On Bounding the Betti Numbers and Computing ... - Semantic Scholar
On Bounding the Betti Numbers and Computing the Euler Characteristic of Semi-algebraic Sets Saugata Basu
Abstract
In this paper we give a new bound on the sum of the Betti numbers of semi-algebraic sets. This extends a well-known bound due to Oleinik and Petrovsky [19], Thom [23] and Milnor [18]. In separate papers they proved that the sum of the Betti numbers of a semi-algebraic set de ned by 1 0 0 ( ) 1 is bounded by ( ( )) Given a semialgebraic set de ned as the intersection of a real variety, = 0 ( ) whose real dimension is with a set de ned by a quanti er-free Boolean formula with atoms of the form, =0 0 0 ( ) 1 we prove that the sum of the Betti numbers of is bounded by ( ( )) In ?thespecial case, when is de ned by = 0 1 0 0 we have a slightly tighter bound of ( ( )) This result generalises the Oleinik-Petrovsky-Thom-Milnor bound in two directions. Firstly, our bound applies to arbitrary semi-algebraic sets, not just for basic semi-algebraic sets. Secondly, the combinatorial part (the part depending on ) in our bound, depends on the dimension of the variety rather than that of the ambient space. It also generalizes a result in [7] where a similar bound is proven for the number of connected components. In the second part of the paper we use the tools developed for the above results, as well as some additional techniques, to give the rst single exponential time algorithm for computing the Euler characteristic of arbitrary semi-algebraic sets. S
P
; : : : ; Ps k
S
; deg Pi
d;
i
s;
O sd
R
Q
k
k R ;
:
; deg Q
d;
0
k ;
Pi
; Pi >
s
k
; Pi
; : : : ; Ps >
;
:
s
1 Introduction Let P = fP1; : : :; Pk g R[X1; : : :; Xk ]; be a family of polynomials whose degrees are bounded by d; and let S be a semi-algebraic set de ned by a quanti er-free Boolean formula, with atoms of the form Pi f>; 1is deg (Pi): Then, for all suciently small > 0; the set de ned by (Q 0) ^1is ((1 ? )Pi + 0); is homotopy equivalent to the set S: Moreover, the above set is bounded by a smooth hypersurface Q = 0; which has a nite number of critical points for the projection map onto the X1 co-ordinate and these critical points are non-degenerate and have distinct X1 co-ordinates. P
Proof: Let S ( ) = (Q 0) \1is ((1 ? )Pi + 0):
We prove that for suciently small > 0; S is a deformation retract of S ( ) which will prove the rst part of the lemma. Since S is compact there exists a constant R such that, x 2 S ) x21d0 + + x2kd0 + 1 < R: For 0 < < 1=R; any point x = (x1; : : :; xk ) satisfying, P1 (x) 0; : : :;0 Ps(x) 0;0 will also satisfy Q 0: This follows directly from the de nition of Q and the fact that x21d + + x2kd + 1 < R: Thus, for every connected component C of S there exists a connected component C 0 of Q 0 such that C C 0 : Moreover, the signs of the polynomials, + (1 ? )Pi ; 1 i s cannot change over C 0; because if one of them became zero Q will be negative at that point. But, since C 0 contains C; and is suciently small, it is clear that + (1 ? )Pi > 0; 1 i s over C 0 : Thus, for small enough, S S ( ): Replacing by a new variable t; consider the set D Rk+1 de ned using the same inequalities as S ( ) with replaced by t: Let : Rk+1 ! R; and x : Rk+1 ! Rk ; denote the projections onto the t and the X co-ordinates, respectively. Then, from the de nition of D; it is clear that x (D \ ?1(0)) = S: Now for suciently small > 0 D \ ?1 (0) is a deformation retract of D \ ?1 ([0; ]); and there exists a retraction, (see [8] for details) : D \ ?1([0; ]) [0; ] ! D \ ?1([0; ]); such that (D \ ?1 ([0; ]); a) = D \ ?1([0; a]); a 2 [0; ]: Thus, we have the map, : S ( ) [0; ] ! S ( ); de ned by, (x; a) = x ( ((x; ); a)); x 2 S ( ); a 2 [0; ]; gives a deformation retract of S ( ) to S: We next show that the set S ( ) is bounded by connected components of the smooth hypersurface de ned by Q = 0: First observe that the set Q 0 is bounded.This follows from the fact that 2d0 > sd and thus the second term in Q dominates the rst as jxj becomes large. Secondly, the