Let us first recall the following definition.

**Definition 3.1:** *A 3-dimensional almost Kenmotsu manifold is called a (**k*, μ, *v*)-almost Kenmotsu manifold if the Reeb vector field satisfies the (*k*, μ, *v*)-nullity condition, that is,
$$\begin{array}{l}R\left(X,\phantom{\rule{thinmathspace}{0ex}}Y\right)\xi =k\left(\eta \left(Y\right)X-\eta \left(X\right)Y\right)+\mu \left(\eta \left(Y\right)hX-\eta \left(X\right)hY\right)+v\left(\eta \left(Y\right){h}^{\prime}X-\eta \left(X\right){h}^{\prime}Y\right)\end{array}$$(11) *for any vector fields X, Y, where k, μ and v are smooth functions*.In the framework of almost Kenmotsu manifolds, some classes of nullity conditions were studied by many authors. We observe that a (*k*, μ, *v*)-nullity condition becomes a

*–*

*k*-nullity condition if *k* is a constant and *μ = v* = 0 (see [10]);

*–*

*generalized **k*-nullity condition if *k* is a function and *μ = v* = 0 (see [11]);

*–*

*(**k, μ*)-nullity condition if *k* and *μ are* constants and v = 0 (see [8]);

*–*

*generalized (**k, μ*)-nullity condition if *k* and *μ are* functions and v = 0 (see [12] and [11]);

*–*

*(**k*, *v*)′-nullity condition if *k* and v are constants and *μ* = 0 (see [8]);

*–*

*generalized (**k*, *v*)′-nullity condition if *k* and *v* are functions and *μ* = 0 (see [12] and [11]).

*Using the above definitions and some results shown in [8] we have***Theorem 3.2:** *Any 3-dimensional non-unimodular Lie group admits a left invariant almost Kenmotsu structure for which the Reeb vector field satisfies the (**k, μ, v*)-nullity condition with k, μ and v being constants.

