CC BY
10
Vladikavkaz Mathematical Journal 2021, Volume 23, Issue 4, P. 89-95
УДК 517.98
DOI 10.46698/c1197-8093-8231-u
BOUNDED ORTHOMORPHISMS BETWEEN LOCALLY SOLID VECTOR LATTICES
R. Sabbagh1 and O. Zabeti1
1 University of Sistan and Baluchestan, P.O. Box 98135-674, Zahedan, Iran E-mail: sabaghrahele@gmail.com, o.zabeti@gmail.com
Abstract. The main aim of the present note is to consider bounded orthomorphisms between locally solid vector lattices. We establish a version of the remarkable Zannen theorem regarding equivalence between orthomorphisms and the underlying vector lattice for the case of all bounded orthomomorphisms. Furthermore, we investigate topological and ordered structures for these classes of orthomorphisms, as well. In particular, we show that each class of bounded orthomorphisms possesses the Levi or the AM-properties if and only if so is the underlying locally solid vector lattice. Moreover, we establish a similar result for the Lebesgue property, as well.
Key words: orthomorphism, bounded orthomorphism, /-algebra, locally solid vector lattice. Mathematical Subject Classification (2010): 46A40, 47B65, 46A32.
For citation: Sabbagh, R. and Zabeti, O. Bounded Orthomorphisms Between Locally Solid Vector Lattices, Vladikavkaz Math. J., 2021, vol. 23, no. 4, pp. 89-95. DOI: 10.46698/c1197-8093-8231-u.
1. Motivation and Introduction
Let us start with some motivation. Suppose X is an Archimedean vector lattice and Orth(X) is the space of all orthomorphisms on X. This space has many important consequences using just the order structure (see [1, Section 2.3]). One of the most remarkable advantages in Orth(X) is the pointwise lattice operations. On the other hand, there are several non-equivalent ways to define bounded operators between locally solid vector lattices; furthermore, these spaces have some ordered and topological structures, as well (see [2, 3] for a detailed exposition). Therefore, it is natural to expect some special properties from bounded orthomorphisms defined on a locally solid vector lattice. This is what our paper is about. We shall consider some topological and ordered structures for different types of bounded orthomorphisms. In particular, we establish a version of the known Zannen theorem [1, Theorem 2.62] for each category of bounded orthomorphisms. Moreover, we investigate topologically and ordered closedness for these classes of operators, as well. Now, let us recall some preliminaries we need in the sequel.
A vector lattice X is called order complete if every non-empty bounded above subset of X has a supremum and X is Archimedean if nx ^ y for each n € N implies that x ^ 0. It is known that every order complete vector lattice is Archimedean. A set S С X is called a solid set if x € X, y € S and |x| ^ |y| imply that x € S. Also, recall that a linear topology т on a vector lattice X is referred to as locally solid if it has a local basis at zero consisting of solid sets.
Suppose X is a locally solid vector lattice. A net (xa) С X is said to be order convergent to x € X if there exists a net (z^) (possibly over a different index set) such that zp l 0 and
© 2021 Sabbagh, R. and Zabeti, O.
for every ,0, there is an a0 with |xa — x| ^ zp for each a ^ a0. A set A C X is called order closed if it contains limits of all order convergent nets which lie in A. Keep in mind that a topology t on a vector lattice X making it a locally solid vector lattice is referred to as Fatou if it has a local basis at zero consisting of solid order closed neighborhoods. In this case, we say that X has the Fatou property. Observe that a locally solid vector lattice (X, t) is said to have the Levi property if every T-bounded upward directed set in X+ has a supremum. Finally, recall that a locally solid vector lattice (X, t) possesses the Lebesgue property if for every net (ua) in X, ua I 0 implies that ua 0.
