Proof. The first statement is merely a consequence of the integrability of \(J\). Since \(\mathfrak{g}\) is 2-step solvable, \(\mathfrak{g}_J'\) is clearly abelian. In order to see \(\mathfrak{g}_J'\) is an ideal of \(\mathfrak{g}'+J\mathfrak{g}'\), we observe that, since \(\mathfrak{g}\) is \(2\)-step solvable, \([\mathfrak{g}_J', \mathfrak{g}']=0\). Hence, it is sufficient to prove \([\mathfrak{g}_J', J \mathfrak{g}']\subseteq \mathfrak{g}_J'\). On the other hand, using again \(2\)-step solvability and integrability of \(J\), for any \(Y\in \mathfrak{g}'\) and \(X\in\mathfrak{g}_J'\) we have \([X, JY]=-J[JX, JY].\) This proves the remaining claims. ◻

Proof. The assertion (2) is clear because the derived series of \(\mathfrak{g}\) is contained in the \(J\)-adapted one. Finally, (1) follows from the results in [47], according to which a nilpotent Lie algebra is never \(J\)-perfect implying that the \(J\)-adapted derived series must terminate with \(0\). ◻

Proof. Denoting with \(\pi\colon \mathfrak{g} \to \mathfrak{g}/\mathfrak{k}\) the canonical projection, we have that \(\pi(\mathfrak{g}^{[j]}(J))=(\mathfrak{g}/\mathfrak{k} )^{[j]}(J)\) vanishes for some \(j\in \mathbb{N}\). Therefore, \(\mathfrak{g}^{[j]}(J)\subseteq \ker\pi=\mathfrak{k}\) and thus \(\mathfrak{g}^{[l+j]}(J)=(\mathfrak{g}^{[j]}(J))^{[l]}(J)\subseteq \mathfrak{k} ^{[l]}(J)=0\), for some \(l\in \mathbb{N}\). Finally, if \(\mathfrak{k}_1,\mathfrak{k}_2\) are \(J\)-solvable ideals it is obvious that \(\mathfrak{k}_1 \cap \mathfrak{k}_2\) is \(J\)-solvable. From the isomorphism \((\mathfrak{k}_1+\mathfrak{k}_2)/\mathfrak{k}_1\simeq \mathfrak{k}_2/(\mathfrak{k} _1\cap \mathfrak{k}_2)\) we note that \((\mathfrak{k}_1+\mathfrak{k}_2)/\mathfrak{k}_1\) is solvable and thus also \(\mathfrak{k}_1+\mathfrak{k}_2\) is. ◻

Proof. To prove (1), we note that if \(J\) is simple it is obviously semisimple. Moreover, \(\mathfrak{g}'+J\mathfrak{g}'\) is a \(J\)-invariant ideal of \(\mathfrak{g}\), hence we either have \(\mathfrak{g}'+J\mathfrak{g}'=0\) or \(\mathfrak{g}'+J\mathfrak{g}'=\mathfrak{g}\). In the first case we would have that \(\mathfrak{g}\) is abelian, which is not possible, deducing that \(\mathfrak{g}\) must be \(J\)-perfect.

As for (2), thanks to (1), and Lemma 4 we know that \(\mathfrak{g}_J'\) is an ideal of \(\mathfrak{g}\). Now, simplicity forces either \(\mathfrak{g}_J'=0\) or \(\mathfrak{g}_J'=\mathfrak{g}\). However, the latter cannot happen since \(\mathfrak{g}\) is solvable. Therefore, \(\mathfrak{g}_J'=0\) implying that \(J\) is abelian and \(\mathfrak{g}=\mathfrak{g}'\rtimes J \mathfrak{g}'\).

Finally, we prove (3). To any \(J\)-solvable ideal \(\mathfrak{k}\) of the quotient \(\mathfrak{g}/ {\rm Rad}(\mathfrak{g})\) corresponds a \(J\)-solvable ideal of \(\mathfrak{g}\). Therefore, the latter ideal must be contained in \({\rm Rad}(\mathfrak{g})\), implying that \(\mathfrak{k}\) is trivial in the quotient. ◻

Proof. Let \(\alpha=d\gamma\) be a non-zero and semi-positive \((1,1)\)-form, and assume by contradiction that \(\mathfrak{g}\) admits a compatible balanced metric, with fundamental form \(\omega\). Then, if \(\dim(\mathfrak{g})=2n\), \[0=d\omega^{n-1}\wedge \gamma =d(\omega^{n-1}\wedge \gamma)+\omega^{n-1}\wedge d\gamma=\omega^{n-1}\wedge \alpha\,,\] where we used that every \((2n-1)\)-form is closed, thanks to unimodularity. This would imply that \(\alpha=0\), against our assumption, hence \(\mathfrak{g}\) cannot carry balanced metrics. ◻

Proof. We have \[{\rm tr}(J{\rm ad}_{[X, JX]})= -{\rm tr}([J, {\rm ad}_{JX}]{\rm ad}_{X})={\rm tr}([J, {\rm ad}_{JX}]J[J,{\rm ad}_{X}])+ {\rm tr}([J, {\rm ad}_{JX}]J{\rm ad}_XJ)\,.\] On the other hand, \[{\rm tr}([J, {\rm ad}_{JX}]J{\rm ad}_XJ)={\rm tr}(-{\rm ad}_{JX}J{\rm ad}_{X}+ {\rm ad}_{JX}{\rm ad}_{X}J )={\rm tr}(J[{\rm ad}_{JX},{\rm ad}_{X}])=-\operatorname{tr}(J\operatorname{ad}_{[X,JX]})\,,\] which in particular, using 1 , gives \[\eta(X,JX)=-\frac{1}{4}{\rm tr}([J, {\rm ad}_{JX}]J[J,{\rm ad}_{X}])=-\frac{1}{4}{\rm tr}([J, {\rm ad}_{JX}]^2)\,,\] as claimed. ◻

Proof. By Proposition 20 we know that \(\rho=d \sigma\). Now, given a balanced metric \(\omega\), we have \[s^{{\rm Ch}}(\omega)\frac{\omega^n}{n!}=\rho\wedge \frac{\omega^{n-1}}{(n-1)!}=d\sigma\wedge \frac{\omega^{n-1}}{(n-1)!}=0\] using that \(d\omega^{n-1}=0\) and unimodularity. ◻

Proof. By simply using Jacobi identity and abelianity of \(J\), we have that \[\operatorname{ad}_{JX}\operatorname{ad}_{JY}=-\operatorname{ad}_{JX}\operatorname{ad}_YJ=-\operatorname{ad}_{[JX, Y]}J-\operatorname{ad}_Y\operatorname{ad}_{JX}J=\operatorname{ad}_{J[JX,Y]}-\operatorname{ad}_Y\operatorname{ad}_X\,,\] as claimed. ◻

Proof. Most of the proposition hinges on the identity ?? . Up to complexification, we can use Lie’s Theorem and assume that \(\operatorname{ad}_Z\) is upper triangular for all \(Z\in\mathfrak{g}\), after choosing a suitable basis of \(\mathfrak{g}\). Now, take \(X \in\mathfrak{n}_J\). Then \(\operatorname{ad}_X\) and \(\operatorname{ad}_{JX}\) are strictly upper triangular. For any \(Y\in \mathfrak{g}\), this forces \(\operatorname{ad}_Y\operatorname{ad}_X\), \(\operatorname{ad}_{JX}\operatorname{ad}_{JY}\), and hence also \(\operatorname{ad}_{J[JX, Y]}\), to be strictly upper triangular, and thus nilpotent. Thanks to this \(J[JX, Y]\in \mathfrak{n}\) and consequently \([JX, Y]\in\mathfrak{n}_J\), proving that \(\mathfrak{n}_J\) is an ideal. The last part of (1) is straightforward.

As for (2), we use ?? again. Indeed, for any \(X,Y\in \mathfrak{g}\), \[B^{1,1}(JX , Y)=\frac{1}{2}{\rm tr}(\operatorname{ad}_{JX}\operatorname{ad}_Y-\operatorname{ad}_{X}\operatorname{ad}_{JY})=-\frac{1}{2}{\rm tr}(\operatorname{ad}_{J[X, Y]})=\frac{1}{2}{\rm tr}(J\operatorname{ad}_{[X, Y]})=-\eta(X,Y)=-d\beta(X,Y)\,,\] where \(\beta\) is the form defined in 3 , which proves the claim.

To show (3), we exploit the closure of \(\eta\). First of all, we notice that \(B^{1,1}(\mathfrak{n}, J \mathfrak{n})=0\), since \(\mathfrak{n} \subseteq \ker B\). Now, let \(X, Y\in\mathfrak{g}\) and \(Z\in\mathfrak{n}\), we have \[0=d\eta(JX,JY, Z)=B^{1,1}(J[JX, JY], Z)-B^{1,1}(J[JX, Z], JY)+B^{1,1}(J[JY, Z], JX)\,.\] Using that \([JX, JY]\in\mathfrak{n}\), we get \[0=-B^{1,1}([JX, Z], Y)+B^{1,1}([JY, Z], X)=B^{1,1}([X, JZ], Y)-B^{1,1}([Y, JZ], X)\,,\] which implies \[B^{1,1}([JZ, X], Y)=B^{1,1}([JZ, Y], X)\,,\] as claimed. On the other hand, \[B^{1,1}([Z, X], Y)=B^{1,1}([JZ, JX], Y)=B^{1,1}([JZ, Y], JX)=B^{1,1}(J[Z, JY], X)\,,\] concluding the proof of (3).

As for (4), we observe that \(\ker B\cap J \ker B\subseteq \ker \eta\,.\) Viceversa, fix \(X\in \ker \eta\) and \(Y\in \mathfrak{n}\). By expanding the relation \(\eta(X, Y)=0\) in terms of \(B\), using part (2), we obtain \(B(X, JY)=0.\) Together with \(\mathfrak{n} \subseteq \ker B\), this implies that \(X\in \ker B\). On the other hand, expanding \(\eta(X, JY)=0\), we get \(B(JX, JY)=0\,,\) which yields \(JX\in \ker B\). Thus \(X \in \ker B \cap J \ker B\), giving \(\ker B \cap J \ker B =\ker \eta\). Finally, to show that \(\ker \eta = \mathfrak{n}_J\), let \(X\in \ker \eta\) and write \(X=Z+JY\), where \(Y, Z\in \mathfrak{n}\). It follows that \(JY\in \ker B\). Thus, we have \[{\rm tr}(\operatorname{ad}_{JY}^2)=0\,.\] We shall prove that \({\rm tr}(\operatorname{ad}_{JY}^{k+1})=0\) for all \(k\ge 2\). Using ?? , we infer \[\begin{align} \operatorname{ad}_{JY}^{k+1}=\operatorname{ad}_{JY}^{k-1}\operatorname{ad}_{J[JY, Y]}-\operatorname{ad}_{JY}^{k-1}\operatorname{ad}_Y^2\,. \end{align}\] Now, \(\operatorname{ad}_Y\) is nilpotent. Hence, as above, \(\operatorname{ad}_{JY}^{k-1}\operatorname{ad}_Y^2\) is nilpotent and therefore \[{\rm tr}(\operatorname{ad}_{JY}^{k+1})={\rm tr}(\operatorname{ad}_{JY}^{k-1}\operatorname{ad}_{J[JY, Y]})\,.\] Applying ?? once again, we obtain \[\operatorname{ad}_{JY}^{k-1}\operatorname{ad}_{J[JY, Y]}=\operatorname{ad}_{JY}^{k-2}\operatorname{ad}_{J[JY,[JY, Y]]}-\operatorname{ad}_{JY}^{k-2}\operatorname{ad}_{[JY, Y]}\operatorname{ad}_Y\,,\] and thus \[{\rm tr}(\operatorname{ad}_{JY}^{k+1}) ={\rm tr}(\operatorname{ad}_{JY}^{k-2}\operatorname{ad}_{J[JY,[JY, Y]]})\,.\] Iterating this process, we eventually obtain \[{\rm tr}(\operatorname{ad}_{JY}^{k+1})={\rm tr}(\operatorname{ad}_{JY}\operatorname{ad}_{J[JY,[JY, \cdots [JY, Y]]\cdots]})=0\,,\] since \(JY\in \ker B\). This ultimately leads to the fact that \(\operatorname{ad}_{JY}\) is nilpotent, and hence \(\operatorname{ad}_{X}\) is nilpotent, proving that \(\ker \eta \subseteq \mathfrak{n}\). On the other hand, since clearly \(J \ker \eta =\ker \eta\), we deduce that \(\ker \eta \subseteq \mathfrak{n}_J\). Conversely, it is well-known that \(\mathfrak{n} \subseteq \ker B\) and so \(\mathfrak{n}_J\subseteq \ker B \cap J \ker B=\ker \eta\).

Finally, we prove (5). Let \(\mathfrak{k}\) be a \(J\)-invariant ideal of \(\mathfrak{g}\). Being \(B^{1,1}\) of type \((1,1)\) guarantees straightforwardly that \(\mathfrak{k} ^{\perp}\) is \(J\)-invariant. Let \(X\in \mathfrak{k}\), \(Y\in \mathfrak{k} ^{\perp}\) and \(Z\in \mathfrak{g}\) and write \(Z=Z'+JZ''\) with \(Z',Z''\in \mathfrak{n}\). Thus, applying part (3), we get \[\begin{align} B^{1,1}([Z,Y ], X)&=B^{1,1}([Z',Y ], X)+B^{1,1}([JZ'',Y], X)=B^{1,1}(J[Z',JX],Y)+B^{1,1}([JZ'', X],Y)=0\,, \end{align}\] since \(\mathfrak{k}\) is a \(J\)-invariant ideal. This concludes the first part of (5). The second part is a trivial consequence of the facts that \(B_{\mathfrak{k} }=B_{\mathfrak{g} }|_{\mathfrak{k}\times \mathfrak{k}}\) and that \(J\mathfrak{k} =\mathfrak{k}\). ◻

Proof. We start with (1). Let \(S\) be a subalgebra of \(A\) and consider \(\mathfrak{aff}(S)\subseteq \mathfrak{aff}(A)\). Since \(S\) is a subalgebra, we have that, for any \(X=(a, b), Y=(a', b')\in\mathfrak{aff}(S)\) with \(a,b,a',b'\in S\), \[[X, Y]=(0, ab'-a'b)\in \mathfrak{aff}(S)\,,\] which implies that \(\mathfrak{aff}(S)\) is a subalgebra of \(\mathfrak{aff}(A)\), which is clearly \(J\)-invariant. The statement for ideals follows from the same arguments.

As for (2), we fix a \(J\)-perfect subalgebra \(\mathfrak{s}\) of \(\mathfrak{aff}(A)\). We consider \(S:=p_2\mathfrak{s}'\subseteq A\), where \(p_2\colon \mathfrak{aff}(A)\to A\) is the projection onto the second factor. Now, fix \(a, b\in S\), then there exist \(X,Y\in \mathfrak{s}'\) such that \(X=(0,a)\) and \(Y=(0,b)\). Then, \[\mathfrak{s}'\ni[JY,X]=(0,ab)\,,\] which shows that \(ab\in S\). We conclude that \(S\) is a subalgebra of \(A\).

Now, we prove (3). Using (1), \(\mathfrak{aff}(S)\) and \(\mathfrak{aff}(I)\) are, respectively, a \(J\)-invariant subalgebra and a \(J\)-invariant ideal of \(\mathfrak{aff}(A)\). The claim then follows from the assumption that \(A=S\oplus I\).

To prove (4), we consider \[\mathfrak{aff}( \varphi):=\varphi \oplus \varphi\colon \mathfrak{aff}(A)\to \mathfrak{aff}(A')\,, \quad \mathfrak{aff}(\varphi)(a, b):=(\varphi(a), \varphi(b))\,.\] Using the definition of the bracket and the complex structure on affine Lie algebras, it is straightforward to check that \(\mathfrak{aff}(\varphi)\) is a holomorphic Lie algebra homomorphism. Now, \(\ker \mathfrak{aff}( \varphi)=\ker \varphi\oplus \ker \varphi= \mathfrak{aff}(\ker \varphi)\), as claimed.

