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Category: geometry

Grothendieck’s functor of points

A comment-thread well worth following while on vacation was Algebraic Geometry without Prime Ideals at the Secret Blogging Seminar. Peter Woit became lyric about it :

My nomination for the all-time highest quality discussion ever held in a blog comment section goes to the comments on this posting at Secret Blogging Seminar, where several of the best (relatively)-young algebraic geometers in the business discuss the foundations of the subject and how it should be taught.

I follow far too few comment-sections to make such a definite statement, but found the contributions by James Borger and David Ben-Zvi of exceptional high quality. They made a case for using Grothendieck’s ‘functor of points’ approach in teaching algebraic geometry instead of the ‘usual’ approach via prime spectra and their structure sheaves.

The text below was written on december 15th of last year, but never posted. As far as I recall it was meant to be part two of the ‘Brave New Geometries’-series starting with the Mumford’s treasure map post. Anyway, it may perhaps serve someone unfamiliar with Grothendieck’s functorial approach to make the first few timid steps in that directions.

Allyn Jackson’s beautiful account of Grothendieck’s life “Comme Appele du Neant, part II” (the first part of the paper can be found here) contains this gem :

“One striking characteristic of Grothendieck’s
mode of thinking is that it seemed to rely so little
on examples. This can be seen in the legend of the
so-called “Grothendieck prime”.

In a mathematical
conversation, someone suggested to Grothendieck
that they should consider a particular prime number.
“You mean an actual number?” Grothendieck
asked. The other person replied, yes, an actual
prime number. Grothendieck suggested, “All right,
take 57.”

But Grothendieck must have known that 57 is not
prime, right? Absolutely not, said David Mumford
of Brown University. “He doesn’t think concretely.””

We have seen before how Mumford’s doodles allow us to depict all ‘points’ of the affine scheme $\mathbf{spec}(\mathbb{Z}[x]) $, that is, all prime ideals of the integral polynomial ring $\mathbb{Z}[x] $.
Perhaps not too surprising, in view of the above story, Alexander Grothendieck pushed the view that one should consider all ideals, rather than just the primes. He achieved this by associating the ‘functor of points’ to an affine scheme.

Consider an arbitrary affine integral scheme $X $ with coordinate ring $\mathbb{Z}[X] = \mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k) $, then any ringmorphism
$\phi~:~\mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k) \rightarrow R $
is determined by an n-tuple of elements $~(r_1,\ldots,r_n) = (\phi(t_1),\ldots,\phi(t_n)) $ from $R $ which must satisfy the polynomial relations $f_i(r_1,\ldots,r_n)=0 $. Thus, Grothendieck argued, one can consider $~(r_1,\ldots,r_n) $ an an ‘$R $-point’ of $X $ and all such tuples form a set $h_X(R) $ called the set of $R $-points of $X $. But then we have a functor

$h_X~:~\mathbf{commutative rings} \rightarrow \mathbf{sets} \qquad R \mapsto h_X(R)=Rings(\mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k),R) $

So, what is this mysterious functor in the special case of interest to us, that is when $X = \mathbf{spec}(\mathbb{Z}[x]) $?
Well, in that case there are no relations to be satisfied so any ringmorphism $\mathbb{Z}[x] \rightarrow R $ is fully determined by the image of $x $ which can be any element $r \in R $. That is, $Ring(\mathbb{Z}[x],R) = R $ and therefore Grothendieck’s functor of points
$h_{\mathbf{spec}(\mathbb{Z}[x]} $ is nothing but the forgetful functor.