Recall that, for an Archimedean vector lattice X, by Orth(X), we mean the space of all orthomorphisms on X; more precisely, an order bounded band preserving operator on X is called an orthomorphism. For more details on this subject, see [1]. Observe that a linear operator T on a vector lattice X is called band preserving if in X implies that T(x)±y. Note that by we mean |x| A |y| = 0. Now, suppose X is a locally solid vector lattice. An orthomorphism T on X is called nb-bounded if there exists a zero neighborhood U C X such that T(U) is bounded in X; T is said to be bb-bounded provided that it maps bounded sets into bounded sets. First, observe that compatible with the different spaces of all bounded operators on X, these notions of bounded orthomorphisms are not equivalent, in general. Consider the identity operator on RN; it is bb-bounded and continuous but not nb-bounded. Note that here we consider RN with the product (coordinate-wise) topology. Suppose X is c00 with the norm topology. Consider the orthomorphism T on X defined via T((xn)) = (nxn); indeed, it is neither bb-bounded nor continuous.
The class of all nb-bounded orthomorphisms on X is denoted by Orthn(X) and is equipped with the topology of uniform convergence on some zero neighborhood, namely, a net (Sa) of nb-bounded orthomorphisms converges to zero on some zero neighborhood U C X if for any zero neighborhood V C X there is an a0 such that Sa(U) C V for each a ^ a0. The class of all bb-bounded orthomorphisms on X is denoted by Orthb(X) and is allocated to the topology of uniform convergence on bounded sets. Recall that a net (Sa) of bb-bounded orthomorphisms uniformly converges to zero on a bounded set B C X if for any zero neighborhood V C X there is an a0 with Sa(B) C V for each a ^ a0.
The class of all continuous orthomorphisms on X is denoted by Orthc(X) and is equipped with the topology of equicontinuous convergence, namely, a net (Sa) of continuous orthomorphisms converges equicontinuously to zero if for each zero neighborhood V C X there is a zero neighborhood U C X such that for every e > 0 there exists an a0 with Sa(U) C eV for each a ^ a0. See [10] for a detailed exposition on these classes of operators. In general, we have Orthn(X) C Orthc(X) C Orthb(X) and when X is locally bounded, they coincide.
Furthermore, suppose X is a locally solid vector lattice. We say that X has the AM -property provided that for every bounded set B C X, Bv is also bounded with the same scalars; namely, given a zero neighborhood V and any positive scalar a with B C aV, we have Bv C aV. Observe that by Bv, we mean the set of all finite suprema of elements of B; for ample information, see [3].
All vector lattices in this note are assumed to be Archimedean. For undefined terminology and related topics, see [1, 4].
2. Main Result
First, we have the following useful facts.
Lemma 1. Suppose X is a locally solid vector lattice. Then Orthn(X), Orthb(X), and Orthc(X) are vector lattices.
< It is enough to prove that, in each case, the modulus of the orthomorphism exists and is also bounded. By [1, Theorem 2.40], the modulus of an orthomorphism T exists and satisfies |T|(x) = |T(x)| for each x € X+. Therefore, it is easy to see that if T is either nb-bounded or bb-bounded or continuous, then so is its modulus. >
Furthermore, we have a domination property, as well.
Proposition 1. Suppose X is a locally solid vector lattice and T, S are linear operators on X such that 0 ^ T ^ S. Then we have the following observations.
(i) If S € Orthra(X) then T € Orth«(X).
(ii) If S € Orthb(X) then T € Orthb(X).
(iii) If S € Orthc(X) then T € Orthc(X).
< (i) There exists a solid zero neighborhood U C X such that S(U) is bounded; moreover S is order bounded and band preserving. We need show that T is also order bounded, band preserving and nb-bounded. Choose arbitrary solid zero neighborhood V C X. Find scalar
Y > 0 with S(U) C 7V. For each x € U+, we have 0 ^ T(x) ^ S(x) € yV so that T(x) € V since V is solid. Since U C U+ — U-, we conclude that T(U) is also bounded. It is clear that T is also order bounded. Now, suppose x,y € X such that x±y. By [1, Theorem 2.36 and Theorem 2.40], we have
|T(x)| A |y| = T(|x|) A |y| < S(|x|) A |y| = |S(x)| A |y| = 0,
so that T(x)±y.
