DOI 10.1007/s10957-017-1102-7 F O RU M
A Note on the Paper “The Algebraic Structure of the Arbitrary-Order Cone”
Xin-He Miao1 · Yen-chi Roger Lin2 · Jein-Shan Chen2
Received: 7 June 2016 / Accepted: 13 March 2017
© Springer Science+Business Media New York 2017
Abstract In this short paper, we look into a conclusion drawn by Alzalg (J Optim Theory Appl 169:32–49,2016). We think the conclusion drawn in the paper is incorrect by pointing out three things. First, we provide a counterexample that the proposed inner product does not satisfy bilinearity. Secondly, we offer an argument why a pth-order cone cannot be self-dual under any reasonable inner product structure onRn. Thirdly, even under the assumption that all elements operator commute, the inner product becomes an official inner product and the arbitrary-order cone can be shown as a symmetric cone, we think this condition is still unreasonable and very stringent so that the result can only be applied to very few cases.
Keywords pth-order cone· Second-order cone · Inner product · Jordan algebras Mathematics Subject Classification 17C10· 52A07
Communicated by Sándor Zoltán Németh.
B
Yen-chi Roger Lin [email protected] Xin-He Miao[email protected] Jein-Shan Chen [email protected]
1 Department of Mathematics, Tianjin University, Tianjin 300072, China
2 Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan
1 Introduction
In the recent paper [1], Alzalg claims that the arbitrary-order cone is a symmetric cone for any order greater than or equal to 1, that is, the cone is homogeneous and self-dual.
Ito and Lourenço [2] showed that the pth-order cone in dimension n ≥ 3 are not homogeneous unless p= 2. In this short note, we show that the arbitrary-order cone is not self-dual for any order other than 2. In particular, we provide a counterexample indicating that the inner product defined in [1] is indeed not an inner product because it does not satisfy the bilinearity. We offer an argument why a pth-order cone cannot be self-dual under any reasonable inner product structure onRn. In addition, Alzalg assumes that all elements operator commute in order to show that the arbitrary-order cone is a symmetric cone. We think this condition is unreasonable and very stringent by elaborating the reason and counterexample. To sum up, we think the conclusion drawn in the paper [1] is very limited.
2 A Type of Inner Product
Alzalg first in [1] defines an inner product as follows:
x, yp:= 1 2xT
Jp(x) + Jp(y)
y, (1)
where x, y ∈ Rn= R × Rn−1, p ∈ [1, ∞], and Jp(·) is the matrix given by
Jp(x) :=
⎧⎪
⎨
⎪⎩
1 0
0 ¯x
2 p
¯x22
In−1
, if ¯x = 0,
In, if ¯x = 0
with x = (x1, ¯x) ∈ R × Rn−1.
Alzalg argues thatx, ypdefined as in (1) is a new inner product onRn, and the pth-order cone becomes symmetric under this new product. The pth-order cone inRn is defined as
Pp:=
x= (x1, ¯x) ∈ R × Rn−1: x1≥ ¯xp
.
Unfortunately, the functionalx, ypfails to be bilinear, i.e., the functionalx, ypis not an inner product onRn. We provide the following counterexample.
Example 2.1 For anyv = (v1, ¯v) ∈ R × R2and p ≥ 1, we denote v(p) := ¯v¯v2p2 2
. Consider x = (0, 1, 0), y = (0, 0, 1) and z = (0, 1, 1) in R3. Then, it follows from (1) that (p) = (p). Moreover, we have
x, zp= y, zp= 1 2
x(p) + z(p) ,
x + y, zp= z, zp= 1 2
z(p) + z(p)
= z(p).
Clearly x + y, zp = x, zp+ y, zp if and only if x(p) = z(p). However,
x(p) = 1 for any p ≥ 1, and z(p) = 2(2/p)−1, which equals 1 only when p= 2.
Hence,x, ypis never a bilinear form onRnexcept for p= 2. In other words, x, yp
is not an inner product onRnwhen p= 2.
3 Why Cannot the pth-Order Cone
Ppbe Self-Dual When p = 2?
In this section, we give a reason why the pth-order conePpcannot be self-dual under any (reasonable) inner product onRnwhen p= 2. By a “reasonable inner product”
onRnwe mean that the standard basis{e1, e2, . . . , en} of Rnis an orthogonal frame under the inner product·, ·, that is,
ei, ej = 0 whenever 1 ≤ i, j ≤ n, i = j. (2)
This inner product can be written as
x, y = xTM y, x, y ∈ Rn, (3)
where M is a diagonal matrix M = diag(λ1, λ2, . . . , λn) with λi > 0 for all i = 1, 2, . . . , n. This is a reasonable assumption on the inner products to make, since we will almost never take arbitrary frames inRnto define a pth-order cone. Also we note that the “inner product”·, ·pappeared in [1] satisfies the property (2).
The following proposition can be proved by standard arguments in convex analysis;
see [3].
Proposition 3.1 IfPpis self-dual under the inner product·, ·, then for each nonzero vector x ∈ ∂Ppthere is a nonzero vector x∈ ∂Ppsuch thatx, x = 0.
