REFERENCE IC/88/404

INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS

S7 WITHOUT ANY CONSTRUCTION OF LIE GROUP

Zhou Jian

and

INTERNATIONAL Xu Senlin ATOMIC ENERGY AGENCY

UNITED NATIONS EDUCATIONAL, SCIENTIFIC AND CULTURAL ORGANIZATION

IC/88/404

International Atomic Energy Agency and United Nations Educational Scientific and Cultural Organization INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS It is known that an n-dimensional Lie group G is a parallelizable man- ifold ( The left invariant vector fields Xx,...,Xn determined by a base Xi{e),..., Xn[e) of unit element e £ G are C™ basic vector fields of G ) and an H-space, where an H-space is a space which admits a continuous product with unit, that is a continuous mapping ft : G x G ~» G with a point e e G such that (i(x,e) = p{e, x) = x for any ieG. In [2], R.Bott T S WITHOUT ANY CONSTRUCTION OF LIE GROUP* and J.Milnor proved that the sphere S" is a parailelizable manifold if and only if n = 1,3 or 7. In [l], J.F.Adams proved that S" is an H-space if and only if n = 0,1,3 or 7. A natural question is whether S" is a Lie group for Zhou Jian n = 0,1,3, or 7? At first we will prove that Sr is not a Lie group and even it is not a topological group, then we will prove that S" is a Lie group (or Department of Mathematics, University of Science and Technology of China, topological group) if and only if n — 0,1, or 3. We give several examples of Hefei, Anhui, People's Republic of China parallelizable manifolds and H-spaces. Example 1 It is easy to prove that S1 = {(ii.zj) € R2\x\ + x\ - and 1} = {xi + ii2 £ C = -Rs|rJ +x| = 1} with a multiplication fi is a Lie group, of course it is a parallelizable manifold and an H-space, where C is Xu Senlin " the complex field and /i : C x C —* C, (i{xi + ix?, j/i + iVs) = {*i + «£:) • International Centre for Theoretical Physics, Trieste, Italy. {yi+iyi) — (ii!/i - i»!/*)+«(xi!h + xiyi). For any {x!,x,) £ 5', set vector ! fieid X(xllX1) = (— ij,ii), then X is a C°° basic vector field on 5 , therefore Sl is a parallelizable manifold. s ABSTRACT Example 2 5 = {(I1,X2,X3,I4) R* +x\ + x\ + x\ = 1} = A ix? +}Xs + kxt 6 Q — R \x\ +x\-tx\ + x\ = 1}, where a quaternion field Q is an algebra of dimension 4 over the field R of real numbers, with a base n In [2], R.Bott and J.Milnor proved that the sphere S is a parallelizable manifold composed of 4 elements 1, i, j, k whose multiplication table is given by the if and only if n = 1,3 or 7. In [l], J.F.Adams proved above result and proved that S" following formulas : is an H-space if and only if n = 0,1,3, or 7. Because a Lie group must necessarily be a 1 - i = i • 1 = .', 1 • j = j • 1 = j, 1 • k = k • 1 = h, i2 - j1 = k2 = -1, parallelizable manifold and also an H-space, naturally ones ask that S" is a Lie group for * • J = ~3 •! - k, i • k —• —k • i = -j, j • k — -k-j = i. 7 n = 0, 1,3, or 7 ? In this paper we prove that S is not a Lie group, and it is not even Let (i : Q x Q -» Q, fi{x1+ix1 + jx3 + kxt,yi + iy2 + J!ta + fcl/4) = (xi + a topologkal group. Therefore, S" is a Lie group (or a topological group) if and only if ix2 + jxB-\- kxt) • (yi+i n - 0,1,3. By simple computation, we know that S3 with multiplication /i is a Lie group (also is a paraltelizable manifold and an H-space). For any 3 MIRAMARE - TRIESTE (x1>x1,xs,xt) £ S , set vector fields X{tl ls Ia ri) = (-ij.ii, —x4,x3) , X a December 1988 (x{1, IaIt) = (-X3,Xi,xu-xi) , X( IIiIaiIaiIl) - {-i^-zs.is.ii) , then {X11X2,X3} is a C°° basic vector field on S3', therefore S3 is a parallelizable manifold. To be submitted for publication. Permanent address: Department of Mathematics, University of Science and Technology of China, Hefei, Anhui, People's Republic of China. r Lemma 2 A connected commutative Lie group G is a topological prod- Example 3 S = {{xu...,xt) e R*\ £f=1s? = 1 } = {xi -f- ix2 + jx3 + i uct of a torus and an Euclidean space. That is if G is a complex Lie group, fci4 + ei5 + ie • xe + je • z7 + ke • xs<= if = R \T,f=i x? = 1}, where the Caley then G and (Cm/Zm + iZm) x C are isomorphic; if G is a real Lie group, number field K = {p + qe = (zi + izj + J'XJ + /cx4) + (jB + i • ze + j • IT + then G and {RmjZm) x Rn are isomorphic. (See Yan Zhida and Xu Yichao k • zs)e\zi € R, I = 1,...,8} is an algebra of dimension 8 over the field R of real numbers with a base composed of 8 elements 1, i, j, k, e, ie, je, ke, [11) p.95, Theorem 1.8.7.) whose multiplication table is given by the following formulas : for any p, Lemma 3 Compact simple Lie groups are of the following several q G Q, note g = i6 + ixe + jx7 + /cxg = is - ii 1; B, = SO{21 + 1,R), dim Bi = 2l2 + t,l> 2; 1 Let K, = (Pi C, = Sp[l), d\TnCl=2l +l,l>3; K x if 2 From D, = SO{2l,R), dim D, = 2l - i, I > 4 it follows that the above multiplication of K is not associative. Therefore, 7 S is an H-space, but the multiplication of K does not give the construction and G2,Fi,Ee,Er,Ea, dim G, = 14, dim Ft = 52, dim E6 - 78, dim of a Lie group. E7 = 133, dim E& = 248. (See Yan Zhida and Xu Yichao [11] pp.158 - 7 162.) For any (rj, ...,xB) £ 5 , set vector fields Lemma 4 Let G be a connected and simple connected Lie group, then any connected normal subgroup of G is closed. (See Yan Zhida and Xu I ( Yichao [11] p.97, Theorem 1,8.8.) Xj = (-Xi,-X3,X1,Xi,~Xg,XT,-Xe,X!,), Lemma 5 The factor space GjH of a Lie group G about its closed X* = (-x ,i ,i7,x ,Xi,-a; -Xs,-i4), s 6 8 !