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Would someone be able to list (or provide a reference to) the simply-connected non-compact irreducible symmetric spaces of rank $\ge 1$(as quotients of Lie groups $G/H$)?

Any help would be appreciated!

Sergio Charles
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    Wolf (2011), p. 286 sq.. –  Jul 26 '18 at 00:25
  • @FrancoisZiegler Question: When it says "duality interchanges compact and non-compact simply connected irreducible symmetric spaces further reducing the classification to the compact irreducible case", does this mean that a decomposition of a, say non-compact, symmetric space must have irreducible factors that are compact simple simply connected Lie groups? – Sergio Charles Jul 26 '18 at 00:47
  • I don’t get how this pertains to your question. You are assuming $M$ irreducible, so there is just one factor, which is noncompact. (Wolf explains the decomposition and duality on pp. 234-237.) –  Jul 26 '18 at 01:37
  • @FrancoisZiegler No, I am assuming the factors of $M$ are irreducible ($M$ is a non-compact reducible simply connected symmetric space $M= \mathbb{R^n}\times\prod M_i \times \prod M_j^* \times \prod N_k \times \prod N_l^*$). I am asking if the irreducible factors of the non-compact space are compact by the so-called "duality". – Sergio Charles Jul 26 '18 at 01:49
  • I was referring to the title question, where $M$ is irreducible. If not, and $M$ is still non-compact simply-connected, then there may (not must) be compact factors. But not “by the duality”, which is only a device reducing the classification of noncompact irreducibles to that of compact irreducibles. Unless I am missing something... –  Jul 26 '18 at 02:10
  • @FrancoisZiegler I see where the confusion is. I thought you meant $M$ is a reducible simply connected symmetric space. What do you mean by "if $M$ is still non-compact simply-connected, then there may (not must) be compact factors"? Sorry for the confusion. Do you mean that if there are irreducible non compact factors, they can be changed to irreducible compact factors? I cannot access pages 234-237. – Sergio Charles Jul 26 '18 at 02:30
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    Yes. Not “changed” in the sense of modifying $M$, only in the sense of bijecting them to something already-classified. As to “may (not must)” it was in reference to the “must” your very first comment above. E.g. $\smash{\mathbf R^2\times\mathrm S^2}$ is noncompact with a compact factor. –  Jul 26 '18 at 02:36

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The reference from which I learned the lists of $G/K$'s of various sorts is S. Helgason's book "Differential Geometry and Symmetric Spaces".

(For my own mathematical purposes, explicit details about at least the "classical groups and domains" is very useful, so I do keep these things in my head.) Possibly dropping overt reference to the anomalous isogenies and edge cases and quibbles about notational conventions...

Type A: $SL_n(\mathbb R)/SO(n,\mathbb R)$, $SL_n(\mathbb C)/SU(n)$, $SL_n(\mathbb H)/Sp^*(n)$, $U(p,q)/U(p)\times U(q)$

Types B,D: $O(p,q,\mathbb R)/O(p)\times O(q)$, $O(n,\mathbb C)/SO(n,\mathbb R)$, $O^*_{2n}/U(n)$.

Type C: $Sp_n(\mathbb C)/Sp^*n$, $Sp_n(\mathbb R)/U(n)$, $Sp^*_{p,q}/Sp^*_p\times Sp^*_q$.

The not-well-known cases are: $Sp^*_{p,q}$ is (modeled by) the group of quaternion matrices preserving a quaternion hermitian form. Provably, these have signatures, much as Sylvester's inertia theorem for quadratic forms. And $O^*_{2n}$ is quaternion matrices preserving a skew-hermitian form.

Note that in all cases the three $\mathbb R$-algebras $\mathbb R, \mathbb C, \mathbb H$ play roles. In fact, a completely parallel thing happens for classical groups over $p$-adic fields and in other cases, as in A. Weil's "Algebras with involutions and classical groups", Indian (not Indiana) J. Math. 1960.

paul garrett
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