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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.

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