# Is there an underlying topological space for ind-schemes?

An ind-scheme over a base scheme $$S$$ can be defined in several ways. For simplicity, lets assume that $$S$$ is the spectrum of an algebraically closed field $$k$$. We can define a $$k$$-ind-scheme as a functor from $$k$$-algebras to sets, such that there exists a sequence of $$k$$-schemes with closed embeddings

$$X_1 \to X_2 \to ...\to X_n \to ...$$

and a natural isomorphism of $$X$$ with the functor $$R \mapsto \lim_{\rightarrow}(X_n(R))$$. Such a functor is a sheaf in the Zariski topology and can be uniquely extended as a sheaf to operate on all $$k$$-schemes.

Is there a sensible way to define the "underlying topological space" of an ind-scheme $$X$$? In other words, is it true that the direct limit of the topological spaces underlying the $$X_n$$ (an increasing union really) is independent of the presentation?

If we can take all $$X_n$$ to be quasi-compact, it seems to me that the answer is positive. since for different (quasi-compact) presentations, we will have closed embeddings $$f_n:X_n\to X'_{m(n)}$$. Am I right?

### Edit:

Even though I wrote "any sensible way to define the underlying topological space", I now realize that I am actually only interested in the "direct limit" topology. The main question is the independence of presentation in the general or some special cases.

## 1 Answer

The underlying set of a functor $X : \mathsf{Alg}(k) \to \mathsf{Set}$ is given by $\mathrm{colim}_{K/k} X(K)$, where $K/k$ runs through all field extensions. The topology is defined using open subfunctors. A reference is the book by Demazure and Gabriel on algebraic groups. I have learned it from a script by Marc Nieper-Wißkirchen. I'm not sure what happens for Ind-schemes, though.

• Nice answer. It gives the topology introduced by Kambayashi, which is not the same as the one given by the inductive limit (see the differences here arxiv.org/abs/1109.4088 ) – Jérémy Blanc Sep 18 '13 at 14:00