Algebraic Topology-Rational Homotopy by Felix Y. (ed.)

By Felix Y. (ed.)

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Gibt mit {3 1= 0, ±1 und (1~{3 ~) E Es sei nun N ein Normalteiler von G = SL(2,K) mit N Zu zeigen ist N = G. Falls es ein {3 E K N, so folgt [( 1 ~{3 Fall, dann gibt es Konjugation mit ~ ) , (~ ~ )] (~ ~) (~1 ~) = F2 E N mit 'Y kann 'Y 1= dll ({32 1= 1)) EN. 1st dies nicht der 0 oder {3 1= O. Nach eventueller 0 angenommen werden, ferner 0' = 0 nach Konjugation mit F12 (-air) und schlieBlich {3 = 1 nach Konjugation mit einer Diagonalmatrix aus G. t, gibt es ein {3 E K mit {34 ist F21 (0') E N fiir alle 0' E K.

Man bestimme nach der Methode aus Aufgabe 5 eine Orthonormal basis des durch die Vektoren (1,2,1,2), (0,1,1,1), (2,1,0,-1) aufgespannten Teilraumes von IR4. 7. Es sei V der IR-Vektorraum der 2 X 2 Hermiteschen Matrizen uber zeige, daB 1 heX, Y) := '2 (det(X + Y) - det X - det Y) (C,-). Man eine nicht-ausgeartete symmetrische Bilinearform auf V ist und bestimme die Matrix von h bez. der Basis von V. 8. Es seien 9 und h symmetrische Bilinearformen auf dem endlich-dimensionalen Vektorraum V, h sei nicht-ausgeartet.

Sind ebenfalls posit iv- bzw. negativ-definit. ' 2) Die Diagonalform [Dp,q] ist genau dann positiv-(negativ-)definit, wenn Dpq = E (Dp,q = -E). 3) 1st h positiv-definit, so ist -h negativ-definit und umgekehrt. h. daf3 heine Hermitesche Form auf V mit der entsprechenden Eigenschaft ist), so ist jeder zu (V, h) isometrische Raum ebenfalls positiv-( negativ-,in-)definit. 46 Kapitel I. Die klassischen Gruppen Beispiel. Die inl. angegebene *-Hermitesche Form heX, Y) = Spur(XY*) auf Mat( n, lK) ist fiir (lK, *) = (IR, id), (C, -) positiv-definitj es gilt niimlich h(X,X) = Li,j l~ijI2, wenn X = (~ij).

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