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Example text

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. 13. Now R = p* so so it follows easily that P contains a totally degenerate subspace of dimension n that contains R. Accordingly, by 1. 14, there is a symplectic base PROOF. R = rad P, i GENERATION THEOREMS 29 for V in which R = Fx I + . . + C Fx I FXr + . . + FXr + . . + FXn / 2 CP. 17 we know that a-(~) inID for some ~ n X ~ n symmetric matrix D.

0; hence k 2x = mkx. and we have (3). So P = 1 and O. 54 T. 2. If k is a hyperbolic transformation in rSp,,( V), then k is in Sp,,( V) only if k is an involution. PROOF. 1. D. 3. If a hyperbolic transformation k in rSp,,( V) stabilizes a line in V, then k is in PSp,,( V). PROOF. 4. D. 4. Let k be a hyperbolic transformation in rSp,,(V), and hence in GSp,,(V), . 2 such that m, E F - F . A) where A is the ~ n X ~ n matrix PROOF. 4 we know that k f/:. 3 we know that k can stabilize no line in V.

So 1= q(x,p) = ma-1q(ax, ap) = ma-1q(ax,p) = ma-I. So a is in SPn( V). (2) Take PI' P2 in P with q(pI' P2) = I. D. 7. Every transvection in fSPn( V) is already in SPn( V). Every projective 53 SYMPLECTIC COLLINEAR TRANSFORMATIONS transvection in prSPn( V) is already in PSPn( V), and its representative transvection is in SPn( V). PROOF. 3. So assume n > 4. If a is a transvection in rSPn( V), then a E GSPn( V), and q(P. P) =1= 0 since dim P > n - 1 > ~ n. 6. A projective transvection in prsPn (V) has the form T = k with T a transvection in SL n( V) and k an element of rSPn( V).

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