> subgroup H is an analytic manifold, such that the natural projection p : G —> GjH is analytic, and if H is normal, then GjH is a Lie group, which is called a factor group of G about its closed subgroup H. (See Yan Zhida and Xu Yichao [11] p.94, Theorem 1.8.3 and see G.Hochschild [6] p.94, then {X1,..., JfT} is a. C™ basic vector field on S7, therefore, S7 is a paral- Corollary 2.2.) lelizable manifold. Lemma 6 (E.Cartan) Every closed subgroup H of a Lie group G is a Lie The basic vector fields in above three examples are given respectively subgroup of G and the topology of H must be the induced topology. (See by the multiplications of C, Q and K. For example, in K X^ = ix, X% = jx, Yan Zhida and Xu Yichao [11] p.40,Theorem 1.4.1 and see G.Hochschiid [6] p.92.) XI = kxtX*=e- x, X\ = [ie) • x, X° = {je) • x, Xl = (ke) • x. To prove that S* is not a Lie group, we need use the following 8 Lemmas. Lemma 7 Let p : E -> B be a fibre mapping, xa e p~'(bo) =fAe B, Lemma 1 A connected compact Lie group G is a multiplicative product then there exists a natural homomorphism of groups d : Wn+i(B,bo) —> ir (F, xo), such that the following sequence is exact : of a center C(G) and a connected simple normal subgroup GU..,,G,, i.e. a ->• TT (E,Xo) ~> * (B,bo) -» T {F,I ) ~* Vn{E,X ) -»-..-+ G = C(G) • Gi • • • Ga, where the subgroup G* = Gt • • • G, of G is a n+l n+l n 0 0 connected semisimple normal subgroup, every G,- is non-commutative and iri{B,b0) ~* xo(F, Xo) -> Jro(S,xo) ~* KO{B, b0) , this decomposition is unique under not counting their order. (See Yan where wn is a homotopy group, n = 0,1,2, • • •- (See R.M.Switzer [10] p,56.) Zhida and Xu Ytchao [11] p.110, Theorem 2.2.7.) Lemma 8 Let H be a closed normal subgroup of the Lie group G, then the natural projection p : G ~* GjH is a fibre mapping. (See N.Stcenroci [9] pp.28-33.) Remark 1 From J.Milnor [7] p.114 we know that //S(G) ^ 0 if G is a We now prove that S7 is not a Lie group. non-abelian compact connected Lie group. (See E.Cartan, "La Topologie Theorem 1 S7 is not a Lie group. des Espaces Representatives des Groups de Lie," Paris, Hermann, 1936). 7 Proof Suppose that S7 is a Lie group. Because S7 is compact and Using the above result, ones can prove that S is not a Lie group. In fact, 7 7 connected, from Lemma 1 we know S7 = C{S7) • G\ • • • G,, where Gi is assume that S is a Lie group. If S is a, non- abelian compact connected 3 7 S 7 connected simple normal subgroup, G' = Gi • • • G. is connected semis im- Lie group, then .ff (S ) ^ 0, which contradicts the fact that H (S ) = 0 7 ple normal subgroups. Since S7 is connected and simple connected, from (See R.Bott and Loring W.Tu [3] pp.5 and 36); if S is an abelian compact 7 7 Lemma 4 it follows that G( and 'G* are both closed subgroups. According connected Lie group, it follows from Lemma 2 that S = T . According to to Lemma 6, Gt and G* are Lie subgroups of G. Moreover, compactness (l) in the proof of Theorem 1 we obtain a contradiction again. of S7 implies compactness of closed subgroups Gi and G'. From Lemma 3 Theorem 2 S" is a Lie group if and only if re = 0,1,3. we know that the compact simple Lie group whose dimension is less than Proof (=*• ) Obviously, 5° = {u, —u} is considered as a 0-ctimensional 7 must be SU(2) = S3 (See C.Chevalley [4] pp.17-18), therefore d = $s. Lie group. By Examples 1 and 2, Sl and S3 are also Lie groups. n Note S -• S7/G\ p : S7 -> S is a natural projection, then 5 = plS7) = (<$=} If S is a Lie group for n > 1, then 5" is a parallelizable manifold, p{C(S7)} is a connected compact commutative Lie group. Hence from from [1] and [2] it follows that n = 1,3 or 7. But Theorem 1 tells us that 7 Lemma 2 it follows that S = Rm/Zm = T™ = S1 x • - • x S1. .S is not a Lie group, therefore n = 1,3. m Theorem 3 5" is a topological group if and only if n — 0,1,3. If S7 is a Lie group, then it must one of the be following cases : Proof (<=) Since S" is a Lie group, it is a topological group for n = T 7 7 0,1,3. (1) G* = {«} (u is a unit element of G), S = T , then S = T and by n p.61 in [9| , p52 in [10] we obtain (=>) Suppose that the manifold S is a topological group for n > 1. 7 7 1 1 1 l 1 From A.M. Gleason |5], D. Montgomery and L. Zippin [8], we know that 0 = Xl(S ) = K^T ) = JT,(S' x S x 5" x S x S x S x S ) == 1 1 1 1 1 1 S" is a Lie group. According to Theorem 2, n — 1,3. MS ) SMS ) © MS ) ©MS ) 6 M^ ) ©MS ) © 7 7 Remark 2 Because K^RP } — Z2 4- 0, RP is not a simple connected, 7 Z(&Z® ZQZ@Z, a contradiction. where RP is an n-dimensional real projective space. Therefore, we can 3 4 not use Lemma 4 to prove that RP1 is not a Lie group. (2) G' = £ , S = 7" , then by Lemmas 7 and 8, there is an exact se- 7 quence : From R.Bott and Loring W.Tu |3] p.78 it follows that H*(RP ) = 0. As in Remark 1 we can obtain that RP7 is not a Lie group. Therefore, it is easy prove that RPn is a Lie group if and only if n = 0, 1,3 and RPn is a topological group if and only if n = 0, 1,3. which implies 7T,(r4) = 0, but Mr4) = M^1 xS'xS1 xS1) = 0, this is a contradiction. (3} G* = S3 • Ss. S = 2" (1 < i < 3). Using the second coordinate we know that the dimension of S3 • S3 is 6 at least. By Lemmas 7 and 8, there is an exact sequence : 0 = MS7) -' M7") -* MS' • Ss) = 0 ' which implies °' tn's 's a contradiction. Acknowledgements References

