# 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)?

• No, for an arbitrary choice of the matrices $X_i$, this is impossible already when $n=3$. Commented May 14, 2022 at 8:09
• Thanks, can you give a reference? I still cannot see why this is impossible, even for $n=3$. I can imagine writing the each of the three products $e^{a_iX_i}$ as a product of three coordinate-plane rotations, with the angles depending on $a_i$. Then I have a product of $9$ coordinate-plane rotations with three parameters, $a_1,a_2,a_3$, to tune the $9$ angles. It seems sufficient to be able to generate an arbitrary rotation since one can generate an arbitrary rotation with just three coordinate-plane rotations. Commented May 14, 2022 at 15:04

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.

• Thanks a lot! I do know some general topology, so I understand the proof. The extension to arbitrary $n$ is straightforward. If I were to ask, given a particular basis of $o(n)$ (or a larger set that spans $o(n)$), how would you tell whether the corresponding $F_{(X_1,\dots,X_N)}$ is surjective, I am guessing that is a much harder question? Commented May 14, 2022 at 23:15
• I am asking since you mentioned that for "generic" tuples $(X_1,\dots,X_N)$ the corresponding $F$ is surjective, but I think that does not (at least immediately) follow from this proof. Commented May 14, 2022 at 23:28
• @3Brown1Blue No, I did not say this and it is actually false. As for the question about particular bases, it looks hard. Commented May 15, 2022 at 2:26
• Sorry, have to learn to read twice before commenting. Thanks again for your help! Commented May 15, 2022 at 4:20
• @3Brown1Blue Not a problem. On the second thought, I can prove that for SO(3) and $N=3$, the map $F$ is onto if and only if we have an orthogonal basis of o(3). Commented May 15, 2022 at 4:36