I need to cover a case of $n$-dimensional locally symmetric Riemannian space of rank $n-1$. Is there a simple proof that there is no such irreducible space ($n>4$)? If I need to cite the Cartan classification for that, which concrete paper/book you suggest?
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1$\begingroup$ Multiply hyperbolic plane (or 2d sphere) by the Euclidean space of suitable dimension. $\endgroup$– Moishe KohanSep 17, 2020 at 3:51
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$\begingroup$ @MoisheKohan yes, but what about irreducible ones? $\endgroup$– ArimakatSep 17, 2020 at 13:05
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$\begingroup$ Just go through the list at en.wikipedia.org/wiki/Symmetric_space and references therein. $\endgroup$– Igor BelegradekSep 17, 2020 at 18:48
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1 Answer
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An irreducible symmetric space $M$ of dimension $n$ and rank $n-1$ has dimension $n=2$. Indeed, the rank is the codimension of a principal orbit of the isotropy group $K$ at $p \in M$ acting on the tangent space $T_pM$. So the $K$-orbits are one dimensional. Now any monoparametric subgroup of $K$ has a one dimensional orbit contained in a $2$-dimensional subspace of $T_pM$ hence $K$ has a one dimensional orbit contained in a $2$-dimensional subspace of $T_pM$. So $K$ has an invariant $2$-dimensional subspace in $T_pM$. But $K$ is the holonomy group of $M$ at $p$ hence $K$ acts irreducibly on $T_pM$. Then $dim(T_p M)=2$.