(ii) It is similar to the part (i); just if necessary, replace a bounded set B with its solid hull which is also bounded and use the inclusion B C B+ — B-.
(iii) Similar to the part (i), we conclude that T is a orthomorphism. Choose arbitrary solid zero neighborhood W C X. Find solid zero neighborhood V C X with V — V C W. There exists a solid zero neighborhood U C X with S(U) C V so that S(x) € V for each positive x € U. This implies that T(x) € V, as well; since V is solid. Therefore, T(U+) C V. We conclude that T(U) C T(U+) — T(U-) C V — V C W, as claimed. >
Proposition 2. Suppose X is a locally solid vector lattice. Then Orthn(X), Orthb(X) and Orthc(X) are locally solid vector lattices.
< For Orthn(X): by Lemma 1, Orthn(X) is an Archimedean vector lattice. By [4, Theorem 2.17], it suffices to prove that the lattice operations in Orthn(X) are uniformly continuous. Suppose (Ta) is a net of nb-bounded orthomorphisms on X which converges uniformly on some zero neighborhood to zero. It is enough to show that |Ta| ^ 0. There exists a zero neighborhood U C X such that for each zero neighborhood V C X there is an a0 such that Ta(U) C V for each a ^ ao. So, for each x € U+, by [1, Theorem 2.40], |Ta|(x) = |Ta(x)| € V for sufficiently large a. Note that U and V are solid so that the proof would be complete.
The proof for Orthb(X) is similar to the case of Orthn(X); just, we may assume that every bounded set B C X is solid otherwise, consider the solid hull of B which is bounded, certainly.
For Orthc(X): by Lemma 1, Orthc(X) is an Archimedean vector lattice. By [4, Theorem 2.17], it suffices to prove that the lattice operations in Orthc(X) are uniformly continuous. Suppose (Ta) is a net of continuous orthomorphisms on X which converges equicontinuously to zero. It is enough to show that |Ta| ^ 0. For every arbitrary solid zero neighborhood
V C X, there exists a solid zero neighborhood U C X such that for each e > 0, there is an a0 with Ta(U) C eV for each a ^ a0. So, for each x € U+, by [1, Theorem 2.40], |Ta|(x) = |Ta(x)| € eV for sufficiently large a. Note that U and V are solid so that the proof would be complete. >
Theorem 1. Suppose X is a locally solid vector lattice. If X possesses the AM -property, then so is Orthn(X), Orthb(X) and Orthc(X).
< For Orthn(X): assume that X possesses the AM-property and D C Orthn(X) is bounded. This means that D is bounded uniformly on some zero neighborhood. So, there is a zero neighborhood U C X such that for arbitrary zero neighborhood V C X, there exists a positive scalar a such that {T(U) : T € D} C aV. In particular, for any x € U, the set Bx = {T(x) : T € D} C aV. This implies that, by the assumption, Bxv is also bounded in X. Again, by [1, Theorem 2.43] and using [5, Lemma 3], (T V ... V Tn)(x) = Ti(x) V ... VTn(x) € aV V ... V aV = aV for each Ti,..., Tra € D. So, (Ti V ... V Tra)(U) C aV. This would complete the proof.
The proof for Orthb(X) is similar to the case of Orthn(X); just, we may assume that every bounded set B C X is solid otherwise, consider the solid hull of B which is bounded, certainly.
For Orthc(X): assume that X possesses the AM-property and D C Orthc(X) is bounded. This means that D is bounded equicontinuously. Therefore, for arbitrary zero neighborhood V C X, there exists a zero neighborhood U C X with T(U) C V for each T € D. In particular, for any x € U, the set Bx = {T(x) : T € D} C V. This implies that, by the assumption, Bxv is also bounded in X. Again, by [1, Theorem 2.43] and using [5, Lemma 3], (Ti V ... V Tra)(x) = Ti(x) V ... V Tra(x) € V V ... V V = V for each Ti,..., Tra € D. So, (t1 V ... V Tn)(U) C V. This would complete the proof. >
Remark 1. Observe that the converse of Theorem 1 is not true, in general. Consider X = l1; it does not have the AM-property. By [1, Theorem 4.75 and Theorem 4.77] , Orthn(X) is an AM-space with unit so that a C(K)-space for some compact Hausdorff space K. Therefore, Orthn(X) possesses the AM-property.
