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Abstract
We prove that a Lorentzian concircular structure (LCS)-manifold does not admit any null hypersurface which is tangential or transversal to its characteristic vector field. Due to the above, we focus on its ascreen null hypersurfaces and show that such hypersurfaces admit a symmetric Ricci tensor. Furthermore, we prove that there are no totally geodesic ascreen null hypersurfaces of a conformally flat (LCS)-manifold.
Since the middle of the twentieth century, Riemannian geometry has had a substantial influence on several main areas of mathematical sciences. Primarily, semi-Riemannian (in particular, Lorentzian [3]) geometry has its roots in Riemannian geometry. On the other hand, the situation is quite different for null manifolds with a degenerate metric, since one fails to use, in the usual way, the theory of non-null geometry. Null submanifolds have numerous applications in mathematical physics and General relativity (see [16] for details). This prompted Duggal-Bejancu [6] and Duggal-Sahin [8] to introduce the geometry of null submanifolds. They introduced a non-degenerate screen distribution to construct a null transversal vector bundle which is non-intersecting to its null tangent bundle and developed local geometry of null curves, hypersurfaces, and submanifolds. Their approach is extrinsic contrary to the intrinsic approach of Kupeli [12]. Based on the above books, many authors picked interest in null geometry, for instance, see [1, 2, 5, 9, 10, 11, 13, 15] and many more references cited therein.
In [17, 18], the geometry of Lorentzian concircular structure (LCS)-manifolds is extensively studied. As these spaces are semi-Riemannian, they naturally admit null subspaces. The aim of this paper is to introduce the geometry of null subspaces of (LCS)-manifolds. In particular, we study the geometry of null hypersurfaces of these spaces. Unlike the null hypersurfaces of Sasakian manifolds, null hypersurfaces (LCS)-manifolds are never tangent to the characteristic vector field of (LCS)-manifolds. Moreover, an ascreen null hypersurface admits a symmetric Ricci tensor. In case of a conformally flat (LCS)-manifolds, we prove that no totally geodesic ascreen null hypersurfaces can be admitted. Moreover, a screen conformal or screen totally umbilic null hypersurface is proper totally umbilic. The paper is arranged as follows; In Section 2, we quote some basic notions on (LCS)-manifolds and null hypersurfaces of semi-Riemannian manifolds. Section 3, we prove several new characterization theorems on null hypersurfaces of (LCS)-manifolds.
2. Preliminaries
An (n+2)-dimensional Lorentzian manifold M̄ is a smooth connected paracompact Hausdorff manifold with a Lorentzian metric ḡ, that is, M̄ admits a smooth symmetric tensor field ḡ of type (0, 2) such that for each point p ∈ M̄, the tensor ḡ: T_{p}M̄×T_{p}M̄ –→ ℝ is a non-degenerate inner product of signature (−,+, . . . , +), where T_{p}M̄ denotes the tangent vector space of M̄ at p and ℝ is the real number space. A non-zero vector v ∈ T_{p}M̄ is said to be timelike (resp., non-spacelike, null, spacelike) if it satisfies ḡ_{p}(v, v) < 0 (resp., ≤ 0,= 0,> 0) [6, 8, 16]. Furthermore, the category to which a given vector falls is called its causal character. From now on, we denote by Γ(E) the module of smooth sections of a vector bundle E over M̄.