Finally, by (4), the projection onto the quotient \(\pi\colon A\to A/I\) induces a holomorphic Lie algebra homomorphism \(\mathfrak{aff}( \pi)\colon \mathfrak{aff}(A)\to \mathfrak{aff}(A/I)\) such that \(\ker \mathfrak{aff}(\pi)=\mathfrak{aff}(\ker \pi)=\mathfrak{aff}(I)\). On the other hand, since \(I\) is an ideal, (1) implies that \(\mathfrak{aff}(I)\) is a \(J\)-invariant ideal of \(\mathfrak{aff}(A)\), hence we may consider the canonical projection onto the quotient \(\pi'\colon \mathfrak{aff}(A)\to \mathfrak{aff}(A)/\mathfrak{aff}(I)\). Since \(\ker \pi'=\ker\mathfrak{aff}( \pi)\) and \(\mathfrak{aff}( \pi)\) is surjective, we can find a holomorphic isomorphism \(f\colon \mathfrak{aff}(A)/\mathfrak{aff}(I)\to \mathfrak{aff}(A/I)\) such that \(\mathfrak{aff}( \pi)=f\pi'\), which proves the claim. ◻

Proof. Assume that (1) holds. By Proposition 24, part (1), we have \(\mathfrak{n}(\mathfrak{g})_J=0\). Set \(\mathfrak{k}=\mathfrak{n}(\mathfrak{g})\oplus J \mathfrak{n}(\mathfrak{g})\) and consider \(\mathfrak{k}^{\perp}\), where \((\cdot)^{\perp}\) denotes the orthogonal complement with respect to \(B^{1,1}\). Clearly, \(\mathfrak{k}^{\perp}\) is \(J\)-invariant. Applying Lemma 2, we obtain \(\mathfrak{k}=\mathfrak{n}(\mathfrak{k})\oplus J\mathfrak{n}(\mathfrak{k})\) and, Proposition 24, part (4), gives \(\mathfrak{k} \cap \mathfrak{k}^{\perp}=\ker B^{1,1}_{\mathfrak{k}}= \mathfrak{n}(\mathfrak{k})_J=0\). Now, since \(d\eta=0\) and \(\mathfrak{k}\) is an ideal, for any \(X, Y \in\mathfrak{k}^{\perp}\) and \(Z\in\mathfrak{k}\) we have \[\label{detauguale0} 0=d\eta(X, Y, Z)=B^{1,1}(J[X, Y], Z)-B^{1,1}(J[X, Z], Y)+B^{1,1}(J[Y, Z], X)=B^{1,1}(J[X, Y], Z)\,,\tag{4}\] which implies \([X, Y]\in\mathfrak{k}^{\perp}\cap \mathfrak{k} =0\). Hence \(\mathfrak{k}^{\perp}\) is abelian. Instead, choosing \(X\in\mathfrak{k}^{\perp}, Y, Z \in\mathfrak{n}(\mathfrak{g})\) and using Proposition 24, part (3), we deduce \[B^{1,1}([JZ, X], Y)=B^{1,1}([JZ, Y], X)=0\,, \quad B^{1,1}([Z, X], Y)=B^{1,1}(J[Z, JY], X)=0\,,\] since \([JZ, Y], J[Z, JY]\in\mathfrak{k}\). This allows us to infer that \([\mathfrak{k}^{\perp}, \mathfrak{k} ]\subseteq \mathfrak{k}^{\perp}\cap \mathfrak{k}=0.\) Using 4 again, but now with \(X\in\mathfrak{k} ^{\perp}\), \(Y\in\mathfrak{g}\) and \(Z\in\mathfrak{k}\), we conclude that \(\mathfrak{k}^{\perp}\) is a \(J\)-invariant abelian ideal of \(\mathfrak{g}\). By semisimplicity, this forces \(\mathfrak{k}^{\perp}=0\). In particular, \(\ker B^{1,1}\subseteq \mathfrak{k}^{\perp}=0\), which implies that \(B^{1,1}\) is non-degenerate. Since \(B^{1,1}_{\mathfrak{k}}\) is also non-degenerate we deduce that \(\mathfrak{g}=\mathfrak{n}(\mathfrak{g})\oplus J\mathfrak{n}(\mathfrak{g})\) and (2) follows.

Assume now that (2) holds. We adapt the proof of [53] to prove (3). If \(\mathfrak{g}\) has no proper \(J\)-invariant ideals, the claim is immediate. Let us assume that \(\mathfrak{g}\) has a proper \(J\)-invariant ideal, hence we can consider the minimal non-zero \(J\)-invariant ideal \(\mathfrak{k}\). By Lemma 2 and the minimality of \(\mathfrak{k}\), we obtain \(\mathfrak{k}=\mathfrak{n}(\mathfrak{k} )\oplus J \mathfrak{n}(\mathfrak{k} )\). By Proposition 24, part (5), we know that \(\mathfrak{k} ^{\perp}\) is a \(J\)-invariant ideal. Moreover, \(\mathfrak{k}\cap \mathfrak{k}^{\perp}=\ker B_{\mathfrak{k}}^{1,1}=\mathfrak{n}(\mathfrak{k})_J=0\), thanks to Proposition 24, part (4), and the fact that \(\mathfrak{n}(\mathfrak{g})_J=0\). In particular, \(\mathfrak{g} =\mathfrak{k} \oplus \mathfrak{k}^{\perp}\) as a Lie algebra. Now observe that, if \(\mathfrak{l} \subseteq \mathfrak{k}\) is a \(J\)-invariant ideal, then \([\mathfrak{l} , \mathfrak{k}^{\perp}]\subseteq \mathfrak{k}\cap \mathfrak{k} ^{\perp}=0\), so \(\mathfrak{l}\) is also a \(J\)-invariant ideal of \(\mathfrak{g}\). By the minimality of \(\mathfrak{k}\), it follows that \(\mathfrak{l}\) is either \(0\) of \(\mathfrak{k}\). Hence, \(\mathfrak{k}\) is \(J\)-simple. For the same reason, any \(J\)-invariant ideal of \(\mathfrak{k}^{\perp}\) is also a \(J\)-invariant ideal of \(\mathfrak{g}\). Therefore, \(\mathfrak{n}(\mathfrak{k}^{\perp})_J=0\) and \(\mathfrak{k}^{\perp}=\mathfrak{n}( \mathfrak{k}^{\perp})\oplus J\mathfrak{n}( \mathfrak{k}^{\perp})\). To see the last equality, fix \(X=N+JN'\in \mathfrak{k}^{\perp}\), for some \(N, N'\in\mathfrak{n}(\mathfrak{g})\) and \(Y\in \mathfrak{n}(\mathfrak{k} )\). Since \(B^{1,1}(\mathfrak{n}(\mathfrak{g}), J\mathfrak{n}(\mathfrak{g} ))=0\), we infer \[0=B^{1,1}(X, Y)=B^{1,1}(N, Y)+ B^{1,1}(JN', Y)=B^{1,1}(N, Y)\,.\] Thus, \(N\in (\mathfrak{n}(\mathfrak{k} )\oplus J\mathfrak{n}(\mathfrak{k}))^{\perp}=\mathfrak{k}^{\perp}\). From this, it follows that \(\mathfrak{k}^{\perp}\subseteq \mathfrak{n}(\mathfrak{k}^{\perp})\oplus J\mathfrak{n}(\mathfrak{k}^{\perp})\). Therefore, \(B^{1,1}_{\mathfrak{k} ^{\perp}}\) is non-degenerate, using Proposition 24 part (1) and (4). We can then repeat the procedure on \(\mathfrak{k}^\perp\) and conclude after finitely many steps.

Finally, assume (3), namely \[\mathfrak{g}=\bigoplus_{i=1}^{k}\mathfrak{k}_i\,,\] where \(\mathfrak{k}_i\) is a \(J\)-simple ideal of \(\mathfrak{g}\). Of course, for any \(i=1, \ldots, k\), the projection \(\pi_i\colon \mathfrak{g} \to \mathfrak{k}_i\) is a holomorphic Lie algebra homomorphism. Hence, if \(\mathfrak{k}\subseteq \mathfrak{g}\) is a \(J\)-invariant ideal of \(\mathfrak{g}\), then \(\pi_i(\mathfrak{k} )\) is a \(J\)-invariant ideal of \(\mathfrak{k}_i\). Therefore, \(\pi_i(\mathfrak{k} )\) is either \(0\) or \(\mathfrak{k}_i\). If \(\pi_i(\mathfrak{k})=\mathfrak{k}_i\) then \(\mathfrak{k}_i\subseteq \mathfrak{k} .\) Indeed, \[\mathfrak{k}_i'=[\mathfrak{k}_i, \mathfrak{k}_i]_{\mathfrak{k}_i}=[\mathfrak{k}_i, \pi_i(\mathfrak{k} )]_{\mathfrak{k}_i}=[\mathfrak{k}_i, \mathfrak{k} ]\subseteq \mathfrak{k}\,.\] Since \(\mathfrak{k}\) is \(J\)-invariant and \(\mathfrak{k}_i\) is \(J\)-simple, hence \(J\)-perfect, we also have \(\mathfrak{k}_i=\mathfrak{k}_i'\rtimes J\mathfrak{k}_i'\subseteq \mathfrak{k}\). Consequently, \[\mathfrak{k}=\bigoplus_{\mathfrak{k}_i\subseteq \mathfrak{k} }\mathfrak{k}_i\,.\] But now, it is clear that \(\mathfrak{k}\) cannot be \(J\)-solvable unless \(\mathfrak{k}=0\).

In addition, \[\mathfrak{g}'=\left[\mathfrak{g}, J\mathfrak{g} \right]=\bigoplus_{i=1}^k[\mathfrak{k}_i, J \mathfrak{k}_i]=\bigoplus_{i=1}^k \mathfrak{k}_i'\,.\] Therefore, \[\mathfrak{g}'\rtimes J \mathfrak{g}'=\bigoplus_{i=1}^k\mathfrak{k}_i'\rtimes J\mathfrak{k}_i'=\mathfrak{g} \,,\] showing that \(\mathfrak{g}\) is \(J\)-perfect. ◻

Proof. It is known that every \(D\in {\rm Der}(\mathfrak{g} )\) satisfies \(D\mathfrak{g}\subseteq \mathfrak{n}\). If \(D\in {\rm Der}(\mathfrak{g} , J):=\{D\in {\rm Der}(\mathfrak{g}) \,\, |\, \, DJ =JD\}\), then \(D\mathfrak{g} \subseteq \mathfrak{n} _J=0\), due to semisimplicity of \(\mathfrak{g}\). Hence, \(D=0\), meaning that \({\rm Der}(\mathfrak{g}, J)=\{0\}\). The claim now follows from the well-known fact that \({\rm Der}(\mathfrak{g}, J)={\rm Lie}({\rm Aut}(\mathfrak{g}, J))\). Finally, if \(X \in \mathfrak{g}\) is a holomorphic vector field, then \(\operatorname{ad}_X\in {\rm Der}(\mathfrak{g}, J)\), which implies \(X\in \mathfrak{z}(\mathfrak{g})\). On the other hand, the centre of \(\mathfrak{g}\) is a \(J\)-invariant abelian ideal, so \(X=0\). ◻

Proof. Let us begin observing that, since \(\mathfrak{g}\) is \(2\)-step solvable, the family \(\{\operatorname{ad}_{JV}|_{\mathfrak{g}'}\}_{V\in \mathfrak{g}'}\) consists of commuting operators. Therefore, we can find \(Z\in\mathfrak{g}'\otimes \mathbb{C}\) which is a common eigenvector of \(\{\operatorname{ad}_{JW}|_{\mathfrak{g}'\otimes \mathbb{C}}\}_{W\in\mathfrak{g}' \otimes \mathbb{C}}\), i.e., for any \(W\in \mathfrak{g}' \otimes \mathbb{C}\), \[\label{common} [JW,Z ]=\lambda(W)Z\,, \qquad \lambda\in(\mathfrak{g}'\otimes \mathbb{C})^*\,.\tag{5}\] Hence, in particular, we obtain \[\label{commonZ} [JZ, Z]=\lambda(Z)Z\,.\tag{6}\] Now, we have two possibilities: either \(Z\in\mathfrak{g}'\) or \(Z\) is genuinely complex. Let us assume that \(Z\in\mathfrak{g}'\). This allows us to infer that \[\mathfrak{k}_1 := \langle Z, JZ\rangle\] is a non-zero \(J\)-invariant ideal of \(\mathfrak{g}\). By hypothesis, \(\mathfrak{k}_1 = \mathfrak{g}.\) Moreover, \(\lambda(Z)\ne 0\) otherwise \(\mathfrak{g}\) would be abelian against simplicity of \(J\). Hence, up to renormalisation, we can suppose that \(\lambda (Z)=1\) which gives straightforwardly that \(\mathfrak{g} \simeq \mathfrak{aff}(\mathbb{R})\). It remains to understand the case in which \(Z\) is complex. Let us write \(Z=X+\sqrt{-1}Y\), with \(X, Y\in\mathfrak{g}'\). From 5 , it follows that, for any \(V\in \mathfrak{g}'\), \[\label{affc} [JV, X]=\lambda_1(V)X-\lambda_2(V)Y\,, \qquad [JV, Y]=\lambda_2(V)X+ \lambda_1(V)Y\,,\tag{7}\] where \(\lambda_1=\mathrm{Re}(\lambda)\) and \(\lambda_2=\operatorname{Im}(\lambda)\). This, in particular, guarantees that \[\mathfrak{k}_2:= \langle X, Y, JX, JY \rangle\] is a non-zero \(J\)-invariant ideal of \(\mathfrak{g}\) forcing \(\mathfrak{k}_2 =\mathfrak{g}\), by \(J\)-simplicity. Next, we show that \(\mathfrak{g}\simeq \mathfrak{aff}(\mathbb{C}).\) Applying 7 to \(X\) and \(Y\), we get \[\label{hiddeneqaffC} \begin{align} \relax [JY, Y]=&\,\lambda_2(Y)X+\lambda_1(Y)Y\,, \qquad [JY, X]=\lambda_1(Y)X- \lambda_2(Y)Y\\ [JX, Y]=&\, \lambda_2(X)X+\lambda_1(X)Y\,, \qquad [JX, X]=\lambda_1(X)X- \lambda_2(X)Y\,. \end{align}\tag{8}\]

Now, since \(J\) is abelian then \([JY, X]=[JX, Y]\), hence \(\lambda_2(X)=\lambda_1(Y)\) and \(\lambda_1(X)=-\lambda_2(Y)\), guaranteeing that \([JY, Y]=-[JX, X].\) Rewriting 6 using these relations shows that \(\lambda(Z)= 2\lambda(X)\). Therefore, \(\lambda(Z)\ne 0\), since otherwise \(\lambda_1(X)=\lambda_2(X)=0\) and \(\mathfrak{g}\) would be abelian, against the fact that \(J\) is simple. Then, up to renormalisation, we can assume \(\lambda(X)=1\). Thus, we can rewrite 8 as \[[JY, Y]=-X\,, \quad [JY, X]=Y\,, \quad [JX, Y]=Y\,, \quad [JX, X]=X\,,\] which yields the desired isomorphism. ◻

Proof. By Theorem 27 and Theorem 31 we immediately infer ?? .
To prove (1), we observe that for \(\mathfrak{aff}(\mathbb{R})\) and \(\mathfrak{aff}(\mathbb{C})\) we may choose \[\begin{align} \qquad \qquad \mathfrak{aff}(\mathbb{R})_+&= \mathfrak{aff}(\mathbb{R}) \,, &\mathfrak{aff}(\mathbb{R})_-&= 0 \,,\qquad \qquad \\ \mathfrak{aff}(\mathbb{C})_+&= \langle X,JX \rangle \,, & \mathfrak{aff}(\mathbb{C})_-&= \langle Y, JY \rangle \,, \end{align}\] where we retain the notations of Example 30. It is easy to check that these subspaces satisfy the desired Lie bracket relations. Hence, thanks to ?? , it suffices to choose \[\mathfrak{g}_+:=\mathfrak{aff}(\mathbb{R})^p\oplus \mathfrak{aff}(\mathbb{C})_+^q\,, \qquad \mathfrak{g}_-:=\mathfrak{aff}(\mathbb{C})_-^q\,.\] Furthermore, we define \[s|_{\mathfrak{g}_+}={\rm Id}_{\mathfrak{g}_+}\,, \quad s|_{\mathfrak{g}_-}=-{\rm Id}_{\mathfrak{g}_-}\,.\] Due to the bracket relations above and that \(J\mathfrak{g}_{\pm}=\mathfrak{g}_{\pm}\), we obtain \(s\in {\rm Aut}(\mathfrak{g}, J)\), and clearly \(s^2={\rm Id}\). Next, it is easy to see that \(B^{1,1}_s\) is of type \((1,1)\), since \([s, J]=0\). Moreover, using again that \([s, J]=0\) and \(s\in \mathrm O(B)\), we have, for any \(X, Y\in\mathfrak{g}\), \[\begin{align} B^{1,1}_s(X, Y)= \frac{1}{2}\left(B(X, sY)+ B(JX, JsY)\right) =\frac{1}{2}\left(B(Y, sX)+ B(JY, JsX)\right)=B^{1,1}_s(Y, X)\,, \end{align}\] giving us that \(B_s^{1,1}\) is symmetric. To see positive definiteness, we note that the decomposition of \(\mathfrak{g}\) in \(J\)-simple ideals is orthogonal with respect to \(B_s^{1,1}\). Hence, it is sufficient to prove the claim on each \(J\)-simple ideal. Since \(B_{\mathfrak{aff}(\mathbb{R})}^{1,1}>0\) and \(s_{\mathfrak{aff}(\mathbb{R})}={\rm Id}\), it is enough to prove the statement on \(\mathfrak{aff}(\mathbb{C})\), for which it is easy to see that \[\qquad \, B^{1,1}_{\mathfrak{aff}(\mathbb{C})}|_{\mathfrak{aff}(\mathbb{C})_+\times \mathfrak{aff}(\mathbb{C})_+}>0\,,\quad B^{1,1}_{\mathfrak{aff}(\mathbb{C})}|_{\mathfrak{aff}(\mathbb{C})_+\times \mathfrak{aff}(\mathbb{C})_-}=0\,, \quad B^{1,1}_{\mathfrak{aff}(\mathbb{C})}|_{\mathfrak{aff}(\mathbb{C})_-\times \mathfrak{aff}(\mathbb{C})_-}<0\,.\] Since \(s|_{\mathfrak{aff}(\mathbb{C})_-}=-{\rm Id}\), the claim follows straightforwardly.

