The Apollonius-Toën circles

Interlude I: Serre’s formula (1957)

Example 3.1

Consider the intersection of the axis in 4-dimensions \(R=\mathbb {C}[x,y,z,w]\), with the diagonal

\[ \textcolor{red}{\text{Axis}:=\begin{cases} xz=0\\ xw=0\\ yz=0\\ yw=0\end{cases}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \textcolor{blue}{\text{Diag}:=\begin{cases} x-z=0\\ y-w=0\end{cases}} \]

Geometrically, we find an intersection of multiplicity 2, as seen through the following picture:

\includegraphics[scale=0.2]{Images/all8circles/all8circles-0.png}

Any small perturbation of the diagonal leads to two intersection points.

Now, the problem is that the algebraic computation, leads to a different number. Indeed, by computing the tensor product we obtain

\[ \textcolor{red}{R/(xz,xw,yz,yw)}\otimes _{R} \textcolor{blue}{R/(x-z, y-w)}\simeq \mathbb {C}[x,y]/(x^2, xy,y^2) \]

which as a \(\mathbb {C}\)-vector space, has dimension 3

\[ = \underbrace{\mathbb {C}\oplus \mathbb {C}.\epsilon _x \oplus \mathbb {C}.\epsilon _y}_{\textcolor{black}{3}\neq \textcolor{purple}{2}}\, \, \, \, \, \, \text{too many!} \]
Explanation 3.8

The reason for the different results lies in redundancies, in the following sense: take the following function \(f\)

\[ \textcolor{teal}{f:=xw-yz=w(x-z)-z(y-w)} \]

and observe that it vanishes for two reasons, because it simultaneously belongs to the two ideals (of the diagonal and the axis)

\[ \textcolor{teal}{f}\in \textcolor{blue}{\underbrace{(x-z,y-w)}_{I_{Diag}}}\cap \textcolor{red}{\underbrace{(xz,xz,yz,yw)}_{I_{Axis}}} \]

but the two reasons are different, in the sense that \(f\) does not live in the product ideal

\[ \textcolor{teal}{f}\notin \textcolor{blue}{I_{Diag}}.\textcolor{red}{I_{Axis}} \]

In fact, if we compute the quotient we find an isomorphism

\[ \frac{\textcolor{blue}{I_{Diag}}\cap \textcolor{red}{I_{Axis}}}{\textcolor{blue}{I_{Diag}}.\textcolor{red}{I_{Axis}}} \simeq \mathbb {C}.[\textcolor{teal}{f}] \]

showing us that \(f\) is in fact very special: it is the only function that vanishes non-trivially for two different reasons. Now, when we computed the tensor product

\[ \textcolor{red}{R/(xz,xw,yz,yw)}\otimes _{R} \textcolor{blue}{R/(x-z, y-w)}\simeq \mathbb {C}[x,y]/(x^2, xy,y^2) \]

algebraically solving the intersection, we discarded such ambiguous outcomes. In order to correct the counting we need to account for such phenomena. More precisely:

\[ \text{Correct counting }=\underbrace{\mathbb {C}\oplus \mathbb {C}.\epsilon _x \oplus \mathbb {C}.\epsilon _y - \mathbb {C}.[ \textcolor{teal}{f}]}_{\textcolor{black}{3}-\textcolor{teal}{1}= \textcolor{purple}{2}}\, \, \, \, \, \, \]
Remark 3.9

More generally, the tensor product

\[ \textcolor{red}{R/(xz,xw,yz,yw)}\otimes _{R} \textcolor{blue}{R/(x-z, y-w)}\simeq \mathbb {C}[x,y]/(x^2, xy,y^2) \]

and the quotient

\[ \frac{\textcolor{blue}{I_{Diag}}\cap \textcolor{red}{I_{Axis}}}{\textcolor{blue}{I_{Diag}}.\textcolor{red}{I_{Axis}}} \simeq \mathbb {C}.[\textcolor{teal}{f}] \]

form the first two layers of an hierarchy of redundancies captured by the homology groups of a chain complex

\[ \textcolor{red}{ R/I_{Diag}}\otimes _R^{\textcolor{purple}{\mathbb {L}}}\textcolor{blue}{R/I_{Axis}} \]

computing the derived tensor product. More precisely, its homology groups encode

\[ H_0=\underbrace{ \textcolor{red}{R/I_{Axis}}\otimes _{R} \textcolor{blue}{R/I_{Diag}}}_{\text{Solves the system}} \]
\[ H_1 =\underbrace{ \frac{\textcolor{blue}{I_{Diag}}\cap \textcolor{red}{I_{Axis}}}{\textcolor{blue}{I_{Diag}}.\textcolor{red}{I_{Axis}}}}_{\text{Counts redundancies}} \]
\[ H_2=\text{ Redundancies between redundancies} \]

etc.

Finally, Serre’s formula for the multiplicity of intersections is given by adding the dimensions of homologies of even degree and subtracting dimensions of odd degrees.