By K. Bleuler, H. R. Petry, A. Reetz

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O. Indeed, if f f. 0, let g::: ~~. Then 9 f. 0; because I(x) -- 0 integrating by parts, J[- +00 ~ f(x)f(x) - f(x)f(x)]dx -00 -00 f. 0. symplectic manifolds and f :M1 --+ M 2 a canonical transformation. Show that Solution. Since or equivalently r: M 1 --+ I is an immersion. M 2 is a canonical transformation it follows that CHAPTER I 24 for each x E M I and Xx, Y x E TxM I . As w is nondegenerate, the equality implies = 0, because WI is nondegenerate. Therefore the for each Y x E TxM I , so Xx tangent map T f : T M I ~ T M 2 must be injective at every point x EM, and hence dimMI :::;dimM2 , so f is an immersion .

G}, i. e. l(x)::: jX f(y)dy [resp. oo g(x)::: jX009(y)dyj. Show that (S(R,R),w) is a weak symplectic vector space. Solution. It is clear that w is skew-symmetric and bilinear. We shall prove only its weak nondegeneracy. For this we must check that if f f. 0, there is agE S(R, R) such that w(l,g) as Ixl --+ 00 f. O. Indeed, if f f. 0, let g::: ~~. Then 9 f. 0; because I(x) -- 0 integrating by parts, J[- +00 ~ f(x)f(x) - f(x)f(x)]dx -00 -00 f. 0. symplectic manifolds and f :M1 --+ M 2 a canonical transformation.

Dx n with two nonvanishing functions a and b which by iM a = iMf3 must have the same sign. We construct a family of diffeomorphisms 4>t satisfying 4>;CXt = a; 4>0 = Id M; 0 S t S 1. f = 4>1 will solve our problem. Since M is compact, connected and orientable, we conclude from iM (f3 - a) = 0, that p- a = d'y, for some (It - 1)- The diffeomorphism form 'Y. This is a special case of the De Rham theorem. Since at is a voluml" fOlm, there is a unique time dependent vector field X t solving the equation ix,at + 'Y = 0; 0 S t S 1.