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**Additional resources for Algebraic Geometry Iv Linear Algebraic Groups Invariant Theory**

**Example text**

Iii)-b If f ki j Ai W Ai ! Ai interchanges the boundary components of Ai , then f ki j Ai is a special twist. This whole chapter is devoted to the proof of the following theorem. 1 (cf. [53, Sect. 3]). (i) Any pseudo-periodic map f W ˙g ! ˙g is isotopic to a pseudo-periodic map in standard form. (ii) Suppose two pseudo-periodic maps in standard form f; f 0 W ˙g ! ˙g are homotopic, then there is a homeomorphism h W ˙g ! ˙g isotopic to the identity, such that f D h 1 f 0 h. 20 2 Standard Form The uniqueness statement (ii) above is stronger than the original one ([53, Sect.

1/ W A1 ! i / W Ai ! A ; @A / ! A ; @A / ! 2/ W A2 ! f j A1 / D f 0 j A1 . @ isotopic to 0Ä 1 W A2 ! A3 f 0 j A2 W A2 ! 3/ W A3 ! h1 / 1 D f 0 j A2 . i / W Ai ! h1 / D f 0 j Ai 1: W Ar ! f j A1 / W A1 ! h1 / 1 D f 0 j Ar W Ar ! , f D h1 1 f 0 h1 ) as asserted. 3 (S PECIALIZATION). A an (orientation-preserving) homeomorphism which interchanges the boundary components. Suppose f j @A W @A ! @A is periodic. @ isotopy f W A ! A; such that f0 D f and 0Ä Ä 1; f1 W A ! A is a special twist. We need a sublemma.

Then (i) m0i D m1i D an even number, (ii) . 0i ; i0 / D . mod i / and 0 Ä ıi < i . ) (iv) . Ai /. Proof. Ai / D Ai . Since Ai is amphidrome, f k interchanges the boundary components of Ai . Ai / D Ai does not interchange the boundary components. This implies (i) m0i D m1i D 2k. Obviously, f k j @0 A W @0 A ! @1 A is equivariant with respect to the actions of f 2k j @0 A on @0 A and f 2k j @1 A on @1 A. This proves (ii) . 0i ; i0 / D . 1i ; i1 /. To prove (iii), consider f k W A i ! Ai as f in sublemma 1 and take a parametrization W Œ0; 1 S 1 !