Let (M̄, ḡ) be a Lorentzian manifold. A vector field V defined by ḡ(X, V) = A(X), for any X ∈ Γ(TM̄), is said to be a concircular [17, 18] vector field if, for any X, Y ∈ Γ(TM̄), we have (∇̄_{X} A)Y = α[ḡ(X, Y) − ω(X) A(Y)], where α is a non-vanishing smooth function and ω is a closed 1-form. Here, ∇̄ denotes the Levi-Civita connection of M̄ with respect to ḡ. Suppose that M̄ admits a unit timelike concircular vector field ζ, called the characteristic vector field of the manifold. Then, we have
$$\overline{g}(\zeta ,\zeta )=-1.$$
Since ζ is a unit concircular vector field, it follows that there exists a non-zero 1-form θ such that for
which follows that φ̄ is a symmetric (1, 1) tensor field called the structure tensor field of the manifold. Thus, the Lorentzian manifold M̄ together with the unit timelike concircular vector field ζ, its associated 1-form θ and a (1, 1) tensor field φ̄ is said to be a Lorentzian concircular structure manifold (briefly, an (LCS)-manifold) [17, 18]. In particular, if α = 1, then we obtain the LP-Sasakian structure of Matsumoto [14]. In an (LCS)-manifold, the following relations hold for all vector fields X, Y ∈ Γ(TM̄):
Let (M, g) be a null hypersurface of M̄. This means that, at each x ∈ M the restriction g = ḡ|_{TxM} is degenerate. That is; there exists a non-zero U ∈ T_{x}M such that g(U, X) = 0, for all X ∈ T_{x}M. Therefore, in null setting, the normal bundle TM^{⊥} of the null hypersurface M is a rank 1 vector subbundle of the tangent bundle TM, contrary to the classical theory of non-degenerate hypersurfaces for which the normal bundle has trivial intersection {0} with the tangent one and plays an important role in the introduction of the main induced geometric objects on M. The approach of [6, 8], which we adopt here, consist of fixing, on the null hypersurface, a geometric data formed by a null section and a screen distribution. By screen distribution on M, we mean a complimentary bundle of TM^{⊥} in TM. It is then a rank n non-degenerate distribution over M. In fact, there are infinitely many possibilities of choices for such a distribution provided the hypersurface M be paracompact, but each of them is canonically isomorphic to the factor vector bundle TM/TM^{⊥} [12]. We denote by S(TM) the screen distribution over M. Then we have the decomposition
$$TM=S(TM)\perp T{M}^{\perp},$$
where ⊥ denotes the orthogonal direct sum. From [6] or [8], it is known that for a null hypersurface equipped with a screen distribution, there exists a unique rank 1 vector subbundle tr(TM) of TM̄ over M, such that for any non-zero section ξ of TM^{⊥} on a coordinate neighborhood , there exists a unique section N of tr(TM) on satisfying
where ⊕ denote the direct (non-orthogal) sum. We call tr(TM) a (null) transversal vector bundle along M. In fact, from (2.11) and (2.12) one shows that, conversely, a choice of a transversal bundle tr(TM) determines uniquely the screen distribution S(TM). A vector field N as in (2.11) is called a null transversal vector field of M. It is then noteworthy that the choice of a null transversal vector field N along M determines both the null transversal vector bundle, the screen distribution S(TM) and a unique radical vector field, say ξ, satisfying (2.11). The name screen distribution is justified as follows; in the case M is a null cone of a 4-dimensional Lorentzian manifold, the integral curves of vector fields in TM^{⊥} are null (lightlike) rays and fibers of S(TM) can be visualized as screen that are transversal to these rays.
Let ∇ and ∇^{*} denote the induced connections on M and S(TM), respectively, and P be the projection of TM onto S(TM), then the local Gauss-Weingarten equations of M and S(TM) are the following [6]
for all X, Y ∈ Γ(TM), ξ ∈ Γ(TM^{⊥}) and N ∈ Γ(tr(TM)). In the above setting, B is the local second fundamental form of M and C is the local second fundamental form on S(TM). A_{N} and ${A}_{\xi}^{*}$ are the shape operators on TM and S(TM) respectively, while τ is a 1-form on TM. The above shape operators are related to their local fundamental forms by
for any X, Y ∈ Γ(TM). Moreover, $\overline{g}({A}_{\xi}^{*}X,N)=0$, and ḡ(A_{N} X, N) = 0, for all X ∈ Γ(TM). From these relations, we notice that ${A}_{\xi}^{*}$ and A_{N} are both screen-valued operators. Let ϑ = ḡ(N, ·) be a 1-form metrically equivalent to N defined on M̄. Take η = i^{*}ϑ to be its restriction on M, where i : M → M̄ is the inclusion map. Then it is easy to show that
for all X, Y, Z ∈ Γ(TM). Consequently, ∇ is generally not a metric connection with respect to g. However, the induced connection ∇^{*} on S(TM) is a metric connection.
Denote by R̄ the curvature tensor of the connection ∇̄ on M̄. Using the Gauss-Weingarten formulae (2.13)–(2.16), we obtain the following curvature relations (see details in [6, 8]).
for all X, Y, Z ∈ Γ(TM), ξ ∈ Γ(TM^{⊥}) and N ∈ Γ(tr(TM)).
