Are left isoclinic rotations a group? Assume the following definitions:


*

*Isoclinic rotations are rotations $\varphi$ in $\mathbb{R}^{2n}$ such that there exists $n$ complementary oriented planes $P_i=\langle x_i,y_i\rangle$ such that $\varphi$ acts as a simple rotation of the same angle $\theta$ on each plane $P_i$, either clockwise or counter-clockwise (here we assume that an orientation of $\mathbb{R}^{2n}$ has been chosen, and that the orientations of the $P_i$ are chosen such that it induces the same orientation on $\mathbb{R}^{2n}$).

*Left isoclinic rotations are isoclinic rotations such that, is one wants all single rotations to go (say) counter-clockwise, one needs to change the orientation of the planes an even number of times. Similarly an isoclinic rotation is right is one needs an odd number of orientation changes.
The second definition is home-made (meaning I haven't found a reference stating it), extrapolating the case of dimension 4 given here. In this case, one can show that the set of left isoclinic rotations is isomorphic to the unit quaternions, and therefore is a group. My question is: "Is the set of left (or right) isoclinic rotations still a group in higher dimension?". Ideally, a geometric picture would be appreciated. A good reference would certainly also be useful.
One possible idea to tackle the problem is to use the fact that unit quaternions are isomorphic to $\mathrm{SU}(2)$. This can be understood by identifying $\mathbb{C}^2$ with $\mathbb{R}^4$ and recalling that if a plane is identified with the complex line, a rotation in the plane is the same as the multiplication by a unit complex (see Andrew D. Hwang's answer here). I assume then that isoclinic rotations can be seen as matrices
$$U\left(\begin{array}{ccc}
\lambda_1 & & \\ & \ddots & \\ & & \lambda_n
\end{array}\right)U^\dagger$$
Where $U\in\mathrm{SU}(n)$ denotes a change of basis and the middle matrix (call it $D$) is diagonal such that all $\lambda_i$ are such that $\lambda_i=z$ or $\overline{z}$ for some $z\in\mathbb{C}$. Here $z$ represents a counter-clockwise rotation by some angle $\theta$, and $\overline{z}$ represents the rotation by the same angle but clockwise. Whether it is a left or right isoclinic rotation depends on the parity of the number of $\overline{z}$'s in $D$. From there I'm not sure whether this is closed under the product. Any idea?
 A: 
Is the set of left isoclinic rotations still a group in higher dimension?

Nope. Since the sets of left/right isoclinic rotations are conjugate to each other via any orientation-reversing element of $O(2n)$, it suffices to assume left isoclinic. The next smallest case to consider after 4D is 6D.
There are six coordinates to $\mathbb{R}^6$. Let $L_1$ be the left-isoclinic $90^{\circ}$ rotation in the coordinates $12,34,56$. That is, it is block-diagonal with three copies of the usual $2\times 2$ rotation $90^{\circ}$ rotation matrix. Then let $L_2$ be the same as $L_1$ but flip the orientation/direction of the rotation in the $34$ and $56$ planes (that is, invert or equivalently transpose the second two of the three $2\times 2$ blocks). Observe the composition $L_1L_2$ has is block diagonal, with first block $-I_2$ and the next two blocks both $I_2$. This is a rotation in a single 2D plane, so it is not isoclinic, and thus left isoclinic rotations are not closed under multiplication.
This same example generalizes to higher dimensions too of course.
In fact, we can describe the set of left-isoclinic rotations geometrically and topologically.
Every left isoclinic rotation $L$ is of the form $L=\exp(\theta J)=\cos\theta+\sin\theta J$ where $J$ is a left-isoclinic rotation in all the same 2D planes except by $90^{\circ}$ instead of $\theta$. (Of course, there is more than one way to decompose $\mathbb{R}^{2n}$ into $n$ 2D invariant subspaces, but $J$ is independent of this choice.) We may parametrize all left-isonclinic rotations using this, where $0<\theta<\pi$ is convex, except for $L=\pm I_{2n}$. In this way, the set $\mathcal{L}$ of left-isoclinic rotations is the topological suspension $\mathcal{L}=S(\mathcal{J})$ where $\mathcal{J}$ is the set of all right-angle left-isoclinic rotations (or in other words, orthogonal complex structures $J$). More geometrically, $\mathcal{L}$ is a union of semicircles between $\pm I_{2n}$ parametrized by elements of $\mathcal{J}$ (which are the midpoints of these semicircles).
Suppose we have a path $\gamma(t)=\exp(\theta(t)J(t))$ through $\mathcal{L}$ with $\theta(0)=0$ so $\gamma(0)=I$. Using Euler's formula with $\cos$ and $\sin$ we can differentiate at $t=0$ to get the tangent matrix of the path at $I$:
$$ \gamma=\cos\theta+\sin\theta\, J \\
 \gamma'=-\dot{\theta}\sin\theta+\dot{\theta}\cos\theta\, J + \sin\theta\, \dot{J} \\
 \gamma'(0)= \dot\theta(0)J(0). $$
Thus, the tangent vectors are rays extending through $J$s in $\mathcal{J}$. This is not closed under addition, unfortunately, as the aforementioned $L_1$ and $L_2$ demonstrate. Thus, not only is $\mathcal{L}$ not closed under multiplication, it's not even a manifold at $I$!
It is noteworthy though that $\mathcal{J}$ is a $\mathrm{SO}(2n)$-orbit under conjugation.  If we write $\mathbb{R}^{2n}=\mathbb{C}^n$ then multiplication-by-$i$ is an element of $\mathcal{J}$ and its stabilizer is the embedded copy of $\mathrm{U}(n)$ in $\mathrm{SO}(2n)$. In particular, for $2n=6$, we have $\mathcal{J}\simeq\mathrm{SO}(6)/\mathrm{U}(3)$. Perhaps you can do something interesting with this using the exceptional isomorphism $\mathrm{SO}(6)\cong\mathrm{SU}(4)/\{\pm I_4\}$. At any rate, we can pick $e_6\in\mathbb{R}^6$, then define $\mathcal{J}\to S^4$ by $J\mapsto Je_6$ (note $Je_6\perp e_6$ so $Je_6\in S^4\subset \mathbb{R}^6$). Each fiber looks like a copy of 4D left-isoclinic rotations, so we have a fiber bundle
$$ S^2\to \mathrm{SO}(6)/\mathrm{U}(3)\to S^4. $$