As for (2), we have that, using Proposition 24, part (3), for any \(X\in \mathfrak{g}_{\pm}\cap \mathfrak{g}'\), \[\begin{align} B^{1,1}_s(\operatorname{ad}_{JX} Y, Z)=&\, B^{1,1}(Y, \operatorname{ad}_{JX}sZ) = B^{1,1}(Y, s\operatorname{ad}_{JsX}Z)=\pm B^{1,1}_s(Y, \operatorname{ad}_{JX}Z)\,, \end{align}\] concluding the proof. ◻

Proof. First of all, we define \(JY:=2\sum_{i=1}^{\dim_{\mathbb{C}}\mathfrak{g}_+}JX_i\) and, using the bracket relations of the vectors in the basis \(\mathcal{B}\), we infer that \(\operatorname{ad}_{JY}|_{\mathfrak{g}'}=2{\rm Id}_{\mathfrak{g}'}\). To prove the first claim, we observe that, for any \(X\in \mathfrak{g}'\), \(B_{s}^{1,1}(JY, X)=0\). Using the fact that \(JY\in\mathfrak{g}_+\) and ?? we deduce \[B_s^{1,1}(JY, JX)=\frac{1}{2} {\rm tr}(\operatorname{ad}_{J[JY, X]})={\rm tr}(\operatorname{ad}_{JX})\,,\] thereby proving the first three statements.

The last claim is proved by straightforward computations once we observe that \(\varphi s_{\mathfrak{g}_1 }\varphi^{-1}=s_{\mathfrak{g}_2 }\), which follows from \(\varphi((\mathfrak{g}_1) _{\pm})=(\mathfrak{g}_2)_{\pm}\). This property can be deduced, for instance, using the basis 9 and the fact that \(\varphi\) is a Lie algebra isomorphism. ◻

Proof. From the inclusion \(\mathfrak{k}_1\subseteq \mathfrak{k}_2\), we deduce that \({\rm Rad}(\mathfrak{k}_2)\cap\mathfrak{k}_1\) is a \(J\)-solvable ideal of \(\mathfrak{k}_1\). Hence, \({\rm Rad}(\mathfrak{k}_2)\cap \mathfrak{k}_1\subseteq {\rm Rad}(\mathfrak{k}_1)\). On the other hand, \[\frac{\mathfrak{k}_1}{{\rm Rad}(\mathfrak{k}_2)\cap \mathfrak{k}_1}\simeq\frac{\mathfrak{k}_1 + {\rm Rad}(\mathfrak{k}_2)}{{\rm Rad}(\mathfrak{k}_2)}\,,\] which is a \(J\)-invariant ideal of \(\mathfrak{k}_2/{\rm Rad}(\mathfrak{k}_2)\) and thus \(J\)-semisimple, due to the assumption \(\mathfrak{g}'_J \subseteq {\rm Rad(\mathfrak{k}_2)}\) and Theorem 27. Therefore, \[0={\rm Rad}\left(\frac{\mathfrak{k}_1}{{\rm Rad}(\mathfrak{k}_2)\cap \mathfrak{k}_1}\right)=\frac{{\rm Rad}(\mathfrak{k}_1)}{{\rm Rad}(\mathfrak{k}_2)\cap \mathfrak{k}_1},\] concluding that \({\rm Rad}(\mathfrak{k}_1 )={\rm Rad}(\mathfrak{k}_2)\cap \mathfrak{k}_1\). As for the second identity, note that the quotient \[\frac{\frac{\mathfrak{k}_2}{{\rm Rad}(\mathfrak{k}_2)}}{\frac{\mathfrak{k}_1 + {\rm Rad}(\mathfrak{k}_2)}{{\rm Rad}(\mathfrak{k}_2)}} \simeq \frac{\mathfrak{k}_2}{\mathfrak{k}_1+{\rm Rad}(\mathfrak{k}_2)}\]is \(J\)-semisimple and abelian, since \(\mathfrak{k}_2'+J\mathfrak{k}_2'\subseteq \mathfrak{k}_1\). This can only happen if \(\mathfrak{k}_1 + {\rm Rad}(\mathfrak{k}_2)=\mathfrak{k}_2\). ◻

Proof. Using [8], we know that \(\mathfrak{g}\simeq \mathfrak{aff}(A)\) for some finite-dimensional, associative and commutative \(\mathbb{R}\)-algebra \(A\). Using the Wedderburn’s Principal Theorem, see [55], we can write \[A=J\oplus S\, ,\] where \(J\) is the Jacobson radical and \(S\) is a semisimple subalgebra of \(A\) such that \(S\simeq A/J\), see [55]. On the other hand, using the finite dimensionality of \(A\) as an \(\mathbb{R}\)-algebra, we have that \(J=N\), where \(N\) is the nilradical of \(A\), i.e. the maximal ideal consisting of nilpotent elements. Therefore, \[A=N \oplus S\,.\] Once we have this, we can use Lemma 26, part (3), to infer \[\mathfrak{g}\simeq \mathfrak{aff}(A)=\mathfrak{aff}(N\oplus S)= \mathfrak{aff}(N)\ltimes \mathfrak{aff}(S)\,.\] Since \(S\) is semisimple, \(\mathfrak{aff}(S)\) is the direct sum of \(J\)-simple ideals by Lemma 26, part (3), and hence \(J\)-semisimple by Theorem 27. On the other hand, using [57], we have that \(\mathfrak{aff}(N)\) is a \(J\)-invariant nilpotent ideal, consequently \(\mathfrak{aff}(N)\subseteq \mathfrak{n}_J\). Since \(\mathfrak{aff}(S)\simeq \mathfrak{aff}(A/N)\) is \(J\)-semisimple, we conclude \(\mathfrak{aff}(N)=\mathfrak{n}_J\).

We now prove the uniqueness part of the statement. Let us assume that there exist two \(J\)-adapted Levi subalgebras \(\mathfrak{h}_1, \mathfrak{h}_2\) of \(\mathfrak{g}\). Using the fact that \(\mathfrak{h}_i\) is \(J\)-perfect and Lemma 26, part (2), we know that there exist a subalgebra \(S_i\subseteq A\) such that \(\mathfrak{h}_i=\mathfrak{aff}(S_i)\), for any \(i=1,2\). Thanks to Lemma 26, part (3), and since \(\mathfrak{aff}(N)=\mathfrak{n}_J\), we deduce that \(A=S_1\oplus N=S_2\oplus N\). Now, we can apply [55] to conclude that there exists \(n\in N\) such that \(S_1=(1-n)S_2(1-n)^{-1}\). We obtain \(S_1=S_2\) by commutativity of \(A\). The claim now follows. ◻

Proof. We prove the existence part of the theorem in a stepwise fashion by successively reducing to simpler cases.

Step 1. We show that we may assume that \(\mathfrak{g}\) has no \(J\)-invariant ideals properly contained in \({\rm Rad}(\mathfrak{g})\). To prove this we work inductively on \(\dim_\mathbb{C}(\mathfrak{g})\). If \(\mathfrak{g}\) is \(1\)-dimensional, it can only be either \(\mathfrak{aff}(\mathbb{R})\) or \(\mathbb{C}\) and in both cases the theorem is clearly true. Thus, let \(\mathfrak{k} \subset {\rm Rad}(\mathfrak{g})\) be a \(J\)-invariant ideal in \(\mathfrak{g}\). Since \(\mathrm{Rad}(\mathfrak{g}/\mathfrak{k})={\rm Rad}(\mathfrak{g})/\mathfrak{k}\), by the inductive assumption we have \(\mathfrak{g}/\mathfrak{k} = \mathfrak{s} \ltimes {\rm Rad}(\mathfrak{g})/\mathfrak{k}\), for some \(J\)-semisimple subalgebra \(\mathfrak{s}\). Consider now the canonical projection onto the quotient \(\pi \colon \mathfrak{g}\to \mathfrak{g}/\mathfrak{k}\), then \(\hat{\mathfrak{s}}:= \pi^{-1}(\mathfrak{s})\) is a proper \(J\)-invariant subalgebra of \(\mathfrak{g}\) and so, by the inductive assumption again, it can be written as \(\hat{\mathfrak{s}}= \mathfrak{h}\ltimes \mathrm{Rad}(\hat{\mathfrak{s}})\) for a \(J\)-semisimple subalgebra \(\mathfrak{h}\). Since \(\hat{\mathfrak{s}}/\mathfrak{k}=\pi(\pi^{-1}(\mathfrak{s}))= \mathfrak{s}\) is \(J\)-semisimple, it follows that \(\mathrm{Rad}(\hat{\mathfrak{s}})\subseteq \mathfrak{k}\), but \(\mathfrak{k}\subseteq {\rm Rad}(\mathfrak{g})\) is \(J\)-solvable and thus we must have \(\mathfrak{k}= \mathrm{Rad}(\hat{\mathfrak{s}})\). It then follows that \(\mathfrak{g}= \mathfrak{h}\ltimes {\rm Rad}(\mathfrak{g})\).

Step 2. We reduce here to the case where \(\mathfrak{g}=\mathfrak{g}'+J\mathfrak{g}'\). Using Lemma 39 with \(\mathfrak{k}_1=\mathfrak{g}'+J\mathfrak{g}'\) and \(\mathfrak{k}_2=\mathfrak{g}\), we have \(\mathfrak{g}=\mathfrak{g}'+J\mathfrak{g}'+{\rm Rad}(\mathfrak{g})\) and \({\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')={\rm Rad}(\mathfrak{g})\cap (\mathfrak{g}'+J\mathfrak{g}')\). The latter is a \(J\)-invariant ideal of \(\mathfrak{g}\) contained in \({\rm Rad}(\mathfrak{g})\). Using Step 1, we may assume either \({\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')=0\) or \({\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')={\rm Rad}(\mathfrak{g})\). In the first case, the Levi–Malcev decomposition follows from \(\mathfrak{g}=(\mathfrak{g}'+J\mathfrak{g}')\oplus{\rm Rad}(\mathfrak{g})\). In the second case we have \({\rm Rad}(\mathfrak{g})\subseteq \mathfrak{g}'+J\mathfrak{g}'\), and hence \(\mathfrak{g}=\mathfrak{g}'+J\mathfrak{g}'+{\rm Rad}(\mathfrak{g})\subseteq \mathfrak{g}'+J\mathfrak{g}'\), as claimed.

Step 3. We show that it may be assumed that \(\mathfrak{z}(\mathfrak{g})=0\). We suppose \(\mathfrak{z}(\mathfrak{g})\neq 0\) and provide the Levi–Malcev decomposition for this case. Since \(J\) is abelian, \(\mathfrak{z}(\mathfrak{g})\) is \(J\)-invariant and thus contained in the radical of \(J\). Therefore, again by Step 1, we can suppose \(\mathfrak{z}(\mathfrak{g})= {\rm Rad}(\mathfrak{g})\). In these assumptions, the adjoint representation \(\operatorname{ad}\colon \mathfrak{g}\to \mathrm{End}(\mathfrak{g})\) descends to a well-defined representation of the quotient \[\operatorname{ad}\colon \frac{\mathfrak{g}}{{\rm Rad}(\mathfrak{g})} \to \mathrm{End}(\mathfrak{g})\,.\] We consider the operator \[\Phi(U):=\frac{1}{2} \left(\operatorname{ad}_{JU}-J\operatorname{ad}_{U}\right) \in \mathrm{End}(\mathfrak{g})\,,\] where \(JU\) is the mean curvature vector of \(\mathfrak{g}/{\rm Rad}(\mathfrak{g})\). It is easy to prove that \(\Phi(U) \in \mathrm{End}(\mathfrak{g}, J)\), namely \([\Phi(U),J]=0\). Moreover, for every \(X\in \mathfrak{g}/{\rm Rad}(\mathfrak{g})\), we have \[\begin{align} \relax[\Phi(U), \operatorname{ad}_X]=&\, \frac{1}{2}(\operatorname{ad}_{JU}\operatorname{ad}_X-J\operatorname{ad}_U\operatorname{ad}_X-\operatorname{ad}_X\operatorname{ad}_{JU}+\operatorname{ad}_XJ\operatorname{ad}_U)\\=&\,\frac{1}{2}(\operatorname{ad}_{JU}\operatorname{ad}_X-\operatorname{ad}_X\operatorname{ad}_{JU}-\operatorname{ad}_{JX}\operatorname{ad}_U+\operatorname{ad}_U\operatorname{ad}_{JX})=\frac{1}{2}(\operatorname{ad}_{[JU, X]+[U, JX]})=0\,, \end{align}\] where we used the fact that, since \(\mathfrak{g}\) is \(2\)-step and \(U\in (\mathfrak{g}/{\rm Rad}(\mathfrak{g}))'\), \(\operatorname{ad}_U\operatorname{ad}_X=\operatorname{ad}_U\operatorname{ad}_{JX}=0.\) As a consequence, \(\ker\Phi(U)\) and \(\operatorname{Im}\Phi(U)\) are \(J\)-invariant ideals of \(\mathfrak{g}\). We now consider the operator \[\Omega^{1,1}_{\mathfrak{g}/ {\rm Rad}(\mathfrak{g})}:=\frac{1}{2} \left(\operatorname{ad}^{\mathfrak{g}/{\rm Rad}(\mathfrak{g})}_{JU}-J\operatorname{ad}^{\mathfrak{g}/{\rm Rad}(\mathfrak{g})}_{U}\right) \in \mathrm{End}\left(\frac{\mathfrak{g}}{{\rm Rad}(\mathfrak{g})}, J\right)\,,\] defined so that \(\Omega^{1,1}_{\mathfrak{g}/ {\rm Rad}(\mathfrak{g})} \circ \pi = \pi \circ \Phi(U)\), where \(\pi \colon \mathfrak{g}\to \mathfrak{g}/ {\rm Rad}(\mathfrak{g})\) is the canonical projection onto the quotient. By Remark 37 \(\Omega^{1,1}_{\mathfrak{g}/ {\rm Rad}(\mathfrak{g})}=\mathrm{Id}_{\mathfrak{g}/ {\rm Rad}(\mathfrak{g})}\), whence \(0=\ker \Omega^{1,1}_{\mathfrak{g}/ {\rm Rad}(\mathfrak{g})}=\ker\Phi(U)/{\rm Rad}(\mathfrak{g})\), i.e. \(\ker\Phi(U)={\rm Rad}(\mathfrak{g})\). We then conclude that \(\mathfrak{g}=\operatorname{Im}\Phi(U)\oplus {\rm Rad}(\mathfrak{g})\), completing the proof of this step.

Step 4. Finally, we conclude the existence part of the theorem. Thanks to the previous steps we may assume \(\mathfrak{g}= \mathfrak{g}' + J \mathfrak{g}'\) and \(\mathfrak{z}(\mathfrak{g})=0\). Since \(\mathfrak{z}(\mathfrak{g})=0\), in particular \(\mathfrak{g}'_J=0\), which puts us in the position to apply Theorem 41 to deduce the desired decomposition of \(\mathfrak{g}\).

