Sums of even powers of k-regulous functions
aa r X i v : . [ m a t h . AG ] J u l SUMS OF SQUARES OF K-REGULOUS FUNCTIONS
JULIUSZ BANECKI AND TOMASZ KOWALCZYK
Abstract.
We prove that the Pythagoras number of the ring of 0-regulous functions R ( R n ) is finite and bounded from above by 2 n .We also show that a k -regulous function which is nonnegative on R n cannot be necessarily written as a sum of squares of k -regulousfunctions, provided that k > n >
1. We then obtain lowerbounds for p ( R k ( R n )) of k -regulous functions on R n , which tendsto infinity as k tends to infinity for a fixed n ≥ Introduction
Consider the following problem: let f be a real valued nonnegativefunction on the space X . It is then natural to ask two questions,whether f can be written as a sum of squares of some functions on X ,and if so, how many squares are needed? In this paper we deal withthe case X = R n and f a k -regulous function for n ≥ k ≥ f is a psd if f ( α ) ≥ α ∈ Sper( A ), where Sper( A ) denotes the real spectrum of aring. Equivalently, a k -regulous functions on R n is psd if it is nonneg-ative. Obviously sos implies psd. For a ring A , we denote by p ( A ) thePythagoras number of A , i.e. the smallest positive integer such thatany sum of squares can be written as a sum of at most p ( A ) squares.If such number does not exist, we put p ( A ) = ∞ . It was known since Hilbert that not every nonnegative polynomialcan be written as a sum of squares of polynomials, hence psd = sos.But, by a famous result of Artin, every nonnegative polynomial can bewritten as a sum of squares of rational functions, and as a consequencepsd=sos holds in the field of rational functions on R n . We will beinterested in this question in case of intermediate rings between poly-nomials and rational functions, namely k -regulous functions. This is a Mathematics Subject Classification.
Primary: 26C15, 14P99.
Key words and phrases.
Pythagoras number, k-regulous function, sums ofsquares. still quite new object, first paper devoted to systematic study of thesefunctions was [10]. We refer to [7, 9] for exposition on this theory.Let A be a ring. In general, computing Pythagoras number of A isa very difficult task. It is hard to even determine whether this numberis finite or not. One of the problems in this manner is to determinethe precise value of p ( R ( x , . . . , x n )). Pfister [13] was first to show thatthis number is finite and bounded from above by 2 n . Lower bound isknown to be n + 2, which in particular gives p ( R ( x , x )) = 4 . Situ-ation is different when we consider polynomials, it was shown in [3]that p ( R [ x , . . . , x n ]) = ∞ , for n ≥
2. There are also some resultsconcerning excellent rings, see [6].In Section 2 we discuss the case k = 0. It was shown in [8] thatpsd=sos holds in this case. We provide some lower and upper boundsof the Pythagoras number of p ( R ( R n )) for n ≥
1. In Section 3 weconsider the case k >
0, and we show that psd = sos. Also, we givelower bounds depending on k such that p ( R k ( R n )) tends to infinity as k tends to infinity, provided that n > k -regulousgeometry. Definition 1.1.
Let n be a positive integer and let X ⊂ R n be anirreducible, nonsingular real affine variety (i.e. X ⊂ R n is an algebraicsubset with structure sheaf of regular functions). Let k ∈ N . We saythat a continuous function f : X → R is k -regulous on X if f is ofclass C k on X and f is a rational function on X , i.e. there exists anon-empty Zariski open subset U ⊂ X such that f | U is regular. Sums of squares of -regulous functions In this section we deal with the k -regulous functions for k = 0. Webegin with the following definitions Definition 2.1.
Let A be a ring. We say that an element f ∈ A istotally positive if there exists elements c, d which are sums of squaresin A such that the following identity holds cf = 1 + d. It is clear that a k -regulous function on R n is totally positive if it isstrictly positive on R n . Definition 2.2.
We say that a ring A is a ring of regular functions ifthe set Σ = { P A } consists of units. It is obvious that for any k, n ∈ N any ring of k -regulous functionson R n is a ring of regular functions. UMS OF SQUARES OF K-REGULOUS FUNCTIONS 3
Before proving our first result we recall the following theorem con-cerning rings of regular functions.
Theorem 2.3. [12, Theorem 7.3]
Let A be a ring of regular functions oftranscendence degree ≤ d over R and let f be a totally positive elementof A . Then f is a sum of at most d squares in A . Theorem 2.4.