polynomials + (1 ? )Pi are all strictly positive over S ( ): Hence, S ( ) muct be bounded by the smooth hypersurface Q = 0: It remains to show that the hypersurface Q = 0 is smooth and has a nite number of critical points for the projection map onto the X1 co-ordinate, and that these critical points are non-degenerate with distinct X1 co-ordinate. Let Qt = 1is (t + (1 ? t)Pi ) + ts+1 (X12d0 + + Xk2d0 + 1); then for t = 1; it is clear that Qt = 0 de nes a smooth hypersurface, with all its critical points degenerate, and having distinct X1 co-ordinates. Moreover, these being stable conditions there is an open interval containing 1; such that for t in this interval, these conditions are satis ed. Now, by quanti er elimination over algebraically closed elds, we have that the set of real t for which the above conditions are not met is either a nite set of points or a complement of a nite set of points. It cannot be latter, since it does not contain an open interval around 1; and hence it must be a nite set of points. Thus, there is an open interval (0; t0) such that for 2 (0; t0) the conditions of the proposition are satis ed. 2
Lemma 1 Let S be any semi-algebraic set. Then, for large enough, S 0 = S \ (X12 + + Xk2 );
has the same homotopy type as S:
13
Proof: We consider the set S (t) Rk+1 de ned by the same formulas as S and the inequality, t(X12 + + Xk2) 1: Let and x denote the projections onto the t and the X co-ordinates respectively. Then, for suciently large ; there exists a retraction (see [8] for details), of S 0(t) \ ?1 ((0; 1 ]) ! 0 S (t) \ ?1 ( 1 ): Now, x (S 0(t) \ ?1 ((0; 1 ])) = S; and x (S 0(t) \ ?1 ( 1 )) = S 0: We de ne a map d : S ! S 0 (t) \ ?1 ((0; 1 ]) by d(x) = (x1; : : :; xk ; min(x21 + + x2k ; )): Then, d composed with the retraction mentioned above and x gives a retraction of S 0 to S:
2
Lemma 2 Let S and S () be as in above. Then, for suciently small > 0; S is a deformation retract of S (): Proof: Let S 0() denote the semi-algebraic set de ned similarly as S () with the di erence that, for every equality in we replace the corresponding conjunct, Pj = 0; by the conjunct, Pj ? ^ Pj :
Let, S (t) and S 0(t) be subsets of Rk+1 de ned by the corresponding formulas for S () and S 0 () with replaced by a new variable t: Then it is clear that, S S () S 0 (). Moreover, for suciently small > 0; there exists a retraction S 0() ! S; given by a map,
: S 0 () [0; ] ! S 0 (); satisfying, (S 0(); t) = S 0(t); t 2 [0; ]: Moreover, S () S 0(t); 0 < t < : Now, consider the restriction of to S () [0; ]: It is clear that de nes a proper retraction of S () to S:
2
Lemma 4 Let S and S ?() be as above. Then, for suciently small > 0; S ?() is a deformation
retract of S:
Proof: Let D Rk+1 denote the set de ned by the same formula as S ?() with replaced by a new
variable t: Let and x denote the projections onto the t and the X co-ordinates respectively. Then for suciently small > 0 there exists a retraction of D \ ?1 ((0; ]) ! D \ ?1 ( ): Moreover, it is clear that x (D \ ?1 ((0; ]) = S; and x(D \ ?1 ( )) = S ? ( ): We now de ne a one-one continuous map, d : S ! D \ ?1 ((0; ]); that composed with the retraction and x will give a retraction of S ! S ? (): The details are messy, but the basic idea is that since S is de ned by an open condition, if a point x 2 S then x 2 S ? (t); for all small enough t > 0: We map x to (x; t0) where t0 is either the maximum t for which x 2 S ? (t); or if this maximum is greater than : For, 1 j L; de ne rj : S ! R; as follows: Without loss of generality let Q1 ; : : :; Qlj be the polynomials appearing in j ; For 1 i lj ; let cij = 1, if j (i) = ?; and cij = 0; else. De ne, cij Qi : rj (x) = 1min ( ? 1) il H j
2i?cij
Since the polynomials Hi 1; it is clear that rj is a continuous, well de ned real valued function. De ne, r(x) = min(max1j L rj (x); ): For x 2 S it is clear that, r(x) > 0; and moreover r(x) is continuous. Now, de ne d(x) = (x; r(x)): It is clear that d de nes a one-one continuous map from S to D \ ?1 ((0; ]) and the lemma follows.