But, surely the forgetful functor cannot give us interesting extra information on Mumford’s drawing?
Well, have a look at the slightly extended drawing below :



What are these ‘smudgy’ lines and ‘spiky’ points? Well, before we come to those let us consider the easier case of identifying the $R $-points in case $R $ is a domain. Then, for any $r \in R $, the inverse image of the zero prime ideal of $R $ under the ringmap $\phi_r~:~\mathbb{Z}[x] \rightarrow R $ must be a prime ideal of $\mathbb{Z}[x] $, that is, something visible in Mumford’s drawing. Let’s consider a few easy cases :

For starters, what are the $\mathbb{Z} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $? Any natural number $n \in \mathbb{Z} $ determines the surjective ringmorphism $\phi_n~:~\mathbb{Z}[x] \rightarrow \mathbb{Z} $ identifying $\mathbb{Z} $ with the quotient $\mathbb{Z}[x]/(x-n) $, identifying the ‘arithmetic line’ $\mathbf{spec}(\mathbb{Z}) = { (2),(3),(5),\ldots,(p),\ldots, (0) } $ with the horizontal line in $\mathbf{spec}(\mathbb{Z}[x]) $ corresponding to the principal ideal $~(x-n) $ (such as the indicated line $~(x) $).

When $\mathbb{Q} $ are the rational numbers, then $\lambda = \frac{m}{n} $ with $m,n $ coprime integers, in which case we have $\phi_{\lambda}^{-1}(0) = (nx-m) $, hence we get again an horizontal line in $\mathbf{spec}(\mathbb{Z}[x]) $. For $ \overline{\mathbb{Q}} $, the algebraic closure of $\mathbb{Q} $ we have for any $\lambda $ that $\phi_{\lambda}^{-1}(0) = (f(x)) $ where $f(x) $ is a minimal integral polynomial for which $\lambda $ is a root.
But what happens when $K = \mathbb{C} $ and $\lambda $ is a trancendental number? Well, in that case the ringmorphism $\phi_{\lambda}~:~\mathbb{Z}[x] \rightarrow \mathbb{C} $ is injective and therefore $\phi_{\lambda}^{-1}(0) = (0) $ so we get the whole arithmetic plane!

In the case of a finite field $\mathbb{F}_{p^n} $ we have seen that there are ‘fat’ points in the arithmetic plane, corresponding to maximal ideals $~(p,f(x)) $ (with $f(x) $ a polynomial of degree $n $ which remains irreducible over $\mathbb{F}_p $), having $\mathbb{F}_{p^n} $ as their residue field. But these are not the only $\mathbb{F}_{p^n} $-points. For, take any element $\lambda \in \mathbb{F}_{p^n} $, then the map $\phi_{\lambda} $ takes $\mathbb{Z}[x] $ to the subfield of $\mathbb{F}_{p^n} $ generated by $\lambda $. That is, the $\mathbb{F}_{p^n} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $ consists of all fat points with residue field $\mathbb{F}_{p^n} $, together with slightly slimmer points having as their residue field $\mathbb{F}_{p^m} $ where $m $ is a divisor of $n $. In all, there are precisely $p^n $ (that is, the number of elements of $\mathbb{F}_{p^n} $) such points, as could be expected.

Things become quickly more interesting when we consider $R $-points for rings containing nilpotent elements.

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Grothendieck’s survival talks

The Grothendieck circle is a great resource to find published as well as unpublished texts by Alexander Grothendieck.

One of the text I was unaware of is his Introduction to Functorial Algebraic Geometry, a set of notes written up by Federico Gaeta based on tape-recordings (!) of an 100-hour course given by Grothendieck in Buffalo, NY in the summer of 1973. The Grothendieck-circle page adds this funny one-line comment: “These are not based on prenotes by Grothendieck and to some extent represent Gaeta’s personal understanding of what was taught there.”.

It is a bit strange that this text is listed among Grothendieck’s unpublished texts as Gaeta writes on page 3 : “GROTHENDIECK himself does not assume any responsability for the publication of these notes”. This is just one of many ‘bracketed’ comments by Gaeta which make these notes a great read. On page 5 he adds :

“Today for many collegues, GROTHENDIECK’s Algebraic Geometry looks like one of the most abstract and unapplicable products of current mathematical thought. This prejudice caused har(‘m’ or ‘ess’, unreadable) even before the students of mathematics within the U.S. were worried about the scarcity of academic positions… . If they ever heard GROTHENDIECK deliver one of his survival talks against modern Science, research, technology, etc., … their worries might become unbearable.”