In order to prove the uniqueness part of the theorem, we proceed by induction on \(\dim_{\mathbb{C}}\mathfrak{g}\). If \(\dim_{\mathbb{C}}\mathfrak{g}=1\), then either \(\mathfrak{g}=\mathbb{C}\) or \(\mathfrak{g}=\mathfrak{aff}(\mathbb{R})\) and the statement is clearly true. Assume the assertion is true for any Lie algebra whose complex dimension is strictly less than \(\dim_{\mathbb{C}}\mathfrak{g}\). Let \(\mathfrak{h}_1, \mathfrak{h}_2\) be two \(J\)-adapted Levi subalgebras of \(\mathfrak{g}.\) Assume first \(\mathfrak{g}\ne \mathfrak{g}' + J\mathfrak{g}'\). Then, \(\mathfrak{h}_1, \mathfrak{h}_2\subseteq \mathfrak{g}'+J\mathfrak{g}'\), because they are both \(J\)-perfect. Furthermore, thanks to Lemma 39, we deduce that \(\mathfrak{h}_1, \mathfrak{h}_2\) are \(J\)-adapted Levi subalgebras of \(\mathfrak{g}'+J\mathfrak{g}'\). We then apply the inductive hypothesis and conclude that they are equal. Then, it remains to treat the case when \(\mathfrak{g}=\mathfrak{g}'+J\mathfrak{g}'\). If \(\mathfrak{z}( \mathfrak{g})=0\), then \(\mathfrak{g}_J'=0\) and we may conclude applying Theorem 41. Hence, we may assume that \(\mathfrak{z}(\mathfrak{g})\ne 0\). We distinguish two cases. Let us assume first that \(\mathfrak{z}(\mathfrak{g})\ne {\rm Rad}(\mathfrak{g})\). Consider the canonical projection onto the quotient \(\pi\colon\mathfrak{g} \to\mathfrak{g}/\mathfrak{z}(\mathfrak{g})\) and observe that \(\pi(\mathfrak{h}_i)\), \(i=1,2\), are \(J\)-adapted Levi subalgebras of \(\mathfrak{g}/\mathfrak{z}(\mathfrak{g})\). By the inductive hypothesis, \(\pi(\mathfrak{h}_1)=\pi(\mathfrak{h}_2)\). This allows to infer that \(\mathfrak{h}_1\subseteq \mathfrak{h}_2+\mathfrak{z}(\mathfrak{g}):=\mathfrak{s}\). Of course, \({\rm Rad}(\mathfrak{s} )=\mathfrak{z}(\mathfrak{g})\), hence \(\mathfrak{h}_1, \mathfrak{h}_2\) are \(J\)-adapted Levi subalgebras of \(\mathfrak{s}\) and \(\dim_{\mathbb{C}}\mathfrak{s}<\dim_{\mathbb{C}}\mathfrak{g}\), since \(\mathfrak{z}(\mathfrak{g})\ne {\rm Rad}(\mathfrak{g})\). We conclude that \(\mathfrak{h}_1=\mathfrak{h}_2\) by the inductive hypothesis. It remains to understand the case \(\mathfrak{z}(\mathfrak{g})={\rm Rad}(\mathfrak{g})\). In terms of the decomposition \(\mathfrak{g}=\mathfrak{h}_1\ltimes \mathfrak{z}(\mathfrak{g})\), we consider the projections \(p_{\mathfrak{h}_1}\) and \(p_{\mathfrak{z}(\mathfrak{g})}\) on \(\mathfrak{h}_1\) and \(\mathfrak{z}(\mathfrak{g})\), respectively. First of all, we observe that, since \(J\mathfrak{h}_1=\mathfrak{h}_1\) and \(J\mathfrak{z}(\mathfrak{g})=\mathfrak{z}(\mathfrak{g})\), \([p_{\mathfrak{h}_1}, J]=[p_{\mathfrak{z}(\mathfrak{g})}, J]=0\). Hence, for any \(Z\in \mathfrak{h}_2\), we can write \[Z=p_{\mathfrak{h}_1}(Z)+ p_{\mathfrak{z}(\mathfrak{g})}(Z)\,.\] It follows that, for any \(Z_1, Z_2\in \mathfrak{h}_2\), \[p_{\mathfrak{h}_1}([Z_1, Z_2])+ p_{\mathfrak{z}(\mathfrak{g})}([Z_1, Z_2])=[Z_1, Z_2]=[p_{\mathfrak{h}_1}(Z_1), p_{\mathfrak{h}_1}(Z_2)]\,,\] showing that \(p_{\mathfrak{h}_1}\) is a holomorphic Lie algebra homomorphism and that \(p_{\mathfrak{z}(\mathfrak{g})}|_{\mathfrak{h}_2'}=0\). Since \(\mathfrak{h}_2=\mathfrak{h}_2'\oplus J\mathfrak{h}_2'\) and \([p_{\mathfrak{z}(\mathfrak{g})}, J]=0\) it follows that \(p_{\mathfrak{z}(\mathfrak{g})}|_{\mathfrak{h}_2}=0\). Thus, for any \(Z\in\mathfrak{h}_2\), \(Z=p_{\mathfrak{h}_1}(Z)\in\mathfrak{h}_1\) and so \(\mathfrak{h}_2\subseteq \mathfrak{h}_1\) forcing \(\mathfrak{h}_2=\mathfrak{h}_1\), since they have the same dimension, thereby completing the proof. ◻

Proof. First of all, for any \(i=1, \ldots, p+q\) and \(j=1, \ldots, q\), using ?? multiple times, we deduce \[\begin{align} \operatorname{ad}_{JX_i}^2&=\operatorname{ad}_{J[JX_i, X_i]}=\operatorname{ad}_{JX_i}\,,\\ \operatorname{ad}_{JY_{p+j}}^3&=\operatorname{ad}_{J[JY_{p+j},Y_{p+j}]}\operatorname{ad}_{JY_{p+j}}=-\operatorname{ad}_{JX_{p+j}}\operatorname{ad}_{JY_{p+j}}=-\operatorname{ad}_{J[JX_{p+j}, Y_{p+j}]}=-\operatorname{ad}_{JY_{p+j}}\,, \end{align}\] which immediately imply that \(\operatorname{ad}_{JX_i}\) and \(\operatorname{ad}_{JY_{p+j}}\) are diagonalisable with \({\rm Spec}(\operatorname{ad}_{JX_i})\subseteq\{0,1\}\) and \({\rm Spec}(\operatorname{ad}_{JY_{p+j}})\subseteq\{0, \sqrt{-1}, -\sqrt{-1}\}\). Finally, since \(J\) is abelian, \(\{\operatorname{ad}_{JX_1}, \ldots, \operatorname{ad}_{JX_{p+q}}, \operatorname{ad}_{JY_{p+1}}, \ldots, \operatorname{ad}_{JY_{p+q}}\}\) is a family of commuting and diagonalisable endomorphisms of \(\mathfrak{g}\) which are, hence, simultaneously diagonalisable. The last statement is now clear. ◻

Proof. Using Corollary 43, we observe that unimodularity of \(\mathfrak{g}'+J\mathfrak{g}'\) would not allow the \(J\)-adapted Levi subalgebra \(\mathfrak{h}\) to be non-zero, proving (1). As for (2), we notice that, again by Corollary 43, if \(\mathfrak{g}\) is of real type then there cannot be any ideal of the \(J\)-adapted Levi subalgebra in the decomposition given by Corollary 33 which is isomorphic to \(\mathfrak{aff}(\mathbb{C})\). ◻

Proof. Of course \(\ker d_{\tau_{JU}}\cap V^2(\mathfrak{h}_{JU}, \tau_{JU})=V^2(\mathfrak{h}_{JU}, \tau_{JU})\), since \(\dim_{\mathbb{R}}\mathfrak{h}_{JU}=2\). On the other hand, \[(d_{\tau_{JU}}L)(JU, U)=\operatorname{ad}_{JU}L(U)-2L(U)=(\operatorname{ad}_{JU}-2{\rm Id})L(U)\,,\] using the fact that \([JU, U]=2U\) and that \(\tau_{JU}(U)=0\). Now, any \(\alpha \in V^2(\mathfrak{h}_{JU}, \tau_{JU})\) is completely determined by \(\alpha(JU, U)\in \mathfrak{g}_J'\). Hence, the condition \(H^2(\mathfrak{g}, JU)=0\) is equivalent to the surjectivity of \((\operatorname{ad}_{JU}-2{\rm Id})|_{\mathfrak{g}_J'}\) which, in turn, is equivalent to \(2\notin {\rm Spec}(\operatorname{ad}_{JU}|_{\mathfrak{g}_J'})\). ◻

Proof. Let \(\hat{\mathfrak{s}}\) be a subalgebra as in the statement. It is apparent that \(\hat{\mathfrak{s}}_J'\subseteq \mathfrak{g}_J'\) and, since \(\hat{\mathfrak{s}}\subseteq \mathcal{O}(\mathfrak{g}_J')\), we have \(\hat{\mathfrak{s}}=\mathcal{O}_{\hat{\mathfrak{s}}}(\hat{\mathfrak{s}}_J'):=\{X\in \hat{\mathfrak{s}}\, \, |\, \, [\operatorname{ad}_X, J]|_{\mathfrak{s}_J'}=0\}\). Furthermore, since \(\hat{\mathfrak{s}}_J'\subseteq \mathfrak{g}_J'\), the inclusion \(\hat{\mathfrak{s}}\to \mathfrak{g}\) induces a Lie algebra homomorphism \[\varphi\colon \frac{\hat{\mathfrak{s}}}{\hat{\mathfrak{s}}_J'}\to \frac{\mathcal{O}(\mathfrak{g}_J')}{\mathfrak{g}_J'}\,, \quad \varphi(X+\hat{\mathfrak{s}}_J')=X+\mathfrak{g}_J'\,.\] Now, by Theorem 42, we have that \[\frac{\hat{\mathfrak{s}}}{\hat{\mathfrak{s}}_J'}=\mathfrak{h}_{\hat{\mathfrak{s}}}\ltimes \frac{{\rm Rad}(\hat{\mathfrak{s}})}{\hat{\mathfrak{s}}_J'}\,,\quad \frac{\mathcal{O}(\mathfrak{g}_J')}{\mathfrak{g}_J'}=\mathfrak{h}_{\mathfrak{g}}\ltimes\frac{{\rm Rad}(\mathcal{O} (\mathfrak{g}_J'))}{\mathfrak{g}_J'}\,.\] Then, since \(\ker(\varphi|_{\mathfrak{h}_{\hat{\mathfrak{s}}}})=\ker \varphi\cap \mathfrak{h}_{\hat{\mathfrak{s}}}=\frac{\mathfrak{g}_J'}{\hat{\mathfrak{s}}_J'}\cap \mathfrak{h}_{\hat{\mathfrak{s}}}=0\), we see that \(\varphi|_{\mathfrak{h}_{\hat{\mathfrak{s}}}}\) is a holomorphic Lie algebra isomorphism onto \({\rm Im}(\varphi|_{{\mathfrak{h}}_{\hat{\mathfrak{s}}}})\). In particular, this implies that \({\rm Im}(\varphi|_{\mathfrak{h}_{\hat{\mathfrak{s}}}})\) is a \(J\)-semisimple Lie subalgebra of \(\mathcal{O}(\mathfrak{g}_J')/\mathfrak{g}_J'\), and hence \({\rm Im}(\varphi|_{\mathfrak{h}_{\hat{\mathfrak{s}}}})\cap \frac{{\rm Rad}(\mathcal{O} (\mathfrak{g}_J'))}{\mathfrak{g}_J'}=0\). On the other hand, \[\mathfrak{h}_{\hat{\mathfrak{s}}}\simeq \frac{\hat{\mathfrak{s}}}{{\rm Rad}(\hat{\mathfrak{s}})}\simeq \frac{\mathcal{O}(\mathfrak{g}_J')}{{\rm Rad}(\mathcal{O}(\mathfrak{g}_J'))},\] which gives that \({\rm Im}(\varphi|_{\mathfrak{h}_{\hat{\mathfrak{s}}}})\) is a \(J\)-adapted Levi subalgebra of \(\mathcal{O}(\mathfrak{g}_J')/\mathfrak{g}_J'\). Using Theorem 42, we conclude that \(\mathfrak{h}_{\mathfrak{g}}= {\rm Im}(\varphi|_{\mathfrak{h}_{\hat{\mathfrak{s}}}})\). Then, from the last statement in Lemma 35 we deduce \[JU_{\hat{\mathfrak{s}}}+ \mathfrak{g}_J'=\varphi(JU_{\hat{\mathfrak{s}}}+ \hat{\mathfrak{s}}_J')=JU_{\mathfrak{g}}+ \mathfrak{g}_J'.\] The claim now follows, since \(\operatorname{ad}_{JU_{\hat{\mathfrak{s}}}}|_{\hat{\mathfrak{s}}_J'}=\operatorname{ad}_{JU_{\mathfrak{g}}}|_{\hat{\mathfrak{s}}_J'}\). ◻

Proof. We have that \(\mathfrak{g}'+J\mathfrak{g}'\) satisfies the hypotheses of Lemma 49, thanks to Corollary 39. The conclusion follows. ◻

Proof. First of all, we note that in our hypotheses \(\mathfrak{g}_J'\) is a \(J\)-invariant ideal of \(\mathfrak{g}\). With this in mind, we divide the proof into steps.

Step 1. We show that we may assume \({\rm Rad}(\mathfrak{g})=\mathfrak{g}_J'\ne 0\) and that \(\mathfrak{g}\) has no \(J\)-invariant ideals properly contained in \({\rm Rad}(\mathfrak{g})\). First, note that we may suppose \(\mathfrak{g}_J'\ne 0\), otherwise we can simply apply Theorem 42 and conclude. To prove that we may reduce ourselves to the case \({\rm Rad}(\mathfrak{g})=\mathfrak{g}_J'\), we work inductively on \(\dim_\mathbb{C}(\mathfrak{g})\). If \(\mathfrak{g}\) is \(1\)-dimensional it can only be either \(\mathfrak{aff}(\mathbb{R})\) or \(\mathbb{C}\) and in both cases the theorem is clearly true. Let us assume that \(\mathfrak{g}_J'\subset{\rm Rad}(\mathfrak{g})\). Then, since \(\mathfrak{g}/ \mathfrak{g}_J'\) has an abelian complex structure, we can apply Theorem 42 and obtain \[\frac{\mathfrak{g}}{\mathfrak{g}_J'}=\mathfrak{s}\ltimes \frac{{\rm Rad}(\mathfrak{g})}{\mathfrak{g}_J'}\,,\] where \(\mathfrak{s}\) is the \(J\)-adapted Levi subalgebra of \(\mathfrak{g} /\mathfrak{g}_J'\). Now consider \(\hat{\mathfrak{s}}:=\pi^{-1}(\mathfrak{s})\), which is a proper \(J\)-invariant subalgebra of \(\mathfrak{g}\). Since \(\hat{\mathfrak{s}}/\mathfrak{g}_J'=\mathfrak{s}\) is \(J\)-semisimple, we can infer, using the same strategy as in Step 1 of Theorem 42, that \(\mathfrak{g}_J'= {\rm Rad}(\hat{\mathfrak{s}})\). We may then apply Lemma 49 to infer that \(H^2(\hat{\mathfrak{s}},JU_{\hat{\mathfrak{s}}})=0\). Hence, by the inductive hypothesis \(\hat{\mathfrak{s}}\) admits a Levi–Malcev decomposition, and therefore so does \(\mathfrak{g}\). Consequently, we can assume \(\mathfrak{g}_J'={\rm Rad}(\mathfrak{g}).\)

Finally, we prove that \(\mathfrak{g}\) has no \(J\)-invariant ideals properly contained in \({\rm Rad}(\mathfrak{g})\), once again by induction on \(\dim_\mathbb{C}\mathfrak{g}\). If \(\dim_\mathbb{C}\mathfrak{g}=1\), the claim is trivial. Let \(\mathfrak{k} \subset \mathfrak{g}_J'\) be a \(J\)-invariant ideal of \(\mathfrak{g}\). Note that \((\mathfrak{g}/\mathfrak{k})_J'=\frac{\mathfrak{g}_J'}{\mathfrak{k}}\). Furthermore \(\frac{\mathfrak{g}/\mathfrak{k}}{\mathfrak{g}_J'/\mathfrak{k}}\simeq \mathfrak{g}/\mathfrak{g}_J'\) via \(\varphi((X+\mathfrak{k})+ \frac{\mathfrak{g}_J'}{\mathfrak{k}})=X+\mathfrak{g}_J'\). On the other hand, using again Lemma 35, we have \(JU_{\mathfrak{g}/\mathfrak{k}}+ \mathfrak{g}'_J=\varphi((JU_{\mathfrak{g}/\mathfrak{k} }+\mathfrak{k})+\frac{\mathfrak{g}'_J}{\mathfrak{k}})=JU_{\mathfrak{g}}+ \mathfrak{g}_J'.\) Denote by \(\pi^{\mathfrak{k} }\colon \mathfrak{g} \to\mathfrak{g}/\mathfrak{k}\) the quotient map, then \[\operatorname{ad}_{JU_{\mathfrak{g}/\mathfrak{k}}+ \mathfrak{k}}^{\mathfrak{g}/\mathfrak{k} }|_{\mathfrak{g}_J'/\mathfrak{k}}=\pi^{\mathfrak{k}}\operatorname{ad}_{JU_{\mathfrak{g}/\mathfrak{k}}}^{\mathfrak{g}}|_{\mathfrak{g}_J'}=\pi^{\mathfrak{k}}\operatorname{ad}_{JU_{\mathfrak{g}}}^{\mathfrak{g}}|_{\mathfrak{g}_J'}\,.\] Hence, since \(2\notin{\rm Spec}(\operatorname{ad}_{JU_{\mathfrak{g}}}^{\mathfrak{g}}|_{\mathfrak{g}_J'})\), the same holds for \(\operatorname{ad}_{JU_{\mathfrak{g}/\mathfrak{k}}+ \mathfrak{k}}^{\mathfrak{g}/\mathfrak{k} }|_{\mathfrak{g}_J'/\mathfrak{k}}\). Furthermore, since \(\mathfrak{g}\) acts holomorphically on \(\mathfrak{g}_J'\), then \(\mathfrak{g}/\mathfrak{k}\) acts holomorphically on \((\mathfrak{g}/\mathfrak{k})_J'\). Therefore, applying the inductive hypothesis we obtain a Levi–Malcev decomposition of \(\mathfrak{g}/ \mathfrak{k}\). The same argument in Step 1 of Theorem 42, together with an argument analogous to that used in the first part of this proof, guarantees the existence of a \(J\)-adapted Levi subalgebra for \(\mathfrak{g}\).