2
14
Lemma 5 Let S be a semi-algebraic set de ned by a conjunct (Q = 0) ^ (1(P ) _ _ L(P )); where Q is a polynomial, and j ; 1 j L; are sign conditions on a family of polynomials P ; such that P P
none of the j contain an equality. Then, i (S ) = 1j L i(Sj ) and (S ) = 1j L (Sj ); where Sj is the set de ned by the conjunct (Q = 0) ^ j (P ):
Proof: For a sign condition j on P ; de ne j (; P ); to be the formula obtained by replacing every inequality in j (P ) of the type Pi > 0 by Pi ; and Pi < 0 by Pi ?: Let S 0 be the set de ned by the conjunct, (Q = 0) ^ (1 (; P ) _ _ L (; P )); and Sj0 to be the set de ned by, (Q = 0) ^ j (; P ) for 1 j L: Then using the same argument as in the proof of lemma 4 it is possible to show that S 0 and S has the same homotopy type. Also, by the arguments used in the proof of lemma 2 one can show that, Sj0 has the same homotopy type as Sj : P Moreover, S 0 is the disjoint union of the compact sets Sj0 : Therefore, i (S 0) = 1j L i (Sj0 ): Hence it follows that, X i (S ) = i (Sj ): 1iL
Since (S ) is the alternating sum of the Betti numbers of S; it immediately follows that (S ) = P 1iL (Sj ):
2
6.1 The Example of a Solid Torus
p1
v1
p2
p3
v2
v3
p4
v4
X1 - axis
Figure 1: S is the solid torus in 3 space Consider the set bounded by the smooth torus in R3 (see gure 1.) There are four critical points p1 ; p2; p3 and p4 corresponding to the critical values v1 ; v2; v3 and v4 respectively, for the projection map onto the X1 co-ordinate. The Morse data at these points are given in the following table: Critical point Index Tangential data Normal data p1 0 (D0 D2 ; ;) ([0; 1]; ;) 1 1 1 1 p2 1 (D D ; @D D ) ([0; 1]; f0g) p3 1 (D1 D1; @D1 D1 ) ([0; 1]; ;) p4 2 (D2 D0; @D2 D0 ) ([0; 1]; f0g)
15
6.2 Proof of Correctness and Complexity Analysis for the Algorithm in the General Case 6.2.1 Proof of Correctness
First note that the set S is compact. We utilize several results from [24]. We assume that the reader is familiar with algebraic decision trees ([24]). Consider the following xed degree algebraic decision tree denoted by T: Let the polynomial associated to the root be the polynomial, P1 = X12 + + Xk2 ? ; The associated polynomial to each node at level i, 2 i s is Pi : The sets associated to each of the leaf nodes correspond to the 3s possible sign-conditions on the family P : Let the set of leaves corresponding to the sign conditions which are non-empty and included in S be denoted by LS : For a leaf l 2 Ls ; let Sl denote the set associated with it. @A); For any set A; following [24], we de ne i0 (A) to be the rank of the i-th homology group Hi(A; where (A) is the closure of A in the topology of Rk and @A = A ? A: P 0 Similarly we de ne, (A) = 0ik (?1)i i0 (A): If A is compact then i0 (A) = i (A); and 0 (A) = (A): We rst prove a preliminary lemma. Lemma 8 Let X be a semi-algebraic set in Rk which is bounded and semi-closed. For some polynomial f 2 R[X1; : : :; Xk ]; let A = X \ (f 0); and B = X ? A: Then, 0(X ) = 0 (A) + 0 (B):
Proof: The proof is quite similar to the proof of lemma 5 in [24]. From the exact sequence,
@X ) ! Hi (X; A [ @X ) ! ; ! Hi(A [ @X; @X ) ! Hi(X; @X ) = (X; A [ @X ) + (A [ @X; @X ): Moreover, i(X; A [ @X ) = i0 (B ); and we have that, (X; 0 0 2 i (A [ @X; @X ) = i (A) (see [24] for a proof). It follows easily that, (X ) = 0 (A) + 0 (B): Next, we prove that, Lemma 9 (S ) = Pl2LS 0(Sl):
Proof: Note that since S is compact, 0(S ) = (S ): The lemma now follows by a simple induction on
