# A curious interrelationship between distinct embeddings of $SO(M+1)$ into $SO(2M+1)$

The following seems to be a property of $$SO(2M+1)$$ for an arbitrary integer $$M$$, although I have not yet been able to prove it. (I can prove it for, e.g., $$M=1$$, and have numerically checked it for other values of $$M$$.)

Notation:

1. I denote the $$2M+1$$ coordinates of $$\mathbb R^{2M+1}$$ as $$x_0,x_1,\cdots,x_{2M}$$.
2. Let us denote the Lie algebra generators of $$SO(2M+1)$$ by $$L_{ab}$$ with $$a< b \in \{0,1,2,\cdots,2M\}$$. Geometrically, we can think of $$L_{ab}$$ as generating a rotation in the $$(x_a,x_b)$$-plane. If one likes to be more concrete, we can identify it with the matrix $$A \in \mathbb R^{(2M+1)\times (2M+1)}$$ whose only nonzero entries are $$\left(A \right)_{a,b} = 1$$ and $$\left(A \right)_{b,a} = -1$$.
3. For a products of matrices, we use the ordering convention $${\prod}_{i=0}^n A_i := A_1 A_2 \cdots A_n$$.

Conjecture: for any vector $$\vec a \in \mathbb R^M$$ there exists a vector $$\vec b \in \mathbb R^M$$ such that $$\boxed{R := e^{\sum_{k=0}^{M-1} b_k L_{M,M+k+1} } \left( \prod_{i=0}^M e^{\sum_{k=0}^{M-1} a_k L_{i,i+k+1} }\right) e^{\sum_{k=0}^{M-1} b_k L_{M-(k+1),M} } }\; (\in SO(2M+1))$$ leaves the middle coordinate of $$\mathbb R^{2M+1}$$ (i.e., $$x_M$$) invariant.

How can one prove this? And can such a proof be constructive?

For $$M=1$$, one can constructively show that the above is true for $$b = - \arctan( \sin a)$$. For $$M>1$$, I have numerically confirmed it for random vectors $$\vec a$$ for up to $$M \approx 20$$.

EDIT: I have been able to simplify the effective problem.

Firstly, note that the orthogonal matrix $$R$$ leaving the central coordinate invariant is equivalent to the component $$R_{M,M} = 1$$.

Secondly, if we define the vector $$\vec v^T = (0,\cdots,0,1,0,\cdots,0)$$ with the entry $$1$$ on index $$M$$, then one can show that $$\vec v^T \cdot e^{\sum_{k=0}^{M-1} b_k L_{M,M+k+1} } = (0,\cdots,0,\cos (\theta),\sin(\theta) n_0,\sin(\theta) n_1 ,\cdots,\sin(\theta) n_{M-1})$$ where $$\theta := |\vec b|$$ and $$\vec n := \vec b / |\vec b|$$. Hence, since $$R_{M,M} = \vec v^T \cdot R \cdot \vec v$$, one can show that $$R_{M,M} = \vec u^T \cdot A \cdot \vec u$$ where $$\vec u^T = (\cos (\theta),\sin(\theta) n_0,\sin(\theta) n_1 ,\cdots,\sin(\theta) n_{M-1})$$ and $$\boxed{ A_{i,j} := \left( \prod_{l=0}^M e^{\sum_{k=0}^{M-1} a_k L_{l,l+k+1} }\right)_{M+i,M-j} }.$$

In conclusion, we conjecture that the above matrix $$A$$ always has an eigenvector with eigenvalue one, no matter the choice of vector $$\vec a \in \mathbb R^M$$. This indeed seems to be the case (based on playing around with examples). In fact, $$A$$ seems to be a Hankel matrix, with moreover the whole region above the anti-diagonal being zero.

SECOND EDIT: The above matrix $$A$$ (in the box) is defined as a product over matrices. Each individual factor can be obtained in closed form: $$e^{\sum_{k=0}^{M-1} a_k L_{l,l+k+1} } = \left( \begin{array}{ccc} I_{l} & 0 & 0 \\ 0 & \left( \begin{array}{cc} \cos(\alpha) & \sin(\alpha) \vec m^T \\ - \sin(\alpha) \vec m & I_M + ( \cos(\alpha)-1 ) \vec m \cdot \vec m^T \end{array} \right) & 0 \\ 0 & 0 & I_{M-l} \end{array} \right),$$ where $$I_d$$ denotes the identity matrix of size $$d\times d$$, and where $$\alpha := |\vec a|$$ and $$\vec m := \vec a / |\vec a|$$. I have not yet been able to obtain a closed form for $$A$$ itself.