Let n > be a positive integer. Then the Pythagorasnumber of -regulous functions on R n is bounded from above by n .Proof. Let f ∈ R ( R n ) be a sum of squares, and let Z = Z ( f ) be thezero set of f . Consider now the function f on a set U = R n \ Z , it is atotally positive element. By the Theorem 2.3 f can be written as f = f + f + · · · + f n where f i ∈ R ( U ) for i = 1 , , . . . , n . Since the zero set of f is nowheredense in R n in the Euclidean topology, each of the functions f i can becontinously extended to the whole R n . This finishes the proof. (cid:3) Remark . It is a straightforward fact that the Pythagoras numberof 0-regulous functions on R n for n > n + 2 ≤ p ( R ( R n )) ≤ n . In particular p ( R ( R )) = 4.Combining the above with the Theorem 2.6. [8, Theorem 6.1]
Let f ∈ R ( R n ) be a non-negative -regulous function. Then f can be written as a sum of squares in R ( R n ) . we get Theorem 2.7.
Let f ∈ R ( R ) be a non-negative -regulous function.Then we have the identity f = f + f + f + f for some f , f , f , f ∈ R ( R n ) .Remark . All of the above can be also applied to the ring of locallybounded rational functions on R n , resulting in the same inequalitiesconcerning Pythagoras number. JULIUSZ BANECKI AND TOMASZ KOWALCZYK Sums of squares of k -regulous functions In this section we study psd and sos elements in the rings of k -regulous functions on R n for k > n >
1. We show that equalitypsd=sos does not hold and we give lower bound for the Pythagorasnumber.Let p ∈ R [ x , x , . . . , x n ] be a polynomial, p can be written in theform p = p d + p d +1 + . . . , with each p k a homogeneous polynomial. Wedefine the initial form of p as In( p ) := p d . We then have ord( p ) = d .We begin with the following lemma Lemma 3.1.
Let f = pq ∈ R ( x , x , . . . , x n ) be a rational function.Assume that ord( p ) = ord( q ) = d > and In( q ) In( p ) . Then f is notcontinuous at the origin.Proof. Rewrite f = In( p )+ p In( q )+ q where p , q are polynomials of order > d .Take a point a = ( a , a , . . . , a n ) ∈ R n such that In( q )( a ) = 0 and q ( a ) = 0. Then f ( a ) = In( p )( a ) + p ( a )In( q )( a ) + q ( q ) = In( p )( a )In( q )( a ) + p ( a )In( q )( a ) q ( a )In( q )( a ) . By the order consideration, the only relevant term on the right handside is
In( p )( a )In( q )( a ) . If we now converge along the sequence a n = ( a n , . . . , a n n ),then the limit as n tends to infinity is precisely In( p )( a )In( q )( a ) . Hence, it isenough to find a point b ∈ R n such that the analogous limit havedifferent value. This clearly follows from the assumption In( q ) In( p )and d > (cid:3) Lemma 3.2.
Let f ∈ R ( x , x , . . . , x n ) . Suppose that f ′ x i is a polyno-mial for every i. Then f is a polynomial itself.Proof. The proof will be carried out by induction on n . The case n = 1is trivial. Integrating f ′ x with respect to x , we can find a polynomial r such that f − r does not depend on x , i.e. f − r ∈ R ( x , . . . , x n ).The result follows from the induction hypothesis applied to f − r . (cid:3) Lemma 3.3.
Let f = pq ∈ R k ( R n ) . Assume that q (0) = 0 and ord( p ) = ord( q ) + k , then In( p ) = F k ( x , x , . . . , x n ) · In( q ) where F k ( x , x , . . . , x n ) is a homogeneous polynomial of degree k .Proof. Lemma 3.1 deals with the case k = 0, so we may inductivelyassume that the Theorem is true for a given k . Take any i = 1 , . . . , n .Suppose ( In( p )In( q ) ) ′ x i is not identically 0. That means (In( p )) ′ x i In( q ) − UMS OF SQUARES OF K-REGULOUS FUNCTIONS 5
In( p )(In( q )) ′ x i neither is 0, so it is obvious that In( p ′ x i q − pq ′ x i ) =(In( p )) ′ x i In( q ) − In( p )(In( q )) ′ x i . Now applying the induction hypoth-esis to the ( k − f ′ x i = p ′ xi q − pq ′ xi q we getIn( q ) | (In( p )) ′ x i In( q ) − In( p )(In( q )) ′ x i and, as a consequence( In( p )In( q ) ) ′ x i = (In( p )) ′ x i In( q ) − In( p )(In( q )) ′ x i In( q ) ∈ R [ x , x , . . . , x n ]. Lemma 3.2 finishes the induction. (cid:3) The main Theorem of this section, useful in providing some coun-terexamples, goes as follows:
Theorem 3.4.