Together with Claude Chevalley and Pierre Cartier, Grothendieck was an editor of “Survivre et Vivre“, the bulletin of the ecological association of the same name which appeared at regular intervals from 1970 to 1973. Scans of all but two of these volumes can be found here. All of this has a strong 60ties feel to it, as does Gaeta’s decription of Grothendieck : “He is a very liberal man and in spite of that he allowed us to use plenty of tape recorders!” (p.5).

On page 11, Gaeta records a little Q&A exchange from one of these legendary ‘survival talks’ by Grothendieck :

Question : We understand your worries about expert knowledge,… by the way, if we try to explain to a layman what algebraic geometry is it seems to me that the title of the old book of ENRIQUES, “Geometrical theory of equations”, is still adequate. What do you think?

GROTHENDIECK : Yes, but your ‘layman’ should know what a sustem of algebraic equations is. This would cost years of study to PLATO.

Question : It should be nice to have a little faith that after two thousand years every good high school graduate can understand what an affine scheme is … What do you think?

GROTHENDIECK : …. ??

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Geometry of the Okubo algebra

Last week, Melanie Raczek gave a talk entitled ‘Cubic forms and Okubo product’ in our Artseminar, based on her paper On ternary cubic forms that determine central simple algebras of degree 3.

I had never heard of this strange non-associative product on 8-dimensional space, but I guess it is an instance of synchronicity that now the Okubo algebra seems to pop-up everywhere.

Yesterday, there was the post the Okubo algebra by John Baez at the n-cafe, telling that Susumu Okubo discovered his algebra while investigating quarks.

I don’t know a thing about the physics, but over the last days I’ve been trying to understand some of the miraculous geometry associated to the Okubo algebra. So, let’s start out by defining the ‘algebra’.

Consider the associative algebra of all 3×3 complex matrices $M_3(\mathbb{C}) $ with the usual matrix-multiplication. In this algebra there is the 8-dimensional subspace of trace zero matrices, usually called the Lie algebra $\mathfrak{sl}_3 $. However, we will not use the Lie-bracket, only matrix-multiplication. Typical elements of $\mathfrak{sl}_3 $ will be written as $X,Y,Z,… $ and their entries will be denoted as

$X = \begin{bmatrix} x_0 & x_1 & x_2 \\ x_3 & x_4 & x_5 \\ x_6 & x_7 & -x_0-x_4 \end{bmatrix} $

For any two elements $X,Y \in \mathfrak{sl}_3 $ one defines their Okubo-product to be the 3×3 matrix

$X \ast Y = \frac{1}{1-\omega}(Y.X-\omega X.Y) – \frac{1}{3}Tr(X.Y) 1_3 $

where $\omega $ is a primitive 3-rd root of unity and $1_3 $ is the identity matrix. Written out in the entries of X and Y this operation looks horribly complicated

$X \ast Y = \frac{1}{1-\omega} \begin{bmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & -p_{11}-p_{22} \end{bmatrix} $