**Proof:** *By [8, Theorem 5.2] we know that on any 3-dimensional non-unimodular Lie group there exists an almost Kenmotsu structure. Next, we recall the proof of this result shown in [8]. Let **G* be a 3-dimensional non-unimodular Lie group, then there exists a left invariant local orthonormal frame fields {*e*_{1}, *e*_{2}, *e*_{3}} satisfying
$$\begin{array}{l}\left[{e}_{1},{e}_{2}\right]=\alpha {e}_{2}+\beta {e}_{3},\left[{e}_{2},{e}_{3}\right]=0,\phantom{\rule{thinmathspace}{0ex}}\left[{e}_{1},{e}_{3}\right]=\gamma {e}_{2}+\delta {e}_{3}\end{array}$$(12) and *a+δ* = 2, where *a, β, γ, δ ∊* ℝ. We define a metric *g* on *G* by *g(e*_{i}, e_{j} ) = δ_{ij} for 1 ≤ *i*, *j* ≤ 3. Also, we denote by ξ *= −e*_{1} and denote by *η* the dual l-form of ξ. Thus, we may define a (1, 1)-type tensor field ϕ by ϕ(ξ) = 0, ϕ(*e*_{2}) = *e*_{3} and *ϕ(e*_{3}) = *−e*_{2}. Then, one can check that (*G, ϕ, ζ, η*, g) admits a left invariant almost Kenmotsu structure. Next, we prove that the Reeb vector field of this almost Kenmotsu structure satisfies the (*k, μ*, v)-nullity condition with *k, μ*, v being constants. Firstly, using the Levi-Civita equation and (12) we obtain
$$\begin{array}{l}{\mathrm{\nabla}}_{\xi}\xi =0,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\nabla}}_{e2}\xi =\alpha {e}_{2}+\frac{1}{2}\left(\beta +\gamma \right){e}_{3},\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\nabla}}_{e3}\xi =\frac{1}{2}\left(\beta +\gamma \right){e}_{2}+\left(2-\alpha \right){e}_{3},\\ {\mathrm{\nabla}}_{\xi}{e}_{2}=\frac{1}{2}\left(\gamma -\beta \right){e}_{3,}{\mathrm{\nabla}}_{e2}{e}_{2}=-\alpha \xi ,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\nabla}}_{e3}{e}_{2}=-\frac{1}{2}\left(\beta +\gamma \right)\xi ,\\ {\mathrm{\nabla}}_{\xi}{e}_{3}=\frac{1}{2}\left(\beta +\gamma \right){e}_{2},{\mathrm{\nabla}}_{e2}{e}_{3}=-\frac{1}{2}\left(\beta +\gamma \right)\xi ,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\nabla}}_{e2}{e}_{3}=\left(\alpha -2\right)\xi .\end{array}$$(13) Using (13), by a straightforward calculation we obtain
$$\begin{array}{l}R\left({e}_{2},{e}_{3}\right)\xi =0,\\ R\left({e}_{2},\phantom{\rule{thinmathspace}{0ex}}\xi \right)\xi =-({\alpha}^{2}+\frac{1}{4}\left(\beta -\gamma \right)\left(3\beta -\gamma \right)){e}_{2}+\left(\beta \left(\alpha -2\right)-\alpha \gamma \right){e}_{3},\\ R\left({e}_{2},\phantom{\rule{thinmathspace}{0ex}}\xi \right)\xi =\left(\beta \left(\alpha -2\right)-\alpha \gamma \right){e}_{2}-({\left(\alpha -2\right)}^{2}+\frac{1}{4}\left(\beta -\gamma \right)\left(3\gamma -\beta \right)){e}_{3}.\end{array}$$(14) In view of (2), it follows from (13) that
$$\begin{array}{l}h{e}_{2}=\left(\alpha -1\right){e}_{3}-\frac{1}{2}\left(\beta +\gamma \right){e}_{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}h{e}_{3}=\frac{1}{2}\left(\beta +\gamma \right){e}_{3}+\left(\alpha -1\right){e}_{2.}\end{array}$$(15) If ξ satisfies the (*k, μ, v*)-nullity condition, it follows from (11) and (15) that
$$\begin{array}{l}R\left({e}_{2},{e}_{3}\right)\xi =0,\\ R\left({e}_{2},\phantom{\rule{thinmathspace}{0ex}}\xi \right)\xi =(k-\frac{1}{2}\mu \left(\beta -\gamma \right)+v\left(\alpha -1\right)){e}_{2}+(\mu \left(\alpha -1\right)+\frac{1}{2}v\left(\beta -\gamma \right)){e}_{3},\\ R\left({e}_{2},\phantom{\rule{thinmathspace}{0ex}}\xi \right)\xi =(\mu \left(\alpha -1\right)+\frac{1}{2}v\left(\beta -\gamma \right)){e}_{2}+(k+\frac{1}{2}\mu \left(\beta -\gamma \right)-v\left(\alpha -1\right)){e}_{3}).\end{array}$$(16) Comparing (14) with (16) we state that there exists a unique solution for *k, μ* and v provided that either *β + γ* ≠ 0 or *α* ≠ 1, namely,
$$\begin{array}{l}k=-{\alpha}^{2}+2\alpha -\frac{1}{4}{\left(\beta -\gamma \right)}^{2}-2,\phantom{\rule{0.056em}{0ex}}\mu =\beta -\gamma \end{array},\phantom{\rule{thinmathspace}{0ex}}v=-2.$$(17) Notice that the (*k, μ*, −2)-nullity condition defined by relation (17) implies that *G* has a non-Kenmotsu almost Kenmotsu structure if we assume that *h ≠* 0 (or equivalently, either *β + γ* ≠ 0 or *α* ≠ 1).Otherwise, if *h* = 0, taking into account (15) we observe that the condition *β + γ* = 0 and *α* = 1 holds. Using *h* = 0 in (2) gives that ▽ξ = id - *η* ⊗ ξ and hence *R*(*X, Y*)ξ *= η(X)Y* − *η(Y)X* for any vector fields *X, Y*. This implies that ξ satisfies the (− 1, 0, 0)-nullity condition and by Proposition 2.1 we see that in this case *G* has a Kenmotsu structure. This completes the proof. ☐The following proposition follows directly from (15) and (17).

