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.

Let $G$ be a (EDIT: connected) Lie group of dimension $n$, and let $\mathfrak{g}$ be the associated Lie algebra. If $x_1,\ldots,x_n$ is a basis for $\mathfrak{g},$ is it necessarily true that the 1-parameter subgroups $e^{tx_1},\ldots,e^{tx_n}$ generate $G$?

Note: It is sufficient to show that the subgroup generated by the 1-parameter subgroups is closed, since it follows that it is a Lie subgroup of dimension $n$. In particular, it must contain a neighborhood of the identity, which generates $G$.

Also, note that it is not always true that the subgroup generated by 1-parameter subgroups is a Lie subgroup; consider the 1-parameter subgroup of the 2-torus that is a line with irrational slope.

share|improve this question
    
$G$ needs to be connected at least. I don't know the answer in general. –  Grumpy Parsnip May 19 '12 at 14:43
    
This should be true although the details seem messy. The basic observation is that for $t$ sufficiently small we have $e^{t(c_1 x_1 + ... + c_n x_n)} \approx e^{t c_1 x_1} ... e^{t c_n x_n}$. Then use the fact that the exponential map is a diffeomorphism on a sufficiently small neighborhood of the identity. –  Qiaochu Yuan May 19 '12 at 15:04
    
@JimConant: Thanks, fixed. –  Rob Silversmith May 19 '12 at 15:16
1  
@QiaochuYuan: That was my basic idea too. I think this works: Let $x=\sum c_ix_i\in\mathfrak{g}.$ Consider the map $f:\mathfrak{g}\to G$ given by $f(x)=e^{c_1x_1}\cdots e^{c_nx_n}$. We have $$df_0(x)=\left.\frac{d}{dt}f(tx)\right|_{t=0}=\frac{d}{dt}(1+tc_1x_1+O(t^2))...‌​(1+tc_nx_n+O(t^2))=\frac{d}{dt}(1+tx+O(t^2))=x.$$ This implies $df$ is surjective in a neighborhood of zero, which implies $f$ is open in a neighborhood of zero. Thus the image of $f$, which is inside the subgroup generated by the 1-parameter subgroups, contains a neighborhood of the identity, and thus is the entire group. –  Rob Silversmith May 19 '12 at 16:12
    
@Rob: Looks good to me - you may want to post this as an answer to the question. –  Jason DeVito May 19 '12 at 16:56

1 Answer 1

Let $x=\sum c_ix_i\in\mathfrak{g}$. Consider the map $f:\mathfrak{g}\to G$ given by $f(x)=e^{c_1x_1}\cdots e^{c_nx_n}$. We have $$df_0=\left.\frac{d}{dt}f(tx)\right|_{t=0}=\frac{d}{dt}(1+tc_1x_1+O(t^2))...(‌1+tc_nx_n+O(t^2))=\frac{d}{dt}(1+tx+O(t^2))=x.$$

This implies $df$ is surjective in a neighborhood of zero, which implies $f$ is open in a neighborhood of zero. Thus the image of $f$, which is inside the subgroup generated by the 1-parameter subgroups, contains a neighborhood of the identity, and thus is the entire group.

share|improve this answer

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.