3. Classification Results
Let us start off with the following important observation. Let (M, g) be a null hypersurface of a (LCS)-manifold (M̄, ḡ). As M̄ is Lorenztian, any screen distribution over M is Riemannian. Next, since ζ is a global vector field of M̄, we decompose it as follows;
and W a smooth section tangent to S(TM). Using (3.1) and (2.1), we have g(W, W)+2ab = −1. Suppose that a or b vanishes, then it follows that g(W, W) = −1. Consequently, W is a unit timelike vector field of S(TM). This is a contradiction as S(TM) is Riemannian. Thus, we have the following.
Lemma 3.1
There exist no null hypersurface of a (LCS)-manifold (M̄, ḡ) such that ζ is tangent or transversal to M.
In the theory of null hypersurfaces of Sasakian manifolds, it is possible to select a screen distribution S(TM) containing φ̄TM^{⊥} and φ̄tr(TM) as subbundles. See [8, 11, 10] for details. This does not hold for null hypersurfaces of a (LCS)-manifold (M̄, ḡ). More precisely, we have the following result.
Theorem 3.2
Let (M, g) be a null hypersurface of a (LCS)-manifold (M̄, ḡ). Then φ̄TM^{⊥} ⊄ S(TM) and φ̄tr(TM) ⊄ S(TM). Moreover, φ̄TM^{⊥} ∩ TM^{⊥} = {0} and φ̄tr(TM) ∩ tr(TM) = {0}.
Proof
First, we not that φ̄ξ or φ̄N are non-zero sections. In fact, suppose that φ̄ξ = 0 then using (2.8) and (3.2), we have 0 = ḡ(φ̄ξ, φ̄ξ) = b^{2}. This, together with (3.1), suggests that ζ is tangent to M. This is a contraction in view of Lemma 3.1. In the same way, one can show that φ̄N ≠ 0. Next, assume that φ̄ξ belongs to S(TM). Then, by (2.6) and (2.10), we have 0 = ḡ(φ̄ξ, ξ) = b^{2}, from which ζ turns out to be tangent to M which is a contradiction by Lemma 3.1. In the same way, if φ̄N belongs to S(TM) we see from (2.11) and (2.6) that 0 = ḡ(φ̄N, N) = a^{2}. In this case (3.1) gives ζ = W + bN. As ζ is timelike, we obtain g(W, W) = −1. This is a contradiction as S(TM) is Riemannian. Therefore, φ̄TM^{⊥} ⊄ S(TM) and φ̄tr(TM) ⊄ S(TM). Next, suppose that φ̄TM^{⊥} ∩ TM^{⊥} ≠ {0}. Then there exist a non-vanishing smooth function β such that φ̄ξ = βξ. Applying φ̄ to this equation and using (2.7), we get (β^{2} − 1)φ̄ξ = bζ. Applying φ̄ to this last relation and considering (2.7), one gets β(β^{2} − 1)ξ = 0, which implies that β = ±1 since β ≠ 0. Now let β = −1 such that φ̄ξ + ξ = 0. Taking the inner product of this relation with ξ and using (2.6), we get b^{2} = 0. This is a contradiction since by Lemma 3.1, ζ can not be tangent to M. On the other hand, if β = 1 we get φ̄ξ = ξ. But (2.6) implies φ̄ξ = ξ + bζ. From these two relations we deduce that bζ = 0. Since ζ is a unit timelike vector field, the last relation gives b = 0, which is a contradiction as before. Thus, φ̄TM^{⊥} ∩ TM^{⊥} = {0}. In the same way, one can show that φ̄tr(TM) ∩ tr(TM) = {0}, which completes the proof.
Using the language of [11], we will say that a null hypersurface (M, g) of a (LCS)-manifold (M̄, ḡ) is ascreen if the characteristic vector field ζ belongs to S(TM)^{⊥}[= TM^{⊥} ⊕ tr(TM)]. Equivalently, M is ascreen if W = 0. On an ascreen null hypersurface, the following holds.
Theorem 3.3
Let (M, g) be a null hypersurface of a (LCS)-manifold M̄. Then M is an ascreen null hypersurface of M̄ if and only if φ̄TM^{⊥} = φ̄tr(TM).
