The Field With One Element

April 17, 2007

Posted by David Corfield

For some grand theory building and an answer to the question ‘What is the field with one element?’, see Nikolai Durov’s New Approach to Arakelov Geometry.

There’s something extremely intriguing about a mathematical entity which has known effects, but which has not been defined. It generates a sense of independent reality. As I mentioned in the Tuesday 8 November entry on my old blog, a vector space over the ‘field with one element’ is a pointed set. Thinking in such terms makes sense of many combinatorial facts, see TWF 187.

Here’s Durov’s answer:

The ‘field with one element’ is the free algebraic monad generated by one constant (p. 26) or the universal generalized ring with zero (p. 33).

This will need some unpacking.

Posted at April 17, 2007 12:02 PM UTC

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17 Comments & 1 Trackback

Re: The Field With One Element

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Hmm,

…a lot of statements in algebraic topology become statements about homological algebra over $𝔽_{1}$ .(p. 56)

Posted by: David Corfield on April 20, 2007 2:32 PM | Permalink | Reply to this

Re: The Field With One Element

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Thanks for pointing out this paper! I’ll have to read it… when I get a little time.

Posted by: John Baez on April 20, 2007 10:08 PM | Permalink | Reply to this

Re: The Field With One Element

It might take more than a little time - that’s a big paper, considering it’s about something any undergraduate can tell you doesn’t exist.

There was a conference on this subject at IHES a few weeks ago
comparing the various approaches and there seem to be quite a few papers popping up here and there. For instance Paul Lescot has some preprints on the IHES server describing a nice elementary approach to absolute linear algebra where the field of one element rather perversely has two elements.

Maybe somebody who attended this workshop might tell us more about the latest developments ?

Posted by: Dan Evans on April 21, 2007 12:11 PM | Permalink | Reply to this

Re: The Field With One Element

I see Durov/Dourov was given two slots in this high quality meeting.

Posted by: David Corfield on April 22, 2007 9:52 AM | Permalink | Reply to this

Re: The Field With One Element

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Subobjects in the 2-topos of categories are (p. 29) pullbacks of the forget functor from pointed sets to sets. That’s $U : F_{1} - Mod \to$ $F_{\emptyset} - Mod$ , generated by the scalar restriction from $F_{\emptyset} \to F_{1}$ .

Posted by: David Corfield on April 21, 2007 5:06 PM | Permalink | Reply to this

Re: The Field With One Element

The field with one element is described as ‘mysterious’ here (p. 39), although it does have a maximal unramified extension.

Posted by: David Corfield on April 26, 2007 12:20 PM | Permalink | Reply to this

Re: The Field With One Element

Having just skimmed the (very clearly written) introduction to Durov’s paper, it doesn’t look as though anything particularly profound is going on as regards the field with one element.

Durov defines a “generalised ring” to be a finitary monad on Set, which certainly is a generalisation of a ring! The field with one element is the algebraic theory with one nullary constant and nothing else, whose algebras are obviously pointed sets.

Presumably the deeper point is that various things that are traditionally done with rings can in fact be done with arbitrary algebraic theories, though I haven’t got far enough to see that yet.

Posted by: Robin on May 3, 2007 10:26 PM | Permalink | Reply to this

Re: The Field With One Element

PS. Later on, he defines a generalised field to be an algebraic theory (sorry, a generalised ring :-)) with no non-trivial quotients. That’s certainly an interesting sense in which the “field with one element” acts like a field. Could that be all there is to it? It would be rather wonderful if so.

Posted by: Robin on May 4, 2007 12:22 AM | Permalink | Reply to this

Re: The Field With One Element

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I’m taking a look at Durov’s paper now, and it seems fascinating — mainly because he tries such a massive generalization of algebraic geometry.

He does seem to make heavy weather of some standard material about algebraic theories and monads, presumably because these concepts are not part of the normal vocabulary of algebraic geometry. But, the idea is simple enough: he’s claiming he can do algebraic geometry with ‘commutative algebraic monads’ instead of commutative rings.