I assume then that isoclinic rotations can be seen as [unitary] matrices.

The reason this works in 4D is because left/right isoclinic rotations commute. Indeed, there is the exceptional isomorphism $\mathrm{SO}(4)=(S^3\times S^3)/\mathbb{Z}_2$, with left/right isoclinic rotations corresponding to the two $S^3$ factors, and $\mathbb{Z}_2$s nontrivial element $(-1,-1)$. However, in 6D there is no (nontrivial) right-isoclinic rotation $R$ that commutes with all left-isoclinic rotations $L$.
To see this, consider left-isoclinic rotations of $\mathbb{C}\oplus\mathbb{H}$ coming from left-multiplication by elements of $S^1\times S^3$. If $R$ commutes with all of these, it commutes with any combination of them. Exercise. Show a combination of them is a 2D rotation in $\mathbb{C}$ and the zero map on $\mathbb{H}$, and thus conclude $R$ must preserve the invariant subspaces $\mathbb{C}$ and $\mathbb{H}$. Then show $R$ cannot commute with a left-isoclinic rotation that does not preserve the 2D subspace $\mathbb{C}$.
Here's a reasonable way to construct a specific example. For simplicity we can consider isoclinic rotations which admit coordinate planes as invariant 2D subspaces. I will use the "cycle notation" $(12)$ to mean the rotation which rotates the positive $x_1$-axis to the positive $x_2$ axis (and acts trivially on the orthogonal complement of the $x_1x_2$-plane). Of course, this means $(12)^{-1}=(21)$.
Consider $L=(12)(34)(56)$ and $R=(23)(45)(61)$. (Check $234561$ is an odd permutation of $123456$, so $R$ is indeed right-isoclinic.) If $L$ and $R$ don't commute as permutations of the six axes, then they can't commute as rotations. To calculate $RLR^{-1}$, simply apply $R$ to each number in the cycle notation of $L$. Thus, conjugating by $R$ turns ther permutation $(12)$ into $(36)$, which is not present in $L$, so $RLR^{-1}\ne L$. That's because $R$ "breaks apart" the axes in the invariant 2D subspaces of $L$. Similarly, for any (nontrivial) right-isoclinic rotation $R$ there is a noncommuting left-isoclinic rotation and vice-versa in six dimensions, and thus beyond if we generalize this idea.
Notice the $2$-involutions (the products of two $2$-cycles) in $S_4$, which make up a copy of the Klein-four group $V_4$, all commute with each other, reflecting the fact all left/right icoclinic rotations commute in 4D (since quaternions are associative). On a related note, just as there is an exceptional isomorphism $S_4\to S_3$ with kernel $V_4$, there is a homomorphism $\mathrm{SO}(4)\to\mathrm{SO}(3)$ with kernel the left-isoclinic rotations!