the levels of T; starting from the leaf nodes and going up, and using lemma 8 at each node. We omit the details. 2
We next prove that, following the notation introduced in the algorithm, that i0(Sj ) = i (Uj ; Vj ); and thus 0 (Sj ) = (Uj ; Vj ): The proof of this appears in [24] (page 621) and is omitted. It follows from the exact sequence,
! Hi(Vj ) ! Hi(Uj ) ! Hi(Uj ; Vj ) ! Hi?1(Vj ) ! ; we immediately deduce that, (Uj ; Vj ) = (Uj ) ? (Vj ): P This in conjunction with lemma 9 shows that, (S ) = 1j m ( (Uj ) ? (Vj )); and this proves the correctness of the algorithm.
6.2.2 Complexity Analysis
The cost of computing all the non-empty sign conditions of the family is P is sk+1 dO(k) (see [5]. ?s Moreover, there can be only k (O(d))k such non-empty sign conditions. For each such sign-condition included in S; we call the algorithm for computing the Euler charateristics of basic semi-algebraic sets 16
twice. The sum of the degrees of the polynomials involved in each such call is O(sd). Thus each call costs (ksd)O(k): Hence, the total complexity of the algorithm is bounded by (ksd)O(k): The bound in the bit model follows easily once we note that bit sizes of the intermediate values are bounded by L(skd)O(k): This proves theorem 2.
17
Abstract
In this paper we give a new bound on the sum of the Betti numbers of semi-algebraic sets. This extends a well-known bound due to Oleinik and Petrovsky [19], Thom [23] and Milnor [18]. In separate papers they proved that the sum of the Betti numbers of a semi-algebraic set de ned by 1 0 0 ( ) 1 is bounded by ( ( )) Given a semialgebraic set de ned as the intersection of a real variety, = 0 ( ) whose real dimension is with a set de ned by a quanti er-free Boolean formula with atoms of the form, =0 0 0 ( ) 1 we prove that the sum of the Betti numbers of is bounded by ( ( )) In ?thespecial case, when is de ned by = 0 1 0 0 we have a slightly tighter bound of ( ( )) This result generalises the Oleinik-Petrovsky-Thom-Milnor bound in two directions. Firstly, our bound applies to arbitrary semi-algebraic sets, not just for basic semi-algebraic sets. Secondly, the combinatorial part (the part depending on ) in our bound, depends on the dimension of the variety rather than that of the ambient space. It also generalizes a result in [7] where a similar bound is proven for the number of connected components. In the second part of the paper we use the tools developed for the above results, as well as some additional techniques, to give the rst single exponential time algorithm for computing the Euler characteristic of arbitrary semi-algebraic sets. S
P
; : : : ; Ps k
S
; deg Pi
d;
i
s;
O sd
R
Q
k
k R ;
:
; deg Q
d;
0
k ;
Pi
; Pi >
s
k
; Pi
; : : : ; Ps >
;
:
s
1 Introduction Let P = fP1; : : :; Pk g R[X1; : : :; Xk ]; be a family of polynomials whose degrees are bounded by d; and let S be a semi-algebraic set de ned by a quanti er-free Boolean formula, with atoms of the form Pi f>; 1is deg (Pi): Then, for all suciently small > 0; the set de ned by (Q 0) ^1is ((1 ? )Pi + 0); is homotopy equivalent to the set S: Moreover, the above set is bounded by a smooth hypersurface Q = 0; which has a nite number of critical points for the projection map onto the X1 co-ordinate and these critical points are non-degenerate and have distinct X1 co-ordinates. P
Proof: Let S ( ) = (Q 0) \1is ((1 ? )Pi + 0):
