The n-Category Café

Fixed. Yes, that’s what I meant.

Cool! Are the groups $G/\Delta_i$ in this tower of approximations nilpotent?

Simon wrote: I was trying to find expressions for braid invariants coming from Vassiliev theory, and Chen’s itereated integrals was the way to do that.

did you succeed?

Here is a lazy question that I’m embarrassed to be confused about, but I’ve been too lazy to deconfuse myself for years.

Let $X$ be a topological space and $A$ the algebra of Sullivan differential forms on $X$. Now:

-If $X$ is 1-connected, then the minimal model $M$ of $A$ can be used to compute the higher rational homotopy groups of $X$ in that $$M^+/(M^+)^2 =Hom(\pi_*(X),Q),$$ where $M^+$ is the positively graded portion of $M$.

-On the other hand, even if $X$ is not 1-connected, the 1-minimal model of $M(1)$ of $A$ can be used for the Malcev completion of $\pi_1(X)$, in that

$$M(1)^+/(M(1)^+)^2 =Hom(Lie[\pi_1(X)\otimes Q],Q)$$

The point I was never able to clarify to myself is:

In general, that is, for $X$ not 1-connected, does $M$ give you the $Q-$nilpotent completion of the whole $\pi_*(X)$?

One could ask the analogous question for the bar complex $B(A)$. Is $H^*(B(A))$ the $Q$-nilpotent completion of $\pi_*(X)$?

Yes. A rational vector space is the same as a divisible torsionfree abelian group. Incidentally, an abelian group is divisible if and only if it is injective in the category of abelian groups, and is torsionfree if and only if it is flat in the category of abelian groups.

Another way to look at this is that the map

$\mathbb{Z} \hookrightarrow \mathbb{Q}$

is epi in $\mathbf{Rings}$. A rational structure on $G$ is a ring map $\mathbb{Q} \to [G,G]$, automatically under $\mathbb{Z}$ since $\mathbb{Z}$ is the initial ring; so the epi-ness says that there’s at most one such map.

So in $\mathbf{Rings}^\mathrm{op}$, the category of affine schemes (spaces, of a sort), $\Spec(\mathbb{Q})$ is a subobject of the terminal object! It’s a truth-value! :-)

…so (sorry, digression), I wish I had a better intuition for categories of schemes, and in particular the rôle played be spectra of fields! I’ve seen just enough to get an intuition of how rich and subtle it can be, but not enough to understand the subtleties much beyond that. For example: $\Spec(\mathbb{Z})$ is the terminal object; but it’s in many ways very far from being a point. Conversely, spectra of fields are “points” — certainly in terms of their underlying topological spaces, but they seem to function geometrically as points as well — but in general, they’re not necessarily subterminal! What’s going on? (Besides my parochial expectation from $\mathbf{Set}$-like categories that “terminal object” and “point” should be similar ideas.)

Seriously, I know a full answer to this question would essentially be EGA together with most algebraic geometry since then, but I’d be very interested to hear what others around here (some of whom certainly know much more AG than I do) can suggest in the way of pithy insights.

Something I am curious about is how “motivic” such constructions are, if mutually compatible realization functors exist which transfer structures between them, and how “anabelian” they are, how much such constructions say about rational points.

Tom wrote:

Something that bothered me for a while about rational homotopy, as an outsider, was this phrase “the homotopy groups are rational vector spaces”. A priori the (higher) homotopy groups are abelian groups. So does this mean that there exists a rational vector space structure? That there exists a unique one? That one is somehow specified?

I think it’s because you’re a category theorist. Lately I’ve been explaining this stuff to various people at UCR, and whenever James Dolan overhears me say “the homotopy groups are rational vector spaces” he chimes in and says “and that’s just a property, right?”

Somehow I just took for granted that it was a property, even before I knew this property was “being a divisible torsion-free abelian group”, as Todd pointed out here.

Psychoanalyzing myself, I think it’s because I could easily see that any abelian group homomorphism between rational vector spaces is also a vector space map. So, the forgetful functor from rational vector spaces to abelian groups is full. And since it’s obviously faithful, this functor only forgets a property — not structure or stuff.

At least I think that’s why I was so nonchalant about saying that some abelian group ‘is’ a rational vector space.

Since James had sensitized me to this issue, I considered saying something like this at the start of week286:

First, $X'$ is a “rational space”: a 1-connected pointed space whose homotopy groups are rational vector spaces. (And in case you’re wondering, this is just a property of an abelian group.)

But I decided this would puzzle more people than it would de-puzzle!

But now I can add your remark to my ‘Addenda’. It is indeed important.

It’s funny how his initial mistake drawing them matched my initial mistake in describing them… which luckily was fixed by the time anyone saw what I wrote!

Kenneth Baker has allowed me to use some of his figures. Here’s an early stage of the construction of the rational circle:

The right edge of the red band is our original circle, drawn in a tricky way to make the whole picture more manageable. The left edge of the red band is homotopic to 2 times the loop traced out by this original circle. The left of the yellow band is homotopic to 6 times it, and the left edge of the green band is homotopic to 24 times it!

If we remove the red band we see how the yellow one wraps around it 3 times:

and if we remove the yellow band we see how the green one wraps around it 4 times:

Here’s a kind of cross-section that reveals more about what’s going on:

You’re probably curious about how Kenneth Baker drew these pictures. Here’s how:

These pictures are done using Rhino 3D. Actually I’m using the beta version of their port to OS X. There’s a function (called Flow) that lets you map a “spine” of an object to another curve to tell it how to deform the object. This is how I went from the chopped open version to the round one. It’s also how I managed to make the orange wrap around the green and the red wrap around the orange.

Wouldn’t that make a delicious cake? Or maybe it’s a speck of pollen?