A few comments about this, mainly just expanding on what Robin already said:

In Chapter 4 he defines an ‘algebraic monad’ to be a monad on $Set$ which preserves filtered colimits. But, as Lawvere showed, this is just a painfully technical way of talking about an ‘algebraic theory’ — roughly speaking, a theory describing gadgets with some specified $n$ -ary operations satisfying some specified equational laws.

An example of the kind of algebraic theory Durov cares about is ‘the theory of $R$ -modules’ for a ring $R$ . In short, he doesn’t really generalize the concept of ring; he really generalizes the concept of $R$ -module, replacing it by ‘model of an algebraic theory’ — that is, a gadget with some $n$ -any operations satisfying some equations.

That’s a truly massive generalization. It runs the risk of vacuity.

But, the really important rings in algebraic geometry are the commutative ones! So, it’s $R$ -modules with $R$ commutative that Durov really needs to generalize. And, he does this using the concept of a ‘commutative’ algebraic monad, which he introduces in Chapter 5. Again it’s easier to think of these in terms of algebraic theories. An algebraic theory is ‘commutative’ if all its operations commute in a suitable sense.

What does it mean for an $n$ -ary operation and an $m$ -ary operation to ‘commute’? Well, for $n = m = 1$ it just means the obvious thing: two unary operations $r$ and $s$ on a set $X$ commute if

$s (r (x)) = r (s (x))$

for all $x \in X$ . If $s$ is binary and $r$ is unary, we instead need

$s (r (x), r (y)) = r (s (x, y)) .$

It’s fun to guess the general pattern from these examples!

Here’s why ‘the theory of $R$ -modules’ is commutative when $R$ is commutative. Given an $R$ -module $X$ for a commutative ring $R$ , each element $r \in R$ gives a unary operation on $X$ , namely ‘multiplication by $r$ ’. All these unary operations commute:

$r s x = s r x .$

Similarly, addition is a binary operation on $X$ , and this commutes with all the above unary operations:

$r (x + y) = r x + r y$

And so on: all the operations commute!

A commutative algebraic theory can have at most one unary operation (or constant), which for the theory of $R$ -modules is just $0$ . The ‘field with one element’, in Durov’s approach, is really just the algebraic theory that has only one operation — a unary operation. Models of this theory are just pointed sets.

This fits nicely with my own intuitions about linear algebra over the field with one element. A pointed set acts like a ‘vector space over the field with one element’; a set acts like a projective space over the field with one element.

Given our discussions elsewhere on this blog, I’d like to read what Durov says about ‘exterior algebras’ over the field with one element!

Posted by: John Baez on May 4, 2007 7:38 AM | Permalink | Reply to this

Re: The Field With One Element

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This is sounding very interesting! Is it possible to push the analogy further?

(1)

pointed set \leftrightarrow vector space over field F with one element

(2)

set \leftrightarrow projective space over field with one element

(3)

G - set \leftrightarrow action of G on projective space

If so, then if we have a $G$ -set $X$ , how are we to think of a projective representation of the groupoid $X / / G$ ?

Posted by: Bruce Bartlett on May 4, 2007 10:03 AM | Permalink | Reply to this

Re: The Field With One Element

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Bruce wrote:

This is sounding very interesting! Is it possible to push the analogy further?

(1) $pointed set \leftrightarrow vector space over field F with one element$

(2) $set \leftrightarrow projective space over field with one element$

(3) $G - set \leftrightarrow action of G on projective space$

Sort of — but it’s even better than that!

If we have any simple Lie group $G$ , we can describe it using algebraic equations which make sense over any field $k$ , getting an ‘algebraic group’ $G_{k}$ .

Physicists are familiar with the real and complex forms of Lie groups, like $G_{ℝ} = GL (n, ℝ)$ and $G_{ℂ} = GL (n, ℂ)$ . This is a big — and well-known — generalization of that.

If you pick any collection of dots in the Dynkin diagram for $G$ , you get, for any field $k$ , an algebraic variety $X_{k}$ on which $G_{k}$ acts. The most familiar of these are ‘Grassmannians’.