Proof
Following the method of [11], suppose that M is ascreen null hypersurface, then (2.1) reduces to ζ = aξ + bN, where a = θ(N) and b = θ(ξ) are non-vanishing smooth functions. Applying φ̄ to this relation and using the fact that φ̄ζ = 0, we get φ̄ξ + bφ̄N = 0. Thus, one gets φ̄ξ = ωφ̄N, where $\omega =-{\scriptstyle \frac{b}{a}}\ne 0$, a non vanishing smooth function. This implies that φ̄TM^{⊥} ∩ φ̄tr(TM) ≠ {0}. Since rank φ̄TM^{⊥} = rank φ̄tr(TM) = 1, it follows that φ̄TM^{⊥} = φ̄tr(TM). Conversely, suppose that φ̄TM^{⊥} = φ̄tr(TM). Then, there exists a non-vanishing smooth function ω such that φ̄ξ = ωφ̄N. Taking the inner product of this relation with respect to φ̄ξ and φ̄N in turn, we get b^{2} = ω(ab + 1) and ωa^{2} = ab + 1, respectively. Since ω ≠ 0, we have a ≠ 0, b ≠ 0 and b^{2} = (ωa)^{2}. The latter gives b = ±ωa. The case b = ωa implies that ab = ωa^{2} = ab + 1, which is a contradiction. Thus b = −ωa, from which 2ab = −1. Since $\omega =-{\scriptstyle \frac{b}{a}}$, a ≠ 0 and φ̄ξ = ωφ̄N, it is easy to see that aφ̄ξ + bφ̄N = 0. Applying φ̄ to this equation, and using b^{2} = ω(ab + 1) together with 2ab = −1, we get ζ = aξ + bN. Therefore, M is ascreen null hypersurface of M̄, which completes the proof.
Now, we will construct an example of this class of hypersurface in (LCS)-manifold.
Example 3.4
Consider a 3-dimensional manifold M̄ = {(x, y, z) ∈ ℝ^{3}}, where (x, y, z) are the standard coordinates in ℝ^{3}. Let {E_{1}, E_{2}, E_{3}} be linearly independent global frame on M̄ given by E_{1} = e^{−}^{z}(∂_{x}+y∂_{y}), E_{2} = e^{−}^{z}∂_{y} and E_{3} = e^{−2}^{z}∂_{z}. Let ḡbe the Lorentzian metric defined by ḡ(E_{1}, E_{2}) = ḡ(E_{1}, E_{3}) = ḡ(E_{2}, E_{3}) = 0, ḡ(E_{1}, E_{1}) = ḡ(E_{2}, E_{2}) = 1 and ḡ(E_{3}, E_{3}) = −1. Let θ be the 1-form defined by θ(X) = ḡ(X, E_{3}), for any X ∈ Γ(TM̄). Let φ̄ be a (1,1) tensor field defined by φ̄E_{1} = E_{1}, φ̄E_{2} = E_{2} and φ̄E_{3} = E_{3}. Then using the linearity of φ̄ and ḡ we have θ(E_{3}) = −1, φ̄^{2}X = X + θ(X)E_{3} and ḡ(φ̄X, φ̄Y) = ḡ(X, Y) + θ(X)θ(Y), for any X, Y ∈ Γ(TM̄). Thus, for ζ = E_{3}, (φ̄, ζ, θ, ḡ) defines a Lorentzian paracontact structure on M̄, which is also a (LCS)-structure. Consequently, (M̄,φ̄, ζ, θ, ḡ) is a (LCS)-manifold with α = e^{−2}^{z} such that Xα = ρθ(X), where ρ = 2e^{−4}^{z} (see [18]). Now, we define a null hypersurface M of (M̄,φ̄, ζ, θ, ḡ) as M = {(x, y, z) ∈ ℝ^{3} : y = z}. Thus, TM is spanned by U_{1} = E_{1} and U_{2} = ∂_{y} + ∂_{z}, in which TM^{⊥} is spanned by ξ = U_{2}. The transversal bundle tr(TM) is spanned by N, where N = (α/2)(∂_{y} − ∂_{z}). By using the definition of φ̄, we have φ̄N = (α/2)φ̄ξ. Hence, by Theorem 3.3 we have φ̄TM^{⊥} = φ̄tr(TM). Next, applying φ̄ to ζ = aξ + bN and considering the previous relations, we get 2a + αb = 0. Since 2ab + 1 = 0, we obtain $\zeta =(\sqrt{\alpha}/2)\xi -(1/\sqrt{\alpha})N$. Therefore, M is an ascreen null hypersurface of M̄.
