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September 26, 2016

Euclidean, Hyperbolic and Elliptic Geometry

Posted by John Baez

There are two famous kinds of non-Euclidean geometry: hyperbolic geometry and elliptic geometry (which almost deserves to be called ‘spherical’ geometry, but not quite because we identify antipodal points on the sphere).

In fact, these two kinds of geometry, together with Euclidean geometry, fit into a unified framework with a parameter ss \in \mathbb{R} that tells you the curvature of space:

  • when s>0s \gt 0 you’re doing elliptic geometry

  • when s=0s = 0 you’re doing Euclidean geometry

  • when s<0s \lt 0 you’re doing hyperbolic geometry.

This is all well-known, but I’m trying to explain it in a course I’m teaching, and there’s something that’s bugging me.

It concerns the precise way in which elliptic and hyperbolic geometry reduce to Euclidean geometry as s0s \to 0. I know this is a problem of deformation theory involving a group contraction, indeed I know all sorts of fancy junk, but my problem is fairly basic and this junk isn’t helping.

Here’s the nice part:

Give 3\mathbb{R}^3 a bilinear form that depends on the parameter ss \in \mathbb{R}:

v sw=v 1w 1+v 2w 2+sv 3w 3 v \cdot_s w = v_1 w_1 + v_2 w_2 + s v_3 w_3

Let SO s(3)SO_s(3) be the group of linear transformations 3\mathbb{R}^3 having determinant 1 that preserve s\cdot_s. Then:

  • when s>0s \gt 0, SO s(3)SO_s(3) is isomorphic to the symmetry group of elliptic geometry,

  • when s=0s = 0, SO s(3)SO_s(3) is isomorphic to the symmetry group of Euclidean geometry,

  • when s<0s \lt 0, SO s(3)SO_s(3) is isomorphic to the symmetry group of hyperbolic geometry.

This is sort of obvious except for s=0s = 0. The cool part is that it’s still true in the case s=0s = 0! The linear transformations having determinant 1 that preserve the bilinear form

v 0w=v 1w 1+v 2w 2 v \cdot_0 w = v_1 w_1 + v_2 w_2

look like this:

(cosθ sinθ 0 sinθ cosθ 0 a b 1)\left( \begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ a & b & 1 \end{array} \right)

And these form a group isomorphic to the Euclidean group — the group of transformations of the plane generated by rotations and translations!

So far, everything sounds pleasantly systematic. But then things get a bit quirky:

  • Elliptic case. When s>0s \gt 0, the space X={v sv=1}X = \{v \cdot_s v = 1\} is an ellipsoid. The 1d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with XX are the points of elliptic geometry. The 2d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with XX are the lines. The group SO s(3)SO_s(3) acts on the space of points and the space of lines, preserving the obvious incidence relation.

Why not just use XX as our space of points? This would give a sphere, and we could use great circles as our lines—but then distinct lines would always intersect in two points, and two points would not determine a unique line. So we want to identify antipodal points on the sphere, and one way is to do what I’ve done.

  • Hyperbolic case. When s<0s \lt 0, the space X={v sv=1}X = \{v \cdot_s v = -1\} is a hyperboloid with two sheets. The 1d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with XX are the points of hyperbolic geometry. The 2d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with X sX_s are the lines. The group SO s(3)SO_s(3) acts on the space of points and the space of lines, preserving the obvious incidence relation.

This time XX is hyperboloid with two sheets, but my procedure identifies antipodal points, leaving us with a single sheet. That’s nice.

But the obnoxious thing is that in the hyperbolic case I took XX to be the set of points with v sv=1v \cdot_s v = -1, instead of v sv=1v \cdot_s v = 1. If I hadn’t switched the sign like that, XX would be the hyperboloid with one sheet. Maybe there’s a version of hyperbolic geometry based on the one-sheeted hyperboloid (with antipodal points identified), but nobody seems to talk about it! Have you heard about it? If not, why not?