with

[tex]\begin{eqalign} \\ p_{11} &= (1-\omega)x_0y_0+x_3y_1+x_6y_2-\omega(x_1y_3+x_2y_6)-\frac{1}{3}T \\ p_{12} &= x_1y_0+x_4y_1+x_7y_2-\omega(x_0y_1+x_1y_4+x_2y_7) \\ p_{13} &= x_2y_0+x_5y_1-x_0y_2-x_4y_2-\omega(x_0y_2+x_1y_5-x_2y_0-x_2y_4) \\
p_{21} &= x_0y_3+x_3y_4+x_6y_5 – \omega(x_3y_0+x_4y_3+x_5y_6) \\
p_{22} &= (1-\omega)x_4y_4+x_1y_3+x_7y_5 – \omega(x_3y_1+x_5y_7) – \frac{1}{3}T \\
p_{23} &= x_2y_3+x_5y_4-x_0y_5-x_4y_5-\omega(x_3y_2+x_4y_5-x_5y_0-x_5y_4) \\
p_{31} &= x_0y_6+x_3y_7-x_6y_0-x_6y_4-\omega(x_6y_0+x_7y_3-x_0y_6-x_4y_6) \\
p_{32} &= x_1y_6+x_4y_7-x_7y_0-x_7y_4 – \omega(x_6y_1+x_7y_4-x_0y_7-x_4y_7) \\
T &= 2x_0y_0+2x_4y_4+x_1y_3+x_2y_6+x_3y_1+x_5y_7+x_6y_2+x_7y_5+x_0y_4+x_4y_0
\end{eqalign}[/tex]

The crucial remark to make is that $X \ast Y $ is again a trace zero matrix. That is, we have defined a new operation on $\mathfrak{sl}_3 $.

$\mathfrak{sl}_3 \times \mathfrak{sl}_3 \rightarrow \mathfrak{sl}_3~\qquad~\qquad~(X,Y) \mapsto X \ast Y $

This Okubo-product is neither a Lie-bracket, nor an associative multiplication. In fact, it is a lot ‘less associative’ than that other 8-dimensional algebra, the octonions. The only noteworthy identity it has is that $X \ast (Y \ast X) = (X \ast Y) \ast X $. So, why should we be interested in this horrible algebra?

Well, let us consider the subset of $\mathfrak{sl}_3 $ consisting of those matrices X satusfying $Tr(X^2)=0 $. That is, with the above notation, all matrices X such that

$x_0^2+x_4^2+x_1x_3+x_2x_6+x_5x_7=0 $

In the 8-dimensional affine space $\mathfrak{sl}_3 $ these matrices form a singular quadric with top the zero-matrix. So, it is better to go projective. That is, any non-zero matrix $X \in \mathfrak{sl}_3 $ determines a point in 7-dimensional projective space $\mathbb{P}^7 $ with homogeneous coordinates

$\overline{X} = [x_0:x_1:x_2:x_3:x_4:x_5:x_6:x_7] \in \mathbb{P}^7 $

and the points $\overline{X} $ corresponding to solutions of $Tr(X^2)=0 $ form a smooth 6-dimensional quadric $Q \subset \mathbb{P}^7 $ with homogeneous equation

$Q = \mathbb{V}(x_0^2+x_4^2+x_1x_3+x_2x_6+x_5x_7) $

6-dimensional quadrics may be quite hard to visualize, so it may help to recall the classic situation of lines on a 2-dimensional quadric (animated gif taken from surfex).

A 2-dimensional quadric contains two families of lines, often called the ‘blue lines’ and the ‘red lines’, each of these lines isomorphic to $\mathbb{P}^1 $. The rules-of-intersection of these are :

  • different red lines are disjoint as are different blue lines
  • any red and any blue line intersect in exactly one point
  • every point of the quadric lies on exactly one red and one blue line

The lines in either family are in one-to-one correspondence with the points on the projective line. We therefore say that there is a $\mathbb{P}^1 $-family of red lines and a $\mathbb{P}^1 $-family of blue lines on a 2-dimensional quadric.

A 6-dimensional quadric $Q \subset \mathbb{P}^7 $ contains two families of ‘3-planes’. That is, there is a family of red $\mathbb{P}^3 $’s contained in Q and a family of blue $\mathbb{P}^3 $’s. Can we determine these red and blue 3-planes explicitly?

