Isometries of Hermitian symmetric spaces

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Proof of Theorem 2 (Eschenburg-Quast-T.) We prove it by induction over the dimension. The beginning of the induction is the observation that an extrinsically symmetric flat torus is a product of Eu- clidean circles (Proposition 4). As induction step, we show an alternative for arbitrary compact symmetric spaces X: Either X is a Riemannian product of Euclidean spheres and possibly flat torus or it contains a certain totally geodesic submanifold , a so called meridian (Proposition 1). 28
Further, a meridian M − of X is extrinsically symmetric, then so is M − (Proposition 2). Thus passing to meridi- ans again and again, we will lower the dimension preserving preserving the maximal torus unless we reach a space which is a Riemannian product of spheres and possibly a torus and whose maximal torus is a Riemannian product of circles. 29
Page 1 and 2: Isometries of Hermitian symmetric s
Page 3 and 4: 1. Introduction A Riemannian symmet
Page 5 and 6: M : a Riemannian manifold τ : an i
Page 7 and 8: 2. Extrinsically symmetric spaces M
Page 9 and 10: P = G/K : a Riemannian symmetric sp
Page 11 and 12: I(M) : the group of isometries of M
Page 13 and 14: 3. Isometries of Hermitian symmetri
Page 15 and 16: Consider the map ι : P → g, p
Page 17 and 18: Proof: P = P1 × · · · × Pr : a
Page 19 and 20: Let f ∈ I(P ). Let m be the midpo
Page 21 and 22: Lemma (I f)∗(Jo) ∈ {±Jo}. Proo
Page 23 and 24: Remark It is a known fact that ever
Page 25 and 26: 4. Another result Theorem 2 (Loos 1
Page 27: Proposition 3 Let M be a compact Ri

Isometries of Hermitian symmetric spaces

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