One of the authors (X.S.) would like to thank Professor Abdus Salam, International Atomic Energy Agency and UNESCO for hospitality at the [l] Adams, J.F., "On the non-existence of elements of liopf invariant International Centre for Theoretical Physics, Trieste. The authors wish one", Ann. Math., T2 (i960), 20-104. to express their sincere thanks to Professors J.Eeils and A.Verjovsky for [2] Bott, R.., and Milnor, J., "On the parcdhHzability of the, spheres", Bull. Amer. Math. Soc, 64 (1958), 87-89. their interest and encouragement. This work is partially supported by the [3) Bott, R., and Tu, Loring W., Differential Forms in Algebraic Topol- Chinese Science Fund. ogy, (Springer-Verlag, New York, 1982). [4] ChevaSley, C, Theory of Lie Groups, (Princeton, University Press, 1946). [5] Gleason, A.M., "Groups without small subgroups, Ann. Math., 56 (1952), 193-212. [6] Hochschild, G., The Structure of Lie Groups, (Holden-Day,. San Francisco, London, Amsterdam, 1965). [7] Milnor, J., Morse Theory, (Princeton University Press, 1963). |8] Montgomery, D., and Zippin, L., "Smalt subgroups of finite-dimensional groups", Ann. Math., 56 (1952), 213-241. [9] Steenrod, N., The Topology of Fibre Bundles, (Princeton University Press, 1951). [10] Switsser,R-.M., Algebraic Topology-Homoiopy and Homology, (Springer- Verlag, Berlin, Heidelberg, New York, 1975). [llj Yan Zhida and Xu Yichao, Lie Group and Lie Aigebra,{ High Edu- cational Press, China, 1985).

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