A null hypersurface (M, g) of a semi-Riemannian manifold (M̄, ḡ) is called totally umbilic [6, p. 106] if there exist a smooth function μ on a coordinate neighborhood such that ${A}_{E}^{*}X=\mu PX$, or equivalently,
In case μ = 0 on , we say that M is totally geodesic otherwise it is proper totally umbilic in M̄. Furthermore, M is screen totally umbilic [6, p. 109] if there exist a smooth function ϱ on a coordinate neighborhood such that A_{N}X = ϱPX, or equivalently,
In case ϱ = 0 on , we say that M is screen totally geodesic otherwise it is proper screen totally umbilic in M̄.
Let (M, g) be a null hypersurface. We say that M is screen conformal [8, p. 60] if there exist a non-vanishing smooth function ψ on a coordinate neighborhood such that ${A}_{N}X=\psi {A}_{E}^{*}X$, or equivalently,
In case ψ is a constant function, we say that M is screen homothetic.
Suppose that M is an ascreen null hypersurface of M̄. In view of (3.1), we have ζ = aξ + bN, where a and b are given by (3.2). Differentiating this relation and using (2.14) and (2.16), we get
In view of (3.3), (3.4), (3.5), (3.7), (3.9) and (3.10), we have the following result.
Theorem 3.5
Let (M, g) be an ascreen null hypersurface of a (LCS)-manifold M̄. Then the following holds;
(1) zero is an eigenvalue of A_{N} with respect to ξ,
(2) M is a screen integrable null hypersurface and locally isometric to, whereis a null curve tangent to TM^{⊥}and M′ is a leaf of S(TM),
(3) a screen conformal or screen totally umbilic M is proper totally umbilic,
(4) a screen totally geodesic M is never totally geodesic,
(5) there exist no M with a or b constant and τ = 0.
Proof
Parts (1), (3), (4) and (5) are obvious. Part (2) follows from (3.9) and similar arguments as in [7], and the proof is completed.
Remark 3.6
In [11], it was proved (see Theorem 3.4 therein) that a Sasakian manifold does not admit any ascreen null hypersurfaces which are screen totally umbilic or screen conformal. We remark, based on Theorem 3.5, that this is not the case with ascreen null hypersurfaces of (LCS)-manifolds. Futhermore, it was shown, by the same author in Theorem 3.5, that S(TM) is never parallel, and if dim M̄ >3 then S(TM) is not integrable. We note once again that this is not the same with ascreen null hypersurfaces of (LCS)-manifolds.
It is well-known that the Ricci tensor of null hypersurface (and generally of null submanifold) is not symmetric. This is because the induced connection is not a metric connection (see relation (2.18)). In line with the above, the authors in [6, p. 99] (also see [8]) proves the following result.
Let (M, g) be a null hypersurface of a semi-Riemannian manifold (M̄, ḡ). Then the Ricci tensor of the induced connection ∇ is symmetric, if and only if, each 1-form τ induced by S(TM) is closed, i.e., dτ = 0, on.
With reference to Theorem 3.7, we have the following result.
Theorem 3.8
The Ricci tensor of an ascreen null hypersurface (M, g) of a (LCS)- manifold (M̄, ḡ) is symmetric.
Proof
By the second relation in (3.7) and the fact that b ≠ 0, we have
Finally, considering (3.16) and Theorem 3.7, we see that the Ricci tensor of M is symmetric. This completes the proof.
Remark 3.9
A similar conclusion, as in Theorem 3.8, can be arrived at if the first relation of (3.7), i.e., τ (X) = [Xa − αη(X) − aαθ(X)]/a is used. In fact, differentiating this relation with respect to Y ∈ Γ(TM) and considering (2.9), (2.13), (2.14) and (2.17), we derive
Observe that the right hand side of (3.21) vanishes on M since S(TM) is integrable (see Theorem 3.5) and that C(ξ, PX) = 0, for all X ∈ Γ(TM), by (3.10). That is dτ = 0 and thus, by Theorem 3.8, M admits a symmetric Ricci tensor.
