In differential geometry, Lie's theorems allow us to integrate any Lie algebra representation to a Lie group representation. The algebraic version of this is more complicated (and I'm not terribly familiar with it), but over a field of characteristic zero we can still get some results from Tannaka duality. This all fails in characteristic $p$, unfortunately. One of the algebraic motivations I've seen for the theory of formal groups is that they rectify the failure of Lie algebras to encode higherdimensional differential information in positive characteristic, with this representation theoreticissue given as an example of that failure. But do formal groups actually fix this particular problem? Do we have an adjunction between algebraic groups and formal groups, and can we use it to integrate (some) representations from formal groups to algebraic groups?

2$\begingroup$ Why do you see the different behaviour in char p as a "failure" to be repaired, rather than an interesting phenomenon to be investigated on its own terms? $\endgroup$– David LoefflerApr 14, 2022 at 6:43

4$\begingroup$ You might like the distribution algebra (sometimes called hyper algebra) of G. In my world (G reductive) it remedies many of the shortcomings of the Lie algebra. $\endgroup$– Geordie WilliamsonApr 14, 2022 at 9:52

3$\begingroup$ The formal group tries to be the simply connected form. The Lang isogeny $G/G(\mathbb F_p)=G$ shows that no positive dimensional group scheme is simply connected as a scheme. Every commutative group scheme has a covering by a group, which has more representations. But I believe that a semisimple group has only finitely many covering homomorphisms and that one that is "simply connected" in the sense of the weight lattice is maximal. You could hope that its finite dimensional representations are the same as those of its formal group. I believe this is true. $\endgroup$– Ben WielandApr 14, 2022 at 17:49
2 Answers
The short answer is that the characteristic $p$ picture genuinely has more depth. In particular it's not correct to think of formal groups as "better" than Lie groups. In fact, they can be put on equal footing with each other.
I think it's helpful to put all the objects you're interested in in the same category. In this case it is the category of group objects in formal schemes, which I will call formal group objects. There is some notational confusion here. By rights, the category of formal group schemes should denote this category (which includes all algebraic groups as well). However, often when people say "formal group scheme" or "formal group" they mean infinitesimal groups, which are formal group schemes with a single point (and other times they mean something else more restrictive, such as formally smooth infinitesimal group schemes, which I don't want to consider separately). Here I'll hopefully avoid the notational confusion and use group objects for group objects in formal schemes and infinitesimal groups for group objects with one point.
Now in characteristic zero, every algebraic group $G$ has a unique infinitesimal subgroup $\hat{G}\subset G$ (as a group object) subject to the condition of having the same tangent space, i.e., the natural map $T_e\hat{G}\to T_e G$ being an isomorphism. In characteristic $p$, this is no longer the case. If $G$ is an algebraic group in characteristic $p$ then there is still a maximal subgroup object $\hat{G}\subset G$ corresponding to the "full" formal group of $G$, but there is another smaller object $$\hat{G}^{(1)}\subset \hat{G},$$ the first Frobenius neighborhood, which also has the same tangent space (there are also objects $\hat{G}^{(2)}, \dots,$ in addition to possible other intermediate subgroups). The group $\hat{G}^{(1)}$ has the nice property that its representations are equivalent to representations of the Lie algebra $\mathfrak{g}$ as a symmetric monoidal category, and this uniquely characterizes $\hat{G}^{(1)}.$ In fact if $\mathfrak{g}$ is an arbitrary Lie algebra in characteristic $p$, it might not integrate to a smooth algebraic group $G$, but it will always integrate to a group object $G^{(1)}$ that looks like a first Frobenius neighborhood.
Now even if you look at the "full" formal neighborhood $\hat{G}\subset G,$ you will not have anything resembling integration of representations. Indeed, if $G\to \text{End}(V)$ is a representation of a smooth algebraic group scheme, you do get induced representations of $\hat{G}$ and $\hat{G}^{(1)}$ (equivalently, $\mathfrak{g}$). But you can neither integrate representations of $\hat{G}^{(1)}$ to representations of $\hat{G}$ nor representations of $\hat{G}$ to representations of $G$ in any reasonable sense, even in onedimensional cases.

$\begingroup$ Sorry, could you please give a reference / a brief summary of the "first Frobenius neighborhood" that you mentioned? Thanks. $\endgroup$– Z. MApr 15, 2022 at 9:49
Onedimensional algebraic groups are essentially multiplicative, elliptic curves, or additive, so their associated formal groups have height $1$, $2$ or $\infty$. There exist onedimensional commutative formal groups of arbitrary height, so most of these cannot be integrated.
(This is a wellknown fact in chromatic homotopy theory, related to the special status of $K$theory and elliptic homology as opposed to Morava $K$theories of height $>2$.)