Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given an analytic group action of $SL(n, \mathbb{R})$ on $\mathbb{R}^m$ fixing the origin, in this article the author then proceeds to "complexify the analytic $SL(n, \mathbb{R})$ action to obtain a local holomorphic action of $SL(n, \mathbb{C})$ on a neighbourhood of the origin in $\mathbb{C}^m$" (pg 139, 1st paragraph of theorem 2.6).

That sounds quite reasonable, given that the action is analytic. I believe I see how one obtains $SL(n, \mathbb{C})$ as the complexification of $SL(n, \mathbb{R})$, but how may I describe the action of an element in $SL(n, \mathbb{C})$ on $\mathbb{C}^m$ given that the group action is the result of a complexified action of $SL(n, \mathbb{R})$?

I've the following for the complexification of $SL(n, \mathbb{R})$: we think of this group as sitting in $SL(n, \mathbb{C})$. The lie sub-algebra $\mathfrak{sl}(n, \mathbb{R})$ is the set of n-by-n matrices over $\mathbb{R}$ with vanishing trace. Looking at $\mathfrak{sl}(n, \mathbb{R})\oplus i \mathfrak{sl}(n, \mathbb{R})$ it is clear that we obtain all n-by-n matrices over $\mathbb{C}$ with vanishing trace, that is $\mathfrak{sl}(n, \mathbb{C})$. The complexification of $SL(n, \mathbb{R})$ is then the image of $\mathfrak{sl}(n, \mathbb{C})$ under the exponential map, which translates to the connected component of $SL(n, \mathbb{C})$ containing the identity. Since $SL(n, \mathbb{C})$ is connected the image is all of $SL(n, \mathbb{C})$.

So taking an element $g\in SL(n, \mathbb{C})$ we can write it as $g=e^{\alpha+i\beta}$ for $\alpha, \beta \in \mathfrak{sl}(n, \mathbb{R})$, but I'm not sure how to put these pieces together to give me the action of $g$ on $\mathbb{C}^m$. Can this reasoning be completed, or is there perhaps a better approach?

share|improve this question
    
Just to be sure, in that first part: Since you specify that it should fix the origin, you don't actually assume the action to be linear? –  Tobias Kildetoft Feb 27 '13 at 16:28
    
Right, the action is not assumed to be linear but only to fix the origin –  Amos Joshua Feb 28 '13 at 10:44

1 Answer 1

up vote 0 down vote accepted

The following is only a partial answer, because I'm working only locally in both the group and the space on which it acts. For a complete answer, one would have to show that the resulting action is (or can be extended to be) well-defined on the whole group.

The given action of $SL(n,\mathbb R)$ on $\mathbb R^n$ is given, in a neighborhood of the identity in $SL(n,\mathbb R)$ and the origin in $\mathbb R^n$, by convergent power series in the entries of matrices in $SL(n,\mathbb R)$ and the coordinates of points in $\mathbb R^n$. Use the same power series with complex matrix entries and complex coordinates to define a local action of $SL(n,\mathbb C)$ on $\mathbb C^n$.

share|improve this answer
    
An analytic action of $SL(n, \mathbb{R})$ means that the map induced by an element $g$ that sends $x\in\mathbb{R}^m$ to $gx$ is given by a convergent power series in the coordinates of $x\in\mathbb{R}^m$, but the series would be different for each element of $SL(n, \mathbb{R})$, right? When you say that we can have one convergent series that describes the action of all elements, that is made possible because we are "working only locally in both the group and the space on which it acts"? Does this correspond to the author's suggestion that we obtain a "local holomorphic" action? –  Amos Joshua Feb 27 '13 at 16:49
    
An analytic action would be given locally by a single power series in the coordinates of both $x$ and $g$. The reason my answer is only local is that the power series might converge only in a neighborhood of the point where $g=I$ and $x=0$. You can similarly get locally convergent power series about any real $g$ and $x$, and they'll converge for nearby complex $g$ and $x$ too, but more work is needed to define the action far from the real part. As I understand it, "local" in the question means you don't need to worry about $x$ far from 0, but you do for $g$ far from $I$. –  Andreas Blass Feb 27 '13 at 16:56
    
that makes sense in the context of the rest of the article (the goal is to conjugate the action to its linear part in a neighborhood of the origin, i.e. locally in $\mathbb{R}^m$, but for all of $SL(n, \mathbb{R}^m)$). Thank you for your help! Are there any common methods one might employ to establish convergence for g far from I? (Not in detail, just the general idea) –  Amos Joshua Feb 28 '13 at 14:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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