Let x̃ = λξ, then it follows that Ñ = (1/λ)N. Moreover, B̃ = B and
This shows that B and τ depend on the section ξ on . By Theorem 3.8 and Poincaré’s lemma we obtain τ (X) = Xf, where f is a smooth function on . Let λ = exp(f) in (3.22), then, τ̃ = 0 on . Thus, we have
Corollary 3.10
Let (M, g) be an ascreen null hypersurface of a (LCS)-manifold (M̄, ḡ). There exist a pair {ξ, N} onsuch that the corresponding 1-form τ from the Weingarten equation vanishes.
A semi-Riemannian manifold (M̄, ḡ) of constant sectional curvature c is called a semi-Riemannian space form (see [16, p. 80]) and denoted by M̄ (c). The curvature tensor field R̄ of M̄(c) is given by
Let M̄(c) be a (LCS)-manifold of constant curvature c. Then by Proposition 3.1 of [18, p. 451], c satisfies c = (α^{2} − ρ). Using this fact together with (3.9) and (3.23), we get the following result.
Theorem 3.11
Let (M, g) be a screen totally geodesic ascreen null hypersurface of a (LCS)-space form M̄(α^{2} − ρ) then α^{2} = ρ. Moreover, M is a proper totally umbilic null hypersurface whose immersion into M̄ is affinely equivalent to the graph immersion of a certain function F : M −→ ℝ.
The notion of quasi-constant curvature was introduced by Chen and Yano [4]. Moreover, it was shown in [18] that a conformally flat (LCS)-manifold (M̄, ḡ) is of quasi-constant curvature and its curvature tensor R̄ satisfies
where c_{1} and c_{2}(≠ 0) are smooth functions. We denote such a manifold by M̄(c_{1}, c_{2}). The following result holds.
Theorem 3.12
Let (M, g) be an ascreen null hypersurface of a conformally flat (LCS)-manifold M̄(c_{1}, c_{2}) of dimension > 3. Then, M is totally umbilic if and only if the umbiliciy factor μ in (3.3) is a solution of the following differential equations
Moreover, if M is totally umbilic then it is a screen totally umbilic null hypersurface with screen umbilicity factor ϱ := −(α+aμ)/b in (3.4) satisfying the following pair of differential equations
for any X, Y, Z ∈ Γ(TM). Replacing X (or Y) by ξ, we get ξμ+μτ(ξ)−μ^{2}−c_{2}b^{2} = 0. Again replacing X, Y by PX, PY, respectively, in (3.27) and using the fact that S(TM) is non-degenerate, we have
Next, suppose that there exist such that PX′μ + μτ(PX′) ≠ 0 at x ∈ M. It then follows from (3.28) that all vectors from the fiber S(TM)_{x} are collinear with (PX′)_{x}. This is a contradiction as dim(S(TM))_{x} > 1, which proves the first part of the theorem.
Observe, from (3.9), that M is also screen totally umbilic with ϱ := −(α+aμ)/b. Following the same steps as above, this time with (2.20), (2.15), (2.18), (3.4), (3.10) and (3.24), we get
for all X, Y, Z ∈ Γ(TM). Replacing X (or Y) by ξ in (3.29) and using the fact b = θ(ξ), we get ξϱ − ϱτ(ξ) − ϱμ − c_{1} − c_{2}ab = 0. As M is ascreen, we have 2ab+1 = 0. Considering these two relations, we have $\xi \varrho -\varrho \tau (\xi )-\mu \varrho -{c}_{1}+{\scriptstyle \frac{{c}_{2}}{2}}=0$. The last differential equation follows as in (3.28) and the following arguments. This completes the proof.
It then follows from Theorem 3.12 that the following result holds.
Corollary 3.13
(1) There exist no totally geodesic ascreen null hypersurfaces of a conformally flat (LCS)-manifold M̄(c_{1}, c_{2}).
(2) There exist no screen totally geodesic ascreen null hypersurfaces of a conformally flat (LCS)-manifold M̄(c_{1}, c_{2}) with c_{1} = 0.
Acknowledgements
The author is very grateful to the reviewer and editor for providing valuable suggestions to improve the quality of this paper. Part of this work was done while the author was still at the University of KwaZulu-Natal.
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