By Buekenhout F., Huybrechts C.
We turn out the lifestyles of a rank 3 geometry admitting the Hall-Janko staff J2 as flag-transitive automorphism workforce and Aut(J2) as complete automorphism crew. This geometry belongs to the diagram (c·L*) and its nontrivial residues are whole graphs of measurement 10 and twin Hermitian unitals of order three.
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Another basis e1 , . . , en determines the same orientation if and only if the matrix transforming one basis into the other has positive determinant. Another (equivalent) method is the choice of an ordered sequence of n + 1 affinely independent points a0 , . . , an ∈ E: we assume that these points determine the same orientation as the basis a1 − a0 , . . , an − a0 of the linear space E. 14) Exercise. Let e1 , . . , en be a basis and let the affinely independent n points a0 , . . , an have the coordinates: ai = j=1 aij ej .
24) Lemma. Let U ⊂ E be a compact subset of a Euclidean space and let f: U → E be a C 1 map. e. the derivative map Dfx0 : E → E is the isomorphism. Then x0 is the isolated zero of f and deg (f, x0 ) = deg (Dfx0 ) = sgn det (A), where A is the matrix representing Dfx0 . In other words the degree at a regular point equals the degree of the derivative map. Proof. The regular point x0 is the isolated zero by the Inverse Function Theorem. It remains to show that deg (f, x0 ) = sgn (det Dfx0 ). 23). We will show that the segment homotopy between the maps f(x) and L(x) = Dfx0 (x − x0 ) has no zeroes on the boundary of a sufficiently small ball K(x0 , δ).
Proof. 13). Now ind (f ×f ) = deg (id ×id’−f ×f ) = deg (id−f) deg (id’−f ) = ind (f)·ind (f ). The fixed point index possesses also a very important property which will enable us to extend its definition to a much larger class of spaces. 11) Lemma (Commutativity Property). Let U ⊂ E, U ⊂ E be open subsets of Euclidean spaces and let f: U → E and g: U → E be continuous maps. Then the composites gf: V = f −1 (U ) → E, fg: V = g−1 (U ) → E have homeomorphic fixed point sets Fix (gf) = Fix (fg). If moreover these sets are compact, then ind (fg) = ind (gf).