Dimension of a Lie group associated to a finite group

Question

Let $G$ be a finite group. Consider—I think I've got my terminology right—the full subcategory $\mathcal C$ of the coslice category $G \downarrow \operatorname{Lie}$ of Lie groups under $G$ whose objects are (morphisms into) connected Lie groups. In other words, consider the category whose objects are morphisms $G \to H$ with $H$ a connected Lie group, and whose morphisms are triangles under $G$ in the obvious sense. (I changed 'over' to 'under', per @QiaochuYuan's comment.) There is nothing inherently category theoretic about my questions, but the language seems to be well suited to it.

(1) Does $\mathcal C$ have an initial object?

(2) Even if the answer to (1) is ‘no’, there is a well defined dimension $$d(G) = \min \{\dim(H) \mathrel: \text{$H$ a connected Lie group and $G$ embeds in $H$}\}.$$ (Note that the set is non-empty; $G$ embeds in an appropriate general linear group via its regular representation.) What purely group theoretic information about $G$ is recorded by $d(G)$?

(3) What changes in (1) or (2) if we replace topologically connected Lie groups by Zariski connected linear algebraic groups over a fixed (not necessarily characteristic 0, not necessarily algebraically closed) field $k$ (and so differentiable maps by algebraic maps)?

Answer 1

Q1: The answer is no already for $H = C_2$.

$C_2$ embeds into exactly one $1$-dimensional connected Lie group, namely $S^1$, and does so uniquely, because there is a unique element of order $2$ in $S^1$. This embedding is weakly initial: if $C_2 \to G$ is any other map from $C_2$ to a connected Lie group $G$ then it extends, not-necessarily-uniquely, to a map $S^1 \to G$. This follows from the fact that $G$ has a maximal compact subgroup $K$, which is connected, together with the fact that the exponential map $\exp : \mathfrak{k} \to K$ is surjective.

Now we need a general fact about weakly initial objects.

Proposition: If $C$ is a category with an initial object $0$, then the weakly initial objects are exactly the objects admitting a morphism to $0$, which is necessarily a split epimorphism.

Proof. If $i$ is a weakly initial object, then by definition it admits a morphism $f : i \to 0$. Since $0$ is initial this is necessary and sufficient for it to admit a morphism to every other object. Moreover there is a unique map $0 \to i$, and the composition $0 \to i \xrightarrow{f} 0$ is necessarily $\text{id}_0$. So $f$ has a section, hence is a split epimorphism. $\Box$

It follows that if the category of maps from $H = C_2$ to connected Lie groups has an initial object then it must admit a split epimorphism from the embedding $C_2 \to S^1$ above. This map must remain a split epimorphism on underlying Lie groups after forgetting the embedding. But the only nontrivial split epimorphism out of $S^1$ is the identity (there is also the zero map to the trivial Lie group but $C_2$ can't embed into that). So if there is an initial object it must be the embedding $C_2 \to S^1$.

But the embedding $C_2 \to S^1$ is not initial, because it has nontrivial endomorphisms. Namely, the map

$$S^1 \ni z \mapsto z^n \in S^1$$

sends $-1$ to $-1$ whenever $n$ is odd.

Q2: This seems to me like both a pretty delicate and a pretty open-ended question so I don't know what could qualify as a complete answer to it, but here are some initial observations.