Let p and q be two homogeneous polynomials such that deg ( p ) − deg ( q ) = 2 d > and pq is a sum of l squares of d -regulousfunctions. Then pq is also a sum of at most l homogeneous polynomialsof degree d , in particular pq is a polynomial.Proof. By the assumption, p/q can be written as pq = l X i =0 h i r with h , . . . , h l , r ∈ R [ x , x , . . . x n ]. Multiplying by denominators andcomparing the lowest degree terms of both sides we getIn( r ) p = q · In( l X i =0 h i ) = q · l X i =0 e h i where for each i either e h i = In( h i ) or e h i is identically zero. If In( r (0)) =0 then each of non-zero e h i is a homogeneous polynomial of degree d . IfIn( r (0)) = 0 then by the above lemma, each e h i = In( r ) · F id where F id isa homogeneous polynomial of degree d . In either case we are left witha representation of pq as a sum of at most l homogeneous polynomialsof degree d , as required. (cid:3) Take the k -regulous function f k = x k x + y ∈ R k ( R n ) with k > n >
1. Consider the idealBad( f k ) := { h ∈ R k ( R n ) : h f k = l X i =1 g i , g i ∈ R k ( R n ) } . JULIUSZ BANECKI AND TOMASZ KOWALCZYK
Obviously, x + y ∈ Bad ( f k ), hence Z (Bad( f k )) ⊂ Z ( x + y ). In orderto prove that this set is nonempty, it is enough to show (because of theweak Nullstellensatz) that 1 / ∈ Bad( f k ). It is though, an immediateconsequence of Theorem 3.4 (since deg( x k +2 ) − deg( x + y ) = 2 k and x k x + y is not a polynomial). As a consequence, we get Theorem 3.5.
The function f k is not a sum of squares in R k ( R n ) .Remark . In Dellzells PhD Thesis [5] he showed that the set ofbad points of a nonnegative polynomial has codimension at least 3.The above result shows that the set of bad points of a nonnegative k -regulous function on R n for k > n > p ( R k ( R n )) dependingon k . In particular, p ( R k ( R n )) if finite, tends to infinity as k tends toinfinity for a fixed n ≥ d be a positive integer, R [ x , . . . , x n ] d denotes the R -vector spaceof forms of degree 2 d . Denote by p ( n, d ) the Pythagoras number of aspace of forms of degree 2 d . In [3] it was shown that p ( R [ x , . . . , x n ]) = ∞ if n ≥
2. As a consequence, the Pythagoras number p ( n, d ) tends toinfinity if d tends to infinity, provided that n ≥
3, see [14] for discussionon lower and upper bounds.
Theorem 3.7.
Let k > and n > be positive integers. Then thePythagoras number of R k ( R n ) satisfies p ( n, k ) ≤ p ( R k ( R n )) . Proof.
By the definition of p ( n, k ) there exists a homogeneous poly-nomial f of degree 2 k being a sum of squares in R [ x , . . . , x n ], whichcannot be represented as a sum of p ( n, k ) − f is a sumof p ( n, k ) − k -regulous functions, Theorem 3.4 yields animmediate contradiction. Since f is a sos in R k ( R n ), we get p ( n, k ) ≤ p ( R k ( R n )) . (cid:3) Remark . It is worth noting that for n = 2 the above Theorem givesus nothing as p (2 , k ) = 2, for any k >
0. However, as a consequenceof [2] and the fact that codimension of the bad set of a nonnegativepolynomial is at least 3, The Motzkin polynomial M ( x, y ) = 1 + x y ( x −
3) + x y can be written as a sum of precisely 4 regular functions, hence4 ≤ p ( R k ( R )) UMS OF SQUARES OF K-REGULOUS FUNCTIONS 7 for k ≥ A be a ring. Consider the following length ⌣ l ( a ) = inf { n | a r +1 is a sum of n squares } and the modified Pythagoras number (as suggested by Pfister in [12]) ⌣ P ( A ) = sup { ⌣ l ( a ) | a is a sum of squares } . We will prove the following
Theorem 3.9.
Let k ≥ , and n ≥ be positive integers. Then themodified Pythagoras number of the ring of k -regulous functions on R n satisfies ⌣ P ( R k ( R n )) ≤ n . Furthermore, for every nonnegative k -reguluos function f we have ⌣ l ( f ) ≤ n . Before we prove the Theorem, recall the Lojasiewicz property
Proposition 3.10. [7, Lemme 5.1]
Let f be a k -regulous functions on R n and g be a k -regulous function on U , where U is the complement ofthe zero set of f . Then there exists a positive integer N such that thefunction f N g is k -regulous on R n .Proof of Theorem 3.9 Let f be a nonnegative k -regulus function on R n and let Z be the zero set of f . Then f is a totally positive unit on U := R n \ Z , hence by Theorem 2.3 we get an equality f | U = P n i =1 f i ,where each f i is a k -regulous function on U . By the Lojasiewicz prop-erty, there exists N a positive integer large enough, such that f i f N isa k -regulous function on R n , for each i . This implies the equality f N +1 = n X i =1 ( f n f i ) on R n , which finishes the proof. Remark . In the paper [11] Kucharz introduced the notion of k -regulous Nash function. It is a straightforward calculation, that theresults of this paper holds also for k -regulous Nash functions on R n . JULIUSZ BANECKI AND TOMASZ KOWALCZYK
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