Lie groups and algebraic groups both correspond with Lie algebras, which are by definition the left invariant vector field. But the topology of Lie groups and algebraic groups are different. Are their Lie algebras different?

More concretely, let $G=GL(n,\mathbb C)$ for $n$ a positive integer. When it is considered as a Lie group, the production is given the product topology. But when it is considered as an algebraic group, it is the Zariski topology that is used. In the first case, $G$ is not connected, while in the second case, $G$ is connected. Both of them have Lie algebra $M(n,\mathbb C)$. Are the two Lie algebras "really" all the same?

If they are all the same, is this always the case when $G$ is an arbitrary group which can be considered both as a Lie group and as an algebraic group?

  • 1
    $\begingroup$ $GL(n,\mathbb{C})$ is connected, whether you use the product topology or the Zariski topology. $\endgroup$ – Ted Nov 15 '12 at 8:56
  • $\begingroup$ @Ted: Thank you for the comment. Is it wrong that $GL(n, \mathbb C)^{\circ}$ consists of the matrices of determinant $>0$, when it is a Lie group? $\endgroup$ – ShinyaSakai Nov 17 '12 at 1:56
  • $\begingroup$ If the field were $\mathbb{R}$, you would get 2 components because $\mathbb{R} - \{0\}$ is disconnected. But $\mathbb{C} - \{0\}$ is connected, so that argument does not apply. In fact, $GL(n,\mathbb{C})$ is connected (with the topology as a subspace of $\mathbb{C}^{n^2}$). $\endgroup$ – Ted Nov 17 '12 at 7:44
  • $\begingroup$ @Ted: I was mistaken. Thank you very much. $\endgroup$ – ShinyaSakai Nov 18 '12 at 7:58

Your statement about "the case when $G$ is an arbitrary group which can be considered both as a Lie group and as an algebraic group" doesn't really make sense on the face of it: Lie groups are not algebraic groups and vice versa. But if $G$ is an algebraic group over $\mathbb{C}$, then the set of its points $G(\mathbb{C})$ is a complex Lie group, and the Lie algebra of $G$ (as an algebraic group) and of $G(\mathbb{C})$ (as a Lie group) coincide. Similarly with $\mathbb{C}$ replaced by $\mathbb{R}$. Here the algebraic group $G$ and the Lie group $G(\mathbb{C})$ are very different things: don't confuse an algebraic variety with the set of its points.

Perhaps it would help to draw a diagram:

$$ \begin{array}{ccc} \text{algebraic groups} & \to & \text{Lie groups}\\ & \searrow & \downarrow\\ & & \text{Lie algebras} \end{array}$$

(This is a diagram of functors between categories, if you like). The diagonal arrow and the downward arrow have very different definitions; they're not just two instances of the same functor applied to topological groups where we regard an algebraic group as a topological group in the Zariski topology. And that, I hope, clears up your confusion about the topologies being different: it doesn't matter, because the functors are different too. :-)

  • $\begingroup$ Thank you very much for the answer. "If $G$ is an algebraic group over $\mathbb C$, then the set of its points $G(\mathbb C)$ is a complex Lie group", since we must have a manifold structure on a Lie group, would you please tell me how is the manifold structure on $G(\mathbb C)$ defined? Thanks a lot. $\endgroup$ – ShinyaSakai Nov 18 '12 at 7:53
  • $\begingroup$ If $X$ is any non-singular algebraic variety over $\mathbb{C}$, then $X(\mathbb{C})$ is a complex manifold by the implicit function theorem. (Moreover, any algebraic group over a field of char 0 is nonsingular.) $\endgroup$ – David Loeffler Nov 18 '12 at 12:00

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.