Next:

  • Euclidean case. When s=0s = 0, the space X={v sv=1}X = \{v \cdot_s v = 1\} is a cylinder. The 1d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with XX are the lines of Euclidean geometry. The 2d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with XX are the points. The group SO s(3)SO_s(3) acts on the space of points and the space of lines, preserving their incidence relation.

Yes, any point (a,b,c)(a,b,c) on the cylinder

X 0={(a,b,c):a 2+b 2=1} X_0 = \{(a,b,c) : \; a^2 + b^2 = 1 \}

determines a line in the Euclidean plane, namely the line

ax+by+c=0 a x + b y + c = 0

and antipodal points on the cylinder determine the same line. I’ll let you figure out the rest, or tell you if you’re curious.

The problem with the Euclidean case is that points and lines are getting switched! Points are corresponding to certain 2d subspaces of 3\mathbb{R}^3, and lines are corresponding to certain 1d subspaces.

You may just tell me I just got the analogy backwards. Indeed, in elliptic geometry every point has a line orthogonal to it, and vice versa. So we can switch what counts as points and what counts as lines in that case, without causing trouble. Unfortunately, it seem for hyperbolic geometry this is not true.

There’s got to be some way to smooth things down and make them nice. I could explain my favorite option, and why it doesn’t quite work, but I shouldn’t pollute your brain with my failed ideas. At least not until you try the exact same ideas.

I’m sure someone has figured this out already, somewhere.

Posted at September 26, 2016 2:25 AM UTC

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Re: Euclidean, Hyperbolic and Elliptic Geometry

Are you sure that it’s the space of all 2d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with XX (which is all of them) that gives you the points of Euclidean geometry?

A geometric way to say “a point (a,b,c)(a,b,c) on the cylinder determines a line ax+by+c=0a x + b y + c = 0” is to take the line from the origin through that point, take the plane through the origin perpendicular to it, and then intersect that plane with the plane z=1z=1 (which we regard as our “Euclidean plane”). The unique line through the origin not intersecting the cylinder is x=y=0x=y=0, which is perpendicular to the plane z=0z=0, which doesn’t intersect the plane z=1z=1.

The corresponding thing for a 2d linear subspace (a plane through the origin) would be to take the line through the origin perpendicular to it and intersect that line with z=1z=1; then a line through the origin lies on a plane through the origin exactly when the corresponding line in z=1z=1 contains the corresponding point in z=1z=1. But there are plenty of planes through the origin whose orthogonal complement doesn’t intersect z=1z=1 at all, namely those containing z=0z=0 — yet these planes do intersect the cylinder, each in a pair of parallel lines (whereas all the other planes intersect the cylinder in an ellipse).

Of course, this is just the usual way to identify z=1z=1 with the finite points of P 2\mathbb{R}P^2.

Posted by: Mike Shulman on September 26, 2016 5:04 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Okay, so instead of

The 2d linear subspaces of 3\mathbb{R}^3 having nonempty intersection with XX are the points.

I think I should say

The 2d linear subspaces of 3\mathbb{R}^3 intersecting XX in an ellipse are the points.

Posted by: John Baez on September 26, 2016 4:37 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

I’m also confused about the hyperboloid when s<0s\lt 0. Wikipedia tells me that an equation of the form

x 2a 2+y 2b 2z 2c 2=1 \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1

is a hyperboloid of one sheet, and that seems to be what you get here (with a=b=1a=b=1 and c=1/|s|c = 1/\sqrt{|s|}. And I can visualize the limiting process better that way too: as s0 +s\to 0^+ the ellipsoid stretches off to infinity in the zz-direction becoming a cylinder, where as s0 s\to 0^- the hyperboloid gets narrower and narrower becoming a cylinder; I don’t see how to obtain a cylinder as a limiting case of a hyperboloid with two sheets.

Posted by: Mike Shulman on September 26, 2016 5:27 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Okay, I got some signs wrong there: to get the hyperboloid with two sheets, I want two minus signs in my bilinear form s\cdot_s, not just one. I’ll take the liberty of fixing that silly mistake: it doesn’t eliminate my problem, it just makes it clearer what the problem really is

Posted by: John Baez on September 26, 2016 6:24 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

I believe you have made a sign error. When s<0, I can try to solve 1 = x 2+y 2z 2x^2 + y^2 - z^2. I find a 1-sheeted hyperboloid. Indeed, I must, since this should be near your cylinder.