**Proposition 3.3:** *The Reeb vector field of the non-Kenmotsu almost Kenmotsu structure defined in Theorem 3.2 satisfies the (k, v)’-nullity condition if and only if ß = γ and either α* ≠ 1 *or β* ≠ 0.From (13) and (15) we obtain the following:

**Proposition 3.4:** *On the non-Kenmotsu almost Kenmotsu structure defined in Theorem 3.2 there holds* ▽_{ξ}*h* = (*ß - γ)h’*.Next, we show that under certain restrictions of *k* and μ the converse of the above Theorem 3.2 is true.

**Theorem 3.5:** *Any 3-dimensional non-Kenmotsu (k, μ,v)-almost Kenmotsu manifold with k a constant and μ invariant along the Reeb vector field is locally isometric to a 3-dimensional non-unimodular Lie group*.

**Proof:** *Let **M*^{3} be a 3-dimensional (*k, μ, v*)-almost Kenmotsu manifold with *h* ≠ 0 and *k* a constant, then 𝒰_{1} is a non-empty subset. It follows from (11) that the Reeb vector field ξ is an eigenvector field of the Ricci operator, i.e. Q *ξ* = 2*kξ*. Using this in Lemma 2.3 we have *λ*^{2} = —*k* — 1 ≠ 0 and hence we get
$$\begin{array}{l}b=c=0.\end{array}$$(18) In view of (8), by a simple computation we obtain that
$$\begin{array}{l}R\left(e,\phantom{\rule{thinmathspace}{0ex}}\xi \right)\xi =-\left({\lambda}^{2}+2\lambda a+1\right)e+2\lambda \varphi e.\end{array}$$(19) Also, it follows from (11) that
$$\begin{array}{l}R\left(e,\phantom{\rule{thinmathspace}{0ex}}\xi \right)\xi =\left(k+\lambda \mu \right)e-\lambda v\varphi e.\end{array}$$(20) Obviously, comparing (19) with (20) we obtain *μ* = —2*a* and *v* = —2. Using (18) in (10) we obtain that *e(a) = ϕe(a)* = 0. In view of the assumption *μ* invariant along ξ, we conclude that *a* is a constant. In this context, it follows from (8) that
$$\begin{array}{l}\left[\xi ,\phantom{\rule{thinmathspace}{0ex}}e\right]=\left(\lambda +a\right)\varphi e-e,\left[e,\phantom{\rule{thinmathspace}{0ex}}\varphi e\right]=0,\phantom{\rule{thinmathspace}{0ex}}\left[\varphi e,\phantom{\rule{thinmathspace}{0ex}}\xi \right]=\left(a-\lambda \right)e+\varphi e.\end{array}$$(21) A Lie group *G* is said to be unimodular if its left-invariant Haar measure is also right-invariant. It is wel1-known a Lie group *G* is unimodular if and only if the endomorphism ad_{X} $\mathfrak{g}\phantom{\rule{thinmathspace}{0ex}}\to \phantom{\rule{thinmathspace}{0ex}}\mathfrak{g}$ given by ad _{X} (*Y*) = [*X, Y*] has trace equal to zero for any *X* ∊ *g*, where *g* denotes the Lie algebra associated to *G*. Following Milnor [13], we state that *M*^{3} is locally isometric to a 3-dimensional non-unimodular Lie group. In fact, from (21) we see that its unimodular kernel {*X ∊ g* : trace(ad_{X}) = 0} is commutative and of 2-dimension and trace(ad_{ξ}) = —2. This completes the proof. ☐If the Reeb vector field ξ satisfies the (*k, μ*, v)-nullity condition, putting *Y* = ξ into (11) gives that
$$\begin{array}{l}l=-k{\varphi}^{2}+\mu h+v{h}^{\prime}.\end{array}$$(22) Using (22) in (3) and (4) we obtain *h*^{2} = (k + 1)ϕ^{2} and hence the following proposition is true.

**Proposition 3.6:** *On any 3-dimensional (k, μ, v)-almostKenmotsu manifold there holds that ▽*_{ξ}*h = μh’ — (ν +* 2)*h*.Note that the above proposition is in fact a generalization of Proposition 3.4.

## Comments (0)