on cohomology \[ H^2_{\mathrm{Lie}}( \widehat{\mathfrak{g}}; \widehat{\mathfrak{g}}) \longrightarrow HL^2 ( \widehat{\mathfrak{h}}; \widehat{\mathfrak{h}}) \] is injective in this case.