Yamaguchi, Kohhei

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Mathematical Journal of OkayamaUniversityVolume 46, Issue 1 2004 Article 30JANUARY 2004Remark on Cup-ProductsKohhei Yamaguchi∗∗University of Electro-CommunicationsCopyright c©2004 by the authors. Mathematical Journal of Okayama University is produced byThe Berkeley Electronic Press (bepress). http://escholarship.lib.okayama-u.ac.jp/mjouMath. J. Okayama Univ. 46 (2004), 115{120REMARK ON CUP-PRODUCTSKohhei YAMAGUCHI1. Introduction.Let L be ¯nite dimensional CW complex and L1 be a subcomplex of Lgiven by(1.1)(L = L1 [® en (® 2 ¼n¡1(L1))L1 = Sq [¡S¸2¤ em(¸)¸¢;where n ¡ 2 ¸ dim em(¸)¸ = m(¸) ¸ q + 2 ¸ 4 are integers (¸ 2 ¤). Letiq : Sq ! L1 and ® 2 ¼n(L;L1) be an inclusion map and the characteristicmap of the top cell en in L. Then it is known ([2], [4]) that8><>:¼k(L1) = 0 if k < q; ¼k(L;L1) = 0 if k < n;¼q(L1) = Z ¢ iq »= Z; ¼n(L;L1) = Z ¢ ® »= Z; ®jSn¡1 = ®;¼n+q¡1(L;L1) = Z ¢ [®; iq]r © ®¤¼n+q¡1(Dn; Sn¡1) »= Z© ¼n+q¡1(Sn);where [ ; ]r denotes the relative Whitehead product.In this paper we study the relation between relative Whitehead productsand cup-products on certain ¯nite dimensional complexes. In particular, weshall prove some generalization of the result obtained by I. M. James in [3]concerning these relations, and it is as follows:Theorem 1.1. Under the above assumption (1.1), let ¯ 2 ¼n+q¡1(L) be anelement such that(1.2) i¤(¯) = m[®; iq]r + ® ± ½ (m 2 Z; ½ 2 ¼n+q¡1(Dn; Sn¡1));where i¤ : ¼n+q¡1(L)! ¼n+q¡1(L;L1) denotes the induced homomorphism.Then if K = L [¯ en+q is the mapping cone of ¯ and ek 2 Hk(K;Z) »= Z(k = q; n; n+ q) denotes the corresponding generator, the following relationholds:(1.3) en ¢ eq = m en+q:Mathematics Subject Classi¯cation. Primary 55Q15; Secondary 55N45, 55Q35.Key words and phrases. CW complexes, coaction map, Whitehead product.This research was partially supported by Grant-in-Aid for Scienti¯c Research (No.13640067 (C)(2) and No. 16540056 (C)(2)), The Ministry of Education, Culture, Sports,Science and Technology, Japan.1151Yamaguchi: Remark on Cup-ProductsProduced by The Berkeley Electronic Press, 2004116 K. YAMAGUCHIThis kind of results was ¯rst obtained by I. M. James [3] for 3-complexesof the forms K = Sp [ eq [ ep+q. Although the above generalization shouldbe well known on the basis of the result due to James, we could not ¯ndany literature and it may be worth-while to write this result. In fact, in thesubsequent paper [5] we shall use this result to study the classi¯cations ofhomotopy types of simply-connected 8 dimensional Poincar¶e complexes.2. Coaction maps.We take M = L _ Sn = (L1 [® en) _ Sn and let ¹0 : L ! M be thecoaction map given by pinching the equator of the top cell en in L. LetjL : L!M; jn : Sn !M; jq : Sq !M jM :M ! (M;L1)be corresponding natural inclusions, and let rL : M ! L be the retraction.Then we have(2.1) rL ± jL = idL ' rL ± ¹0:Lemma 2.1. Let ¹0¤ : ¼¤(L;L1) ! ¼¤(M;L1) be the induced homomor-phism.(i) ¹0¤(®) = jL¤(®) + jM ¤(jn).(ii) ¹0¤([®; iq]r) = jL¤([®; iq]r) + [jM ¤(jn); iq]r.(iii) ¹0¤(® ± ½) = jL¤(® ± ½) + jM ± jn ± ½.Proof. (i) The assertion (i) is clear.