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I would like to work a theorem on a article who deals with the rank one symmetric spaces.

i looked up the definition of symmetric spaces of rank one, but I did not find a satisfactory definition then what is the meaning of rank, intuitively and mathematically? please if anybody already worked with rank one symmetric spaces..

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  • $\begingroup$ What is unsatisfactory with the definition on Wikipedia? $\endgroup$
    – Yemon Choi
    Commented Mar 19, 2012 at 0:51
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    $\begingroup$ the definition on Wiki is a technical definition: "is the maximum dimension of a subspace of the tangent space (to any point) on which the curvature is identically zero", i need more meaningful definition (intuitively meaning) deeper, maybe another equivalence of that definition. $\endgroup$
    – Abdelmajid Khadari
    Commented Mar 19, 2012 at 1:02
  • $\begingroup$ i feel like this definition is not, very practical. $\endgroup$
    – Abdelmajid Khadari
    Commented Mar 19, 2012 at 1:04
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    $\begingroup$ A minor addition to Alex's answer: The same geometric characterization goes through in the case of compact symmetric spaces, only the assertion about (totally-geodesic) flat submanifolds becomes local and the curvature of the manifold is $\ge 0$, with curvature $>0$ iff the symmetric space has rank 1. $\endgroup$
    – Misha
    Commented Mar 20, 2012 at 11:28

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First the algebraic definition. A (non-compact) symmetric space is of the form $G/K$, where $G$ is a (non-compact) semisimple Lie Group defined over $\mathbb{R}$, and $K$ is a maximal compact subgroup of $G$.

Then the rank of a symmetric space is the dimension of the "maximal $\mathbb{R}$-split torus", i.e. the maximal dimension of an abelian diagonalizable over $\mathbb{R}$ subgroup of $G$.

The geometric meaning is that the rank is the dimension of the maximal flat submanifold of the symmetric space. If the rank is $1$, then the maximal flats are geodesics, and the symmetric space turns out to be negatively curved.

If the rank is larger then one, then the symmetric space is only non-positively curved. However, higher-rank symmetric spaces have spectacular rigidity properties (e.g. Margulis superrigidity, arithmeticity and the normal subgroup property come to mind).

There are only three families of rank 1 symmetric spaces,

1) hyperbolic $n$-space, corresponding to the Lie group $SO(n,1)$.

2) complex hyperbolic $n$-space, corresponding to the Lie group $SU(n,1)$.

3) quaternionic hyperbolic $n$-space, corresponding to the Lie group $Sp(n,1)$.

There is also one exceptional example:

4) the Cayley upper half plane, corresponding to the Lie group $F_4^{-20}$.

The spaces 3) and 4) have some but not all of the rigidity properties of higher rank (in particular, superrigidity and arithmeticity, but not the normal subgroup property).

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    $\begingroup$ thank you very much Pr. Alex Eskin, i find this answer very helpful, and i accept your answer. $\endgroup$
    – Abdelmajid Khadari
    Commented Mar 19, 2012 at 1:13
  • $\begingroup$ @Eskin wrote that rank 1 implies negative curvature, but projective space is positively curved and is rank 1, I am confused. $\endgroup$
    – wishcow
    Commented Jan 11, 2017 at 9:40
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    $\begingroup$ @wishcow It's because in the first line the manifolds were restricted to non-compact ones. $\endgroup$
    – Pratyush Sarkar
    Commented Apr 6, 2018 at 21:06
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This question reminds me of when I was a graduate student. At some point Gelfand asked me "What is the rank of a symmetric space" and I just spat back the usual definition, something like what Matrix found in Wikipedia. Gelfand shook his head as if I had said something really stupid and proceeded to explain:

Euclidean space, hyperbolic space, complex projective space (and so on) are rank one. Why? Because if you have two pairs of points and the distance between them is the same, then there is an isometry that takes one pair of points to the other. ONE invariant is all you need to determine whether two pairs of points are the same up to isometries.

The Grassmannian of two-planes in ${\mathbb R}^4$ has rank two : you need two invariants to determine if two pairs of points are equivalent up to isometry. Take two planes in four-space passing through the origin. Draw a circle with center zero in one plane. Project it orthogonally onto the second plane. You get an ellipse, but you cannot compare it to the circle because it lives on a different plane so project it back to the first plane. The minor and major axes of your ellipse (with respect to the circle) are two invariants that are preserved by any isometry of the pair of planes. Conversely if you have two pairs of planes that have the same two invariants, then there is an isometry of the Grassmannian that takes one pair of planes to the other.

I went back home and the uninsightful book I was reading on symmetric spaces went back to the library the next day.

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    $\begingroup$ Wow. But where do we find beautiful definitions or explanations like this? $\endgroup$
    – Deane Yang
    Commented Mar 19, 2012 at 10:55
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    $\begingroup$ To be fair, though, this interpretation of rank is explained carefully and precisely in Helgason's "Differential Geometry, Lie Groups, and Symmetric Spaces", where he explains, in terms of Weyl chambers, exactly how the invariants arise. $\endgroup$
    – Robert Bryant
    Commented Mar 19, 2012 at 15:23
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    $\begingroup$ @Robert: Thanks for the reference! If I remember correctly, Helgason is more interested in developing Cartan's theory of symmetric spaces so it is natural that it does not give you this sort of insight up front. However, is there any place where the theory is developed from a metric viewpoint? There are a few pages in Busemann's Geometry of Geodesics, but really very little. $\endgroup$
    – alvarezpaiva
    Commented Mar 19, 2012 at 18:24
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    $\begingroup$ @alvarezpavia: Try Eberlein's book "Geometry of nonpositively curved manifolds" and/or Ballmann, Gromov, Schroeder "Manifolds of nonpositive curvature". What Gelfand told you is just the Cartan decomposition: Your symmetric space is $X=G/K$, where $G$ is a reductive group and $K$ maximal compact. Cartan decomposition: $G=KA_+K$, where $A_+$ is Weyl chamber. Thus, if $o\in X$ is fixed by $K$, then every orbit $Kx$ in $X$ intersects $A_+$ exactly once, at a point $y=c(x)$. Thus, the pairs of points $(o,x)$ are parameterized by $y\in A_+$. Dimension of $A_+$ is the rank of the space $X$. QED $\endgroup$
    – Misha
    Commented Mar 20, 2012 at 7:50
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    $\begingroup$ @alvarezpavia: By the way, if you think of the 2-point invariant of pairs of points $(p,x)$ in rank $n$ symmetric space that Gelfand told you about, as "vector-valued distance" on $X$ (which is the ordinary distance if $n=1$) then you obtain some interesting and useful geometry on $X$ with "triangle inequalities" that generalize the ordinary ones, etc. $\endgroup$
    – Misha
    Commented Mar 20, 2012 at 7:55
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If $M=G/K$ has noncompact type and ${\rm rank}(M)=r\ge 1$, then the space of invariant differential operators on $M$ is generated by a $r$-tuple. One can choose the first element of this $r$-tuple to be the Laplace/Casimir operator, so for rank $1$ symmetric spaces it is always generated by the Laplacian.

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    $\begingroup$ If you have a question like this, then you should probably ask it as a separate question, rather than leaving it as part of an answer to someone else's 7-year-old question. You can always link to this one in your question as background or motivation. $\endgroup$
    – LSpice
    Commented May 10, 2019 at 1:43

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