1
$\begingroup$

Consider the inclusion $j$ of ${\mathbb{R}}$ as the real axis of ${\mathbb{C}}$. On ${\mathbb{C}}$ I have a real polynomial algebra ${\mathbb{R}}[x,\bar{x}]$, where $\bar{x}$ denotes conjugation. Restricting to real values for $x$, this algebra collapses to the usual real polynomial algebra ${\mathbb{R}}[x]$ on ${\mathbb{R}}$. (This is the reason why I consider real algebra ${\mathbb{R}}[x,\bar{x}]$ instead of complex algebra ${\mathbb{C}}[x,\bar{x}]$.

Now I'm trying to be generalize this to the inclusion of a real variety to its corresponding complex variety defined by the same equations. The first step to do this is to realize the above construction in terms of direct images and inverse images of sheaves. That means given the real polynomial algebra ${\mathbb{R}}[x]$, I wish to express ${\mathbb{R}}[x,\bar{x}]$ via some direct image or inverse image operations. I've tried taking the real part $\mathrm{Re}:{\mathbb{C}}\to{\mathbb{R}}$ too, but it doesn't work.

Any suggestions? Or am I trying to force something unnatural?

$\endgroup$
2
  • $\begingroup$ I don't understand what ${\mathbb{R}}[x,\bar{x}]$ means. $\endgroup$ Commented Sep 23, 2010 at 5:36
  • $\begingroup$ I meant polynomials in two variables $x$ and $\bar{x}$. This real algebra has an involution which takes the polynomial $x+\bar{x}^2$ to $\bar{x}+x^2$, for example. $\endgroup$
    – user2529
    Commented Sep 23, 2010 at 7:36

1 Answer 1

3
$\begingroup$

I'm going to make a bold guess about what you are doing. You want to view $\mathbb{C}$ as the real points of two dimensional affine space over $\mathbb{R}$, and its ring of functions can be seen as $\mathbb{R}[u,v]$, where $u$ is the coordinate along the real axis, and $v$ is the coordinate along the imaginary axis. The real line is then the real points of $\operatorname{Spec} \mathbb{R}[u]$, and complex conjugation on $\mathbb{C}$ is the automorphism that fixes $u$ and takes $v$ to $-v$. We can make new variables $x = u+v$ and $\bar{x} = u-v$, so that complex conjugation switches $x$ with $\bar{x}$. The real line in these new coordinates is given by the real points of $\operatorname{Spec} \mathbb{R}[x+\bar{x}]$.

I don't think the operation you want to do has much relation with sheaf operations, at least in the coherent sense. However, there is a way to view it in terms of sheaves of sets. If you want to express $\mathbb{R}[x,\bar{x}]$ using $\mathbb{R}[x]$, you can base change the real line to the complex numbers to get $\mathbb{C}[u]$ (i.e., pull back along the map $\operatorname{Spec} \mathbb{C} \to \operatorname{Spec} \mathbb{R}$), then do a Weil restriction (i.e., push forward) along the same map. Since the two operations are adjoint, you get a canonical "unit" map that embeds the real axis into the complex numbers.

$\endgroup$
1
  • $\begingroup$ Originally I had ${\mathbb{C}}$ as a complex manfold and $\bar{{\mathbb{C}}}$ as its conjugate manifold. (That is the transition functions are now antiholomorphic). Then the product manifold ${\mathbb{C}}\times \bar{\mathbb{C}}$ has a sheaf of rings formed ${\mathbb{C}}[x]\otimes \mathbb{C}[\bar{y}]$ by tensoring the original two coordinate sheaves. Restricting along the diagonal $(x,\bar{x})$ gives the ${\mathbb{C}}[x,\bar{x}]$ $\endgroup$
    – user2529
    Commented Sep 23, 2010 at 7:43

You must log in to answer this question.