(ii) ¹0¤([®; iq]r) = [¹0¤(®); iq]r = [jL¤(®) + jM ¤(jn); iq]r (by (i))= [jL¤(®); iq]r + [jM ¤(jn); iq]r= jL¤([®; iq]r) + [jM ¤(jn); iq]r:(iii) Since 2n¡ 3 ¸ n+ q ¡ 1, ½ 2 ¼n+q¡1(Dn; Sn¡1) »= ¼n+q¡2(Sn¡1) =E¼n+q¡3(Sn¡2) and we have¹0¤(® ± ½) = ¹0¤(®) ± ½ = (jL¤(®) + jM ¤(jn)) ± ½ (by (i))= jL¤(® ± ½) + jM ¤(jn) ± ½:¤Lemma 2.2. Under the same assumption as Theorem 1.1,(2.2) ¹0¤(¯) = jL¤(¯) +m[jn; jq] + jn ± ½ in ¼n+q¡1(M);where ¹0¤ : ¼n+q¡1(L)! ¼n+q¡1(M) denotes the induced homomorphism.2Mathematical Journal of Okayama University, Vol. 46 [2004], Iss. 1, Art. 30http://escholarship.lib.okayama-u.ac.jp/mjou/vol46/iss1/30REMARK ON CUP-PRODUCTS 117Proof. Consider the commutative diagram(2.3)¼r(L)¹0¤¡¡¡¡! ¼r(M) rL¤¡¡¡¡! ¼r(L) jL¤¡¡¡¡! ¼r(M)i¤??y j¤??y i¤??y j¤??y¼r(L;L1)¹0¤¡¡¡¡! ¼r(M;L1) rL¤¡¡¡¡! ¼r(L;L1) jL¤¡¡¡¡! ¼r(M;L1):We take µ = jL¤(¯) +m[jn; jq] + jn ± ½ ¡ ¹0¤(¯): It su±ces to show µ = 0.For this purpose, it is su±cient to prove the following two equations:(2.4) j¤(µ) = 0; and rL¤(µ) = 0:In fact, let iL1 : L1 ! L be an inclusion and consider the exact sequence¼n+q¡1(L1)(jL±iL1 )¤¡¡¡¡¡¡! ¼n+q¡1(M) j¤¡¡¡¡! ¼n+q¡1(M;L1):Because j¤(µ) = 0, there exists some element Á 2 ¼n+q¡1(L1) such thatµ = (jL ± iL1)¤(Á). Hence, using rL¤(µ) = 0, we haveµ = (jL ± iL1)¤(Á) = jL ± (rL ± jL) ± iL1¤(Á) (by (2:1))= jL ± rL ± (jL ± iL1)¤(Á) = jL ± rL¤(µ) = 0:So it remains to prove (2.4). First, we show j¤(µ) = 0. We note thatj¤¹0¤(¯) = ¹0¤(i¤(¯)) (by (2:3))= ¹0¤(m[®; iq]r + ® ± ½) (by (1.2))= m¹0¤([®; iq]r) + ¹0 ± ® ± ½= m(jL¤([®; iq]r + [jM ± jn; iq]r))+(jL ± ® ± ½+ jM ± jn ± ½) (by Lemma 2.1)= jL¤(m[®; iq]r + ® ± ½) +m[jM ± jn; iq]r + jM ± jn ± ½= jL¤(i¤(¯)) +m[jM ± jn; iq]r + jM ± jn ± ½ (by (1.2)).Because [jM ±jn; iq]r = j¤([jn; jq]) and jM ±jn ±½ = j¤(jn ±½), we can rewritej¤¹0¤(¯) = j¤(jL¤(¯)) +mj¤([jn; jq]) + j¤(jn ± ½):Hence,j¤(µ) = j¤(jL¤(¯) +m[jn; jq] + jn ± ½¡ ¹0¤(¯)) = 0:Next we prove rL¤(µ) = 0. Since rL : M ! L is a retraction, rL ± jn = 0and we have(2.5)(rL¤(jn ± ½) = (rL ± jn) ± ½ = 0;rL¤([jn; jq]) = [rL ± jn; rL ± jq] = 0:3Yamaguchi: Remark on Cup-ProductsProduced by The Berkeley Electronic Press, 2004118 K. YAMAGUCHIHence,rL¤(µ) = rL¤(jL¤(¯) +m[jn; jq] + jn ± ½¡ ¹0¤(¯))= rL¤jL¤(¯) +m ¢ rL¤([jn; jq]) + rL¤(jn ± ½)¡ rL¤¹0¤(¯)= rL¤jL¤(¯)¡ rL¤¹0¤(¯) (by (2.5))= ¯ ¡ ¯ (by (2:1))= 0:¤3. Proof of Theorem 1.1.First, recall the function ¯rst de¯ned by I. M. James [3].De¯nition. Let p; q ¸ 2 be integers and let X be a ¯nite dimensional CWcomplex with dimX · p+ q ¡ 1 such that Hp+q¡1(X;Z) is a ¯nite group.We ¯x the elements 0 6= ek 2 Hk(X;Z) (k = p; q). Since dimX · p+q¡1,ep ¢ eq = 0 2 Hp+q(X;Z). We de¯ne the function h : ¼p+q¡1(X) ! Z asfollows. For any element ¸ 2 ¼p+q¡1(X), we take X¤ = X [¸ ep+q. Wedenote by e0p+q the generator of Hp+q(X¤; X;Z) »= Z corresponding to thetop cell in X¤. Let ~i : X ! X¤ be an inclusion and let e0k 2 Hk(X¤;Z)(k = p; q) be corresponding elements such that ~i¤(e0k) = ek.In this situation, consider the exact sequenceHp+q(X¤; X;Z)~j¤¡¡¡¡! Hp+q(X¤;Z) ~i¤¡¡¡¡! Hp+q(X;Z):Since 0 = ep ¢ eq = ~i¤(e0p) ¢~i¤(e0q) = ~i¤(e0p ¢ e0q), there exists an integer m0 2 Zsuch that e0p ¢ e0q = m0e0p+q. Then we de¯ne the