Any element of $SO(n)$ decomposes as a product of $n(n-1)/2$ elements, each an exponential of mutually linearly independent generators? Physics person here, so this might be a simple question that has a straightforward answer in some subfield of math that I am not aware of. Thanks in advance!
We are given an arbitrary set of $n(n-1)/2$ skew-symmetric matrices $X_i$ that span the real vector space of the skew-symmetric matrices. For any $T \in SO(n)$, can one decompose $T$ as $T = \Pi_{i=1}^{n(n-1)/2} e^{a_i X_i}$, for some $a_i \in \mathbb{R}$?
I have not been able to find a reason why this should not be the case for an arbitrary basis (or, at least, an orthogonal basis) of the space of skew-symmetric matrices. For example, if the $X_i$ are orthogonal (wrt Frobenius inner product) and each generates a plane (2D) rotation, then there is a way to decompose $T$ as claimed - though the decompositions I am aware of specify particular orders of the compositions of the $e^{a_i X_i}$.
More generally, for a finite dimensional matrix Lie group, can one decompose any element as a product of exponentials of scaled basis elements of the corresponding Lie algebra for any basis (with the product length equal to the number of parameters describing the group)?
 A: I will give a proof that the answer to your question is negative for $n=3$, but, with a bit more work, this proof extends to all $n\ge 3$.
I will prove that for every $N$, there exist vectors $X_1,...,X_N$ spanning the Lie algebra $o(3)$ such that the map
$$
F=F_{(X_1,...,X_N)}: (a_1,...,a_N)\mapsto \exp(a_1 X_1) ... \exp(a_N x_N),$$
$$
(a_1,...,a_N)\in {\mathbb R}^N,
$$
is not onto $SO(3)$, i.e. the image of this map is not the entire group $SO(3)$. (You are asking about the case when $N=3$.) In other words, there exists a matrix $T\in SO(3)$ which cannot be written as a product
$$
\exp(a_1 X_1) ... \exp(a_N X_N). 
$$
To prove this, first note that, by rescaling the vectors $X_i$, it suffices to consider the case when they have unit norm with respect to the standard metric on the Lie algebra $o(3)$.
For each unit vector $X\in o(3)$, the map
$$
a\mapsto \exp(aX)
$$
is periodic, with period $P=2\sqrt{2}\pi$. Thus, it suffices to limit ourselves to $N$-tuples $(a_1,...,a_N)$ satisfying
$$
0\le a_i\le P, i=1,...,N. 
$$
The key is that the domain of the map $F$ is now compact. The map
$$
(a_1,...,a_N, X_1,...,X_N)\mapsto \exp(a_1 X_1) ... \exp(a_N x_N),$$
is continuous. It follows that the image of the map $F$ depends continuously with the $N$-tuples $(X_1,...,X_N)$. This is intuitively clear and I can write a proof, provided you are familiar with some basic multivariable real analysis (like the notion of compactness), or, better, general topology. But, let's continue. Start with the $N$-tuple of unit vectors in $o(3)$ of the form
$$
(X,...,X). 
$$
For such tuple, the image of the map $F=F_{(X,...,X)}$ equals $\exp({\mathbb R}X)\cong SO(2)$ (a circle inside $SO(3)$). Therefore, by continuity noted above, if an $N$-tuple $(X_1,...,X_N)$ is close to $(X,...,X)$, then the image of the corresponding map $F=F_{(X_1,...,X_N)}$ is close to that circle.
A bit more precisely, if $U$ is a neighborhood of that circle in $SO(3)$) (think of a small tube around the circle), then for all $(X_1,...,X_N)$ sufficiently close to $(X,...,X)$, the image of the map  $F=F_{(X_1,...,X_N)}$ lies in $U$.
Hence, the map  $F=F_{(X_1,...,X_N)}$ is not onto.
Lastly, note that "generic" tuples $(X_1,...,X_N)$ will span the entire Lie algebra $o(3)$. Such generic tuples one can find arbitrarily close to $(X,...,X)$. This proves the claim.
So, what is the difference between these examples and the statement about representing each element of $SO(3)$ as a product of three matrices from the given 1-parameter subgroups $\exp({\mathbb R}X_i), i=1,2,3$? In that statement, the triple $(X_1, X_2, X_3)$ forms an orthogonal basis of $o(3)$. In contrast, in my examples, the angles between the vectors $X_1,...,X_N\in o(3)$ will be close to zero.
