Let $k$ be a field. A space with functions over $k$ is topological space X together with a family $O_X$ of k-subalgebras $O_X(U)\subseteq Map(U,k)$ for every open set $U$ that satisfy

a) If $U\subseteq V$ are two open sets and $f\in O_X(V)$, then $f$ restricted to $U$ is in $O_X(U)$

b) (Axiom of gluing) Given open sets $U_i \subseteq X$ for $i\in I$ and $f_i \in O_X(U_i)$ for $i\in I$ with $$f\restriction_{U_i\bigcap U_j}=f_j\restriction_{U_i\bigcap U_j}$$ for any $i,j \in I$, then there exists a unique function $f: \bigcup _i U_i \to k$ with $f\restriction_{U_i}=f_i$ for any $i\in I$ and $f\in O_X(\bigcup _i U_i)$.

Since I had a nice introduction in commutative algebra, I worked through the first chapter of Wedhorn and Görtz's book about algebraic geometry. Unfortunately I haven't any descent knowledge in geometry, so I can't really imagine what's really about the notion of spaces of functions. I read in the book, that the notion is motivated by the notion of a manifold and the realed-valued continous maps on its opens sets. I suppose with the k-valued maps we can analyse the geometric structure of the topological space, although I can't really imagine how that really should work.

Can you give me some examples, especially from "classical" algbraic geometry (i.e. where X is a algebraic variety)?

new (more precisely) questions:

For a geometric object (such as a manifold or variety), can be the properties of the functions on the open sets interpreted as geometric properties of the object and vice versa?

Are geometric properties of a object even formulated in the terms of functions on the open sets?

Maybe a example for the first question, which would also fit (in some way) the idea in the second question: A point on algebraic variety is non-singular iff the localization of affine coordinate ring at the corresponding maximal ideal of the point is regular.

  • 2
    $\begingroup$ The definition you give is of a "space with functions", which is more commonly called a "sheaf of functions on a space". This is different from a "space of functions", in which the points of the space are themselves functions. $\endgroup$ – Alex Kruckman Jul 3 '14 at 23:09
  • $\begingroup$ Do you know how to evaluate an element of a commutative ring in a prime ideal. This is the essence in considering arbitrary rings as function spaces. For the classical case, it is exactly what you have said, you just pick the function on the variety exactly as in the case of smooth functions or holomorphic functions. The motivation is that there exists an anti equivalence between commutative rings and affine schemes. So looking at the sheaf of regular functions is the same as looking to the space. You can try making this correspondence for continuous functions on a compact spaces. $\endgroup$ – user40276 Jul 4 '14 at 1:38
  • $\begingroup$ I fixed the wrong name of the definition. Yes, I have some (but only very little) knowledge about "abstract" algebraic geometry (e.g. For a ring $R$, $Spec(R)$ will be the topological space...) But my question is much more elementary. How is the geometric structure of a object (e.g. manifold or algebraic variety) encoded on the functions on its open sets? For example, how can I see just by looking at the functions on the open sets, if the object has a genus, many edges, a "peak" point etc.? Can you give me some examples of such objects and the connection to the functions on the open sets? $\endgroup$ – bjn Jul 4 '14 at 22:51

Yes, geometric properties of objects can be captured in the sheaves of functions on them. For example, consider Castelnuovo's criterion, which gives a condition for when a curve on a smooth algebraic surface to be "contracted" to a point, to produce a new smooth surface. The conditions are purely geometric, but if you look at the proof given in Ch. V of Hartshorne's book, the method involves a detailed consideration of the sheaf of functions, and sheaves of modules over it.

Some local properties are easy to encode in this way, e.g. as you note (non-)singularity at a point can be detected by algebraic properties of the associated local ring.

But the encoding of more global properties (e.g. intersection theory) is more subtle. Cohomology typically plays an important role.

It's a little like topology: if you look at the definition of a topological space for the first time, you might quite legitimately doubt that this could be an appropriate language for exploring a concrete problem such as whether a disk can be retracted onto its boundary circle. And indeed, it takes quite a lot of effort to build up the tools necessary for proving such a statement using the language of topological spaces. Similarly, it takes some effort in the foundations of algebraic geometry to build up enough language and tools to express geometric properties in the language of sheaves of rings and modules. This is the goal of Hartshorne's text: the foundations of Ch. II and III serve to establish the necessary tools for investigating the geometry of Ch. IV and V.

  • $\begingroup$ Can you give an example and counterexample (in classical algbraic geometry) of such curve on a smooth algebraic surface, which can be contracted to point to produce a new smooth surface. I can't figure this geometrically out what such point should look like. $\endgroup$ – bjn Jul 24 '14 at 19:44
  • $\begingroup$ @bjn: Consider the case of smooth cubic surfaces in $\mathbb P^3$. These have $27$ lines, each of which is a $\mathbb P^1$ with self-intersection $-1$. You can always find six lines that are disjoint, and then these can be blown down to product a $\mathbb P^2$; in other words, each smooth cubic surface is the blow-up of $\mathbb P^2$ at six points. $\endgroup$ – user160609 Jul 24 '14 at 20:01

This is a special case of the notion of a ringed space, which should be seen as an abstract framework for doing geometry. It is motivated by many specific examples. Any topological space has its ring of continuous functions (say, real-valued), actually for every open subset. One can easily check that a),b) are satisfied, namely continuous functions can be glued. Another example: Any smooth manifold is not just a topological space, but also comes equipped with its rings smooth of smooth functions on its open subsets. Again it is easy to check that a),b) hold. Now, if we take a variety in the sense of classical algebraic geometry, then we have the notion of a regular function. Locally these are fractions of polynomial functions. Because their definition is local, it is again trivial to check a),b).

Here is an example: Consider the projective line $\mathbb{P}^1$ (in variables $x,y$) with the two standard open subsets $\{x \neq 0\}$ and $\{y \neq 0\}$. These are copies of the affine line. Their rings of regular functions are $k[\frac{y}{x}]$ and $k[\frac{x}{y}]$. We have regular functions $y/x$ on $\{x \neq 0\}$ and $x/y$ on $\{y \neq 0\}$. These don't coincide on the intersection, hence don't glue a regular function on $\mathbb{P}^1$. In fact, every regular function on $\mathbb{P}^1$ is constant. This is because $k[\frac{y}{x}] \cap k[\frac{x}{y}] = k$ with intersection taken in the function field $k(\frac{x}{y})$ on $\mathbb{P}^1$.

  • $\begingroup$ Thanks for your answer. Can be the property "every regular function on the projective line is constant" interpreted geometrically? What does this say about the projective line? For more general questions in the same spirit, see my opening post. $\endgroup$ – bjn Jul 17 '14 at 20:42

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