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This is a follow-up (refinement?) of this question.

In learning some algebraic topology, I've learned to think of an affine scheme as spec $R$. (I've been told that this is a legitimate use of terminology because it provides an embedding of $Rings^{op}$ into a larger category.) One construction we can do, for example, is create the "tangent scheme", which is obtained by localizing and completing. The examples I've been looking at are $\hat{\mathbb{G}}_a$, $\hat{\mathbb{G}}_m$, $(\mathbb{Z}[[t]],F)$ (FGLs (1-dimensional, commutative) more generally), $T_1C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_m)\cong C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_a)$ ("commutative $k$-variate FGLs satisfying the 2-cocycle condition"), etc. We then have exponential maps, which e.g. in the last case is $exp:C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_a) \rightarrow C^k(\hat{\mathbb{G}}_a,\hat{\mathbb{G}}_m)$ given by $g\mapsto 1+g$.

Just to test the waters, here is my question (although please feel free to push it further or in a different direction). Presumably these exponential maps are not always injective. Once we apply this picture to a particular ring (or perhaps even before?), can we translate differential-geometric ideas like conjugate points, geodesics, "$exp$ is a local isomorphism", etc. into algebro-geometric language? When we can, which theorems for manifolds carry over to schemes and which must we discard?

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Your question is of such a braodness, that I wonder how one can answer this but pointing you to a textbook about schemes? (In any case, what you are calling "differential-geometric ideas" involve Riemannian structures, and the schemes you are handling don't have anything similar to that...) –  Mariano Suárez-Alvarez Nov 6 '10 at 18:17
Great! What's a good book? I'd welcome just a small sampling of analogies, too, if you have any favorites. Also, what other differential-geometric structures were you thinking of? Riemannian was just the first one that came to mind. –  Aaron Mazel-Gee Nov 6 '10 at 20:58
Oops, I totally misread that. Yeah, I wouldn't expect intuition for smooth geometry to be immediately interpretable in an algebraic setting. But I've heard that there's still a decent notion of curvature on a scheme, for instance? –  Aaron Mazel-Gee Nov 7 '10 at 1:46
@Aaron: Dear Aaron, There is no notion of curvature for a scheme. In the case of varieties over the complex numbers, there is a deep relationship between certain algebro-geometric properties (Kodaira dimension, structure of the canonical bundle, ...) and differential geometric properties of the underlying complex analytic manifold (existence of metrics with various curvature properties); the most famous is probably the theory of Calabi--Yau varieties. But this story is not a part of scheme theory; rather, it is a part of complex algebraic and analytic geometry. –  Matt E Dec 26 '10 at 15:20
Hey, check it out: mathoverflow.net/questions/19308/… –  Kevin H. Lin Jan 7 '11 at 10:23
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I detect that you're really asking about Lie groups and Lie algebras.

Much of the theory of Lie groups and Lie algebras can be done algebraically or formally. Analysis only really comes in when you start to talk about convergence -- such as that of $\operatorname{exp} : \mathfrak{g} \to G$. If you don't care about convergence, you can still treat $\operatorname{exp}$ formally. Or, if $\mathfrak{g}$ is nice enough, say nilpotent, then everything works out -- you get convergence of $\operatorname{exp}$ for free.

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I've been wondering about Lie theory over fields other than $\mathbb{R}$ and $\mathbb{C}$, yes, but actually Matt E's comment alludes to the sorts of things I was hoping to get out of this question. In any case, what do people use this "generalized" Lie theory for? Or where does it show up? –  Aaron Mazel-Gee Dec 27 '10 at 8:49
It is certainly possible and very natural and very interesting to consider algebraic groups like $GL_n$ or $SL_n$ over any field or even any ring (with identity)... Otherwise, there's a theorem (I forget the precise statement) which relates formal group laws over a ring R and cohomology theories (in the sense of Eilenberg-Mac Lane) with values in R-modules... –  Kevin H. Lin Jan 7 '11 at 10:27
Probably this is the Landweber exact functor theorem? This gives sufficient conditions for a formal group law determined by $MU_* \rightarrow R$ to give rise to a complex-oriented homology theory $A_*(X) = MU_*(X) \otimes_{MU_*}R$ with that prescribed formal group law. (Flatness is too strong a condition to be useful; this one is more delicate and involves the coaction of the homology cooperations $MU_*MU$.) –  Aaron Mazel-Gee Sep 25 '11 at 21:38
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$\def\A{\mathbb{A}} \def\spec{\mbox{Spec}} \def\R{\mathbb{R}} \def\Q{\mathbb{Q}} \def\Z{\mathbb{Z}}$

Here is an example of the sort of thing that I'm hoping to learn about from this math.SE question. This is copied directly from an email that I sent a while back, hence the tone.


TIL (today I learned)...


