By Jean-Paul Pier

ISBN-10: 0471893900

ISBN-13: 9780471893905

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**Example text**

The equation = (J) then makes [3 = I; similarly [3' = 1; but then)f = f); even this is absurd. Therefore, in all cases, the assumption that) fixes all lines in P leads to a contradiction. So step (2) is established. (3) Given any line L I in PI' there are elements (J2 and (J3 with Aa2 = a3 such that )(J a2 E (J2 res a3 E ~ not big dilation (J2 = 2 R2 c: P (J3 ~I not big dilation I <: res LI (J3 <: 2 c: R3 c: PI' To see this, pick T L , and) as in step (2). Then)P = P and)L =1= L for some line L in P, by step (2).

D. 28 O. T. 15. i4) V, and let (J be the in I i n X i n matrix over F. Then (I) (J is in SPn(V) if and only if A is symmetric. (2) If (J is in SPn( V), it is hyperbolic if and only if A is alternating. PROOF. 17. Now suppose that SPn( V). So A = (a i) is symmetric. If (J is hyperbolic, then ajj (y;, '±2 = q (J is in aijXi ) 1=1 = q(yP lryJ = 0, so A is alternating. D. , that (J is hyperbolic. 16. Let char F = 2. i4) where A' is invertible and diagonal if (J with A = in SPn (V), there is a (40) not hyperbolic, and A' has the form uL 01 1 + 1 I 0 If (J is hyperbolic.

A with residual line Fa are the transvections Ta a'A as a runs through F. 3. REMARK. Let .! be a symplectic base for V. 17 that a is an element of SPn( V). If we now derive S' from S by (I) adding a multiple of one column to another and then doing the same thing with the corresponding rows, or (2) interchanging two columns and then interchanging the corresponding rows, then the linear transformation a' with a' - ( 47) in .! is still in SPn( V) since S' is still symmetric. But in fact a and a' are conjugate in SPn(V), To see this observe that in either case S' has the form S' = 'TST for some Tin GLn/iF).

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