Let *M*^{3} be three-dimensional almost Kenmotsu manifold. We shall first consider the case *h* = 0, then by Proposition 2.1 we see that *M*^{3} is Kenmotsu and ▽_{ξ}*h* = 0 holds trivially. If the curvature tensor of *M*^{3} is harmonic, following Proposition 4.1 we observe that the scalar curvature of *M*^{3} is a constant. Notice that Inoguchi in [16, Proposition 3.1] proved that a three-dimensional Kenmotsu manifold of constant scalar curvature is of constant sectional curvature –1. This yields that a three-dimensional Kenmotsu manifolds with harmonic curvature tensor is of constant sectional curvature – 1.

Now let us consider a three-dimensional almost Kenmotsu manifold *M*^{3} satisfying *h* ≠ 0, then 𝒰_{1} is a nonempty subset. By applying Lemma 2.2 and a direct calculation we obtain
$$\begin{array}{l}({\mathrm{\nabla}}_{\xi}h)e=\xi (\lambda )e+2a\lambda \varphi e\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}({\mathrm{\nabla}}_{\xi}h)\varphi e=-\xi (\lambda )\varphi e+2a\lambda e.\end{array}$$(25)
By the assumption condition ▽_{ξ}*h* = 0 and (25) we have
$$\xi (\lambda )=a={0}_{;}$$(26)
where we have used that *λ* is positive on 𝒰_{1}. In what follows, we denote *f* by
$$f=e(c)+\varphi e(b)+{b}^{2}+{c}^{2}+2.$$(27)
Then, using (26) and (27) we obtain from Lemmas 2.2 and 2.3 that
$$({\mathrm{\nabla}}_{\xi}Q)\xi =-\xi (\varphi e(\lambda )+2\lambda b)e-\xi (e(\lambda )+2\lambda c)\varphi e.$$(28)
$$({\mathrm{\nabla}}_{\xi}Q)e=-\xi (\varphi e(\lambda )+2\lambda b)\xi -\xi (f)e.$$(29)
$$({\mathrm{\nabla}}_{\xi}Q)\varphi e=-\xi (e(\lambda )+2\lambda c)\xi -\xi (f)\varphi e.$$(30)
$$\begin{array}{l}({\mathrm{\nabla}}_{e}Q)\xi =2(\varphi e(\lambda )-3\lambda e(\lambda )+2\lambda b-2{\lambda}^{2}c)\xi \\ \begin{array}{ccc}& & +(f-2-e(\varphi e(\lambda )+2\lambda b)-b(e(\lambda )+2\lambda c))e\end{array}\\ \begin{array}{ccc}& & +(2{\lambda}^{3}+b(\varphi e(\lambda )+2\lambda b)-e(e(\lambda )+2\lambda c)-\lambda f)\varphi e.\end{array}\end{array}$$(31)
$$\begin{array}{l}({\mathrm{\nabla}}_{e}Q)e=(f-2-e(\varphi e(\lambda )+2\lambda b)-b(e(\lambda )+2\lambda c))\xi \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-(e(f)+2\varphi e(\lambda ))e+(e(\lambda )+\lambda \varphi e(\lambda )+2{\lambda}^{2}b-2\lambda c)\varphi e.\end{array}$$(32)
$$\begin{array}{c}\left({\mathrm{\nabla}}_{e}Q\right)\varphi e\phantom{\rule{thinmathspace}{0ex}}=\left(2{\lambda}^{3}\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}f\lambda \phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}b\left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda b\right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}e\left(e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda c\right)\right)\phantom{\rule{thinmathspace}{0ex}}\xi \\ +\left(e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\lambda \varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}2\lambda c\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2{\lambda}^{2}b\right)e\\ +\left(2\lambda \left(e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda c\right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}e\left(f\right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}4\lambda b\right)\varphi e\phantom{\rule{thinmathspace}{0ex}}.\end{array}$$(33)
$$\begin{array}{l}\left({{\mathrm{\nabla}}_{\varphi}}_{e}Q\right)\xi \phantom{\rule{thinmathspace}{0ex}}=2\left(e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}\lambda \varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+2\lambda c\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}2{\lambda}^{2}b\right)\xi \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\left(2{\lambda}^{3}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}c\left(e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda c\right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\varphi e\left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda b\right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\lambda f\right)e\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\left(f\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\varphi e\left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda c\right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}c\left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda b\right)\right)\varphi e.\end{array}$$(34)