...an incredibly neat way of computing the algebraic de Rham cohomology of certain affine spaces $\A^n_R=\spec(R[x_1,...,x_n])$ (and their formal analogs, if you know what that means -- the story is almost exactly the same). This beautifully illustrates the interplay of differential geometry and algebraic geometry: on the one hand, algebraic geometry imports ideas from differential geometry all the time (although occasionally the reverse is true!), but then the much more general algebraic setting affords new and exciting techniques and results. Of course, the analogies extend much deeper: in the algebro-geometric setting one can define vector fields, connections, curvature, integrability, etc.; you name it, Grothendieck's probably already done it.

~ ~ ~ ~ ~


Recall that the sheaf $\Omega^1$ of Kähler differentials of $\A^n_R$ (relative to $R$ -- this essentially means that we're treating $R$ as the "constants") is supposed to be the "sheaf of 1-forms on $\A^n_R$". As a quasicoherent sheaf, this is determined by an $R[x_1,...,x_n]$-module, namely the free module generated by the symbols $\{dx_i\}$. More generally, the sheaf of $n$-forms on $\A^n_R$ is just defined to be the $n^{th}$ alternating power $\Lambda^n(\Omega^1)$ (as an $R[x_1,...,x_n]$-module), as it should be; in particular, $\Omega^0=\Lambda^0(\Omega^1)=R[x_1,...,x_n]$, the functions on $\A^n_R$. Of course. These all fit together into a chain complex

$$ 0\rightarrow \Omega^0\rightarrow \Omega^1\rightarrow \Omega^2\rightarrow \cdots, $$

where the differential is given by the usual formula

$$d(f \cdot dx_{i_1} \wedge ... \wedge dx_{i_k}) = \sum_j \frac{\partial f}{\partial x_j} \cdot dx_j \wedge dx_{i_1} \wedge ... \wedge dx_{i_k},$$

and the cohomology of this complex is by definition the de Rham cohomology of our scheme. Excellent.

~ ~ ~ ~ ~


Now, suppose that $R$ is torsion-free, and let $K=R\otimes \Q$. Recall the Poincare lemma, a standard differential-geometric fact, which says that $\R^n$ has the de Rham cohomology of a point. (This is what makes all of differential de Rham cohomology tick.) You will note from the wikipedia link that all that's needed to define the necessary chain-homotopy is a notion of integral. But actually this is entirely formal, and in our setting where all our functions are polynomials, we merely need an appropriate notion of integration of polynomials: this is precisely what tensoring with $\Q$ buys us. That is to say,

$$ \begin{array}{ll} H^0(\A^n_K)=K & \mbox{ (given by the constant functions),} \\ H^i(A^n_K)=0 & \mbox{ for }i>0. \end{array} $$

Now, here we go. I'll stick to 1-dimensional cohomology for simplicity. Suppose that $\omega \in \Omega^1_{\A^n_R}$ is a closed 1-form. This means that $d\omega = 0 \in \Omega^2_{\A^n_R}$. However, we can also consider $\omega \in \Omega^1_{\A^n_K}$; it's still closed of course, but $\A^n_K$ has no 1-dimensional cohomology, so every closed form is exact, so $\omega=df$ for some $f \in \Omega^0_{\A^n_K}=K[x_1,...,x_n]$, and f is unique up to adding a constant. Thus we have a canonical isomorphism

$$H^1(\A^n_R) = \frac{\{ f \in K[x_1,...,x_n] : df \in \Omega^1_{\A^n_R}\mbox{ and }f(0)=0\}}{\{f \in R[x_1,...,x_n] : f(0)=0\}}.$$

This is fantastic!

To illustrate, let's start with the simplest case $R=\Z$, $n=1$. We're looking for polynomials $f(x) \in \Q[x]$ such that $f(0)=0$ and $df$ is an integral 1-form, and we're hoping that $f(x)$ itself isn't integral. So, write $f(x)=\sum a_k x^k$. Then $df = \sum k\cdot a_k x^{k-1} dx$. So we're demanding that $k\cdot a_k \in \Z$, but allowing that $a_k \notin \Z$. Of course, we can choose these $a_k$ distinctly since everything in sight is linear. Moreover, any choice $a_k \in \Z$ is considered trivial. Thus, the first de Rham cohomology group of the affine line over $\Z$ is

$$ H^1(\A^1_\Z) = \Z/1\Z \oplus \Z/2\Z \oplus \Z/3\Z \oplus \Z/4\Z \oplus \Z/5\Z \oplus \ldots !$$

Now, if you extend to multivariable functions and write $f(x) = \sum_{J} a_J x^J$ (using multiindex notation), then you'll get that each $a_J \in \Q$ must have denominator a divisor of all the $j_1,\ldots,j_n$, so $a_J$ can be chosen precisely from $\Z[1/\mbox{gcd}(j_k)]$, and hence

$$H^1(\A^n_\Z) = \bigoplus_J \Z/\mbox{gcd}(j_k)\Z,$$

where $J=(j_1,...,j_n)$ runs over all length-$n$ multiindices (and every number divides $0$, of course).

...all of which is the cutest thing I've seen all week. Maybe I need to get out more.

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