Step 2. First of all, thanks to Step 1, we may assume \(0\neq \mathfrak{g}'_J={\rm Rad}(\mathfrak{g})\) and \(\mathfrak{g}\) has no \(J\)-invariant ideals properly contained in \({\rm Rad}(\mathfrak{g})\). Moreover, \(\mathfrak{g}/\mathfrak{g}_J'\) is \(J\)-semisimple with an abelian complex structure, hence, it is \(J\)-perfect by Theorem 27. As in Step 2 of Theorem 42, we can apply Lemma 39 to conclude that \(\mathfrak{g}\) is \(J\)-perfect. In this setting, we consider \[\Phi\colon \mathfrak{g} \to {\rm End}(\mathfrak{g} )\,,\quad \Phi(X)=\frac{1}{2}(\operatorname{ad}_{JX}-J\operatorname{ad}_X)\,, \quad X \in\mathfrak{g}\,.\] First of all, using integrability of \(J\), for any \(X\in \mathfrak{g}\), we deduce \[\label{cazzettiinano} [\Phi(X), J]=0\,, \quad \Phi(JX)=-\Phi(X)J\,.\tag{10}\] By ?? and 10 , we know that \(\mathfrak{g}_J'\subseteq \ker \Phi\) and \(J\ker \Phi=\ker \Phi\). Furthermore, as in Lemma 4, we observe that \(\mathfrak{g}\) acts holomorphically on \(\ker \Phi\). Therefore, \([\mathfrak{g}, \ker \Phi]\subseteq \mathfrak{g}_J'\), giving that \(\ker \Phi\) is a \(J\)-solvable ideal of \(\mathfrak{g}\). Thus, \(\mathfrak{g}_J'= \ker \Phi\). In view of this, \(\Phi\) descends to the quotient, i.e. \[\Phi\colon \frac{\mathfrak{g}}{ \mathfrak{g}_J'}\to {\rm End}(\mathfrak{g}).\] Moreover, we observe that, if \(X\in\mathfrak{g}/\mathfrak{g}_J'\) and \(Y\in\mathfrak{g}_J'\), then \[\label{giovannibuco} \Phi(X)\operatorname{ad}_Y=\operatorname{ad}_{\Phi(X)Y}\,.\tag{11}\] Indeed, using \(2\)-step solvability and that \(X\) acts holomorphically on \(\mathfrak{g}_J'=\ker \Phi\), we have \[\begin{align} \Phi(X)\operatorname{ad}_Y=&\, \frac{1}{2}(\operatorname{ad}_{JX}\operatorname{ad}_Y-J\operatorname{ad}_X\operatorname{ad}_Y)= \frac{1}{2}(\operatorname{ad}_{[JX, Y]}-\operatorname{ad}_XJ\operatorname{ad}_Y)=\frac{1}{2}(\operatorname{ad}_{[JX, Y]}-\operatorname{ad}_X\operatorname{ad}_{JY})\\ =&\, \frac{1}{2}\operatorname{ad}_{[JX, Y]-[X, JY]}=\operatorname{ad}_{\Phi(X)Y}\,, \end{align}\] as claimed. We will only be interested in \(\Phi(U)\), where \(JU\in J(\mathfrak{g} /\mathfrak{g}_J')'\) is the mean curvature vector of \(\mathfrak{g} /\mathfrak{g}_J'\), introduced in Definition 34. We extend the endomorphism \(\Phi(U)\) by \(\mathbb{C}\)-linearity on \(\mathfrak{g}^{\mathbb{C}}:=\mathfrak{g}\otimes \mathbb{C}\). By abuse of notation, this extension will again be denoted \(\Phi(U)\). Since \(\mathfrak{g}_J'\) is an ideal of \(\mathfrak{g}\), and by 10 , \([\Phi(U), J]=0\), we deduce \[\Phi(U)((\mathfrak{g}_J')^{1,0})\subseteq (\mathfrak{g}_J')^{1,0}\,,\qquad \Phi(U)((\mathfrak{g}_J')^{0,1})\subseteq (\mathfrak{g}_J')^{0,1}\,.\] Consequently, we can find \(\lambda \in \mathbb{C}\) such that \(\ker \bigl(\left(\Phi(U)-\lambda {\rm Id}\right)|_{(\mathfrak{g}_J')^{1,0}}\bigr)\ne 0\). By simply exploiting \(\mathbb{C}\)-linearity, from 11 , we have \[\left(\Phi(U)-\lambda {\rm Id}\right)\operatorname{ad}^{\mathfrak{g}^{\mathbb{C}}}_Y=\operatorname{ad}^{\mathfrak{g}^{\mathbb{C}}}_{(\Phi(U)-\lambda {\rm Id})Y}\,, \qquad Y \in (\mathfrak{g}_J')^{1,0}\,.\] Hence, by the above and the fact that \(\mathfrak{g}\) acts holomorphically on \(\mathfrak{g}_J'\), we infer that \(\ker \bigl(\left(\Phi(U)-\lambda {\rm Id}\right)|_{(\mathfrak{g}_J')^{1,0}}\bigr)\) and \(\ker \bigl(\left(\Phi(U)-\bar \lambda {\rm Id}\right)|_{(\mathfrak{g}_J')^{0,1}}\bigr)\) are \(J\)-invariant ideals of \(\mathfrak{g}^{\mathbb{C}}\) contained in \(\mathfrak{g}_J'\otimes \mathbb{C}\). We then consider \[\mathfrak{k}:=\ker \left(\left(\Phi(U)-\lambda {\rm Id}\right)|_{(\mathfrak{g}_J')^{1,0}}\right)\oplus \ker \left(\left(\Phi(U)-\bar \lambda {\rm Id}\right)|_{(\mathfrak{g}_J')^{0,1}}\right)\subseteq \mathfrak{g}_J'\otimes \mathbb{C}\] which is a non-zero \(J\)-invariant ideal of \(\mathfrak{g}^{\mathbb{C}}\) which is stable under conjugation. Thus, \(\mathfrak{k}=(\mathfrak{k} \cap \mathfrak{g})\otimes \mathbb{C}\,,\) where \(\mathfrak{k} \cap \mathfrak{g}\subseteq \mathfrak{g}_J'\) is a non-zero \(J\)-invariant ideal of \(\mathfrak{g}.\) As a consequence, using the first step, we may assume \(\mathfrak{k}\cap \mathfrak{g}= \mathfrak{g}_J'\) and, thus, \(\mathfrak{k}= \mathfrak{g}_J'\otimes \mathbb{C}\). In particular, we conclude that \[\Phi(U)|_{(\mathfrak{g}_J')^{1,0}}=\lambda {\rm Id}_{(\mathfrak{g}_J')^{1,0}}\,, \quad\Phi(U)|_{(\mathfrak{g}_J')^{0,1}}=\bar \lambda {\rm Id}_{(\mathfrak{g}_J')^{0,1}}\,.\] Denote \(a:={\rm Re}(\lambda)\) and \(b:={\rm Im}(\lambda)\). Then we infer that \[\label{baumiao} \Phi(U)|_{\mathfrak{g}_J'}=a{\rm Id}_{\mathfrak{g}_J'}+ b J|_{\mathfrak{g}_J'}\,.\tag{12}\] Now, since \(H^2(\mathfrak{g}, JU)=0\) and by Lemma 48, we deduce that \((a, b)\ne (1,0)\). Therefore, from 12 follows that the map \[\tilde{\Psi}:=\Phi(U)-a{\rm Id}-b J\colon \mathfrak{g} \to \mathfrak{g}\] descends to a well-defined map on the quotient, namely \[\Psi\colon \frac{\mathfrak{g}}{ \mathfrak{g}_J'}\to \mathfrak{g}\,.\] The first thing to observe is that \(\Psi\) is injective. Indeed, recalling Definition 36 and Remark 37, \[\pi \tilde{\Psi}=\pi\Phi(U)-a\pi-bJ\pi=\Omega_{\mathfrak{g}/ \mathfrak{g}_J'}\pi-a\pi- bJ \pi=((1-a){\rm Id}- bJ)\pi\,.\] Hence, if \(X\in \ker \tilde{\Psi}\), then \((1-a)\pi(X)=bJ\pi(X)\), which is possible if and only if \(\pi(X)=0\), since \((a, b)\ne (1, 0)\), therefore \(X\in \mathfrak{g}_J'\). On the other hand, \(\operatorname{Im}(\Psi)\cap \mathfrak{g}_J'=0\). Indeed, if \(\Psi(\pi (X))\in\mathfrak{g}_J'\), \(X\in\mathfrak{g}\), then again \[0=\pi \Psi(\pi(X))=\pi \tilde{\Psi}(X)=((1-a)\pi(X)-bJ\pi(X))\] implying that \(\pi(X)=0\), as claimed. Moreover, \({\rm Im}\Psi\) is \(J\)-invariant, since \([\Psi, J]=0\).

It remains to prove that \(\operatorname{Im}\Psi\) is a subalgebra of \(\mathfrak{g}\). To do so, we fix \(X, Y\in (\mathfrak{g}/\mathfrak{g}_J')'\) and compute \([\Psi(X),\Psi(Y)]\), \([\Psi(JX),\Psi(JY)]\) and \([\Psi(X),\Psi(JY)]\). Let us start by observing that \[\Psi(X)=\frac{1}{2}[JU,X]-aX-bJX\,, \qquad \Psi(Y)=\frac{1}{2}[JU,Y]-aY-bJY\,.\] Then, using \(2\)-step solvability, Jacobi identity and integrability of \(J\), we get \[\begin{align} [\Psi(X),\Psi(Y)]&=b\left( -\frac{1}{2}[[JU,X],JY]+a[X,JY]-\frac{1}{2}[JX,[JU,Y]]+a[JX,Y]+b[JX,JY]\right)\\ &=b\left( -\frac{1}{2}[JU,[X,JY]+[JX,Y]]+a[X,JY]+a[JX,Y]+b[JX,JY]\right)\\ &=b\left( \frac{1}{2}[JU,J[JX,JY]]-aJ[JX,JY]+b[JX,JY]\right)=b\Psi(J[JX,JY])=0\,, \end{align}\] since \(J[JX, JY]\in \mathfrak{g}_J'\). On the other hand, using integrability, we have \[\Psi(JX)=\frac{1}{2}J[JU,X]-aJX+bX\,, \qquad \Psi(JY)=\frac{1}{2}J[JU,Y]-aJY+bY\,,\] thus \[\begin{align} [\Psi(JX),\Psi(JY)]&=\frac{1}{4}[J[JU,X],J[JU,Y]]-\frac{a}{2}[J[JU,X],JY]+\frac{b}{2}[J[JU,X],Y]-\frac{a}{2}[JX,J[JU,Y]]\\&\quad +a^2[JX,JY]-ab[JX,Y]+\frac{b}{2}[X,J[JU,Y]]-ab[X,JY]\,. \end{align}\] We treat the first term as follows. Using integrability, \(2\)-step solvability and Jacobi identity, we have \[\begin{align} \relax [J[JU, X], J[JU, Y]] &=J([J[JU, X], [JU,Y]]+[[JU, X], J[JU, Y]])\\ &=J\left([JU,[J[JU,X],Y]]+ [JU,[X, J[JU, Y]]]\right)\,. \end{align}\] Also, by integrability \[[J[JU,X],JY]+[JX,J[JU,Y]]=J[J[JU,X],Y]+J[[JU,X],JY]+J[JX,[JU,Y]]+J[X,J[JU,Y]]\,,\] whence \[\begin{align} [\Psi(JX),\Psi(JY)]&=\frac{1}{2}J\Psi\left([J[JU,X],Y]+[X,J[JU,Y]]\right)\\ &\quad-\frac{a}{2}J[[JU,X],JY]-\frac{a}{2}J[JX,[JU,Y]]+a^2[JX,JY]-ab[JX,Y]-ab[X,JY]\\ &=\frac{1}{2}J\Psi\left([J[JU,X],Y]+[X,J[JU,Y]]\right)-\frac{a}{2}J[JU,[X,JY]]-\frac{a}{2}J[JU,[JX,Y]]\\ &\quad +a^2J[JX,Y]+a^2J[X,JY]-ab[JX,Y]-ab[X,JY]\\ &=\Psi\left(\frac{1}{2}J[J[JU,X],Y]+\frac{1}{2}J[X,J[JU,Y]]-aJ[X,JY]-aJ[JX,Y]\right)=0\,, \end{align}\] since the argument of \(\Psi\) lies in \(\mathfrak{g}_J'\). Finally, we compute \[\begin{align} [\Psi(X),\Psi(JY)]&=\frac{1}{4}[[JU,X],J[JU,Y]]-\frac{a}{2}[[JU,X],JY]-\frac{a}{2}[X,J[JU,Y]]+a^2[X,JY]\\ &\quad -\frac{b}{2}[JX,J[JU,Y]]+ab[JX,JY]-b^2[JX,Y]\\ &=\frac{1}{4}[JU,[X,J[JU,Y]]]-\frac{a}{2}[JU,[X,JY]]-\frac{a}{2}[X,J[JU,Y]]+a^2[X,JY]\\ &\quad -\frac{b}{2}J[X,J[JU,Y]]-\frac{b}{2}J[JX,[JU,Y]]+abJ[X,JY]+abJ[JX,Y]-b^2[JX,Y]\\ &=\frac{1}{2}\Psi([X,J[JU,Y]])-a\Psi([X,JY])-bJ\left(\frac{1}{2}[JU,[JX,Y]]-a[JX,Y]-bJ[JX,Y]\right)\\ &=\Psi\left(\frac{1}{2}[X,J[JU,Y]]-a[X,JY]-bJ[JX,Y]\right)\,. \end{align}\] It follows that \([\operatorname{Im}\Psi, \operatorname{Im}\Psi]\subseteq \operatorname{Im}\Psi\), and hence \(\operatorname{Im}\Psi\) is a subalgebra, as desired. ◻

Proof. We start by observing that the hypothesis \(H^2(\mathfrak{g}, JU)=0\) and Corollary 50 put us in the position to apply Theorem 51 and infer that \[\mathfrak{g}'+J\mathfrak{g}'=\mathfrak{h}\ltimes {\rm Rad}(\mathfrak{g}'+ J \mathfrak{g}')\,, \qquad \mathfrak{h}\simeq \frac{\mathfrak{g}'+J\mathfrak{g}'}{{\rm Rad}(\mathfrak{g}'+ J\mathfrak{g}')}.\] We first prove that if \([X,Z],[X,Z']\in {\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\), then \([X,S],[X,T] \in {\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\). Surely \([X,S],[X,T]\in \mathfrak{g}'+J\mathfrak{g}'\) and we can decompose \[[X,S]=\sum_{i=1}^{\dim \mathfrak{h}'} \alpha_i Y_i + W_1\,, \qquad [X,T]=\sum_{i=1}^{\dim \mathfrak{h}'} \beta_i Y_i+ W_2 \,,\] where \(\{Y_1, \ldots, Y_{\dim \mathfrak{h}'}\}\) is a basis of \(\mathfrak{h}'\) and \(W_1, W_2\in {\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\). Then, since \(S\in\mathfrak{g}'\) and \({\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\) is an ideal in \(\mathfrak{g}'+J\mathfrak{g}'\), \(2\)-step solvability and ?? imply \[\begin{align} 0=&\, [[X,JU],S]=[[X,S],JU]+[X,[JU,S]] =\sum_{i=1}^{\dim \mathfrak{h}'}\left(-2\alpha_i+a\alpha_i+b\beta_i\right)Y_i+W_3\,,\\ 0=&\, [[X,JU],T]=[[X,T],JU]+[X,[JU,T]]=\sum_{i=1}^{\dim \mathfrak{h}'} \left(-2\beta_i-b\alpha_i+a\beta_i\right)Y_i+W_4\,, \end{align}\] where \(W_3, W_4\in {\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\). Hence, for all \(i=1, \ldots, \dim \mathfrak{h}'\), we obtain the system of equations \[\label{sefattolastoria} \begin{cases} (a-2)\alpha_i+b\beta_i=0\,, \\ (a-2)\beta_i-b\alpha_i=0\,. \end{cases}\tag{16}\] Assume by contradiction that \([X,S]\notin \mathrm{Rad}(\mathfrak{g}'+J\mathfrak{g}')\), then there exists \(i=1, \ldots, \dim \mathfrak{h}'\) such that \(\alpha_i\ne 0\). Therefore, from the second equation in 16 we have \(b=(a-2)\frac{\beta_i}{\alpha_i}\), which, plugged in the first equation, forces \(a=2\) and consequently \(b=0\) which is against our assumptions. We conclude similarly if \([X,T]\notin \mathrm{Rad}(\mathfrak{g}'+J\mathfrak{g}')\).