And, all this stuff works when $k = 𝔽_{1}$ is the field with one element!

Even better, when we let $k = 𝔽_{q}$ be the finite field with $q$ elements, everything about the action of $G_{𝔽_{q}}$ on $X_{𝔽_{q}}$ is a $q$ -deformation of the action of $G_{𝔽_{1}}$ on $X_{𝔽_{1}}$ .

I explained this stuff in week185, week186, and week187. But, I didn’t have a formal theory of ‘the field with one element’. It will be nice to see what having such a theory lets us do.

I don’t really understand your other question very well… but, we definitely want to study the weak quotients $X_{k} / / G_{k}$ in a systematic way.

Posted by: John Baez on May 4, 2007 4:50 PM | Permalink | Reply to this

Re: The Field With One Element

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Thanks John, this is great. I’m certainly going to go through week185, week186 and week187 now.

By my last question, I just meant the following. Suppose we have a left $G$ -set $X$ , and a $U (1)$ -valued equivariant 2-cocycle on $X$ .

That just means that $ϕ$ is a function which sends composable pairs of morphisms in the action groupoid $X / / G$ to elements of $U (1)$ , which we’ll write as

(1)

ϕ_{x} (g_{2}, g_{1}) \equiv ϕ (g_{2} g_{1} \cdot x \overset{g_{2}}{\leftarrow} g_{1} \cdot x \overset{g_{1}}{\leftarrow} x) \in U (1),

satisfying the 2-cocycle equation

(2)

ϕ_{x} (g_{3}, g_{2} g_{1}) ϕ_{x} (g_{2}, g_{1}) = ϕ_{x} (g_{3} g_{2}, g_{1}) ϕ_{g_{1} \cdot x} (g_{3}, g_{2}) .

One thinks of an equivariant 2-cocycle over $X$ as an ‘equivariant line bundle over $X$ ’.

In that way, a twisted equivariant vector bundle $σ$ over $X$ is defined as a projective representation of the groupoid $X / / G$ . In other words, it consists of a $ℂ$ -vector space $σ (x)$ for each $x \in X$ , together with a map

(3)

σ (\overset{g}{\leftarrow} x) : σ (x) \to σ (g \cdot x)

for each $x \in X$ and $g \in G$ , which lifts the action of $G$ on $X$ up to the 2-cocycle $ϕ$ . In other words, we have

(4)

σ (\overset{g_{2}}{\leftarrow} g_{1} \cdot x) σ (\overset{g_{1}}{\leftarrow} x) = ϕ_{x} (g_{2}, g_{1}) σ (\overset{g_{2} g_{1}}{\leftarrow} x)

for all $g_{1}, g_{2} \in G$ and $x \in X$ .

Okay, now you understand my notation. So a twisted equivariant vector bundle over $X$ is the same thing as a projective representation of the groupoid $X / / G$ .

So here’s the amusing ‘double-up’ of the word ‘projective’, which led me to ask my question above.

Namely, let $F$ be the field with one element. Heuristically, of course! Now, if a $G$ -set $X$ can be thought of as an action of $G$ on the projective space $ℙ F^{n}$ ,

(5)

G - set X \leftrightarrow action of G on ℙ F^{n},

then how are we to think of a projective representation of the groupoid $X / / G$ ? We get a phenomenon

(6)

projective \to {projective}^{2}

if you see what I mean $pic$ !

Posted by: Bruce Bartlett on May 4, 2007 6:05 PM | Permalink | Reply to this

Re: The Field With One Element

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Okay, now I get what you mean… but I still don’t have anything intelligent to say. So, I’ll say something unintelligent — but make it so hard to understand that everyone thinks it’s intelligent.

In algebraic geometry we often think about projective space ${ℂℙ}^{n}$ , or nice spaces sitting inside projective space… so an algebraic geometer studying ‘twisted equivariant vector bundles’ will probably like to think about these things over projective space.

