Assume the following definitions:
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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 $\mathbb{R}^{2n}$).
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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?
Best Answer
Short answer: no.
The simplest counterexample to my mind comes from letting $A$ and $B$ be right-angle rotations in the oriented planes formed from consecutive pairs of coordinates ($x_1x_2$ and $x_3x_4$ and so on up to $x_{2n-1}x_{2n}$), but which go in opposite directions in all but the first two planes. Then the product $AB$ will be act as $-I_4$ on the first four coordinates, but act as $+I_{n-4}$ on the rest, which will not be isoclinic beyond four dimensions. (We will refer back to this $A$ and $B$ when discussing how isoclinic rotations don't form a manifold at $\pm I$ by looking at tangent vectors).
Speaking of "isoclinic," let's talk about what that word means.
For any real vector space we can speak of the collection of all possible ordered bases. There is a natural way to topologize this (as a subset of a product of spaces), and then the collection has two connected components. We can refer to these components, or to equivalence classes of bases, as the two orientations of space. If we choose one in particular, we can call it the positive orientation, which makes the other negative. In the case of a real inner product space, we can use orthonormal frames instead of ordered bases.
A real version of the spectral theorem guarantees that any orthogonal transformation of a real inner product space can be represented, with respect to some positive orthonormal frame as a basis, as a block-diagonal matrix with $2\times2$ rotation matrices for blocks, and possibly a $\pm1$ in the corner if the dimension is odd (corresponding to having determinant $\pm1$ respectively). We can pick an orientation on each of the 2D planes (corresponding to the blocks) so that the 2D rotations are all by convex angles ($0^\circ$-$180^\circ$). If the dimension is even and none of the angles are straight ($0^\circ$ and $180^\circ$, corresponding to $\pm I$ blocks), then the planes' orientations are well-defined and can be "added up" to get an orientation of space (by taking an orthonormal frame for each plane, and putting them together we get a frame for the whole space and the resulting orientation is independent of what order the frames are put together in). Assume even dimensions now.
If the convex angle of rotation for a 2D plane is unique in a given spectral decomposition of an orthogonal transformation, then that 2D plane is itself unique. This uniqueness disappears when a convex angle appears with multiplicity. At the opposite extreme, an isoclinic rotation is when all the convex angles are the same. If the angles are not straight, we can add up the orientations, and if the result is positive/negative we get a left/right isoclinic rotation respectively. We consider $\pm I$ both left and right isoclinic.
It is true that any isoclinic rotation is unitary with respect to some complex structure, but not necessarily the same complex structure as other isoclinic rotations. Indeed, if $R$ is isoclinic with strictly convex angle $\theta$, it is expressible as $\exp(\theta J)=\cos(\theta)I+\sin(\theta)J$ where $J$ is a $90^\circ$ in each of the same invariants planes as $R$. Notice $J$ is in both the special orthogonal group and in its lie algebra, and is a complex structure. Then, $R$ is unitary with respect to the complex structure induced from $J$.
This setting also helps us understand how much we are deviating from uniqueness in our spectral decompositions. The sets of orthogonal invariant planes we can use for a spectral decomposition of $R$ correspond bijectively to the sets of complex unitary bases we can use for our complex inner product space (using $J$ as a complex structure, again). In other words, the possible spectral decompositions of $R$ form a space topologically equivalent to a complex projective space $\mathbb{CP}^{n-1}$.
It's also worth noting something else that is special about 4D, besides the fact that left and right isoclinic rotations forms subgroups: they commute. Indeed, $\mathrm{SO}(4)$ is isomorphic to the central product $S^3\times_{S^0}S^3$, where the two copies of $S^3$ correspond to multiplying the quaternion algebra $\mathbb{H}$ (a 4D real inner product space!) on the left/right by unit quaternions, corresponding to left/right isoclinic rotations respectively (whence the terms left and right).
In 6D and beyond, there is no left isoclinic that commutes with all right isoclinic rotations or vice-versa. Let's give a proof sketch. Suppose to the contrary $R$ is right isoclinic and commutes with all left isoclinic rotations $L$. Thus, if we say our vector space is $\mathbb{C}\times\mathbb{H}$, it in particular commutes with all rotations coming from left-multiplication by $S^1\times S^3$, so it commutes with all linear combinations of these rotations, so it commutes with an operator which is a 2D rotation on $\mathbb{H}$ but annihilates $\mathbb{H}$ (exercise!), so $\mathbb{C}$ and $\mathbb{H}$ are invariant, but we can find a left isoclinic rotation $R$ which doesn't preserve $\mathbb{C}$ in which case $L$ and $R$ cannot commute.
Or, we could argue with a particular example. Since $\exp(\theta_1 J_1)$ and $\exp(\theta_2 J_2)$ commute iff $J_1$ and $J_2$ do, it suffices to consider complex structures for our isoclinic rotations. All complex structures are conjugate and conjugation preserves commutativity, so it suffices to pick a particular complex structure $J_1$ and find another $J_2$ which doesn't commute.
