3
$\begingroup$

In Jacobson's Basic Algebra I, in Kaplansky's Linear algebra and geometry and in Artin's Geometric algebra, a hyperbolic plane is defined to be a two-dimensional, nondegenerate inner product space with nonzero null vectors. (I also noticed that Lang went out of his way to define these for alternating forms only, and if I remember right, one of the three other authors did it for symmetric forms only, but I'm not sure at the moment. This may be beside the point.)

When I first heard of these, I thought "oh, there must be a model of synthetic hyperbolic geometry in there somewhere." I thought this was a good guess because, after all, if you fix a point in the synthetic Euclidean plane and look at the transformations that fix it, it looks like the geometry of the regular dot product on $\Bbb R^2$, and the distance even matches the one extracted from the dot product.

But now I'm not so sure the analogous hyperbolic picture is so direct. It's hard to see why the "hyperbolic inner product plane" defined above would be the same as what is going on around a fixed point in hyperbolic geometry. In fact, I have a friend trying to convince me that the two are unrelated. One of his reasons is "in hyperbolic geometry, there aren't two distinguished null lines through points like there are in the inner product space." Finally, it's totally unclear (to me) if there is any connection between the inner product and the distance in hyperbolic geometry.

The fact that both orthogonal and symplectic geometries can have hyperbolic planes adds another layer of complexity that I don't know what to do with.

I'm anticipating one of these types of answers:

Of course there's a useful model of hyperbolic geometry in the hyperbolic plane. You just have to take this subset with these lines and here's how the inner product matches up to the distance like so...

or

Yeah, they're related, you just need to think of hyperbolic geometry and Minkowskian geometry this way...

or

No, they're not really directly related. You should just just think of the inner product version separately as Minkowskian geometry...

Of course, I may not be anticipating the best answer possible: I hope to be surprised! :)


Note: I am aware that you can take $\Bbb R^3$ with a metric signature $++-$ and view it projectively to make a model of the synthetic hyperbolic geometry complete with a distance function, but that is not the direction of this question.

I also considered taking the subset of the hyperbolic space consisting of just the points with positive (or just negative) lengths. I wasn't able to convince myself that it was "the right picture," but maybe someone else can convince me.

$\endgroup$
3
$\begingroup$

In algebraic books, especially those dealing with quadratic forms, there is a usage in English for "hyperbolic plane" that has nothing to do with non-Euclidean Geometry. From Rational Quadratic Forms by Cassels, it means a 2-dimensional quadratic space $U,\phi$ with a basis $u_1,u_2$ such that $$ \phi(u_1) = \phi(u_2) =0, \; \; \; \phi(u_1, u_2) = 1. $$ This is a natural definition, given Witt Cancellation and the decomposition, theorem 18 in Kaplansky's little book, of a quadratic space into a sum of hyperbolic planes and a non-isotropic subspace.

The English usage follows Ernst Witt, Theorie der quadratischen Formen in beliebign Körpern, J. reine angew. Math., vol. 176, pages 31-44, (1936). I do not know the German term Witt used...I found the article online, if I find the correct phrase I will edit that here.

EDIT: it appears that Witt gave no name to such things. I do not see an obvious use of the word isotropic either, so we have a standard terminology problem, we know where the concepts began and in what language, we know what they are now called in English, what are the intermediate steps?

$\endgroup$
2
$\begingroup$

I think that this is just a matter of a strange dimension shift when comparing the terminology of the algebraic authors with the terminology of the geometry authors.

Although you say that your "Note" is not the direction of your questions, nonetheless it is regarded as being the answer to your question. You can see this discussed very explicitly as the "hyperboloid model of hyperbolic geometry" in for example, Thurston's book The Geometry and Topology of 3-manifolds, and in the later published work Three-dimensional Geometry and Topology. This model of the hyperbolic plane need not be thought of projectively. It can instead be thought of as the "sphere of radius $i$", namely the hyperboloid $x^2 + y^2 - t^2 = -1$ ($t > 0$). By restricting the inner product to the tangent spaces of that hyperboloid, you get a Riemannian metric on the hyperboloid which is an exact model of the hyperbolic plane $\mathbb{H}^2$.

The 2-dimensional "hyperbolic plane" discussed in your first paragraph, with signature $(+1,-1)$ can be similarly used to obtain a model of the hyperbolic line $\mathbb{H}^1$, by restricting the inner product to the hyperbola $x^2 - t^2 = -1$ ($t>1$). Of course, $\mathbb{H}^1$ is not particularly special, being isometric to $\mathbb{R}$.

$\endgroup$
  • $\begingroup$ As for the second paragraph, it is not the answer to my question although it's the best answer to a similar question. I'm asking very narrowly if there is a relationship between the 2-d space and the hyperbolic plane, and modifications to that are dodging the narrow question. In recent conversations I've had, it seems the correct answer to my question is "no such direct relationship." $\endgroup$ – rschwieb Aug 23 '14 at 14:35
  • $\begingroup$ @rschwieb I will second the final sentence of your comment as the correct answer, and will point to my third paragraph as providing the underlying reason behind that answer: it is related to $\mathbb{H}^1$, not to $\mathbb{H}^2$. In general I like to write my answers in a wide context even when they dodge the narrow question. $\endgroup$ – Lee Mosher Aug 23 '14 at 14:40
  • $\begingroup$ I'm deliberately omitting the projective versions because I already understand them well, and i don't think the term originates there. Will Jagy mentioned it might have come down to us through German papers. And sure, I understand your reasons for including it in the solution. Certainly I don't disagree with anything else you said, besides my one objection :) $\endgroup$ – rschwieb Aug 23 '14 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.