Nevertheless, when we consider a locally solid f-algebra X, we can have the converse, as well. Before this, we have two useful facts. Recall that by an f-algebra X, we mean a Riesz algebra such that given x,y € X with x A y = 0, we have zx A y = xz A y = 0 for each positive z € X. A locally solid Riesz space which is an f-algebra at the same time is called a locally solid f-algebra. Observe that a locally solid f-algebra is a topological algebra in its own nature so that the multiplication is continuous in this case. For a comprehensive context, see [6, 7].
Theorem 2. Suppose X is a locally solid f-algebra with a multiplication unit e. Then there is an f-algebra isomorphism homeomorphism from X onto Orthb(X).
< By [1, Theorem 2.62], there is an f-algebra isomorphism from X onto Orth(X) defined by u ^ Tu such that Tu(x) = ux. Now, consider this mapping from X into Orthb(X); note that each Tu is bb-bounded using [7, Proposition 2.1]. Furthermore, another using of [1, Theorem 2.62], convinces that this mapping is also onto and an f-isomorphism. So, it is enough to show that it is a homeomorphism. Suppose (ua) is a null net in X. We need show that TUa is null in the topology of uniform convergence on bounded sets. Fix a bounded set B C X and choose arbitrary zero neighborhood V C X. By another application of [7, Proposition 2.1], we find a zero neighborhood U C X such that UB C V. There exists an a0 with ua € U for each a ^ a0 so that TUa (B) = uaB C V for sufficiently large a. For the converse, suppose (TUa) is a null net in Orthb(X). This means that for each x € X, TUa (x) ^ 0 in X. Put x = e, the multiplication unit of X. We see that ua ^ 0, as claimed. >
Theorem 3. Suppose X is a locally solid f-algebra with a multiplication unit e. Then there is an f-algebra isomorphism homeomorphism from X onto Orthc(X).
< The proof has the same line as in the proof of Theorem 2. By [1, Theorem 2.62], there is an f -algebra isomorphism from X onto Orth(X) defined by u ^ Tu such that Tu(x) = ux. Now, consider this mapping from X into Orthc(X); note that each Tu is continuous since the multiplication is continuous in X. Furthermore, another using of [1, Theorem 2.62], convinces that this mapping is also onto and an f-isomorphism. So, it is enough to show that it is a homeomorphism. Suppose (ua) is a null net in X. We need show that TUa is null equicontinuously. Choose arbitrary zero neighborhood V C X. there exists a zero neighborhood U C X such that UU C V. For each e > 0, there exists an a0 with ua € eU for each a ^ a0 so that TUa (U) = uaU C eUU C eV for sufficiently large a. For the converse, suppose (TUa) is a null net in Orthc(X). This means that for each x € X, TUa(x) ^ 0 in X. Put x = e, the multiplication unit of X. We see that ua ^ 0, as claimed. >
Corollary 1. By considering [1, Theorem 2.62] and also Theorem 2 and Theorem 3, we conclude that when X is a locally solid f -algebra with unit, we have, Orthb(X) = Orthc(X) = Orth(X) = X. Therefore, in this case, we can transfer the Lebesgue property or the Fatou property between X and different classes of bounded orthomorphisms.
Remark 2. Note that we can expect Corollary 1 for Orthn(X), in general. Consider X = Rn; it is a locally solid f-algebra with unit. By Corollary 1, Orthb(X) = Orthc(X) = Orth(X) = X = Rn. But we claim that Orth«(X) = {0}. For if 0 = T € Orth«(X), then by [1, Theorem 2.62], it should be of the form T = Tu(x) = ux for some 0 = u € X. Nevertheless, in this case, T can not be nb-bounded since X is not locally bounded.