We prove that for suciently small > 0; S is a deformation retract of S ( ) which will prove the rst part of the lemma. Since S is compact there exists a constant R such that, x 2 S ) x21d0 + + x2kd0 + 1 < R: For 0 < < 1=R; any point x = (x1; : : :; xk ) satisfying, P1 (x) 0; : : :;0 Ps(x) 0;0 will also satisfy Q 0: This follows directly from the de nition of Q and the fact that x21d + + x2kd + 1 < R: Thus, for every connected component C of S there exists a connected component C 0 of Q 0 such that C C 0 : Moreover, the signs of the polynomials, + (1 ? )Pi ; 1 i s cannot change over C 0; because if one of them became zero Q will be negative at that point. But, since C 0 contains C; and is suciently small, it is clear that + (1 ? )Pi > 0; 1 i s over C 0 : Thus, for small enough, S S ( ): Replacing by a new variable t; consider the set D Rk+1 de ned using the same inequalities as S ( ) with replaced by t: Let : Rk+1 ! R; and x : Rk+1 ! Rk ; denote the projections onto the t and the X co-ordinates, respectively. Then, from the de nition of D; it is clear that x (D \ ?1(0)) = S: Now for suciently small > 0 D \ ?1 (0) is a deformation retract of D \ ?1 ([0; ]); and there exists a retraction, (see [8] for details) : D \ ?1([0; ]) [0; ] ! D \ ?1([0; ]); such that (D \ ?1 ([0; ]); a) = D \ ?1([0; a]); a 2 [0; ]: Thus, we have the map, : S ( ) [0; ] ! S ( ); de ned by, (x; a) = x ( ((x; ); a)); x 2 S ( ); a 2 [0; ]; gives a deformation retract of S ( ) to S: We next show that the set S ( ) is bounded by connected components of the smooth hypersurface de ned by Q = 0: First observe that the set Q 0 is bounded.This follows from the fact that 2d0 > sd and thus the second term in Q dominates the rst as jxj becomes large. Secondly, the polynomials + (1 ? )Pi are all strictly positive over S ( ): Hence, S ( ) muct be bounded by the smooth hypersurface Q = 0: It remains to show that the hypersurface Q = 0 is smooth and has a nite number of critical points for the projection map onto the X1 co-ordinate, and that these critical points are non-degenerate with distinct X1 co-ordinate. Let Qt = 1is (t + (1 ? t)Pi ) + ts+1 (X12d0 + + Xk2d0 + 1); then for t = 1; it is clear that Qt = 0 de nes a smooth hypersurface, with all its critical points degenerate, and having distinct X1 co-ordinates. Moreover, these being stable conditions there is an open interval containing 1; such that for t in this interval, these conditions are satis ed. Now, by quanti er elimination over algebraically closed elds, we have that the set of real t for which the above conditions are not met is either a nite set of points or a complement of a nite set of points. It cannot be latter, since it does not contain an open interval around 1; and hence it must be a nite set of points. Thus, there is an open interval (0; t0) such that for 2 (0; t0) the conditions of the proposition are satis ed. 2
Lemma 1 Let S be any semi-algebraic set. Then, for large enough, S 0 = S \ (X12 + + Xk2 );
has the same homotopy type as S:
13
Proof: We consider the set S (t) Rk+1 de ned by the same formulas as S and the inequality, t(X12 + + Xk2) 1: Let and x denote the projections onto the t and the X co-ordinates respectively. Then, for suciently large ; there exists a retraction (see [8] for details), of S 0(t) \ ?1 ((0; 1 ]) ! 0 S (t) \ ?1 ( 1 ): Now, x (S 0(t) \ ?1 ((0; 1 ])) = S; and x (S 0(t) \ ?1 ( 1 )) = S 0: We de ne a map d : S ! S 0 (t) \ ?1 ((0; 1 ]) by d(x) = (x1; : : :; xk ; min(x21 + + x2k ; )): Then, d composed with the retraction mentioned above and x gives a retraction of S 0 to S:
2
Lemma 2 Let S and S () be as in above. Then, for suciently small > 0; S is a deformation retract of S (): Proof: Let S 0() denote the semi-algebraic set de ned similarly as S () with the di erence that, for every equality in we replace the corresponding conjunct, Pj = 0; by the conjunct, Pj ? ^ Pj :
Let, S (t) and S 0(t) be subsets of Rk+1 de ned by the corresponding formulas for S () and S 0 () with replaced by a new variable t: Then it is clear that, S S () S 0 (). Moreover, for suciently small > 0; there exists a retraction S 0() ! S; given by a map,
: S 0 () [0; ] ! S 0 (); satisfying, (S 0(); t) = S 0(t); t 2 [0; ]: Moreover, S () S 0(t); 0 < t < : Now, consider the restriction of to S () [0; ]: It is clear that de nes a proper retraction of S () to S:
2
Lemma 4 Let S and S ?() be as above. Then, for suciently small > 0; S ?() is a deformation
retract of S:
Proof: Let D Rk+1 denote the set de ned by the same formula as S ?() with replaced by a new
variable t: Let and x denote the projections onto the t and the X co-ordinates respectively. Then for suciently small > 0 there exists a retraction of D \ ?1 ((0; ]) ! D \ ?1 ( ): Moreover, it is clear that x (D \ ?1 ((0; ]) = S; and x(D \ ?1 ( )) = S ? ( ): We now de ne a one-one continuous map, d : S ! D \ ?1 ((0; ]); that composed with the retraction and x will give a retraction of S ! S ? (): The details are messy, but the basic idea is that since S is de ned by an open condition, if a point x 2 S then x 2 S ? (t); for all small enough t > 0: We map x to (x; t0) where t0 is either the maximum t for which x 2 S ? (t); or if this maximum is greater than : For, 1 j L; de ne rj : S ! R; as follows: Without loss of generality let Q1 ; : : :; Qlj be the polynomials appearing in j ; For 1 i lj ; let cij = 1, if j (i) = ?; and cij = 0; else. De ne, cij Qi : rj (x) = 1min ( ? 1) il H j
2i?cij
Since the polynomials Hi 1; it is clear that rj is a continuous, well de ned real valued function. De ne, r(x) = min(max1j L rj (x); ): For x 2 S it is clear that, r(x) > 0; and moreover r(x) is continuous. Now, de ne d(x) = (x; r(x)): It is clear that d de nes a one-one continuous map from S to D \ ?1 ((0; ]) and the lemma follows.
2
14
Lemma 5 Let S be a semi-algebraic set de ned by a conjunct (Q = 0) ^ (1(P ) _ _ L(P )); where Q is a polynomial, and j ; 1 j L; are sign conditions on a family of polynomials P ; such that P P
none of the j contain an equality. Then, i (S ) = 1j L i(Sj ) and (S ) = 1j L (Sj ); where Sj is the set de ned by the conjunct (Q = 0) ^ j (P ):
Proof: For a sign condition j on P ; de ne j (; P ); to be the formula obtained by replacing every inequality in j (P ) of the type Pi > 0 by Pi ; and Pi < 0 by Pi ?: Let S 0 be the set de ned by the conjunct, (Q = 0) ^ (1 (; P ) _ _ L (; P )); and Sj0 to be the set de ned by, (Q = 0) ^ j (; P ) for 1 j L: Then using the same argument as in the proof of lemma 4 it is possible to show that S 0 and S has the same homotopy type. Also, by the arguments used in the proof of lemma 2 one can show that, Sj0 has the same homotopy type as Sj : P Moreover, S 0 is the disjoint union of the compact sets Sj0 : Therefore, i (S 0) = 1j L i (Sj0 ): Hence it follows that, X i (S ) = i (Sj ): 1iL
Since (S ) is the alternating sum of the Betti numbers of S; it immediately follows that (S ) = P 1iL (Sj ):
2
6.1 The Example of a Solid Torus
p1
v1
p2
p3
v2
v3
p4
v4
X1 - axis