function h : ¼p+q¡1(X)! Zby h(¸) = m0 (cf. [3], page 378).Lemma 3.1 ([3]). h : ¼p+q¡1(X)! Z is a homomorphism.Proof. This follows from Theorem 4.1 of [3]. ¤Lemma 3.2. Under the same assumption as Theorem 1.1, we denote byh : ¼p+q¡1(M)! Z the linear function given as above. Then we have(i) h(jL¤(¯)) = h(jn ± ½) = 0(ii) h([jn; jq]) = 1Proof. (i) Let us consider the mapping cone M¤1 = M [jL¤(¯) en+q. SinceM¤1 »= (L [¯ en+q) _ Sn, clearly a0 ¢ b0 = 0, where a0 2 Hn(M¤1 ;Z) »= Z andb0 2 Hq(M¤1 ;Z) »= Z denote the generators corresponding to Sn and Sq.Hence, h(jL¤(¯)) = 0. Similar method also shows h(jn ± ½) = 0.(ii) Consider the mapping cone M¤2 = M [[jn;jq ] en+q. Then M¤2 = (L _Sn) [[jn;jq ] en+q contains the subcomplex Sq _ Sn [[iq ;in] en+q = Sq £ Sn.4Mathematical Journal of Okayama University, Vol. 46 [2004], Iss. 1, Art. 30http://escholarship.lib.okayama-u.ac.jp/mjou/vol46/iss1/30REMARK ON CUP-PRODUCTS 119So if a0 2 Hn(M¤2 ;Z) »= Z and b0 2 Hq(M¤2 ;Z) »= Z denote the generatorscorresponding to Sn and Sq, the product a0 ¢ b0 represents the generator ofHn+q(M¤2 ;Z) and we have h([jn; jq]) = 1. ¤Proof of Theorem 1.1. Let N be a mapping cone of ¹0¤(¯) given by N =M [¹0¤(¯) en+q = (L_Sn)[¹0¤(¯) en+q. De¯ne the map f : K = L[¯ en+q !M [¹0¤(¯) en+q = N by(f jL = ¹0 : L!Mf jen+q = degree one map on the top cell en+q:If f¤ : H¤(N;Z) ! H¤(K;Z) denotes the induced homomorphism and lete0k 2 Hk(N;Z) »= Z (k = q; n; n+ q) be the generators corresponding to cellsSq, Sn or en+q, then f¤(e0k) = ek for k = q; n or n + q. Now consider thehomomorphism h : ¼n+q¡1(M)! Z. Then we have(3.1) e0n ¢ e0q = h(¹0¤(¯))e0n+q:Hence,en ¢ eq = f¤(e0n) ¢ f¤(e0q) = f¤(e0n ¢ e0q) = f¤(h(¹0¤(¯))e0n+q)= h(¹0¤(¯))f¤(e0n+q) = h(¹0¤(¯))en+q:So it remains to show that h(¹0¤(¯)) = m. Since h is a homomorphism, itfollows from Lemma 2.2 and 3.2 thath(¹0¤(¯)) = h(jL¤(¯) +m[jn; jq] + jn ± ½)= h(jL¤(¯)) +m ¢ h([jn; jq]) + h(jn ± ½)= 0 +m ¢ 1 + 0 = mand this completes the proof of Theorem 1.1. ¤Acknowledgements. The author is indebted to Professor J. Mukai for nu-merous helpful conversations concerning relative Whitehead products andunstable homotopy theory.References[1] A. L. Blakers and W. S. Massey, Products in homotopy theory, Annals of Math.58(1953) 295{324.[2] I. M. James, On the homotopy groups of certain pairs and triads, Quart. J. Math.Oxford 5(1954) 260{270.[3] I. M. James, Note on cup-products, Proc. Amer. Math. Soc. 8(1957) 374{383.[4] W. S. Massey, Exact couples in algebraic topology (Parts I and II), Annals of Math.56(1952) 363{396.[5] J. Mukai and K. Yamaguchi, Homotopy classi¯cation of twisted complex projectivespaces of dimension 4, to appear in J. Math. Soc. Japan.5Yamaguchi: Remark on Cup-ProductsProduced by The Berkeley Electronic Press, 2004120 K. YAMAGUCHIKohhei YamaguchiDepartment of Information MathematicsUniversity of Electro-CommunicationsChofugaoka, Chofu, Tokyo 182-8585 Japane-mail address: kohhei@im.uec.ac.jp(Received September 30, 2003 )6Mathematical Journal of Okayama University, Vol. 46 [2004], Iss. 1, Art. 30http://escholarship.lib.okayama-u.ac.jp/mjou/vol46/iss1/30

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