Theorem 4. Suppose X is a locally solid vector lattice. If X possesses the Levi property, then so are Orthn(X), Orthb(X), and Orthc(X).
< We prove the result for Orthn(X); the proofs for other cases are similar. Assume that (Ta) is a bounded increasing net in Orthn(X)+. The general idea follows from the proof of [3, Theorem 2.15]. This implies that there is a zero neighborhood U C X such that (Ta(U)) is uniformly bounded for each a. So, for each x € X+, the net (Ta(x)) is bounded and increasing in X+ so that it has a supremum, namely, ax. Define T : X+ ^ X+ via T(x) = ax. It is an additive map; it is easy to see that ax+y ^ ax + ay. For the converse, fix any a0. For each a ^ a0, we have Ta(x) ^ ax+y — Ta(y) ^ ax+y — Tao (y) so that ax ^ ax+y — Tao (y). Since a0 was arbitrary, we conclude that ax + ay ^ ax+y. By [1, Theorem 1.10], it extends to a positive operator T : X ^ X. It is clear that T is also order bounded. We need show that it is band preserving. By [1, Theorem 2.36], it is sufficient to prove that x±y implies that Tx±y. Note that each Ta is band preserving so that
|T(x)| A |y| = ( V |Ta(x)|) A |y| = V(|T«(x)| A |y|) = 0.
^ a ' a
Suppose V is an arbitrary order closed zero neighborhood in X. There is a positive scalar Y with Ta(U) C yV. This means that T(U) C yV since V is order closed. >
Remark 3. Observe that the converse of Theorem 4 is not true, in general. Consider the Banach lattice E consists of all piecewise linear continuous functions on [0,1]. By [8, Theorem 6], Orthn(E) = Orth(E) is the space {a/ : a € R}, where I denotes the identity operator on E. Clearly Orthn(E) possesses the Levi property but E fails to have the Levi property (it is not even order complete). Moreover, this example also shows that the Lebesgue property does not transfer from the space of all bounded orthomorphisms into the underlaying space, as well.
Note that a linear operator T between vector lattices X and Y is said to be disjoint preserving provided that for each x,y € X with x±y, we have T(x)±T(y). Moreover, recall
that a linear operator T on a vector lattice X is called band preserving if given x, y € X, x±y, implies that T(x)±y. For more information, see [1].
Lemma 2. Suppose X, Y are locally solid vector lattices such that Y is Hausdorff and (Ta) is a net of disjoint preserving operators from X into Y which is convergent pointwise to the operator T. Then T is also disjoint preserving.
< Choose arbitrary solid zero neighborhood W C Y; there exists a zero neighborhood V C Y with V+V C W. Consider x, y € X with x±y. Find an index a0 with (Tao —T)(x) € V and (Tao — T)(y) € V. We claim that T(x)±T(y). By the Birkhoff's inequality, we have
0 < ||T (x)| A |T (y)|-|Tao (x)|A|Tao (y) ||
= ||T (x)| A |T (y)| - |Tao (x)| A |T (y)| + |Tao (x) A |T (y)| - |Tao (x)| A |Tao (y)||
< |Tao (x) - T(x)| + |Tao (y) - T(y)| € V + V = W.
Since Tao is disjoint preserving, |Tao (x)| A |Tao (y)| = 0. Since W is solid, we conclude that |T(x)| A |T(y)| € W. This happens for each arbitrary solid zero neighborhood W C Y. Therefore, |T(x)| A |T(y)| =0 as claimed. >
Lemma 3. Suppose X is a Hausdorff locally solid vector lattice and (Ta) is a net of band preserving operators on X which is convergent pointwise to the operator T. Then T is also band preserving.
