There is a well-known duality between compact symmetric spaces and symmetric spaces of noncompact type. Basically it goes as follows: If $$G/K$$ is a symmetric space of noncompact type, $$g=k+p$$ the corresponding Lie-algebra decomposition, then $$g'=k+ip$$ (as a subalgebra of the complexification of $g$) is the Lie algebra of a simply connected Lie group $G'$ and $$G'/K$$ is by definition the dual symmetric space of $G/K$. ($G'/K$ is compact because the Killing form is negative definite on $k$ and positive definite on $p$, hence it is negative definite on $k+ip$, which implies the Lie group $G'$ is compact.)

A special instance is the duality between hyperbolic and spherical geometry: For the hyperbolic 3-space $$G/K=SL(2,C)/SU(2)$$ we obtain $k+ip=su(2)+su(2)$, hence $$G'/K=SU(2)\times SU(2)/SU(2)=SU(2),$$ which is the 3-sphere. The tangent space of $G/K$ was $p$, the tangent space of $G'/K$ is $ip$. Thus one can say that all infinitesimal computations in $G'/K$ are obtained from those in $G/K$ by multiplying the argument by $i$.

I have read in several places (e.g. Helgason) that this duality together with the formula $$\cosh(x)=\cos(ix)$$ would explain the analogies between spherical and hyperbolic geometry, for example (for rectangular triangles with edges $a,b,c$) between the formula $$\cos c=\cos a \cos b$$ in spherical and $$\cosh a = \cosh b \cosh c$$ in hyperbolic geometry.

Finally my question. How can one actually use the duality, that is, the fact that infinitesimally arguments are just multiplied by i, to derive e.g. the hyperbolic from the spherical cosine theorem or vice versa?

This seems not so obvious as one might think: given a hyperbolic triangle one can lift it to a triangle in $p$ (i.e take the preimage under the exponential map), which of course corresponds to a triangle in $ip$ that projects to a spherical triangle. But the edge-lengths of these triangles seem not directly related (and certainly they are not related via multiplication by $i=\sqrt{-1}$ :-)), so I am wondering how duality actually may help to explain such analogous formulae as $\cos c = \cos a \cos b$ vs. $\cosh a = \cosh b \cosh c$. (Formally of course one is obtained from the other by multiplying the lengths with $i$. But this does not seem to have a geometric interpretation.)

  • $\begingroup$ Have you tried infinitesimal triangles? This would explain the identities on the level of Taylor expansions. $\endgroup$
    – S. Carnahan
    Jan 20, 2012 at 7:05
  • $\begingroup$ Taylor expansions of what? The length of the three edges? $\endgroup$
    – ThiKu
    Jan 20, 2012 at 9:10


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