2015
DOI: 10.21099/tkbjm/1429103724
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Realizations of globally exceptional $\mathbf{Z}_2 \times \mathbf{Z}_2$-symmetric spaces

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Cited by 2 publications
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“…Here, the spinor groups Spin (6) and Spin (10) above are respectively realized as the subgroup (F 4 ) E 1 ,E 2 ,E 3 ,F 1 (e k ),k=0,1 and the subgroup (E 8 ) σ ′ 4 ,so (6) of (E 8 ) σ ′ 4 , where the definitions or the details of (F 4 ) E 1 ,E 2 ,E 3 ,F 1 (e k ),k=0,1 and (E 8 ) σ ′ 4 ,so (6) are shown later. This amounts to the globally realization of one of seven types with an automorphism of order 4 on E 8 .…”
Section: Introductionmentioning
confidence: 99%
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“…Here, the spinor groups Spin (6) and Spin (10) above are respectively realized as the subgroup (F 4 ) E 1 ,E 2 ,E 3 ,F 1 (e k ),k=0,1 and the subgroup (E 8 ) σ ′ 4 ,so (6) of (E 8 ) σ ′ 4 , where the definitions or the details of (F 4 ) E 1 ,E 2 ,E 3 ,F 1 (e k ),k=0,1 and (E 8 ) σ ′ 4 ,so (6) are shown later. This amounts to the globally realization of one of seven types with an automorphism of order 4 on E 8 .…”
Section: Introductionmentioning
confidence: 99%
“…Again we would like to state about the group (E 8 ) σ ′ 4 ,so (6) . The essential part to prove the isomorphism as a group is to show the connectedness of the group (E 8 ) σ ′ 4 ,so (6) . In order to obtain this end, we need to treat the complex case as follows: Spin (10, C),…”
Section: Introductionmentioning
confidence: 99%
“…Jiménez, there exist seven compact simply connected Riemannian 4-symmetric spaces G/H in the case where G is of type E 8 . In the present article, we give the explicit form of automorphismsw 4υ4 and µ 4 of order four on E 8 induced by the C-linear transformations w 4 , υ 4 and µ 4 of the 248-dimensional vector space e 8 C , respectively. Further, we determine the structure of these fixed points subgroups (E 8 ) w 4 , (E 8 ) υ 4 and (E 8 ) µ 4 of E 8 .…”
mentioning
confidence: 99%
“…Case hγ H = G γ 1 so(6) ⊕ so(10)σ 4 (Spin(6) × Spin(10))/Z 4 2 iR ⊕ su(8)w 4 (U(1) × SU(8))/Z 24 3 iR ⊕ e 7υ 4 (U(1) × E 7 )/Z 2 4 su(2) ⊕ su(8)μ 4 (SU(2) × SU(8))/Z 4 5 su(2) ⊕ iR ⊕ e 6ω 4 (SU(2) × U(1) × E 6 )/(Z 2 × Z 3 )…”
mentioning
confidence: 99%
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