If you just want to get the space of points correct, you could instead use the quadratic form s(x 2+y 2)+z 2s(x^2+y^2) + z^2. The locus of norm 1 is then an ellipsoid degenerating to two planes at z=±1z=\pm 1, then becoming a 2-sheeted hyperboloid. Wrong symmetry group, though.

Posted by: Theo Johnson-Freyd on September 26, 2016 5:45 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Okay, I’ve fixed my sign error, with no attempt at doing it in a pretty way.

As you note, defining

v sw=s(v 1w 1+v 2w 2)+v 3w 3v \cdot_s w = s(v_1 w_1 + v_2 w_2) + v_3 w_3

would make

X={v sv=1} X = \{ v \cdot_s v = 1 \}

into an ellipsoid for s>0s \gt 0, two planes for s=0s = 0, and a hyperboloid with two sheets for s<0s \lt 0. In every case X/{vv}X/\{v \sim -v\} is exactly the right space of points.

All this sounds very nice, and indeed I considered it. The reason I didn’t use this is that now the group of determinant-one transformations preserving the bilinear form s\cdot_s is no longer the Euclidean group when s=0s = 0. I guess that’s what you meant by “wrong symmetry group though”.

The symmetry group contains the Euclidean group, but much more. The dimension of the symmetry group ‘jumps up’ at s=0s = 0. I could just artificially decree I don’t want all those symmetries, but this seems obnoxious to me.

Posted by: John Baez on September 26, 2016 7:22 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Your link to “Group contraction” points again to deformation theory, but I think it should point to group contraction.

Posted by: L Spice on September 26, 2016 2:35 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Thanks! Fixed!

Posted by: John Baez on September 26, 2016 4:25 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

David Speyer writes:

I seem to be locked out of comments at the nn-Café. Just a quick remark:

You’ve heard of geometry on the hyperboloid with one sheet — it’s called de Sitter space! The point is that, if you restrict your signature (2,1) metric and restrict it to the tangent space to a point on the hyperboloid with one sheet, you get a (1,1) metric, so you are doing relativity, not Riemmannian geometry.

Posted by: John Baez on September 26, 2016 4:31 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Duh, of course! Thanks. So then we’re doing Lorentzian geometry with an SO(2,1)SO(2,1) symmetry group.

Posted by: John Baez on September 26, 2016 4:33 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Maybe this is more complicated than you’re going for, but one context in which to make sense of all this is Cayley-Klein geometry. This may be some of the “fancy junk” that you know, but in case you don’t, and for other readers who don’t, I’ll describe it quickly.

The idea is that we fix a conic in P 2\mathbb{R}P^2 and define a “geometry” on its complement (at least we get a distance measurement, an angle measurement, and a symmetry group – the projective transformations fixing the conic). For different types of conic you get different types of geometries, including hyperbolic, elliptic, Euclidean, and Lorentzian.

The sneaky bit is that you have to be very general about what you mean by a “conic”. All ordinary conics in P 2\mathbb{R}P^2 are equivalent, having a homogeneous equation like x 2+y 2z 2=0x^2+y^2-z^2=0. When points of P 2\mathbb{R}P^2 are regarded as 1d subspaces of 3\mathbb{R}^3, then this conic is a double cone through the origin, and its complement is disconnected into the “inside”, which is the lines intersecting the two-sheeted hyperboloid (in a pair of antipodal points), and the “outside”, which is the lines intersecting the one-sheeted hyperboloid (also in a pair of antipodal points). The inside geometry is hyperbolic, while the outside geometry is hyperbolic along some lines (those that intersect the conic) and elliptic along others (those that don’t).

