# Is a basis for the Lie algebra of a Lie group also a set of infinitesimal generators for the Lie group?

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.

-
$G$ 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
@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

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.