• The existence of maximal compact subgroups implies that we can restrict our attention to embeddings into compact connected Lie groups. In particular the only $1$-dimensional compact connected Lie group is $S^1$ and the only $2$-dimensional compact connected Lie group is $S^1 \times S^1 = T^2$.
• If $H = C_n$ is a cyclic group then it embeds into $S^1$ and this is clearly best possible so $d(C_n) = 1$. Conversely the finite subgroups of $S^1$ are exactly the finite cyclic groups so $d(H) = 1$ characterizes the finite cyclic groups.
• If $d(H) = 2$ then $H$ is not finite cyclic but embeds into $T^2$. By considering the projections to each factor this means $H$ embeds into a product $C_n \times C_m$ of two finite cyclic groups. Working one prime at a time this means that $H$ is itself the product of two finite cyclic groups; see, for example, Hagen von Eitzen's answer here.
• It's tempting to conjecture that if $H$ is finite abelian then the minimal embedding is an embedding of $H$ into some torus (of dimension the minimal $k$ such that $H$ can be written as a product of $k$ cyclic groups) but I don't see how to prove it. It's not true in general that a finite abelian subgroup of a compact connected Lie group $K$ is contained in a maximal torus; for example, as Daniel Fischer describes here, $C_2 \times C_2$ embeds into $SO(3)$, whose maximal torus is $SO(2) \cong S^1$.
• In any case, by considering embeddings into a torus we get that if $H$ is finite abelian then $d(H) \le k$ where $k$ is minimal such that $H$ embeds into a product of $k$ finite cyclic groups. Working one prime at a time, this is equivalently the supremum over all primes $p$ of the minimal number of generators of the localization $H_{(p)}$ of $H$ at $p$, which I believe implies that it's the minimal number of generators of $H$ (sometimes called the rank). Alternatively, embeddings $H \to T^k$ are Pontryagin dual to quotients $\mathbb{Z}^k \to H^{\ast}$, and every finite abelian group is Pontryagin self-dual.
• From this point on we'll only consider nonabelian $H$, which have $d(H) \ge 3$. If $d(H) = 3$ and $H$ is nonabelian then $H$ is a finite subgroup of either $SU(2)$ or $SO(3)$ and these are famously completely classified although I don't know what source to point you to for the details; this MO question might be a start. For $SO(3)$ the nonabelian groups are the dihedral groups $D_n$, the tetrahedral group $A_4$, the octahedral group $S_4$, and the icosahedral group $A_5$. For $SU(2)$ the nonabelian groups are central extensions of these; the binary dihedral groups, the binary tetrahedral group (same order as but not isomorphic to $S_4$), the binary octahedral group, and the binary icosahedral group (same order as but not isomorphic to $S_5$). So for all other nonabelian groups we have $d(H) \ge 4$.

In general the exact value of $d(H)$ is sensitive to the classification of compact connected Lie groups so, for example, there may be large finite groups with unusually small values of $d(H)$ because they have tricky embeddings into some exceptional Lie group. It seems hard to be too precise here. One of the only general results I know is that for a bounded value of $d(H)$, $H$ must have a normal abelian subgroup of bounded index (by the Jordan-Schur theorem), but I think the bound is quite bad.

Q3: This also seems pretty delicate and pretty open-ended. For embeddings of a finite group $H$ into the group of $k$-points $G(k)$ of a reductive group $G$ it's possible to write down explicit and tight bounds on the sizes of the Sylow subgroups of $H$ in terms of $G$ and $k$; see for example Serre's Bounds for the orders of the finite subgroups of $G(k)$. To give the flavor of these results here is the bound for $GL_n(\mathbb{Q})$, which is due to Minkowski: the Sylow $\ell$-subgroup of a finite group $H \hookrightarrow GL_n(\mathbb{Q})$ can have order at most $\ell^{M(n, \ell)}$ where

$$M(n, \ell) = \left\lfloor \frac{n}{\ell-1} \right\rfloor + \left\lfloor \frac{n}{\ell(\ell-1)} \right\rfloor + \left\lfloor \frac{n}{\ell^2(\ell-1)} \right\rfloor + \dots \le \frac{n \ell}{(\ell - 1)^2}.$$

This bound is tight, meaning $GL_n(\mathbb{Q})$ has a finite $\ell$-subgroup of this order, which acts as a Sylow $\ell$-subgroup in the sense that every finite $\ell$-subgroup embeds into it up to conjugacy. (Note the analogy to $S_n$ where the corresponding exponent is given by Legendre's formula, which is the above formula with $\ell-1$ replaced by $\ell$. Since $S_n \hookrightarrow GL_n(\mathbb{Q})$ this is a lower bound on the above as expected.) It follows that the order of a finite subgroup $H$ of $GL_n(\mathbb{Q})$ divides $M(n) = \prod_{\ell} \ell^{M(n, \ell)}$ (this bound is not tight).

But minimizing over all embeddings into algebraic groups makes the answer sensitive to the classification of algebraic groups as above so I don't know how to easily say anything about it. If $k$ has characteristic $0$ then finite groups can't embed nontrivially into unipotent groups over $k$ so I believe that Levi decomposition implies that we can restrict our attention to connected reductive groups, but I'm not too familiar with the classification of reductive groups. And if $k$ has characteristic $p$ then $p$-groups embed into unipotent groups over $k$ so those will have exceptional behavior.

Generally, considering the special case of embeddings of finite abelian groups into algebraic tori shows that the answer is sensitive to which roots of unity exist over $k$ or over low-degree extensions of $k$, and the bounds given by Serre in the paper above validate this idea.