In addition to ordinary conics, however, we also allow complex conics without any real points, typified by the homogeneous equation x 2+y 2+z 2=0x^2+y^2+z^2=0, which has no solutions in P 2\mathbb{R}P^2 (since (0,0,0)(0,0,0) is not a homogeneous coordinate) but has plenty in P 2\mathbb{C}P^2. We still however consider the geometry as defined on the complement of the conic in P 2\mathbb{R}P^2, not P 2\mathbb{C}P^2, so in this case we have all the points of P 2\mathbb{R}P^2, and we get elliptic geometry.

We also allow degenerate conics; but we have to be extra sneaky about what we mean by those. A nondegenerate conic determines, and is determined by, its “dual conic” consisting of all the lines tangent to it. However, for degenerate conics this relationship breaks down, and it turns out that for the purposes of Cayley-Klein geometry we should consider “a conic” to be a pair consisting of a “point-conic” and a “line-conic” that are “related like duals”. In the nondegenerate case, each determines the other, but not in the degenerate case.

A degenerate point-conic can be a pair of lines (x 2y 2=0x^2-y^2=0) or a single “doubled” line (x 2=0x^2=0). Dually, a degenerate line-conic can be a pair of points or a single “doubled” point. And all of these could be either real or imaginary. It turns out that the pair consisting of a single real “doubled” line and two imaginary points on that line gives rise to Euclidean geometry. The space of points is the complement of one line in P 2\mathbb{R}P^2, where the missing line is of course “at infinity”. Unlike in the hyperbolic and elliptic cases, in this case (sometimes called “parabolic”) we can’t define an absolute distance, but with a limiting process we can compare relative distances. Accordingly, the transformation group preserving the line at infinity and the two fixed imaginary points on it is the group of similarity transformations of Euclidean geometry.

Posted by: Mike Shulman on September 26, 2016 7:09 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Thanks! Here’s a preliminary comment just to indicate what I’m trying to do.

I’m teaching a course on linear algebraic groups, and I want to do it from a perspective that emphasizes projective geometry and lots of examples. For example, I want to motivate Borel and parabolic subgroups using the general concept of flag, which arises naturally in incidence geometry.

So, first I need to explain a bit about Klein geometry and the Erlangen program. But to motivate that, I want to introduce the classic examples: Euclidean, hyperbolic and elliptic geometry and their ‘unification’ in projective geometry. For each kind of geometry we have a group GG, and for each type of geometrical figure in that geometry we have a subgroup HGH \subseteq G.

To do this beautifully, I wanted to present Euclidean, hyperbolic and elliptic geometry in a way that makes them look as similar as possible. I want them to all be described by the same algebraic setup, with single parameter ss that you can adjust to get the 3 different cases.

In more fancy terms, which I may not be using quite correctly, I’m hoping for a real linear algebraic group GG and a subgroup HH ‘fibered over the real affine line’, whose fibers over the point ss are groups G sG_s and H sH_s such that G s/H sG_s/H_s is the Euclidean plane, hyperbolic plane or elliptic plane depending on whether s=0s = 0, s>0s \gt 0 or s<0s \lt 0. This should secretly be an example of Grothendieck’s relative point of view. But I want to go through this very swiftly and painlessly, leaving a lot of these ideas implicit.

I will now read the rest of what you said, and see if it helps.

Posted by: John Baez on September 26, 2016 9:27 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Section 2 of http://arxiv.org/pdf/math-ph/9910041v1.pdf has a description of all nine geometries as G/H where G sits in a 2-parameter family of 3-dimensional Lie groups and H is 1-dimensional subgroup. It doesn’t really touch on the projective-space view of the geometries, though.

Posted by: Layra on September 26, 2016 10:33 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Thanks! I know how the G/HG/H description works individually for each of the three geometries I’m interested in; the issue is treating all three in one blow by treating GG and HH as depending on a parameter ss. Ideally I want GG and HH to be affine algebraic varieties ‘fibered over the real affine line’, meaning that GG and HH depend in a purely algebraic way on the parameter ss.