Now we prove the lemma. Suppose \([X,Z],[X,Z'] \in \mathfrak{q}\) then, by the first part of the proof, we know that \([X,S],[X, T]\in {\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\) and we may write \[[X,S]=P+Z_1\,,\qquad [X,T]=Q+Z_2\,,\] with \(P,Q\in V\), for some complement \(V\) of \(\mathfrak{g}_J'\) in \({\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\), and \(Z_1, Z_2\in\mathfrak{g}_J'\). By repeating the computation done above, we get \[\begin{cases} [JU,P]= aP+bQ + [X,Z]+Z_3\,,\\ [JU,Q]= -bP+aQ +[X,Z']+Z_4\,,\\ \end{cases}\] for some \(Z_3, Z_4\in\mathfrak{g}_J'\). Applying \(\operatorname{ad}_{JU}\) to these equations and using the assumption that \([X,Z],[X,Z']\in \mathfrak{q}\), we get \[\label{pezzodistoria} \begin{cases} [JU, [JU,P]]= a[JU,P]+b[JU,Q] +Z_5\,,\\ [JU, [JU,Q]]= -b[JU,P]+a[JU,Q] +Z_6\,,\\ \end{cases}\tag{17}\] with \(Z_5,Z_6\in \mathfrak{g}'_J\). Now, either \([JU,P],[JU,Q] \in \mathfrak{g}'_J\), meaning \([X,S],[X,T]\in \mathfrak{q}\), or \(\lambda=a+\sqrt{-1}b\) is an eigenvalue of \(\operatorname{ad}_{JU} \vert_{\frac{{\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')}{\mathfrak{g}'_J}}\) with eigenvector \([JU,P]-\sqrt{-1}[JU,Q]\). Thus, by 14 we must have \(\lambda\in \{0,2\}\). On the other hand, since \((a,b)\neq (2,0)\), we necessarily have \(a=b=0\). Using 13 , we conclude from 17 that \([JU,P],[JU,Q] \in \mathfrak{g}'_J\), as desired. ◻

Proof. First, we aim to prove that \(\mathfrak{q}\) is an ideal, where \(\mathfrak{q}\) is defined in 15 . Notice that, by 2-step solvability we have \[\operatorname{ad}_{JU}\vert_{(\mathfrak{g}'_J)^{1,0}}=\operatorname{ad}_{JU}\vert_{(\mathfrak{g}'_J)^{1,0}}+\sqrt{-1}\operatorname{ad}_{U}\vert_{(\mathfrak{g}'_J)^{1,0}}=\operatorname{ad}_{JU+\sqrt{-1}U}\vert_{(\mathfrak{g}'_J)^{1,0}}\] and so, by Lie’s theorem, there exists a basis \(\{W_1,\dots,W_s\}\) of \((\mathfrak{g}'_J)^{1,0}\) with respect to which \(\operatorname{ad}_{JU}\vert_{(\mathfrak{g}'_J)^{1,0}}\) is lower triangular. In particular, the eigenvalues \(\lambda_i\in \mathbb{C}\) of \(\operatorname{ad}_{JU}\vert_{(\mathfrak{g}'_J)^{1,0}}\) satisfy \[[JU, W_k]=\lambda_k W_k+ V_k\,, \qquad V_k\in \bigoplus_{i<k} \langle W_i\rangle \qquad k=1,\dots,s\,.\] Writing \(a_k:=\mathrm{Re}(\lambda_k)\), \(b_k:=\operatorname{Im}(\lambda_k)\), and \(Z_k:=\mathrm{Re}(W_k)\), this is equivalent to \[\begin{cases} [JU, Z_k]= a_kZ_k+b_kJZ_k+ \tilde{Z}_k\,, & \tilde{Z}_k\in \bigoplus_{i<k} \langle Z_i,JZ_i \rangle\,, \\ [JU,JZ_k]= -b_kZ_k+a_kJZ_k+\tilde{Z}'_k\,, & \tilde{Z}'_k\in \bigoplus_{i<k} \langle Z_i,JZ_i\rangle \,,\\ \end{cases} \qquad k=1,\dots,s\,.\] Note that \(\mathfrak{g}'_J \subseteq \mathfrak{q}\). We first prove that \([\mathfrak{g},\mathfrak{g}'_J]\subseteq \mathfrak{q}\). In order to do this, we use induction on \(k\), and show that \([\mathfrak{g},Z_k],[\mathfrak{g},JZ_k]\subseteq \mathrm{Rad}(\mathfrak{g}'+J\mathfrak{g}')\). We observe that the condition \(H^2(\mathfrak{g},JU)=0\) entails that \((a_k,b_k)\neq (2,0)\), for all \(k=1,\dots,s\). For \(k=1\), using Lemma 53 with \(S=Z_1, T=JZ_1, Z=Z'=0\), we see that we have \([\mathfrak{g},Z_1],[\mathfrak{g},JZ_1]\subseteq \mathfrak{q}\). Now, suppose that \([\mathfrak{g},Z_{i}], [\mathfrak{g},JZ_{i}]\subseteq \mathfrak{q}\), for all \(i=1,\dots,k-1\), and let us prove the claim for \(k\). By the inductive assumption, we can again apply Lemma 53 with \(S=Z_k\), \(T=JZ_k\), \(Z=\tilde{Z}_k\), and \(Z'=\tilde{Z}'_k\). Hence, we conclude that \([\mathfrak{g},Z_k],[\mathfrak{g},JZ_k]\in \mathfrak{q}\) for all \(k\), meaning that \([\mathfrak{g},\mathfrak{g}'_J]\subseteq \mathfrak{q}\). Knowing this, it now follows immediately from Lemma 53 that \([\mathfrak{g},\mathfrak{q}]\subseteq \mathfrak{q}\).

Let \(\mathfrak{k}\) be the ideal of \(\mathfrak{g}\) generated by \(\mathfrak{g}'_J\), i.e. \[\mathfrak{k}=\mathfrak{g}'_J+[\mathfrak{g},\mathfrak{g}'_J]+[\mathfrak{g},[\mathfrak{g},\mathfrak{g}'_J]]+\dots\] Since \(\mathfrak{q}\) is an ideal, we have \(\mathfrak{k} \subseteq \mathfrak{q}\subseteq \mathrm{Rad}(\mathfrak{g}'+J\mathfrak{g}')\cap \mathfrak{g}'\). Note that, for every \(X\in \mathfrak{g}\), \(Y\in \mathfrak{k}\) we have \([X,JY]+[JX,Y]\in \mathfrak{g}'_J\) and thus \(\mathfrak{k}+ J \mathfrak{k}\) is an ideal of \(\mathfrak{g}\) contained in \(\mathrm{Rad}(\mathfrak{g}'+J\mathfrak{g}')\). In particular, \(\mathfrak{k} + J\mathfrak{k}\) is a \(J\)-solvable ideal of \(\mathfrak{g}\) which contains \(\mathfrak{g}_J'\), implying that \(\mathfrak{g}_J'\subseteq {\rm Rad}(\mathfrak{g})\). ◻

Proof. Combining Proposition 54 and Lemma 39 immediately gives the first part. In turn, if \(J\) is semisimple, this yields \(\mathfrak{g}'_J\subseteq {\rm Rad}(\mathcal{O}(\mathfrak{g}_J'))={\rm Rad}(\mathfrak{g})\cap \mathcal{O}(\mathfrak{g}_J')=0\). ◻

Proof. Thanks to Corollary 55, we know that \({\rm Rad}(\mathcal{O}(\mathfrak{g}'_J)) \subseteq {\rm Rad}(\mathfrak{g})\). Suppose \({\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))={\rm Rad}(\mathfrak{g})\), which implies \({\rm Rad}(\mathfrak{g})\subseteq \mathcal{O}(\mathfrak{g}'_J)\). Similarly to Step 2 of Theorem 42 we deduce that \(\mathfrak{g}=\mathcal{O}(\mathfrak{g}'_J)\). We are now in the position to apply Theorem 51 and conclude this case.

Therefore we only need to deal with the case \({\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))\subset {\rm Rad}(\mathfrak{g})\). We know that \(\mathfrak{g}/{\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))\) has an abelian complex structure. By Theorem 42, there exists a \(J\)-semisimple subalgebra \(\mathfrak{s}\) such that \[\frac{\mathfrak{g}}{{\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))}=\mathfrak{s}\ltimes \frac{{\rm Rad}(\mathfrak{g})}{{\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))}\,, \quad \mathfrak{s}\simeq \frac{\mathfrak{g}}{{\rm Rad }(\mathfrak{g})}\,.\] Consider the \(J\)-invariant subalgebra \(\hat{\mathfrak{s}}=\pi^{-1}(\mathfrak{s})\) of \(\mathfrak{g}\), where \(\pi\colon \mathfrak{g} \to \frac{\mathfrak{g}}{{\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))}\) is the canonical projection onto the quotient. As in Step 1 of Theorem 42, we deduce \({\rm Rad}(\hat{\mathfrak{s}})={\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))\). This, together with the fact that \(\mathfrak{s}=\hat{\mathfrak{s}}/{\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))\) is \(J\)-perfect, implies that \(\hat{\mathfrak{s}}\subseteq \mathcal{O}(\mathfrak{g}'_J)\). Using this last fact and the fact that \(\hat{\mathfrak{s}}\) is a subalgebra, we see that \(\hat{ \mathfrak{s}}\) acts holomorphically on \(\hat{\mathfrak{s}}_J'\). Now, by Corollary 55, we observe that \[\begin{align} \frac{\hat{\mathfrak{s}}}{{\rm Rad}(\hat{\mathfrak{s}})}\simeq \frac{\mathfrak{g}}{{\rm Rad}(\mathfrak{g})}=\frac{\mathcal{O}(\mathfrak{g}'_J)+{\rm Rad}(\mathfrak{g})}{{\rm Rad}(\mathfrak{g})}\simeq \frac{\mathcal{O}(\mathfrak{g}'_J)}{{\rm Rad}(\mathfrak{g})\cap \mathcal{O}(\mathfrak{g}'_J)}=\frac{\mathcal{O}(\mathfrak{g}'_J)}{{\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))} \end{align}\] Thus, we are in the position to apply Lemma 49 to conclude that \(H^2(\hat{\mathfrak{s}}, JU_{\hat{\mathfrak{s}}})=0\). Hence, \(\hat{\mathfrak{s}}\) satisfies the hypotheses of Theorem 51, which allows us to write \(\hat{\mathfrak{s}}=\mathfrak{h}\ltimes {\rm Rad}(\mathcal{O}(\mathfrak{g}'_J)).\) It follows that \(\mathfrak{g}=\mathfrak{h}\ltimes {\rm Rad}(\mathfrak{g})\), which concludes the existence part of the theorem.

The only thing left to prove is uniqueness of the \(J\)-adapted Levi subalgebra. The argument is by induction on \(\dim_{\mathbb{C}}\mathfrak{g}\) and goes as in the proof of Theorem 42, replacing \(\mathfrak{z}(\mathfrak{g})\) with \(\mathfrak{g}'_J\). The only case that is slightly different is when \(\mathfrak{g}\) is \(J\)-perfect and \(0\neq \mathfrak{g}'_J={\rm Rad}(\mathfrak{g})\), so we provide some details. Let \(\mathfrak{h}_1,\mathfrak{h}_2\) be \(J\)-adapted Levi subalgebras of \(\mathfrak{g}\). Let \(p_{\mathfrak{h}_1}\) and \(p_{\mathfrak{g}'_J}\) be the holomorphic projections of \(\mathfrak{g}\) onto \(\mathfrak{h}_1\) and \(\mathfrak{g}'_J\), respectively. For any \(Z_1, Z_2\in \mathfrak{h}_2\), we have \[p_{\mathfrak{h}_1}([Z_1, Z_2])+ p_{\mathfrak{g}'_J}([Z_1, Z_2])=[Z_1, Z_2]=[p_{\mathfrak{h}_1}(Z_1), p_{\mathfrak{h}_1}(Z_2)]+[p_{\mathfrak{h}_1}(Z_1), p_{\mathfrak{g}'_J}(Z_2)]+[p_{\mathfrak{g}'_J}(Z_1), p_{\mathfrak{h}_1}(Z_2)]\,,\] and thus \(p_{\mathfrak{h}_1}\) is a holomorphic Lie algebra homomorphism and \[\label{boloseipazzo} p_{\mathfrak{g}'_J}([Z_1, Z_2])=[p_{\mathfrak{h}_1}(Z_1), p_{\mathfrak{g}'_J}(Z_2)]+[p_{\mathfrak{g}'_J}(Z_1), p_{\mathfrak{h}_1}(Z_2)]\,.\tag{18}\] Now, choosing \(Z_1=JU_{\mathfrak{h}_2}\) and \(Z_2\in \mathfrak{h}_2'\), we get from 18 \[2p_{\mathfrak{g}'_J}(Z_2)=[JU_{\mathfrak{h}_1}, p_{\mathfrak{g}'_J}(Z_2)]\,,\] where we used \(2\)-step solvability and the fact that \(p_{\mathfrak{h}_1}(JU_{\mathfrak{h}_2})=JU_{\mathfrak{h}_1}\), thanks to Lemma 35, since \(p_{\mathfrak{h}_1}\vert_{\mathfrak{h}_2}\) is a holomorphic Lie algebra isomorphism. But then \(p_{\mathfrak{g}'_J}(Z_2)\) must vanish, since otherwise \(2\) would be an eigenvalue for \(\operatorname{ad}_{JU_{\mathfrak{h}_1}}\vert_{\mathfrak{g}'_J}\), which is prohibited by our assumptions. As \(\mathfrak{h}_2\) is \(J\)-perfect we infer \(p_{\mathfrak{g}'_J}|_{\mathfrak{h}_2}=0\) and thus \(\mathfrak{h}_2=\mathfrak{h}_1\), as desired. ◻

Proof. We know that \(\mathcal{O}(\mathfrak{g}_J')\) is a \(J\)-invariant ideal of \(\mathfrak{g}\) which inherits an SKT metric. Since \(\mathfrak{g}_J'\) is a \(J\)-invariant ideal of \(\mathcal{O}(\mathfrak{g}'_J)\), we can use Theorem 42 to infer that \[\frac{\mathcal{O}(\mathfrak{g}'_J)}{\mathfrak{g}_J'}=\mathfrak{h}\ltimes \frac{{\rm Rad}(\mathcal{O}(\mathfrak{g}'_J))}{\mathfrak{g}_J'}\,, \quad \mathfrak{h}\simeq \frac{\mathcal{O} (\mathfrak{g}_J')}{{\rm Rad}(\mathcal{O}(\mathfrak{g}_J'))}\,.\] We consider the subalgebra \(\mathfrak{h}_U=\langle U, JU \rangle\) of \(\mathfrak{h}\) and its preimage \(\pi^{-1}(\mathfrak{h}_U)\) via the canonical projection onto the quotient \(\pi\colon \mathcal{O}(\mathfrak{g}'_J) \to \mathcal{O}(\mathfrak{g}'_J)/ \mathfrak{g}_J'\). Observe that \(\pi^{-1}(\mathfrak{h}_U)\) is an almost-abelian \(J\)-invariant subalgebra of \(\mathfrak{g}\). Indeed, \(\mathfrak{a}:=\pi^{-1}(\langle U\rangle)=\pi^{-1}(\mathfrak{h}_U')\) is an abelian ideal of \(\pi^{-1}(\mathfrak{h}_U)\), since \(U\in (\mathcal{O}(\mathfrak{g}'_J)/\mathfrak{g}_J')'=\mathcal{O}(\mathfrak{g}'_J)'/\mathfrak{g}_J'\) and \(\mathfrak{g}\) is \(2\)-step solvable. Since \(\mathcal{O}(\mathfrak{g}'_J)\) is SKT, \(\pi^{-1}(\mathfrak{h}_U)\) admits an SKT metric \(g\). We consider an orthogonal complement \(\langle U', JU'\rangle\) of \(\mathfrak{g}_J'\) in \(\pi^{-1}(\mathfrak{h}_U)\) with respect to \(g\). Clearly \(U'=bU+X\), for some \(X\in\mathfrak{g} _J'\) and \(b\in \mathbb{R}\). Therefore, we have \([JU',U' ]=2bU'+ v,\) for some \(v\in\mathfrak{g}_J'\), and thus \(\operatorname{ad}_{JU'}|_{\mathfrak{g}_J'}=b\operatorname{ad}_{JU}|_{\mathfrak{g}_J'}\). We can further suppose, up to scaling, that \(|U'|_g=1\). Now, we are in the position to apply [58] or [59] with \(a=2b\), to infer that the real parts of the eigenvalues of \(\operatorname{ad}_{JU}|_{\mathfrak{g}_J'}\) are either \(0\) or \(-1\). This concludes the proof of the first part.