Yes we can, using the Okubo algebra-product on $\mathfrak{sl}_3 $. Take $X \in \mathfrak{sl}_3 $ defining the point $\overline{X} \in Q $ (that is, $Tr(X^2)=0 $). then all 3×3 matrices one obtains by taking the Okubo-product with left X-factor form a 4-dimensional linear subspace in $\mathfrak{sl}_3 $

$L_X = { X \ast Y~|~Y \in \mathfrak{sl}_3 } \simeq \mathbb{C}^4 \subset \mathfrak{sl}_3 $

so its non-zero matrices determine a 3-plane in $\mathbb{P}^7 $ (consisting of all points with homogeneous coordinates $[p_{11}:p_{12}:p_{13}:p_{21}:p_{22}:p_{23}:p_{31}:p_{32}] $, using the above formulas) which actually lies entirely in the quadric Q. These are precisely the bLue 3-planes in Q. That is, the family of all bLue 3-planes consists precisely of the 3-planes

$\mathbb{P}(L_X) $ with $X \in \mathfrak{sl}_3 $ satisfying $Tr(X^2)=0 $

Phrased differently, any point $\overline{X} \in Q $ determines a blue 3-plane $\mathbb{P}(L_X) $.

Similarly, any point $\overline{X} \in Q $ determines a Red 3-plane by taking Okubo-products with Right X-factor, that is, $\mathbb{P}(R_X) $ is a 3-plane for Q where

$R_X = { Y \ast X~|~Y \in \mathfrak{sl}_3 } \simeq \mathbb{C}^4 \subset \mathfrak{sl}_3 $

and all Red 3-planes for Q are of this form. But, this is not all… these correspondences are unique! That is, any point on the quadric defines a unique red and a unique blue 3-plane, or, phrased differently, there is a Q-family of red 3-planes and a Q-family of blue 3-planes in Q. This is a consequence of triality.

To see this, note that the automorphism group of a 6-dimensional smooth quadric is the rotation group $SO_8(\mathbb{C}) $ and this group has Dynkin diagram $D_4 $, the most symmetrical of them all!

In general, every node in a Dynkin diagram has an interesting projective variety associated to it, a so called homogeneous space. I’ll just mention what these spaces are corresponding to the 4 nodes of $D_4 $. Full details can be found in chapter 23 of Fulton and Harris’ Representation theory, a first course.

The left-most node corresponds to the orthogonal Grassmannian of isotropic 1-planes in $\mathbb{C}^8 $ which is just a fancy way of viewing our quadric Q. The two right-most nodes correspond to the two connected components of the Grassmannians of isotropic 4-planes in $\mathbb{C}^8 $, which are our red resp. blue families of 3-planes on the quadric. Now, as the corresponding dotted Dynkin diagrams are isomorphic



there corresponding homogeneous spaces are also isomorphic. Thus indeed, there is a one-to-one correspondence between points of the quadric Q and red 3-planes on Q (and similarly with blue 3-planes on Q).

Okay, so the Okubo-product allows us to associate to a point on the 6-dimensional quadric Q a unique red 3-plane and a unique blue 3-plane (much as any point on a 2-dimensional quadric determines a unique red and blue line). Do these families of red and blue 3-planes also satisfy ‘rules-of-intersection’?

Yes they do and, once again, the Okubo-product clarifies them. Here they are :

  • two different red 3-planes intersect in a unique line (as do different blue 3-planes)
  • the bLue 3-plane $\mathbb{P}(L_X) $ intersects the Red 3-plane $\mathbb{P}(R_Y) $ in a unique point if and only if the Okubo-product $X \ast Y \not= 0 $
  • the bLue 3-plane $\mathbb{P}(L_X) $ intersects the Red 3-plane $\mathbb{P}(R_Y) $ in a unique 2-plane if and only if the Okubo-product $X \ast Y = 0 $

That is, Right and Left Okubo-products determine the Red and bLue families of 3-planes on the 6-dimensional quadric as well as their intersections!

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