Let $G$ be a simply connected algebraic group defined over $C$.

Note : I see the definition of simply connected as "Every isogeny to $G$ is an isomorphism" as given in Hochschild's "Basic Theory of Algebraic Groups and Lie algebras".

$C$ is an algebraically closed field.

Please help me realise the follwoing things provided they are true:

  1. $G_C$ (the $C$-points of $G$) is simply connected as well.

  2. $Lie(G) \cong Lie(G_C)$

Thanks !

  • 1
    $\begingroup$ Do you mean $G(\mathbb C)$, the complex points with their natural complex topology, or just the subset $G(C)$ of closed points in the Zariski topology? If the latter, the definition of simply connected and of Lie algebra should be the same as for $G$ itself (namely, an algebraic definition, rather than a topological one). $\endgroup$ – Remy Feb 18 '16 at 5:19
  • $\begingroup$ @Remy I meant the latter one , that is, $G$ is a variety having a group structure with appropriate morphisms and the defining equations are over an algebraically closed field $C$ and $G_C$ means the points in $C^n$ satisfying the defining equations. $\endgroup$ – Jagdeep Singh Feb 18 '16 at 5:22
  • $\begingroup$ That is to say, if you're interested in the Zariski topology on $G(C)$, then you're just using the equivalence between separated, integral schemes of finite type over $C$ and varieties over $C$, cf. Hartshorne Prop II.2.6. $\endgroup$ – Remy Feb 18 '16 at 5:22
  • $\begingroup$ @Remy The question is certainly referring to the relationship between $G$ and $G^\text{an}$. $\endgroup$ – Alex Youcis Feb 18 '16 at 13:24
  • $\begingroup$ @AlexYoucis Now since $Lie(G) \cong Lie(G_C)$, doesn't that imply that $G$ and $G_C$ are isogeneous. But since both are simply connected, we don't have any isogenies. Does that mean they are isomorphic. All this is not making sense to me. Please clarify. Thank You ! $\endgroup$ – Jagdeep Singh Feb 20 '16 at 9:51

The fact about Lie algebras is clear if you think explicitly in terms of an embedding $G\hookrightarrow\text{GL_n}$. Then, you can, 'by hand', see what the Lie group is both sides. The first question is slightly more fidgety.

Namely, I think one method of attack is as follows, at least in the case of semisimple $G$. One can show that for an algebraic group $G/\overline{k}$ that $G$ is simply-connected (in the sense of having no non-trivial connected central isogenies) if and only if the coweight lattice equals the lattice generated by the coroots. Now, the exact same statement holds true for (complex) Lie groups, and by standard theory (the fact that all maps between the analytification of reductive groups are algebraic) implies that the coweight lattices/coroots are the same. Thus, it's true that $G^\text{an}$ is (topologically) simply connected.

Another, simpler, approach, which I have not really sussed out totally, is the following. Suppose that $\pi_1^\text{ét}(G)\ne 0$. Then, there exists some connected finite étale cover $G'\to G$. But, it's standard that any such cover $G'$ can be given the structure of an algebraic group such that $G'\to G$ is a central isogeny. Thus, if $G$ is simply-connected in the sense you mentioned, then $\pi_1^{é\text{t}}(G)=0$. So,


But, $\pi_1^\text{top}(H)$ is a finitely generated abelian group for any Lie group $H$. Thus, we see that $\widehat{\pi_1^\text{top}(G^\text{an})}=0$ implies that $\pi_1^\text{top}(G^\text{an})=0$ as desired.


Let me give some more explanation of how one can make the identification


for $G/\mathbb{C}$ a linear algebraic group, and $G^\text{an}$ the analytification.

By standard theory, we may as well assume that we've been handed $G$ as a matrix group—that we have a fixed (closed) embedding $G\hookrightarrow \text{GL}_n$ (I don't want to write $\text{GL}_{n,\mathbb{C}}$. In particular, let's assume that $G$ is defined in $\text{GL}_n$ by the Hopf ideal $I=(f_1,\ldots,f_m)$.

Recall that one can define $\text{Lie}(G)$ as

$$\ker(G(\mathbb{C}[\varepsilon])\to G(\mathbb{C}))$$

where $\mathbb{C}[\varepsilon]:=\mathbb{C}[x]/(x^2)$ are the ring of dual numbers of $\mathbb{C}$, and the map $G(\mathbb{C}[\varepsilon])\to G(\mathbb{C})$ comes from the natural quotient map $\mathbb{C}[\varepsilon]\to\mathbb{C}$ given by $\varepsilon\mapsto 0$. Thus, we see that we can identify $\text{Lie}(G)$ in our case with

$$\ker\left(\{(a_{i,j}+b_{i,j}\varepsilon)\in\text{GL}_n(\mathbb{C}[\varepsilon]): f_\ell(a_{i,j}+b_{i,j}\varepsilon)=0\}\to \{(c_{i,j})\in\text{GL}_n(\mathbb{C}): f_\ell(c_{i,j})=0\}\right)$$

which takes a matrix $(a_{i,j}+b_{i,j}\varepsilon)$ to $(a_{i,j})$.

But, note that this kernel is just

$$\{I+(b_{i,j})\varepsilon:(b_{i,j})\in\text{Mat}_n(\mathbb{C}),\text{ and } f_\ell(1+b_{i,j}\varepsilon)=0\}$$

But, note that $f_\ell(1+b_{i,j}\varepsilon)=0$ if and only if $\varphi_\ell(b_{i,j})=0$, where $\varphi_\ell$ is the linear part of of $f_\ell$ expanded in a Taylor series around the identity matrix. Thus


which as the standard bracket structure (i.e. it's a Lie subalgebra of $\mathfrak{gl}_n$.

Now, I leave it to you to check that the EXACT SAME Lie algebra is what one gets if they compute $\text{Lie}(G_\mathbb{C})$ which still has the embedding into $\text{GL}_n(\mathbb{C})$—it's the vanishing set $V(f_1,\ldots,f_m)$.

EDIT EDIT: As to your questions above

Doesn't that mean that $G$ and $G_\mathbb{C}$ are isogenous?

I don't know what that means. I think you need to be careful here which, depending on your taste of algebraic geometry, may not be super obvious. Namely, $G$ is a linear algebraic group over $\mathbb{C}$ and $G^\text{an}$ (or what you're calling $G_\mathbb{C}$) is a complex Lie group. They don't even live in the same category—one is just the image of the other under the analytification (or GAGA) functor. Asking if they're isogenous doesn't really make sense.

"A homomorphism $G \to H$ of connected affine algebraic groups is an isogeny if and only if $\text{Lie}(G) \to \text{Lie}(H)$ is an isomorphism. So, does this mean that if 2 connected algebraic groups $G$ and $H$ have same Lie algebra, then there will "always" exist an isogeny between them?

Over $\mathbb{C}$ this is true—think analytically, and use the fact that they have the same universal covers.

It's not true in positive characteristic. The groups $\mu_p$ and $\alpha_p$ both have abelian $1$-dimensional Lie algebras. There are no non-constant maps $\mu_p\to\alpha_p$ or vice-versa though

  • $\begingroup$ Please explain "Then, you can, 'by hand', see what the Lie group is both sides. " I understand that given $G$ to be a linear algebraic group, $G$ can be embedded in $GL(n,K)$. I would appreciate if you could elaborate how the lie algebra of the $C-points$ of $G$ is same as the lie algebra of $G$. $\endgroup$ – Jagdeep Singh Feb 20 '16 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.