smooth manifolds as real algebraic set (continued)

Question

There are several ways of producing manifolds,say:

1.orbits space of group action

2.connected sum of manifolds

3.underlying topological space of nonsingular algebraic set

....

here,i am interested in the 3rd one.

A well known theorem due to Nash and Tognoli says that Every compact smooth manifold is diffeomorphic to a nonsingular real algebraic set.

My question is: Given a manifold $M$,How to find an algebraic set EXPLICITLY whose underlying topological space is diffeomorphic to $M$?

For example, can we find a specific real algebraic set corresponding to an orientable riemann surface of genus $g$?

---

Thanks for the very useful answers and comments.

I am a student interested in the topology of manifolds and just came to realize so many manifolds could come from algebraic varieties,so the topology of algebraic varieties would be an interesting topic.

A naive idea could be: The topology of algebraic varieties should be determined by the algebra,i.e. the polynomials which generate the ideal corresponding to the variety.

In practice,i guess,it would be quite a hard problem to read topology from algebra,and in many cases,the other direction is also of interest,i.e.to get algebraic information of the variety by studying the topology of the underlying space.

In the very interesting survey on the topology of real and algebraic varieties given by Janos Kollar

https://web.math.princeton.edu/~kollar/FromMyHomePage/tanig.ps

He calls these two directions as Realization problem and Recognition Problem respectively.

To be more precise,realization problem studies which topological space could be realized as the underlying space of a projective algebraic variety,and recognition problem studies which algebraic properties are determined by its underlying topological space.

These two problems clearly summaries the interplay of algebra and topology in the algebraic variety.

Now my question is:

What's the status of the realization problem,i.e. How much topology could be read from the algebra and how?

Answer 1

It isn't too easy to find good equations for surfaces as algebraic subspaces of $\mathbb{R}^3$, but if you are willing to use $\mathbb{R}^n$ for larger $n$ then the picture is clearer. There are standard ways in algebraic geometry to produce surfaces as complex subvarieties $X\subset\mathbb{C}P^2$. If we let $U$ denote the space of $3\times 3$ hermitian matrices over $\mathbb{C}$ then we can define an embedding $j:\mathbb{C}P^2\to U$ by $$j([z])_{pq}=z_p\overline{z_q}/\sum_r|z_r|^2$$ (Equivalently, this sends a line $L<\mathbb{C}^3$ to the orthogonal projection onto $L$.) It is not hard to see that $j(X)$ is a real algebraic set in $U\simeq\mathbb{R}^9$. This generalises in an obvious way to express any complex projective variety as a real algebraic set.

Answer 2

Speaking of compact orientable survaces, this is very simple, indeed: hyperelliptic curves in $CP^2$ given by equations $w^2=P_n(z)$, where $P$ is a rational function with $2n$ zeros and poles, all zeros and poles distinct. of degree $n$, are smooth embedded surfaces with Euler characteristic $4-2n$.

EDIT. A gap in the previous argument was pointed in the comment of BS: only for genera of the form $g=(d-1)(d-2)/2$ a smooth curve exist in $P^2$, and hyperelliptic curves usually have singular points. But they can be embedded to other spaces.

EDIT 2. Here is a way to construct a smooth compact orientable algebraic surface in $R^3$ of any genus. $(x,y,t)$ are the coordinates in $R^3$. $$t^2=(R^2-x^2-y^2)\prod_{j=1}^n((x-x_j)^2+(y-y_j)^2-r^2),$$ where $R$ is large and the discs $(x-x_j)^2+(y-y_j)^2<r^2$ are disjoint and contained in the disc of radius $R$ centered at the origin.

Answer 3

Check out Jordan and Steiner, Compact surfaces as configuration spaces of mechanical linkages. (available for free, linked provided for convenience).