Let us suppose that *ξ* is of Jacobi type. Then ∇_{X}∇_{X}*ξ* + *R*(*ξ*, *X*)*X* = 0 for any *X* tangent to *M*.

Take an orthonormal basis {*e*_{1}, …, *e*_{2m−1}} of vector fields tangent to *M*. As *ξ* is of Jacobi type, $\begin{array}{}\sum _{i=1}^{2m-1}\end{array}$ ∇_{ei}∇_{ei}*ξ* + *Ric*(*ξ*) = 0. That is,

$$\begin{array}{}\sum _{i=1}^{2m-1}{\mathrm{\nabla}}_{{e}_{i}}\varphi S{e}_{i}+Ric(\xi )=0\end{array}$$(11)

From (9) *Ric*(*ξ*) = 2(*m* − 2)*ξ* + *η*(*Bξ*)*Bξ* −*ρ*(*ξ*)*ϕ* *Bξ* + *η*(*Bξ*)*Bξ* + (*trace S*)*Sξ* − *S*^{2}*ξ*. As *ρ*(*ξ*) = 0 and *η*(*Bξ*) = *g*(*Aξ*, *ξ*) we obtain

$$\begin{array}{}Ric(\xi )=2(m-2)\xi +2g(A\xi ,\xi )B\xi +(traceS)S\xi -{S}^{2}\xi .\end{array}$$(12)

From (11) and (12) we get $\begin{array}{}\sum _{i=1}^{2m-1}\end{array}$ ∇_{ei}*ϕ* *Se*_{i} + 2(*m* − 2)*ξ* + 2*g*(*Aξ*, *ξ*)*Bξ* + (*trace S*)*Sξ* −*S*^{2}*ξ* = 0. Taking its scalar product with *ξ* and bearing in mind that *g*(*Bξ*, *ξ*) = *g*(*Aξ*, *ξ*) we obtain

#### Lemma 4.1

*Let* *M* *be a real hypersurface in* *Q*^{m}, *m* ≥ 3, *such that* *ξ* *is of Jacobi type*. *Then*

$$\begin{array}{}{\displaystyle -trace{S}^{2}+2(m-2)+2g(A\xi ,\xi {)}^{2}+(traceS)\eta (S\xi )=0}\end{array}$$

Now we compute

$$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel \varphi S-S\varphi {\parallel}^{2}=\sum _{i=1}^{2m-1}g((\varphi S-S\varphi ){e}_{i},(\varphi S-S\varphi ){e}_{i})\\ \phantom{\rule{1em}{0ex}}=\sum _{i,j=1}^{2m-1}g((\varphi S-S\varphi ){e}_{i},{e}_{j})g((\varphi S-S\varphi ){e}_{i},{e}_{j})\\ =2\sum _{i,j=1}^{2m-1}g(\varphi S{e}_{j},{e}_{i})g(\varphi S{e}_{j},{e}_{i})+2\sum _{i,j=1}^{2m-1}g(\varphi S{e}_{j},{e}_{i})g(\varphi S{e}_{i},{e}_{j})\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=-2\sum _{j=1}^{2m-1}g(\varphi S{e}_{j},S\varphi {e}_{j})+2\sum _{j=1}^{2m-1}g(\varphi S{e}_{j},\varphi S{e}_{j})\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=2trace{S}^{2}-2g({S}^{2}\xi ,\xi )-2\sum _{j=1}^{2m-1}g({\mathrm{\nabla}}_{{e}_{j}}\xi ,S\varphi {e}_{j})\end{array}$$(13)

where we have used (4).

Take now *U* = ∇_{ξ}*ξ* = *ϕ* *Sξ*. Then we have

$$\begin{array}{c}div(U)=\sum _{i=1}^{2m-1}g({\mathrm{\nabla}}_{{e}_{i}}U,{e}_{i})=\sum _{i=1}^{2m-1}g({\mathrm{\nabla}}_{{e}_{i}}\varphi S\xi ,{e}_{i})\\ =\sum _{i=1}^{2m-1}g(({\mathrm{\nabla}}_{{e}_{i}}\varphi )S\xi ,{e}_{i})+\sum _{i=1}^{2m-1}g(\varphi {\mathrm{\nabla}}_{{e}_{i}}S\xi ,{e}_{i})\\ =\sum _{i=1}^{2m-1}g(\eta (S\xi )S{e}_{i}-g(S{e}_{i},S\xi )\xi ,{e}_{i})-\sum _{i=1}^{2m-1}g({\mathrm{\nabla}}_{{e}_{i}}S\xi ,\varphi {e}_{i}),\end{array}$$(14)

that is

#### Lemma 4.2

*Let* *M* *be a real hypersurface in* *Q*^{m}, *m* ≥ 3, *and* *U* = *ϕ* *Sξ*. *Then*

$$\begin{array}{c}div(U)=(traceS)\eta (S\xi )-\eta ({S}^{2}\xi )-\sum _{i=1}^{2m-1}g({\mathrm{\nabla}}_{{e}_{i}}S\xi ,\varphi {e}_{i}).\end{array}$$