I think I may see how to do this using the trick Theo mentioned, and working ‘for all values of ss at once’ instead of ‘one value at a time’, to avoid problems caused by how the dimension of the group preserving a quadratic form jumps up when that form becomes degenerate. Whether I can present it in a slick polished way by 11 am Tuesday is another matter.

Posted by: John Baez on September 26, 2016 11:06 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Oh! So the paper you referred to really seems to treat GG and HH as depending on a parameter in the way I seek… except that it uses two parameters instead of one. They define a group G=SO κ 1,κ 2(3)G = SO_{\kappa_1,\kappa_2}(3) where the two parameters describe the quadratic form, and define two kinds of lines whose stabilizers are H 1=SO κ 1(2)H_1 = SO_{\kappa_1}(2) and H 2=SO κ 2(2)H_2 = SO_{\kappa_2}(2), respectively. Interesting:

Posted by: John Baez on September 26, 2016 11:11 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Yeah, the two parameters roughly are curvature and signature, so that we get the full set of CK geometries. This allows us to also do stuff with duality; elliptic space is self-dual, but Euclidean and hyperbolic are not, so we can examine their duals. Duality swaps the two parameters.

Posted by: Layra on September 27, 2016 3:29 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Here is a guess about how this point of view might help. Consider the pair consisting of both bilinear forms, the one that has the right points and the one that has the right symmetry group, with one acting on 3\mathbb{R}^3 and one on ( 3) *(\mathbb{R}^3)^\ast:

x sy=sx 1y 1+sx 2y 2+x 3y 3 x \cdot_s y = s x_1 y_1 + s x_2 y_2 + x_3 y_3

a sb=a 1b 1+a 2b 2+sa 3b 3 a \cdot^s b = a_1 b_1 + a_2 b_2 + s a_3 b_3

A linear transformation LL of 3\mathbb{R}^3 induces a transposed linear transformation L L^\top of ( 3) *(\mathbb{R}^3)^\ast; let SO s(3)SO_s(3) be the group of determinant-1 transformations such that LL preserves s\cdot_s and L L^\top preserves s\cdot^s. Then we can define the points to be the 1d subspaces of 3\mathbb{R}^3 having nonempty intersection with x sx=1x \cdot_s x = 1, and the lines to be the 2d subspaces of 3\mathbb{R}^3 that can be written as a(x)=0a(x) = 0 for some a 3a\in \mathbb{R}^3 with a sa=1a\cdot^s a = 1. I haven’t checked all the details, but the “sneaky notion of degenerate conic” approach suggests to me that something like this ought to work.

Posted by: Mike Shulman on September 26, 2016 11:37 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Oops, I meant

the lines to be the 2d subspaces of 3\mathbb{R}^3 that can be written as a(x)=0a(x) = 0 for some a( 3) *a\in (\mathbb{R}^3)^\ast with a sa=1a\cdot^s a = 1.

Posted by: Mike Shulman on September 26, 2016 11:38 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

This sounds promising to me. Indeed in the Euclidean case it struck me that transformations of the form

(cosθ sinθ 0 sinθ cosθ 0 a b 1)\left( \begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ a & b & 1 \end{array} \right)

preserve the quadratic form x 2+y 2x^2 + y^2, and thus preserve the cylinder x 2+y 2=1x^2 + y^2 = 1 in 3\mathbb{R}^3 that has a 2-1 map onto the space of lines in Euclidean plane geometry, while their transposes

(cosθ sinθ a sinθ cosθ b 0 0 1)\left( \begin{array}{ccc} \cos \theta & \sin \theta & a \\ -\sin \theta & \cos \theta & b \\ 0 & 0 & 1 \end{array} \right)

preserve the quadratic form z 2z^2, and thus preserve the surface z 2=1z^2 = 1 that has a 2-1 map onto the space of points in Euclidean plane geometry. The first kind of matrix describes how the Euclidean group acts on lines, while the second describes how it acts on points.