< Choose arbitrary solid zero neighborhood W C Y. Consider x, y € X with x±y. Find an index a0 with (Tao - T)(x) € W. We claim that T(x)±y. By the Birkhoff's inequality, we have
0 < ||T(x)| A |y| - |Tao(x)| A |y|| < ||Tao(x)| - |T(x)|| < |Tao(x) - T(x)| € W.
Since Tao is band preserving, |Tao (x)|A|y| = 0. Since W is solid, we conclude that |T(x)|A|y| € W. This happens for each arbitrary solid zero neighborhood W C X. Therefore, |T(x)|A|y| = 0 as desired. >
Observe that, in general, the uniform limit of a sequence of order bounded operators need not be order bounded (see [1, Example 5.6], due to Krengle); note that, in the example, each Kn is order bounded and Kn — T uniformly, nevertheless, we can choose the coefficients (an) such that T is not even order bounded (the modulus does not exist). Furthermore, Kn t T, but T is not order bounded, as mentioned. Consider this point that in the example, the underlying space E, is not an AM-space. Now, we focus on locally solid vector lattices. Compatible with Lemma 3, [3, Corollary 2.7] and [9, Lemma 3.1 and Lemma 3.2], we have the following. Observe that by a topologically complete topological vector space we mean a topological vector space in which every Cauchy net is convergent.
Corollary 2. Suppose X is a topologically complete Hausdorff locally solid vector lattice with the Levi and AM -properties. Then, Orthb(X) and Orthc(X), with respect to the assumed topologies, are topologically complete. Moreover, they form bands in Orth(X).
Remark 4. Note that we are able to consider some results presented for Orthb(X) and Orthc(X), for Orthn(X), as well. Just, observe that Orthn(X) is not topologically complete as shown in [10, Example 2.22].
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Received December 22, 2020
Raheleh Sabbagh
University of Sistan and Baluchestan,
Department of Mathematics, Faculty of Mathematics,
Statistics, and Computer Science,
P.O. Box 98135-674, Zahedan, Iran
PhD Student
E-mail: sabaghrahele@gmail.com
https://orcid.org/0000-0003-2722-7160
Omid Zabeti
University of Sistan and Baluchestan,
Department of Mathematics, Faculty of Mathematics,
Statistics, and Computer Science,
P.O. Box 98135-674, Zahedan, Iran
Associate Professor
E-mail: o.zabeti@gmail.com
https://orcid.org/0000-0001-6875-2418
Владикавказский математический журнал 2021, Том 23, Выпуск 4, С. 89-95
ОГРАНИЧЕННЫЕ ОРТОМОРФИЗМЫ МЕЖДУ ЛОКАЛЬНО СОЛИДНЫМИ ВЕКТОРНЫМИ РЕШЕТКАМИ
Саббах Р.1, Забети О.1
1 Университет Систана и Белуджистана, Иран, Захедан, P.O. Box 98135-674 E-mail: sabaghrahele@gmail.com, o.zabeti@gmail.com
Аннотация. Цель настоящей заметки — изучение понятия ограниченного ортоморфизма между локально солидными векторными решетками. Устанавливается вариант замечательной теоремы Цаа-нена об изоморфизме между ортоморфизмами и векторными решетками, в которых они действуют, для различных типов ограниченных ортоморфизмов. Кроме того, рассматриваются топологическая и порядковая структура этих классов ортоморфизмов. В частности, показано, что каждый класс ортоморфизмов обладает свойством Леви или AM-свойством в том и только в том случае, когда этим свойством обладает соответствующая локально солидная векторная решетка. Аналогичный результата получен и для свойства Лебега.
Ключевые слова: ортоморфизм, ограниченный ортоморфизм, f-алгебра, локально солидная векторная решетка.
Mathematical Subject Classification (2010): 46A40, 47B65, 46A32.
Образец цитирования: Sabbagh, R. and Zabeti, O. Bounded Orthomorphisms Between Locally Solid Vector Lattices // Владикавк. мат. журн.—2021.—Т. 23, № 4.—C. 89-95 (in English). DOI: 10.46698/c1197-8093-8231-u.
