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Real algebraic geometry, at least to start with, traditionally studies the zero-sets of real polynomials in a given set of variables. But treating, say, the Euclidean plane as an uncoordinatized metric space, one may still consider the ring of functions generated by all functions $D_p(\cdot)=d(\cdot,p)^2$, and also constant functions. Then one has really lost nothing, because (bring back coordinates $x,y$ just for the moment), $x=(D_{(0,0)}-D_{(1,0)} +1)/2$ and $y=(D_{(0,0)}-D_{(0,1)} +1)/2$.

Since the ring definition here makes sense over any metric space, the possibility of a generalized real algebraic geometry arises.

Questions: Does this line of thought occur in the literature? Independently of the geometric optic, does this sort of ring arise anywhere in the literature (other than in the motivating example)? Do complexification and projectivization have well-behaved analogues in this context?

I'm particularly interested in the properties of "real algebraic curves" in the hyperbolic plane. I have narrower questions, but I'll save them until I see response to this.

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It seems that for "most" metric spaces (and hyperbolic space is one of the "most") distance functions (or their squares) will span an infinite-dimensional real vector space (most likely, of uncountable dimension). Thus, doing algebraic geometry in this setting will be somewhat difficult. Note that the map $i: x\mapsto d_x$, where $d_x(y)=d(x,y)$, determines an isometric embedding of any metric space $X$ to $L_\infty(X)$, so you should not expect image of $i$ to be contained in a finite-dimensional affine subspace. On the other hand, hyperbolic plane is a 1-sheeted hyperboloid in $R^3$... –  Misha Jun 2 '12 at 5:03
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... so you have a natural supply of linear functions. For instance, if $x, y$ belong to the upper sheet of the hyperboloid, then their Lorentzian inner product $x\cdot y$ equals $-cosh d(x,y)$, where $d$ is the hyperbolic distance. Thus, provided you replace your square with $-cosh$, you get linear functions $-cosh d(x, )$ on ${\mathbb H}^2$. But then you just get "boring" traditional algebraic geometry. –  Misha Jun 2 '12 at 5:08
    
@Misha: You comments is really an answer. I think that you should rewrite it as an answer. –  André Henriques Jun 17 '12 at 19:54
    
I suppose that I don't understand "Thus, doing algebraic geometry in this setting will be somewhat difficult." Why not "different" instead of "difficult"? (Of course I always find mathematics difficult.) Relations make algebraic objects smaller, but not necessarily more approachable. –  David Feldman Jun 17 '12 at 22:54
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up vote 5 down vote accepted

Upon Andre's request, I am rewriting my comments as an answer, even though I am on a somewhat shaky ground since my experience with infinite-dimensional algebraic geometry is very limited.

It seems that for "most" metric spaces (and hyperbolic space is one of the "most") distance functions (or their squares) will span an infinite-dimensional real vector space. I checked this in the case of distance functions on the hyperbolic space (of dimension $\ge 2$, of course) and the same should hold for the distances squared. The point is that for every metric space $X$, the map $i=i_X: x\mapsto d_x$, where $d_x: X\to {\mathbb R}$ is the function $d_x(y)=d(x,y)$, determines an isometric embedding from $X$ to a linear subspace $V$ in $C(X)$ (space of continuous functions on $X$) with the sup-norm on $V$, where $V$ is the linear span of the image of $i$. On the other hand, hyperbolic space (of dimension $\ge 2$) does not embed isometrically in any finite-dimensional Banach space. Thus, doing algebraic geometry on the hyperbolic space becomes (in my mind) a somewhat daunting task since you have to consider the ring of polynomial functions ${\mathbb R}[V]$ on the infinite-dimensional vector space $V$. In particular, you loose the Noetherian property which makes life difficult. The only context where I have seen infinite-dimensional algebraic geometry is the ind-schemes. Ind-schemes appear naturally when one works with, say, affine Grassmannians and which I had to do exactly once in my life (I mean, thinking of affine buildings in algebro-geometric terms). As far as I can tell, dealing with the ind-scheme based on the ${\mathbb R}[V]$ for the hyperbolic space, would amount to considering geometry of finite configurations of points in ${\mathbb H}^n$ (and, occasionally, lines). I could be mistaken, but for the union $Y$ of two distinct geodesic segments in the hyperbolic space, linear span of the image of $i_Y$ (where we use the restriction of the distance function from ${\mathbb H}^n$) will be infinite-dimensional, so you are not allowed to use more than one geodesic. While geometry of finite subsets of hyperbolic spaces has some uses (see the discussion of the sets $K_m$ below), it strikes me (I am a hyperbolic geometer) as mostly boring. (Maybe logicians can add something interesting here since considering finite subsets in ${\mathbb H}^n$ we are dealing with the elementary theory of the hyperbolic space.)

Gromov (see e.g. his book "Metric Structures for Riemannian and Non-Riemannian Spaces") introduces, for every metric space $X$, the collection of subsets $K_m(X)\subset {\mathbb R}^N$, $N=\frac{m(m-1)}{2}$. The set $K_m(X)$ consists of $N$-tuples of pairwise distances between various $m$-tuples of points in $X$. Then $K_3(X)$ (under some mild assumptions on $X$, e.g., $X$ is an unbounded geodesic metric space) is defined by triangle inequalities and nothing else. The set $K_4(X)$ is quite interesting, since all the "curvature" conditions on metric spaces are defined via quadruples of points. However, Gromov could not come up with any interesting uses for $K_m(X), m\ge 5$, and I do not know what to make of these sets either. Thus, the ind-scheme approach to the algebraic geometry of the hyperbolic space might yield something geometrically interesting for quadruples of points, beyond which algebraic geometry is likely to get disconnected from (geo)metric geometry.

Lastly, MO discussion at Is there an algebraic approach to metric spaces? is related to David's question.

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interesting, this works just as well for spheres, and even $S^1$. –  Will Sawin Jun 23 '12 at 4:57
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