We can now apply Theorem 56 to obtain \[\mathfrak{g}=\mathfrak{h}\ltimes {\rm Rad}(\mathfrak{g})\,,\] where \(\mathfrak{h}\) is a \(J\)-semisimple subalgebra. Using Corollary 33, we have that \(\mathfrak{h}\) is a direct sum of \(p\) copies of \(\mathfrak{aff}(\mathbb{R})\) and \(q\) copies of \(\mathfrak{aff}(\mathbb{C})\). On the other hand, \(\mathfrak{h}\) is again SKT and hence \(q=0\), since \(\mathfrak{aff}(\mathbb{C})\) does not admit any SKT metric. ◻

Proof. Firstly, we prove that \(\mathfrak{g}'\subseteq \mathfrak{a}\). Since \(\mathfrak{a}\) is an ideal, it is sufficient to prove that \([J\mathfrak{a},J\mathfrak{a}]\subseteq \mathfrak{a}\). Fix \(X, Y\in\mathfrak{a}\), exploiting integrability, we have \[\mathfrak{g}'\ni [JX, JY]=J([JX, Y]+[X, JY])\in J \mathfrak{g}'\,,\] concluding that \([JX, JY]\in\mathfrak{g}_J'\subseteq \mathfrak{a}_J\). Similarly to the proof of Lemma 4 we deduce that \(\mathfrak{a}_J\) is an ideal. In addition, using that \(\mathfrak{a}_J\) is abelian, for every \(X\in \mathfrak{a}_J\) and \(Y\in \mathfrak{g}\) we have \(\operatorname{Im}(J\operatorname{ad}_{[X, Y]})\subseteq \mathfrak{a}_J\) and \(J\operatorname{ad}_{[X, Y]}|_{\mathfrak{a}_J}=0\). Consequently \[\eta(X, Y)=-\frac{1}{2}{\rm tr}(J\operatorname{ad}_{[X,Y]})=0\,, \qquad X\in \mathfrak{a}_J\,,\,\,Y \in\mathfrak{g}\,,\] showing that \(\mathfrak{a}_J \subseteq \ker \eta\). Consider a Hermitian metric \(g\) and fix \(Y\in (\mathfrak{a}_J)^{\perp}\cap \mathfrak{a}\), where the orthogonal is with respect to \(g\). Since \(\mathfrak{a}\) is an abelian ideal, \([Y, JY]\in\mathfrak{a}\) and we have that \(\operatorname{ad}_{[Y, JY]}\vert_{\mathfrak{a}_J}=0\). Thus \[\begin{align} \eta(Y, JY)&=-\frac{1}{2} \operatorname{tr}(J\operatorname{ad}_{[Y,JY]} \vert_{\mathfrak{a}_J})- \frac{1}{2}\operatorname{tr}(J \operatorname{ad}_{[Y,JY]}\vert_{(\mathfrak{a}_J)^\perp})=-\frac{1}{2}\operatorname{tr}(J\operatorname{ad}_{[Y,JY]^{\perp}}\vert_{ (\mathfrak{a}_J)^{\perp}})\,, \end{align}\] where \([Y,JY]^{\perp}\) is the orthogonal projection on \((\mathfrak{a}_J)^{\perp}\). Thus, we obtain that \[\eta(Y, JY) =-\frac{1}{2}{\rm tr}^{\mathfrak{g}/\mathfrak{a}_J}(J\operatorname{ad}_{[\pi(Y), J\pi(Y)]})\,.\] The claim that \(\eta=\pi^*\eta_{\mathfrak{g}/\mathfrak{a}_J}\) follows from the above equation and from the inclusion \(\mathfrak{a}_J\subseteq \ker \eta\).

Furthermore, we note that \(\ker \eta=\pi^{-1}(\ker\eta_{\mathfrak{g}/\mathfrak{a}_J})=\pi^{-1}(\mathfrak{n}_J)\), where \(\mathfrak{n}_J\) is the maximal nilpotent \(J\)-invariant ideal of \(\mathfrak{g}/\mathfrak{a}_J\). Using that \(\pi\) is a holomorphic Lie algebra homomorphism, we get that \(\ker \eta\) is a \(J\)-invariant ideal of \(\mathfrak{g}\). Finally, by Lemma 12 \(\ker\eta\) is also \(J\)-solvable. Thus, \(\ker \eta\subseteq {\rm Rad}(\mathfrak{g})\).

Next, we claim that \(\mathfrak{n}_J={\rm Rad}(\mathfrak{g}/ \mathfrak{a}_J)\). Indeed, since \(\mathfrak{g}_J'\subseteq \mathfrak{a}_J\), we have that \(\mathfrak{g}/\mathfrak{a}_J\) has an abelian complex structure \(J\). Moreover, clearly, \(\mathfrak{g}/\mathfrak{a}_J=\mathfrak{a}/\mathfrak{a}_J\oplus J \mathfrak{a}/\mathfrak{a}_J.\) In addition, since \(\mathfrak{g}'\subseteq \mathfrak{a}\), we obtain \((\mathfrak{g}/\mathfrak{a}_J)'_J=(\mathfrak{g}'/\mathfrak{a}_J)\cap J(\mathfrak{g}'/\mathfrak{a}_J)\subseteq (\mathfrak{a}/\mathfrak{a}_J)\cap J(\mathfrak{a}/\mathfrak{a}_J)=0\). Then, using Theorem 41, we conclude the proof of the claim. In particular, since \[\frac{\mathfrak{g}}{\ker \eta}\simeq \frac{\mathfrak{g}/\mathfrak{a}_J}{\pi^{-1}(\mathfrak{n}_J)/\mathfrak{a}_J}\simeq \frac{\mathfrak{g}/\mathfrak{a}_J}{\mathfrak{n}_J}\,,\]this gives that \(\mathfrak{g}/ \ker \eta\) is \(J\)-semisimple and hence \({\rm Rad}(\mathfrak{g})=\ker \eta\), which concludes the proof. ◻

Proof. It is easy to see that \(\eta|_{\mathfrak{\mathfrak{g}}'+J\mathfrak{g}' }=\eta_{\mathfrak{g}'+J\mathfrak{g}' }\,\). We observe that, since \(\mathfrak{g}\) is a \(2\)-step solvable Lie algebra, \(\mathfrak{g}'\) is an abelian ideal of \(\mathfrak{g}'+J\mathfrak{g}'\). We are in the position to apply Proposition 58 to \(\mathfrak{g}'+J\mathfrak{g}'\), and obtain, assuming \(\eta=0\), that \(\mathfrak{g}'+J\mathfrak{g}'={\rm Rad}(\mathfrak{g}'+J\mathfrak{g}')\), which clearly gives that \(\mathfrak{g}\) is \(J\)-solvable.

Viceversa, let us suppose that \(J\) is solvable. In particular, by Proposition 58, \(\mathfrak{g}'+J\mathfrak{g}'\) is \(J\)-solvable and equal to \(\ker \eta_{\mathfrak{\mathfrak{g}}'+J\mathfrak{g}'}=\pi^{-1}(\mathfrak{n}_J)\), where \(\pi\colon \mathfrak{g}'+J\mathfrak{g}' \to (\mathfrak{g}'+J\mathfrak{g}')/\mathfrak{g}_J'\) is the canonical projection onto the quotient and \(\mathfrak{n}_J\) is the maximal \(J\)-invariant nilpotent ideal of \((\mathfrak{g}'+J\mathfrak{g}')/\mathfrak{g}_J'\). Thus, \(( \mathfrak{g}'+J\mathfrak{g}')/\mathfrak{g}_J'=\mathfrak{n}_J\) is nilpotent. We will use this piece of information to conclude that \(\eta=0\). We know that \(\eta=0\) is equivalent to require that \(\beta|_{\mathfrak{g}'}=0\), where \(\beta\) is the potential \(1\)-form of \(\eta\), see 3 . Since \(\mathfrak{g}_J'\) is an abelian ideal in \(\mathfrak{g}'+J\mathfrak{g}'\) by Lemma 4, we observe that \(\mathfrak{g}'_J \subseteq \ker \beta\). Now, let \(X\in\mathfrak{g}'\), we have \[\beta(X)=\frac{1}{2}{\rm tr}(J\operatorname{ad}_X|_{\mathfrak{g}'+J\mathfrak{g}'})=\frac{1}{2}{\rm tr}^{\mathfrak{n}_J}(J\operatorname{ad}_{\pi(X)})+\frac{1}{2}{\rm tr}(J\operatorname{ad}_X|_{\mathfrak{g}'_J})=\pi^*\beta_{\mathfrak{n}_J}(X)\,,\] where we used that \(\operatorname{ad}_X|_{\mathfrak{g}_J'}=0\), because \(X\in \mathfrak{g}'\). Now, thanks to [60], we know that \(\beta_{\mathfrak{n}_J}=0\), since \(\mathfrak{n}_J\) is nilpotent. This concludes the proof. ◻

Proof. Applying Theorem 57 we immediately know that \(\mathfrak{g}=\mathfrak{aff}(\mathbb{R})^p\ltimes {\rm Rad}(\mathfrak{g}),\) for some \(p\ge 0\). Suppose by contradiction that \(p\neq 0\) and consider the holomorphic quotient map \(\pi \colon \mathfrak{g}\to \mathfrak{g}/{\rm Rad}(\mathfrak{g})\simeq \mathfrak{aff}(\mathbb{R})^p\). We know that \(\mathfrak{aff}(\mathbb{R})^p\) is endowed with an exact Kähler form \(-\eta\). Therefore, \(-\pi^*\eta\) gives rise to a non-zero, semi-positive \((1,1)\)-form which is exact. We can now use Corollary 18 to conclude \(p=0\) and thus \(\mathfrak{g}\) must be \(J\)-solvable. In particular, Proposition 59 tells us that \(\rho=0\). In this scenario we can then infer that there exists a Kähler metric invoking [36]. ◻

Proof. By Proposition 58 and Theorem 41, we know that \(\eta=\pi^*\eta_{\mathfrak{g}/\mathfrak{g}'_J}\) and \(\mathfrak{g}/\mathfrak{g}'_J=\mathfrak{h}\ltimes \mathfrak{n}_J\), where \(\mathfrak{n}_J=\ker \eta_{\mathfrak{g}/\mathfrak{g}'_J}\) and \(\mathfrak{h}\simeq \mathfrak{g}/\ker \eta\). Since \(\eta_{\mathfrak{g}/\mathfrak{g}'_J}\) makes the \(J\)-simple ideals of \(\mathfrak{h}\) orthogonal, the conclusion follows from Corollary 43. ◻

Proof. Using Proposition 59, we have that \(\eta=0\), which is equivalent to \(\beta|_{\mathfrak{g}'}=0\). The claim then follows from the abelianity of \(J\) and the observation that, for any \(X\in\mathfrak{g}'\), \(\beta(X)=-\frac{1}{2}{\rm tr}(\operatorname{ad}_{JX})=0.\) ◻

Proof. Using the same argument as in the proof of Proposition 58, we infer that \(\mathfrak{a}\subseteq \ker \rho.\) Now, we decompose \(\mathfrak{g}=\mathfrak{a}^{\perp}\oplus \mathfrak{a}\), where the orthogonal complement is taken with respect to \(g\). As a consequence, for the claim to be true it suffices to prove that \(\rho(X, JX)\le 0\), for all \(X\in\mathfrak{a}^{\perp}\). For any such \(X\), since \(\mathfrak{a} \subseteq C(\mathfrak{g}')\) and \(\mathfrak{a}\) is an ideal, we have \[\begin{align} \rho(X, JX)=&\, -\frac{1}{2}{\rm tr}(J\operatorname{ad}_{[X, JX]}|_{\mathfrak{a}^{\perp}})-\frac{1}{2}{\rm tr}(J\operatorname{ad}_{[X, JX]}|_{\mathfrak{a}}) = -\frac{1}{2}{\rm tr}(J\operatorname{ad}_{[X, JX]}|_{\mathfrak{a}^{\perp}})\\ =&\, -\frac{1}{2}{\rm tr}(J\operatorname{ad}_{[X, JX]^{\perp}}|_{\mathfrak{a}^{\perp}})=-\frac{1}{2}{\rm tr}^{\mathfrak{g}/\mathfrak{a}}(J\operatorname{ad}_{\pi([X, JX])})\,, \end{align}\] where \(\pi \colon \mathfrak{g}\to \mathfrak{g}/\mathfrak{a}\) is the canonical projection onto the quotient and \([X, JX]^{\perp}\) is the orthogonal projection of \([X, JX]\) onto \(\mathfrak{a}^{\perp}\). Exploiting the fact that \(\mathfrak{g}/\mathfrak{a}\) has an abelian complex structure, since \(\mathfrak{g}_J'\subseteq \mathfrak{a}\), we infer \[\begin{align} \rho(X, JX)=&\, -\frac{1}{2}{\rm tr}^{\mathfrak{g}/\mathfrak{a}}(J\operatorname{ad}_{\pi([X, JX])})=\frac{1}{2}{\rm tr}^{\mathfrak{g}/\mathfrak{a}}(\operatorname{ad}_{\pi(J[X, JX])}) =\frac{1}{2}{\rm tr}(\operatorname{ad}_{J[X, JX]}|_{\mathfrak{a}^{\perp}})=-\frac{1}{2}{\rm tr}(\operatorname{ad}_{J[X, JX]}|_{\mathfrak{a}})\,, \end{align}\] where we used once again that \(\mathfrak{a}\) is an abelian ideal of \(\mathfrak{g}\) and unimodularity. Now, we apply Lemma 19 choosing \(Y\in \mathfrak{a}\), and obtain \[\begin{align} 0=&\,g([J[X, JX], JY], JY)+g([J[X, JX], Y],Y) +g([J[X, Y], X], JY)-g([J[X, JY], X], Y)\\ &\, +g([J[JX, Y], JX], JY) -g([J[JX, JY], JX], Y) -|[X, Y]|^2-|[X, JY]|^2\\ &\, - |[JX, Y]|^2-|[JX, JY]|^2\,. \end{align}\] Summing over an orthonormal basis of \(\mathfrak{a}\) with respect to \(g\), we have \[\begin{align} {\rm tr}(\operatorname{ad}_{J[X, JX]}|_{\mathfrak{a}})=- {\rm tr}((J\operatorname{ad}_X)^2|_{\mathfrak{a}})-{\rm tr}((J{\operatorname{ad}_{JX}})^2|_{\mathfrak{a}})+|\operatorname{ad}_X|_{\mathfrak{a}}|^2+|\operatorname{ad}_{JX}|_{\mathfrak{a}}|^2\,. \end{align}\] On the other hand, by integrability of \(J\), we obtain \[\begin{align} 0&=\mathrm{tr}(J\operatorname{ad}_{[X,JX]}|_{\mathfrak{a}}) =\mathrm{tr}(J\operatorname{ad}_X\operatorname{ad}_{JX}|_{\mathfrak{a}})-\mathrm{tr}(J\operatorname{ad}_{JX}\operatorname{ad}_X|_{\mathfrak{a}})\\ &=-\mathrm{tr}(J\operatorname{ad}_X\operatorname{ad}_{X}J|_{\mathfrak{a}})+\mathrm{tr}(J\operatorname{ad}_XJ\operatorname{ad}_{X}|_{\mathfrak{a}})-\mathrm{tr}(J\operatorname{ad}_XJ\operatorname{ad}_{JX}J|_{\mathfrak{a}})-\mathrm{tr}(J\operatorname{ad}_{JX}\operatorname{ad}_X|_{\mathfrak{a}}) \\ &=\mathrm{tr}\left((\operatorname{ad}_X)^2|_{\mathfrak{a}}\right)+\mathrm{tr}\left((J\operatorname{ad}_X)^2|_{\mathfrak{a}}\right)\,. \end{align}\] Hence, using the above equation, we get \[\begin{align} \rho(X, JX)=-\frac{1}{2}{\rm tr}(\operatorname{ad}_{J[X, JX]}|_{\mathfrak{a}})=&\, -\frac{1}{2}\left( {\rm tr}((\operatorname{ad}_X)^2|_{\mathfrak{a}})+{\rm tr}(({\operatorname{ad}_{JX}})^2|_{\mathfrak{a}})+|\operatorname{ad}_X|_{\mathfrak{a}}|^2+|\operatorname{ad}_{JX}|_{\mathfrak{a}}|^2\right)\\ =&\, -2(|{\rm Sym}(\operatorname{ad}_X|_{\mathfrak{a}})|^2+|{\rm Sym}(\operatorname{ad}_{JX}|_{\mathfrak{a}})|^2)\le 0\,. \end{align}\] If \(\rho=0\), then clearly \(\operatorname{ad}_X|_{\mathfrak{a}}\in \mathfrak{so}(\mathfrak{a}, g)\). The only thing remained to check is that \([\operatorname{ad}_X, J]|_{\mathfrak{a}}=0\), for any \(X\in\mathfrak{a}^{\perp}\). Recalling the proof of Lemma 21, we have \[\frac{1}{2}{\rm tr}([\operatorname{ad}_X|_{\mathfrak{a}}, J|_{\mathfrak{a}}]^2 )=\frac{1}{2}{\rm tr}([\operatorname{ad}_X, J]^2 |_{\mathfrak{a}})={\rm tr}(J\operatorname{ad}_{[X, JX]}|_{\mathfrak{a}})=0\,.\] On the other hand, since \(\operatorname{ad}_{X}|_{\mathfrak{a}}, J\in\mathfrak{so}(\mathfrak{a}, g)\), then \([\operatorname{ad}_X, J]|_{\mathfrak{a}}\in \mathfrak{so}(\mathfrak{a}, g)\). Hence, \({\rm tr}([\operatorname{ad}_X, J]^2|_{\mathfrak{a}})=0\) if and only if \([\operatorname{ad}_X, J]|_{\mathfrak{a}}=0\). Finally, if we assume that \(\mathfrak{g}\) is completely solvable, then \(\operatorname{ad}_X|_{\mathfrak{a}}=0\), for any \(X\in\mathfrak{g}\), forcing \(\mathfrak{g}_J'\subseteq\mathfrak{a}\subseteq \mathfrak{z}(\mathfrak{g}).\) Now, if \(\mathfrak{g}\) is \(2\)-step with \(\rho=0\), then \(J\) is solvable and \((\mathfrak{g}'+J\mathfrak{g}')/\mathfrak{g}_J'\) is nilpotent, by Proposition 59. This, together with the fact that \(\mathfrak{g}_J'\) is central, gives that \(\mathfrak{g}'+J \mathfrak{g}'\) is nilpotent and hence \(2\)-step since it is SKT, see [50]. ◻