$$\begin{array}{l}\left({{\mathrm{\nabla}}_{\varphi}}_{e}Q\right)e\phantom{\rule{thinmathspace}{0ex}}=\left(2{\lambda}^{3}\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}f\lambda \phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}c\left(e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda c\right)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\varphi e\left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\lambda b\right)\right)\xi \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\left(\varphi e\left(f\right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}4\lambda c\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\lambda \left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+2\lambda b\right)\right)e\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\lambda e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2{\lambda}^{2}c\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}2\lambda b\right)\varphi e.\end{array}$$(35)
$$\begin{array}{l}\left({{\mathrm{\nabla}}_{\varphi}}_{e}Q\right)\varphi e\phantom{\rule{thickmathspace}{0ex}}=\left(f\phantom{\rule{thickmathspace}{0ex}}-2\phantom{\rule{thickmathspace}{0ex}}\varphi e\left(e\left(\lambda \right)\phantom{\rule{thickmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}2\lambda c\right)\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}c\left(\varphi e\left(\lambda \right)\phantom{\rule{thickmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}2\lambda b\right)\right)\xi \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}\left(\varphi e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\lambda e\left(\lambda \right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2{\lambda}^{2}c\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}2\lambda b\right)e\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\left(\varphi e\left(f\right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2e\left(\lambda \right)\right)\varphi e.\end{array}$$(36)
If we replace *X* and *Y* in (23) by *e* and ξ, respectively, then we have from (29) and (31) that
$$\left\{\begin{array}{l}2(\varphi e(\lambda )-3\lambda e(\lambda )+2\lambda b-2{\lambda}^{2}c)+\xi (\varphi e(\lambda )+2\lambda b)={0}_{,}\\ f-2-e(\varphi e(\lambda )+2\lambda b)-b(e(\lambda )+2\lambda c)+\xi (f)={0}_{,}\\ 2{\lambda}^{3}+b(\varphi e(\lambda )+2\lambda b)-e(e(\lambda )+2\lambda c)-\lambda f=0.\end{array}\right.$$(37) Similarly, if we replace *X* and *Y* in (23) by *ϕ*e and ξ, respectively, we obtain from (30) and (34) that$$\left\{\begin{array}{l}2(e(\lambda )-3\lambda \varphi e(\lambda )+2\lambda c-2{\lambda}^{2}b)+\xi (e(\lambda )+2\lambda c)={0}_{,}\\ f-2-\varphi e(e(\lambda )+2\lambda c)-c(\varphi e(\lambda )+2\lambda b)+\xi (f)={0}_{,}\\ 2{\lambda}^{3}+c(e(\lambda )+2\lambda c)-\varphi e(\varphi e(\lambda )+2\lambda b)-\lambda f=0.\end{array}\right.$$(38) Similarly, if we replace *X* and *Y* in (23) by *e* and *ϕβ*, respectively, we obtain from (33) and (35) that
$$\left\{\begin{array}{l}c(e(\lambda )+2\lambda c)-\varphi e(\varphi e(\lambda )+2\lambda b)+e(e(\lambda )+2\lambda c)=b(\varphi e(\lambda )+2\lambda b{)}_{,}\\ e(\lambda )-\lambda \varphi e(\lambda )+2\lambda c-2{\lambda}^{2}b+\varphi e(f)={0}_{,}\\ \lambda e(\lambda )-\varphi e(\lambda )+2{\lambda}^{2}c-2\lambda b-e(f)=0.\end{array}\right.$$(39) Moreover, it follows from Lemma 2.3 that
$$r=-2({\lambda}^{2}+1)-2f.$$(40)
Applying again the well known formula div $Q\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{1}{2}$ grad(*r*) we have from equations (28), (32) and (36) that
$$\left\{\begin{array}{l}-\xi (\varphi e(\lambda )+2\lambda b)+3\lambda e(\lambda )-\varphi e(\lambda )+2{\lambda}^{2}c-2\lambda b={0}_{,}\\ -\xi (e(\lambda )+2\lambda c)+3\lambda \varphi e(\lambda )-e(\lambda )+2{\lambda}^{2}b-2\lambda c={0}_{,}\end{array}\right.$$(41)
where we have used the scalar curvature *r* = constant, equation (40), the second terms of relations (37) and (38).