So now let us assume thatPpinRnis self-dual under the inner product (3). For each k= 2, 3, . . . , n, we consider the inner product between uk = e1+ekandvk= e1−ek
to see that
0≤ uk, vk = λ1− λk,
that is,λ1≥ λkfor all k= 2, 3, . . . , n. On the other hand, if we apply Proposition3.1 to x = uk, then there is a nonzero vector x= (x1, x2, . . . , xn) ∈ ∂Ppsuch that
0= uk, x = λ1x1+ λkxk.
Sinceλ1 ≥ λk > 0 and |xk| ≤ x1, it is necessary that xk = −x1andλ1= λk; this holds for all k= 2, 3, . . . , n. Therefore, M must be a positive multiple of the identity
matrix, and the inner product·, · is a scalar multiple of the standard Euclidean inner product onRn. Under this structure, the dual cone ofPpisPq, where 1/p + 1/q = 1 (cf. [4]); hence,Pp can be self-dual only when p = 2 (or the trivial case n = 2).
In any case, a pth-order cone cannot be self-dual under any reasonable inner product structure onRnunless p= 2 or n = 2.
4 Operator Commutativity
In order to show the arbitrary-order cone being a symmetric cone, Alzalg further assumes (on page 36 in [1]) that
Without loss of generality, throughout this paper, we assume that all elements operator commute. If elements do not operator commute, we can scale the under- lying optimization problem so that the scaled elements operator commute [5].
We argue that the condition that all elements operator commute is too harsh and not very useful. Recall that for any x, y in a Euclidean Jordan algebra Rn, if the elements x and y operator commute, it implies that x and y have the same spectral decomposition (see [6]), i.e.,
x= λ1(x)e1+ λ2(x)e2, y = μ1(y)e1+ μ2(y)e2,
where{e1, e2} are a Jordan frame in Rn. Indeed, under this condition, the considered pth-order cone inRn becomes the pth-order cone in the subspace generated by e1 and e2inRn. In other words, the pth-order cone is restricted to a set which behaves similarly to the second-order cone because under transformation Jp(x), the pth-order cone can be recast as the second-order cone. It means the assumption leads to a very special subcase, and this is not what we want for real pth-order cone.
From the above-quoted paragraph, it seems that Alzalg thought that all elements can be made operator commute after rescaling. However, as noted by one of the reviewers, the only Euclidean Jordan algebraJ where all elements operator commute is Rn(with the usual inner product and componentwise Jordan product). This can be seen by noting that any two primitive idempotents operator commute in this algebraJ and hence (via simultaneous spectral decompositions) are either identical or orthogonal. This implies that there is only one Jordan frame. (If there is a primitive idempotent outside a given Jordan frame, then it is orthogonal to all the elements of the Jordan frame and hence orthogonal to the unit element.) Via the spectral decomposition theorem, the algebra becomesRn.
In fact, based on the matrix Jp(x) in [1], we can establish the relationship between the pth-order cone and second-order cone. We first define the matrix ¯Jp(x) for any x= (x1, ¯x) ∈ Rnas bellow:
¯Jp(x) :=
⎧⎪
⎨
⎪⎩
1 0
0 ¯x ¯x2
pIn−1
, if ¯x = 0,
In, if ¯x = 0.
Theorem 4.1 For any x= (x1, ¯x) ∈ Rnand p≥ 1, we have (a) x ∈ P2 ⇒ ¯Jp(x)x ∈ Pp;
(b) x∈ Pp ⇒ [ ¯Jp(x)]−1x ∈ P2.
Proof For ¯x = 0, the results of (a) and (b) are obvious. Thus, we only consider the case ¯x = 0.
(a) For any x ∈ P2, we have ¯x2 ≤ x1. Moreover, we know that ¯Jp(x)x = (x1, ¯x ¯x2p¯x) ∈ Rn. Hence, it follows that (¯x2/ ¯xp) · ¯x
p = ¯x2 ≤ x1, i.e., ¯Jp(x)x ∈ Pp.
(b) With the similar arguments, for any x∈ Pp, we obtain that ¯xp≤ x1. Note that ¯Jp(x)−1
x =
x1, ¯x ¯xp2 ¯x
∈ Rn. Therefore, we have (¯xp/ ¯x2) · ¯x
2 =
¯xp≤ x1, which says ¯Jp(x)−1
x∈ P2. The proof is complete.
5 Conclusions
In this short paper, we show that the conclusion that the arbitrary-order cone is a symmetric cone for any order greater than or equal to 1 drawn in a recent paper by Alzalg is invalid. First, we provide a counterexample to show that the inner product proposed therein does not satisfy bilinearity. Secondly, we offer an argument why a pth-order cone cannot be self-dual under any reasonable inner product structure onRn. Thirdly, we show that the assumption of all elements operator commute is unreasonable and very stringent so that the result can only be applied to very few cases.
Acknowledgements The first author’s work is supported by National Natural Science Foundation of China (No. 11471241). The third author’s work is supported by Ministry of Science and Technology, Taiwan. The paragraph about that the only Euclidean Jordan algebra where all elements operator commute isRn, suggested by one of the reviewers, is so valuable that the authors decide to include it. The authors would like to thank anonymous reviewers for suggestions, which improved the paper.
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