From (13) and Lemma 4.2 we obtain

$$\begin{array}{c}div(U)-\frac{1}{2}\parallel \varphi S-S\varphi {\parallel}^{2}=-trace{S}^{2}+\eta (S\xi )(traceS)-\sum _{i=1}^{2m-1}g(({\mathrm{\nabla}}_{{e}_{i}}S)\xi ,\varphi {e}_{i}).\end{array}$$(15)

Then

$$\begin{array}{c}\sum _{i=1}^{2m-1}(g(({\mathrm{\nabla}}_{\xi}S){e}_{i},\varphi {e}_{i})=-\sum _{i=1}^{2m-1}g(\varphi ({\mathrm{\nabla}}_{\xi}S){e}_{i},{e}_{i})\\ =-trace(\varphi ({\mathrm{\nabla}}_{\xi}S))=-trace(({\mathrm{\nabla}}_{\xi}S)\varphi )=-\sum _{i=1}^{2m-1}g(({\mathrm{\nabla}}_{\xi}S)\varphi {e}_{i},{e}_{i})\\ =-\sum _{i=1}^{2m-1}g(({\mathrm{\nabla}}_{\xi}S){e}_{i},\varphi {e}_{i}).\end{array}$$(16)

Thus we conclude

$$\begin{array}{c}\sum _{i=1}^{2m-1}g(({\mathrm{\nabla}}_{\xi}S){e}_{i},\varphi {e}_{i})=0.\end{array}$$(17)

Bearing in mind (17) Codazzi equation yields

$$\begin{array}{c}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sum _{i=1}^{2m-1}g(({\mathrm{\nabla}}_{{e}_{i}}S)\xi ,\varphi {e}_{i})=-\sum _{i=1}^{2m-1}g(\varphi {e}_{i},\varphi {e}_{i})\\ +\sum _{i=1}^{2m-1}g({e}_{i},AN)g(A\xi ,\varphi {e}_{i})+\sum _{i=1}^{2m-1}g({e}_{i},A\xi )g(JA\xi ,\varphi {e}_{i})\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\sum _{i=1}^{2m-1}g(\xi ,A\xi )g(JA{e}_{i},\varphi {e}_{i})\\ =2(m-2)-g(AN,N{)}^{2}+g(\xi ,A\xi )g(AN,N)-g(\xi ,A\xi )(traceA).\end{array}$$(18)

From (6) *g*(*AN*, *N*) = *cos*(2*t*) = −*g*(*Aξ*, *ξ*). Moreover, as {*e*_{1}, …*e*_{2m−1}, *N*} is an orthonormal basis of vectors tangent to *Q*^{m} at any point of *M*, {*Je*_{1}, …, *Je*_{2m−1}, *JN*} is also an orthonormal basis. Then *trace* *A* = $\begin{array}{}\sum _{i=1}^{2m-1}\end{array}$ *g*(*AJe*_{i}, *Je*_{i}) + *g*(*AJN*, *JN*) = −$\begin{array}{}\sum _{i=1}^{2m-1}\end{array}$ *g*(*JAe*_{i}, *Je*_{i}) − *g*(*JAN*, *AN*) = −$\begin{array}{}\sum _{i=1}^{2m-1}\end{array}$ *g*(*Ae*_{i}, *e*_{i}) − *g*(*AN*, *N*) = −*trace A*. Thus *trace A* = 0 and (18) becomes

$$\begin{array}{}\sum _{i=1}^{2m-1}g(({\mathrm{\nabla}}_{{e}_{i}}S)\xi ,\varphi {e}_{i})=-2(m-2)-2g(A\xi ,\xi {)}^{2}.\end{array}$$(19)

From this, Lemma 4.1, Lemma 4.2 and (17) we get

$$\begin{array}{}div(U)=\frac{1}{2}\parallel \varphi S-S\varphi {\parallel}^{2}.\end{array}$$(20)

Now if *M* is Hopf, *U* = 0 and then *ϕ* *S* − *S**ϕ* = 0.

If *M* is compact, $\begin{array}{}\frac{1}{2}\end{array}$ *∫*_{M} ∥*ϕ* *S* − *S**ϕ*∥^{2}*dV* = 0. Thus again *ϕ* *S* − *S**ϕ* = 0.

In both cases as (𝔏_{ξ} *g*)(*X*, *Y*) = *g*((*ϕ* *S* − *S**ϕ*)*X*, *Y*), for any *X*, *Y* ∈ *TM*, we conclude 𝔏_{ξ} *g* = 0 and *ξ* is Killing, obtaining our Theorem.

As *ϕ S* = *Sϕ* we have, [2], that *m* = 2*k* and *M* must be locally congruent to a tube around a totally geodesic ℂ*P*^{k} in *Q*^{m}.

Bearing in mind the expression of the shape operator *S* of such a real hypersurface, [2], it is immediate to see that its structure vector field is of Jacobi type and we conclude the proof of our Corollary.

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