But, I don’t see anything interesting happening, unless you figure out a way to connect the action of $ℂ^{*}$ on $ℂ^{n}$ (which gives $ℂ^{n} / ℂ^{*} = {ℂℙ}^{n - 1}$ ) to some $ℂ^{*}$ -valued cocycle in your twisted equivariant vector bundle. It seems to be a kind of ‘fiber versus base’ business.

(Since you’re a physicist you like $U (1)$ ; algebraic geometers prefer $ℂ^{*}$ , but it’s not a big deal, since the latter is just the complexification of the former. Algebraic geometers like to make things complex, while physicists like ‘keepin’ it real’.)

We did have some fun, in our old Klein 2-geometry discussions, thinking about ‘categorified projective space’. We’d get this by weakly modding out things like $ℂ^{n}$ by the action of $ℂ^{*}$ , getting a groupoid $ℂ^{n} / / ℂ^{*}$ . Maybe you could try getting this idea into the act! Or maybe algebraic geometers should think about this…

Or, maybe they already have. I guess they’d call $ℂ^{n} / / ℂ^{*}$ a ‘stack’, but maybe it’s too trivial of a stack to be worth discussing.

Posted by: John Baez on May 4, 2007 6:31 PM | Permalink | Reply to this

Re: The Field With One Element

John wrote:

The ‘field with one element’, in Durov’s approach, is really just the algebraic theory that has only one operation — a unary operation.

In fact it’s a 0-ary operation.

Posted by: Squark on May 4, 2007 10:56 AM | Permalink | Reply to this

Re: The Field With One Element

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Squark wrote:

The ‘field with one element’, in Durov’s approach, is really just the algebraic theory that has only one operation — a unary operation.

In fact it’s a 0-ary operation.

Whoops! You’re right.

For readers struggling to follow:

A 0-ary operation is often called a ‘nullary’ operation, or more commonly just a ‘constant’. In this particular example, by sheer coincidence, the constant goes by the name of ‘0’. For any ring $R$ , an $R$ -module has an element called ‘0’. But, when $R$ is the mysterious ‘field with one element’ (not really a ring), this is all an $R$ -module has!

So, these technicalities really amount to far less than you might fear. A ‘vector space over the field with 1 element’ is just a fancy name of talking about a set equipped with an element called ‘0’.

It turns out to be useful way of thinking, though!

Posted by: John Baez on May 4, 2007 4:33 PM | Permalink | Reply to this

Re: The Field With One Element

John wrote:

Given our discussions elsewhere on this blog, I’d like to read what Durov says about ‘exterior algebras’ over the field with one element!

Exterior tensor powers / alternative morphisms appear to be not that straightforward in this formalism. Durov discusses them only over the field F_±1 (i.e. for generalized rings with 0 and -). If we want these tensor powers to behave like in usual algebra we need to impose additional conditions on the generalized ring - so-called “alternativity”.

It appears to me one can consider exterior powers / alternative morphisms over F₁ as well. We can define a morphism Φ to be alternative when

Φ(x, x) = 0

without using either the unary - or the binary +. This yields the correct exterior algebra for a pointed set S namely we get

(Λ S) \ {0} = 2^S\{0}

It is interesting that when defining these notions one can use either 0:

[1] Φ(x, x) = 0

or unary -:

[2] Φ(x, y) = -Φ(y, x)

or 0 and binary +:

[3] Φ(x, y) + Φ(y, x) = 0

Durov prefers the combination of [1] and [2]. This excludes F₁; worse, classical semi-rings are also excluded. It would appear that the later are best dealt with using [1] and [3].

Posted by: Squark on May 5, 2007 12:01 PM | Permalink | Reply to this

Re: The Field With One Element

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Interesting! It’s nice to have you at the Café, helping unravel the mysteries of nature. A bit like the good old days at sci.physics.research… but even better, I think.

As you can see, I helped pretty up your Greek letters.

Posted by: John Baez on May 6, 2007 4:21 AM | Permalink | Reply to this

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April 17, 2007