Posted by: John Baez on September 27, 2016 7:37 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

David Speyer wrote:

(I’m still locked out of comments)

You may have already figured this out, but here is Mike Shulman’s remark in explicit matrices:

If QQ is a symmetric matrix, then the group of transformations that preserve the bilinear form QQ is

{L:LQL T=Q} \{L : L Q L^T=Q \}

If QQ is invertible, this equation is equivalent to

L TQ 1L=Q 1. L^T Q^{-1} L = Q^{-1}.

(Invert both sides to get L TQ 1L 1=Q 1L^{-T} Q^{-1} L^{-1} = Q^{-1}, and then move the LL’s over.)

Now, consider the family of matrices Q(s)=diag(1,1,s)Q(s) = diag(1,1,s). For ss nonzero, the equations

Ldiag(1,1,s)L T=diag(1,1,s)L diag(1,1,s) L^T = diag(1,1,s)

and

L Tdiag(1,1,s 1)L=diag(1,1,s 1) L^T diag(1,1,s^{-1}) L = diag(1,1,s^{-1})

are equivalent. Clear out denominators in the second equation to rewrite this as

Ldiag(1,1,s)L T=diag(1,1,s) L\, \diag(1,1,s) L^T = \diag(1,1,s)

and

L Tdiag(s,s,1)L=diag(s,s,1). L^T \diag(s,s,1) L = \diag(s,s,1).

For ss nonzero, this is two equivalent equations. When s=0s=0, the limits are

Ldiag(1,1,0)L T=diag(1,1,0) L \,\diag(1,1,0) L^T = \diag(1,1,0) and

L Tdiag(0,0,1)L=diag(0,0,1), L^T \diag(0,0,1) L = \diag(0,0,1),

which are no longer equivalent.

The plane symmetry group is the matrices with det=1det = 1 obeying both these equations.

Posted by: John Baez on September 30, 2016 7:11 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

Somehow David’s matrix-ridden approach makes it very clear to me what’s going on — probably because I’ve spent some time doing calculations with such matrices. I’m pretty happy now!

Posted by: John Baez on October 1, 2016 1:02 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

David Speyer wrote:

This suggests that there is a more degenerate form of geometry: Consider the limiting family

Ldiag(1,s,s 2)L T=diag(1,s,s 2)L\, \diag(1,s,s^2) L^T = \diag(1,s,s^2)

to get

Ldiag(1,0,0)L T=diag(0,0,1)L \,\diag(1,0,0) L^T = \diag(0,0,1)

and

L Tdiag(0,0,1)L=diag(0,0,1). L^T \diag (0,0,1) L = \diag(0,0,1) .

Indeed, combined with the determinant 1 condition, I believe this is nothing other than the Heisenberg group

(1 * * 0 1 * 0 0 1) \left(\begin{array}{ccc} 1 &* &\ast \\ 0 &1 &\ast \\ 0 &0 &1 \end{array} \right)

I’d always wondered if there was a conceptual explanation for why maximal compact subgroups of split Lie groups have the same dimension as the unipotent radical of their Borels. Maybe there is a general story like this, where there is some family linking together the different forms that include the maximal compact form, with the unipotent radicals in degenerate cases.

Posted by: John Baez on October 1, 2016 1:04 AM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

I don’t have time to think about it myself right now, but is section 3.2 of this paper possibly relevant?

Posted by: Mark Meckes on September 27, 2016 2:47 PM | Permalink | Reply to this

Re: Euclidean, Hyperbolic and Elliptic Geometry

A quick point: The group of matrices preserving a singly-degenerate quadratic form on 3\mathbb{R}^3 is four-dimensional, containing not only the xzxz-shears, the yzyz-shears, and the xyxy-rotations, but also the zz-scaling.

The group preserving a doubly-degenerate quadratic form is six-dimensional, containing the expected zxzx- and zyzy-shears, as well as the xyxy- and yxyx-shears, the xx-scaling, and the yy-scaling.

These symmetries can be mapped to the other model, where they all affect the metric. However, scaling in the zz-direction is projectively equivalent to isotropic scaling in the xyxy-plane, so that particular combination does not affect the metric.

Posted by: Eyal Minsky-Fenick on November 7, 2016 6:49 PM | Permalink | Reply to this

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