Next, using the first equation of (41) and the second equation of (41) in the first terms of (37) and (38), respectively, we obtain
$$\left\{\begin{array}{l}3\lambda e(\lambda )-\varphi e(\lambda )+2{\lambda}^{2}c-2\lambda b={0}_{,}\\ 3\lambda \varphi e(\lambda )-e(\lambda )+2{\lambda}^{2}b-2\lambda c={0}_{,}\end{array}\right.$$(42) and
$$\left\{\begin{array}{l}\xi (\varphi e(\lambda ))+2\lambda \xi (b)={0}_{,}\\ \xi (e(\lambda ))+2\lambda \xi (c)=0.\end{array}\right.$$(43)
Taking the covariant differentiation of relation (42) and using (43) and (26) we have
$$\xi (b)=\xi (c)={0}_{,}$$(44)
where we have used that λ is a positive function. Using (44) and (26) in (10) gives that
$$\left\{\begin{array}{l}e(\lambda )=b-c{\lambda}_{,}\\ \varphi e(\lambda )=c-b\lambda .\end{array}\right.$$(45)
Putting (45) into (42) yields that
$$\left\{\begin{array}{l}2b\lambda -c{\lambda}^{2}-c={0}_{,}\\ 2c\lambda -b{\lambda}^{2}-b=0.\end{array}\right.$$(46)
It follows from (46) that (λ^{2} + 1)(*b*^{2} – *c*^{2}) = 0 and hence we get either *b* – *c* = 0 or *b* + *c* = 0. We continue the discussion with the following two cases.

**Case i**. Using *b = c* in (46) we have either *b* = *c* = 0 or λ = 1. Now we assume that *b* = *c* = 0 holds and using this in (10) we see that λ is a positive constant. By applying *a* = *b* = *c* = 0 in (8) we obtain the following
$$\begin{array}{l}\left[\xi ,\phantom{\rule{thinmathspace}{0ex}}e\right]=\lambda \varphi \mathrm{e}-e,\phantom{\rule{thinmathspace}{0ex}}\left[e,\phantom{\rule{thinmathspace}{0ex}}\varphi e\right]=0,\phantom{\rule{thinmathspace}{0ex}}\left[\varphi e,\phantom{\rule{thinmathspace}{0ex}}\xi \right]=-\lambda e+\varphi e.\end{array}$$

According to J. Milnor [13], we now conclude that *M*^{3} is locally isometric to a three-dimensional non-unimodular Lie group equipped with a left invariant non-Kenmotsu almost Kenmotsu structure. Moreover, using *a* = *b* = 0 in equations (28)–(36) we get

$$\mathrm{\nabla}Q=0.$$

Notice that the above relation holds on a three-dimensional Riemannian manifold if and only if the curvature tensor is parallel, i.e. the manifold is locally symmetric. We also observe that Wang [1, Theorem 3.4] and [2, Theorem 5] proved that a locally symmetric three-dimensional non-Kenmotsu almost Kenmotsu manifold is locally isometric to the Riemannian product ℍ^{2}(−4) × ℝ.

Otherwise, if λ = 1 we obtain from (46) again *b* = *c* and using this in the last two equations of (37) we obtain
$$e(b)=0.$$

Moreover, using λ = 1 and *b* = *c* in the last two equations of (38) we obtain

$$\varphi e(b)=0.$$

In view of (44) we observe that both *b* and *c* are constants. Then it follows from the second equation of (37) that *f* = 2 + 2*b*^{2}. Finally, using this in (28)–(36) gives that ▽*Q* = 0 and this is equivalent to the local symmetry. Then the proof follows from Wang [1, Theorem 3.4] or [2, Theorem 5].

**Case ii**. Now we consider the other case: *b* + *c* = 0. Using this in (46) gives that *b* = *c* = 0, where we have used that λ is positive. Moreover, putting *b* = *c* = 0 in (45) and applying (26) we see that λ is a constant. Therefore, the proof follows from **Case i**. ☐

On a three-dimensional locally symmetric almost Kenmotsu manifold, applying the local symmetry condition we obtain that ▽ξ*l* = 0. Substituting *X* with ξ in (6) implies that ▽_{ξ}ϕ = 0. Then, by taking the covariant differentiation of (3) along ξ we obtain ▽_{ξ}*h*^{2} = ▽_{ξ}*h* o *h + h* o ▽_{ξ}*h* = 0. It follows directly that (▽_{ξ}▽_{ξ}*h*) o *h +* 2(▽_{ξ}*h*)^{2} + *h* o (▽_{ξ}▽_{ξ}*h*) = 0. Also, using ▽_{ξ}*l* = ▽_{ξ}*h* = 0 and taking the covariant differentiation of (4) along ξ we obtain ▽_{ξ}▽_{ξ}*h* = -2▽_{ξ}*h*. Therefore, it is easily seen that (▽_{ξ}*h*)^{2} = 0 and hence we get ▽_{ξ}*h* = 0. Then the following corollary follows directly from Theorem 4.2 and can also be regarded as a generalization of Wang [1, Theorem 3.4] and [2, Theorem 5].

## Comments (0)