Figure 1: S is the solid torus in 3 space Consider the set bounded by the smooth torus in R3 (see gure 1.) There are four critical points p1 ; p2; p3 and p4 corresponding to the critical values v1 ; v2; v3 and v4 respectively, for the projection map onto the X1 co-ordinate. The Morse data at these points are given in the following table: Critical point Index Tangential data Normal data p1 0 (D0 D2 ; ;) ([0; 1]; ;) 1 1 1 1 p2 1 (D D ; @D D ) ([0; 1]; f0g) p3 1 (D1 D1; @D1 D1 ) ([0; 1]; ;) p4 2 (D2 D0; @D2 D0 ) ([0; 1]; f0g)
15
6.2 Proof of Correctness and Complexity Analysis for the Algorithm in the General Case 6.2.1 Proof of Correctness
First note that the set S is compact. We utilize several results from [24]. We assume that the reader is familiar with algebraic decision trees ([24]). Consider the following xed degree algebraic decision tree denoted by T: Let the polynomial associated to the root be the polynomial, P1 = X12 + + Xk2 ? ; The associated polynomial to each node at level i, 2 i s is Pi : The sets associated to each of the leaf nodes correspond to the 3s possible sign-conditions on the family P : Let the set of leaves corresponding to the sign conditions which are non-empty and included in S be denoted by LS : For a leaf l 2 Ls ; let Sl denote the set associated with it. @A); For any set A; following [24], we de ne i0 (A) to be the rank of the i-th homology group Hi(A; where (A) is the closure of A in the topology of Rk and @A = A ? A: P 0 Similarly we de ne, (A) = 0ik (?1)i i0 (A): If A is compact then i0 (A) = i (A); and 0 (A) = (A): We rst prove a preliminary lemma. Lemma 8 Let X be a semi-algebraic set in Rk which is bounded and semi-closed. For some polynomial f 2 R[X1; : : :; Xk ]; let A = X \ (f 0); and B = X ? A: Then, 0(X ) = 0 (A) + 0 (B):
Proof: The proof is quite similar to the proof of lemma 5 in [24]. From the exact sequence,
@X ) ! Hi (X; A [ @X ) ! ; ! Hi(A [ @X; @X ) ! Hi(X; @X ) = (X; A [ @X ) + (A [ @X; @X ): Moreover, i(X; A [ @X ) = i0 (B ); and we have that, (X; 0 0 2 i (A [ @X; @X ) = i (A) (see [24] for a proof). It follows easily that, (X ) = 0 (A) + 0 (B): Next, we prove that, Lemma 9 (S ) = Pl2LS 0(Sl):
Proof: Note that since S is compact, 0(S ) = (S ): The lemma now follows by a simple induction on
the levels of T; starting from the leaf nodes and going up, and using lemma 8 at each node. We omit the details. 2
We next prove that, following the notation introduced in the algorithm, that i0(Sj ) = i (Uj ; Vj ); and thus 0 (Sj ) = (Uj ; Vj ): The proof of this appears in [24] (page 621) and is omitted. It follows from the exact sequence,
! Hi(Vj ) ! Hi(Uj ) ! Hi(Uj ; Vj ) ! Hi?1(Vj ) ! ; we immediately deduce that, (Uj ; Vj ) = (Uj ) ? (Vj ): P This in conjunction with lemma 9 shows that, (S ) = 1j m ( (Uj ) ? (Vj )); and this proves the correctness of the algorithm.
6.2.2 Complexity Analysis
The cost of computing all the non-empty sign conditions of the family is P is sk+1 dO(k) (see [5]. ?s Moreover, there can be only k (O(d))k such non-empty sign conditions. For each such sign-condition included in S; we call the algorithm for computing the Euler charateristics of basic semi-algebraic sets 16
twice. The sum of the degrees of the polynomials involved in each such call is O(sd). Thus each call costs (ksd)O(k): Hence, the total complexity of the algorithm is bounded by (ksd)O(k): The bound in the bit model follows easily once we note that bit sizes of the intermediate values are bounded by L(skd)O(k): This proves theorem 2.
17