Proof. By Corollary 62, we know that \(\mathfrak{g}'+J\mathfrak{g}'\) is unimodular. Using this fact and Proposition 24, part (2), we deduce that \(B^{1,1}=0\), thereby proving the first statement. It follows that \(\ker B\) is \(J\)-invariant ideal, since \(\mathfrak{n} \subseteq \ker B\). Let us now fix \(X\in \mathfrak{n}\) and we prove that \({\rm tr}(\operatorname{ad}_{JX}^k)=0\), for any \(k\ge 2\).

For \(k=2\), we have that \[0=B(X, X)=-B(JX, JX)=-{\rm tr}(\operatorname{ad}_{JX}^2)\,.\] Now, let us suppose that \(k\ge 3\). The strategy is the same as in the proof of Proposition 24, part (4). Using ?? and Lie’s Theorem, we can conclude, iterating \(k-1\) times the argument, that \[{\rm tr}(\operatorname{ad}_{JX}^k)={\rm tr}(\operatorname{ad}_{JX}^{k-2}\operatorname{ad}_{J[JX, X]})={\rm tr}(\operatorname{ad}_{J[JX, [JX, \ldots, [JX, X]]\ldots]})=0\,,\] by unimodularity of \(\mathfrak{g}'+J\mathfrak{g}'\). Hence, \({\rm tr}(\operatorname{ad}_{JX}^k)=0\), for any \(k\ge 2\), which implies \(JX\in \mathfrak{n}\), and thus forces \(\mathfrak{n}\) to be \(J\)-invariant. ◻

Proof. We prove (1). The proof of (2) is similar once one observes that Lie algebras of rigid type are always unimodular, hence \(J\)-solvable by Corollary 44, part (1). Applying Lemma 65, we know that the Killing form is of type \((2,0)+(0,2)\). On the other hand, by complete solvability, we have \[0\le {\rm tr}(\operatorname{ad}_X^2)=B(X, X)=-B(JX, JX)=-{\rm tr}(\operatorname{ad}_{JX}^2)\le 0\,,\] and hence \({\rm tr}(\operatorname{ad}^2_X)=0\), for any \(X\in\mathfrak{g}\). Using again complete solvability, \({\rm tr}(\operatorname{ad}_X^2)=0\) implies that \(\operatorname{ad}_X\) is nilpotent, for any \(X\in\mathfrak{g}\), concluding the proof. ◻

Proof. As for the first claim, by Remark 64, the normaliser \(N_{\mathfrak{g}}\) of \(\mathfrak{g}_J'\) in \(\mathfrak{g}\) satisfies the hypotheses of Proposition 63. Furthermore \(N_{\mathfrak{g}}\) is \(J\)-solvable and completely solvable. Applying the last part of Proposition 63, we have that \(\mathfrak{g}_J'\subseteq \mathfrak{z}(N_{\mathfrak{g}})\cap (\mathfrak{g}'+J \mathfrak{g}')\subseteq \mathfrak{z}(\mathfrak{g}'+J \mathfrak{g}')\). Since \(J\) is solvable, we know that \((\mathfrak{g}'+J\mathfrak{g}')/ \mathfrak{g}_J'\) is nilpotent and \(\mathfrak{g}_J'\subseteq \mathfrak{z}(\mathfrak{g}'+J \mathfrak{g}')\), implying that \(\mathfrak{g}'+J \mathfrak{g}'\) is nilpotent and SKT. Using [51], we obtain that \(\mathfrak{z}(\mathfrak{g}'+J \mathfrak{g}')\) is \(J\)-invariant. Hence, we are in the position to apply Proposition 63 with \(\mathfrak{a}=\mathfrak{z}(\mathfrak{g}'+J \mathfrak{g}' )\), which allows us to infer that \(\mathfrak{g}_J'\subseteq\mathfrak{z}(\mathfrak{g}'+J\mathfrak{g}')\subseteq \mathfrak{z}(\mathfrak{g})\). Therefore, taking the quotient by \(\mathfrak{g}_J'\), we obtain a completely solvable Lie algebra with a solvable abelian complex structure. Now, we can use Corollary 66, part (1), to conclude that \(\mathfrak{g}/\mathfrak{g}_J'\) is nilpotent, which establishes the claim, using that \(\mathfrak{g}_J'\subseteq \mathfrak{z}(\mathfrak{g}).\)

The second claim is now a straightforward consequence of Theorem 57, together with the first part of the proof applied to \({\rm Rad}(\mathfrak{g})\). ◻

Proof. We will again make use of Remark 64. We know that \(N_{\mathfrak{g}}\) is a \(J\)-invariant ideal of \(\mathfrak{g}\) that contains \(\mathfrak{g}'+J \mathfrak{g}'\). In addition, \(\mathfrak{g}'_J\) is an ideal of \(N_{\mathfrak{g}}\) that satisfies the hypotheses of Proposition 63. This allows us to infer that \(\mathfrak{g}'+J\mathfrak{g}'\) acts skew-symmetrically on \(\mathfrak{g}_J'\). On the other hand, being \(J\) solvable, we know that \(\frac{\mathfrak{g}'+J\mathfrak{g}'}{\mathfrak{g}_J'}\) is nilpotent. Now, using the fact that \(\mathfrak{g}\) is of real type, we can infer that \(\mathfrak{g}'+J\mathfrak{g}'\) is nilpotent and \(\mathfrak{g}_J'\subseteq \mathfrak{z}(\mathfrak{g} '+J \mathfrak{g}')\). Finally, applying [50], we conclude that \(\mathfrak{g}'+J \mathfrak{g}'\) is \(2\)-step nilpotent. ◻

Proof. Fix an element \(X\in \mathfrak{g}\) and consider the adjoint representation \[\operatorname{ad}_X-\sqrt{-1}\operatorname{ad}_{JX}=\operatorname{ad}_{X^{1,0}}=\begin{pmatrix} A_X & *\\ 0 & * \end{pmatrix}\,,\] where the matrix is written with respect to the decomposition \(\mathfrak{g}_\mathbb{C}=\mathfrak{g}^{1,0}\oplus \mathfrak{g}^{0,1}\). By Lie’s Theorem we may pick a basis \(\{Z_1,\dots,Z_{n}\}\) of \(\mathfrak{g}^{1,0}\) with respect to which the block \(A_X\) is upper triangular. Set \(Z_j=\frac{1}{\sqrt{2}}(e_{2j-1}-\sqrt{-1}e_{2j})\) so that \(e_{2j}=Je_{2j-1}\). Choose a Hermitian metric \(g\) on \(\mathfrak{g}\) having \(\{e_1,\dots,e_{2n}\}\) as an orthonormal basis and extend it bilinearly to \(\mathfrak{g}_\mathbb{C}\) so that \(\{Z_1,\dots,Z_{n}\}\) is a unitary basis. We then compute \[\begin{align} \mathrm{tr}(A_X)&=\sum_{j=1}^{n}g(A_XZ_j,\bar Z_j)=\frac{1}{2}\sum_{j=1}^{n}g\left([X-\sqrt{-1}JX,e_{2j-1}-\sqrt{-1}e_{2j}], e_{2j-1}+\sqrt{-1}e_{2j}\right)\\ &=\frac{1}{2}\left( \mathrm{tr}(\operatorname{ad}_X)-\mathrm{tr}(J\operatorname{ad}_{JX}) - \sqrt{-1}\, \mathrm{tr}(\operatorname{ad}_{JX})-\sqrt{-1}\,\mathrm{tr}(J\operatorname{ad}_X) \right)\,. \end{align}\] Choosing \(X\in J\mathfrak{n}\) we have that \(\operatorname{ad}_{JX}\) is nilpotent, hence the eigenvalues of \(\operatorname{ad}_{X^{1,0}}\) are those of \(\operatorname{ad}_X\).

In particular, if \(\mathfrak{g}\) is of rigid type, then \(\mathrm{tr}(A_X)\) is purely imaginary, and it follows that \(\mathrm{tr}(J\operatorname{ad}_{JX})=0\), for all \(X\in J\mathfrak{n}\), because \(\mathfrak{g}\) is unimodular. From this we immediately deduce \[\rho(X,Y)=-\frac{1}{2}\mathrm{tr}(J\operatorname{ad}_{[X,Y]})+\frac{1}{2}\mathrm{tr}(\operatorname{ad}_{J[X,Y]})=0\,, \qquad X,Y\in \mathfrak{g}\,,\] which proves (2).

Assume instead that \(\mathfrak{g}\) is completely solvable. Then \(\mathrm{tr}(A_X)\) is real and so \(\mathrm{tr}(J\operatorname{ad}_X)=0\), for all \(X\in J\mathfrak{n}\). This implies that \(\sigma\) vanishes on \(J\mathfrak{g}'\). On the other hand \(\rho=d\sigma=-\sigma \vert_{\mathfrak{g}'}\) therefore Chern–Ricci flatness implies that \(\sigma\) vanishes on \(\mathfrak{g}'+J\mathfrak{g}'\). Now, we have \(0=\sigma\vert_{\mathfrak{g}'+J\mathfrak{g}'}=\sigma_{\mathfrak{g}'+J\mathfrak{g}'}\) hence there exists a closed, nowhere vanishing \((k,0)\)-form \(\Psi\) on \(\mathfrak{g}'+J\mathfrak{g}'\), where \(k=\dim_\mathbb{C}(\mathfrak{g}'+J\mathfrak{g}')\), see [61]. Pick an orthogonal complement \(\mathfrak{k}\) of \(\mathfrak{g}'+J\mathfrak{g}'\) in \(\mathfrak{g}\) and a basis \(z_1,\dots,z_{n-k}\) of \(\mathfrak{k}^{1,0}\). Let \(\varphi^1,\dots,\varphi^{n-k}\) be the dual \((1,0)\)-forms. Since \(\mathfrak{g}' \subseteq \mathfrak{k}^\perp\) the form \(\varphi^1\wedge \dots \wedge \varphi^{n-k}\wedge \Psi\) is closed and nowhere vanishing on \(\mathfrak{g}\) which then implies \(\sigma=0\), again by [61]. ◻

Proof. If the canonical bundle of \(M\) is holomorphically trivial clearly \(\rho=0\). By Proposition 69 we deduce \(\sigma=0\) which implies, together with [61], that there exists a left-invariant holomorphic volume form on \(G\), which then descends to \(M\), concluding the proof. ◻

Proof. We prove the corollary by induction on the solvability step \(s\) of \(J\). If \(s=2\) then the result follows from Proposition 59. Let us assume that the result is true for \(s-1\) and let us prove it for \(s\). If \(\mathfrak{g}\) is \(s\)-step \(J\)-solvable the ideal \(\mathfrak{g}' + J \mathfrak{g}'\) is \((s-1)\)-step \(J\)-solvable, thus, by the inductive assumption, \(\rho_{\mathfrak{g}'+J\mathfrak{g}'}=0\). But then, from Proposition 69 we infer that \(\sigma_{\mathfrak{g}'+J\mathfrak{g}'}=0\). In particular \(\rho=\sigma \vert_{\mathfrak{g}'}=0\). ◻

Proof. We will prove the statement using induction on the step of solvability of \(J\). Let \(J\) be a \(s\)-step solvable complex structure. If \(s=1\), then \(\mathfrak{g}'+J\mathfrak{g}'\) is abelian, hence \(\mathfrak{g}\) is \(2\)-step solvable. We can use Theorem 67 to conclude. Let us assume that the statement is true for complex structures which are \((s-1)\)-step solvable and assume that \(J\) is \(s\)-step solvable. We can then consider \(\mathfrak{k}:=\mathfrak{g}'+J\mathfrak{g}'\) which is unimodular, SKT and completely solvable so that \(\mathfrak{k}_J'\subseteq \mathfrak{g}_J'\), hence abelian. Clearly, the complex structure on \(\mathfrak{k}\) is \((s-1)\)-step solvable. Hence, by the inductive hypothesis, \(\mathfrak{k}\) is \(2\)-step nilpotent. Now, thanks to [51] and [50], we know respectively that \(\mathfrak{z}(\mathfrak{k})\) is a \(J\)-invariant abelian ideal of \(\mathfrak{g}\) and that \(\mathfrak{g}_J'\subseteq \mathfrak{z}(\mathfrak{k})\). Since \(J\) is solvable, we can now apply Corollary 71 and infer that \(\rho =0\). On the other hand, \(\mathfrak{z}(\mathfrak{k})\) can be chosen as the abelian \(J\)-invariant ideal of \(\mathfrak{g}\) in Proposition 63, which implies that \(\mathfrak{g}\) acts skew-symmetrically on \(\mathfrak{z}(\mathfrak{k}).\) We are in the position to use complete solvability and infer that \(\mathfrak{g}_J'\subseteq \mathfrak{z}(\mathfrak{k})\subseteq \mathfrak{z}(\mathfrak{g}).\) In order to conclude, we just need to observe that \(\mathfrak{g}/ \mathfrak{z}(\mathfrak{g})\) is a completely solvable Lie algebra with a solvable abelian complex structure. By Corollary 66, we conclude that \(\mathfrak{g}/\mathfrak{z}(\mathfrak{g})\) is nilpotent and hence \(\mathfrak{g}\) is nilpotent as well, as claimed. ◻

Proof. Assume that \(J\) is not solvable. Then \(\mathfrak{h}:=\mathfrak{g}/ {\rm Rad}(\mathfrak{g})\) is a completely solvable Lie algebra with abelian semisimple complex structure \(J\). Using Proposition 24, part \((2)\), we have that \(-\eta_{\mathfrak{h}}\) is \(d\)-exact and non-negative, thanks to the fact that \(\mathfrak{g}\) is completely solvable. Let \(\pi\colon \mathfrak{g} \to \mathfrak{h}\) be the canonical projection onto \(\mathfrak{h}\). Thus, \(-\pi^*\eta_\mathfrak{h}\) is a \(d\)-exact non-negative \((1,1)\)-form, hence it must vanish by Corollary 18. This is equivalent to \(\eta_\mathfrak{h}=0\), and therefore \(B_{\mathfrak{h}}^{1,1}=0\), contradicting the \(J\)-semisimplicity of \(\mathfrak{h}\), by Theorem 27. Using Corollary 71, we conclude that \(\rho=0\). The final claim follows from [36]. ◻