Let's construct an example in 6D. Use the transposition notation $(ij)$ to be a 2D rotation in the $x_ix_j$-plane from the positive $x_i$ direction to the positive $x_j$ direction, which also means $(ij)^{-1}=(ji)$ and $(ij)(k\ell)=(k\ell)(ij)$ if $i,j\ne k,\ell$. Let $L=(12)(34)(56)$ and $R=(23)(45)(61)$. We can calculate $RLR^{-1}$ by applying $R$ to each axis used in $L$'s "cycle notation" to get $RLR^{-1}=(14)(36)(52)\ne L$. Thus, $R$ and $L$ don't commute, because each "breaks apart" the axes from the other's invariant planes. This example also works beyond 6D: simply append $(78)$ and so on at the end of both $L$ and $R$.
Little fun fact related to this "cycle notation." Notice how there are homomorphisms $\mathrm{SO}(4)\to\mathrm{SO}(3)$ and $S_4\to S_3$ whose kernels are the left-isoclinic rotations and the double transpositions respectively. The actions are given by having $\mathrm{SO}(4)$ act on the 3D real inner product space of (scalar multiples of) left isoclinic complex structures by conjugation, and $S_4$ act on its own double transpositions by conjugation. I wonder if there's some kind of field-with-one-element ($\mathbb{F}_1$) stuff behind this coincidence?
We can also characterize the space of left/right isoclinic rotations topologically.
Let $\mathcal{J}=\mathcal{J}_L\sqcup\mathcal{J}_R$ be the space of all complex structures, which has two components corresponding to being left or right isoclinic, and let $\mathcal{L}$ and $\mathcal{R}$ be the sets of left/right isoclinic rotations. From the formula $\exp(\theta J)=\cos(\theta)+\sin(\theta)J$ for complex structures $J$ we find that $\mathcal{L}$ and $\mathcal{R}$ are topological suspensions of $\mathcal{J}_L$ and $\mathcal{J}_R$ respectively; they can be partitioned into $180^\circ$ arcs all glued at $\pm I$, with $\mathcal{J}_L$ and $\mathcal{J}_R$ parametrizing them (each complex structure $J$ being stationed at the $90^\circ$ midpoint of the arc $\exp(\theta J)$ made from convex $\theta$).
Note $\mathcal{L}$ and $\mathcal{R}$ are not manifolds at $\pm I$. Indeed, their tangent spaces (the set of tangent vectors) there are not flat. They are the union of rays (i.e. the topological cone) through $\mathcal{J}_L$ and $\mathcal{J}_R$, but these spaces are not closed under addition (consider $A+B$ where $A,B$ are defined at the beginning of my answer), so $\mathcal{L}$ and $\mathcal{R}$ could be a bit "spiky" where they meet at $\pm I$. For a visual aid, I like to pretend $\mathcal{J}_L$ and $\mathcal{J}_R$ are both cirlces, so $\mathcal{L}\cup\mathcal{R}$ is a doubly-pinched torus, with $\pm I$ the pinch points and $\mathcal{L}$ and $\mathcal{R}$ the two "halves."
How one-parameter circle subgroups $\exp(\theta J)$ "wrap around" $\mathcal{L}\sqcup\mathcal{R}$ for nonconvex $\theta$ depends on whether the dimension is $0$ or $2$ mod $4$. In the former case, once $\theta$ reaches $180^\circ$ and $\exp(\theta J)$ reaches $-I$, we wrap back around the same half ("left" or "right") we started on. In the latter case, we wrap around the other half. Thus if $n\equiv2$ mod $4$, one-parameter subgroups are split in half between $\mathcal{L}$ and $\mathcal{R}$. This is reflected in the fact that multiplying by $-I$ preserves $\mathcal{L}$ and $\mathcal{R}$ if $n\equiv0$ mod $4$ but swaps them if $n\equiv2$ mod $4$. Technically this means $\mathcal{L}$ and $\mathcal{R}$ are only "spiky" at $\pm I$ if $n\equiv2$ mod $4$; if $n\equiv0$ mod $4$ it's still not "flat" at $\pm I$, it's more like "wavy," just wavy in a way that is symmetric around $\pm I$. Unfortunately the smallest dimension this "wavy but not flat or spiky" case occurs is $8$, so not easy to visualize.
Note $\mathcal{L}\simeq\mathcal{R}$ are homeomorphic (diffeomorphic outside of $\pm I$) since $\mathcal{J}_L\simeq\mathcal{J}_R$ are. The orthogonal group acts on $\mathcal{L}\cup\mathcal{R}$ by conjugation, always preserving the convex angle. Rotations will preserve $\mathcal{J}_L$ and $\mathcal{J}_R$, whereas rotoreflections swap them (thus establishing they are geometrically identical manifolds). Indeed, we have $\mathcal{J}_L\simeq\mathcal{J}_R\simeq\mathrm{SO}(2k)/\mathrm{U}(k)$ and $\mathcal{J}\simeq\mathrm{O}(2k)/\mathrm{U}(k)$ where $n=2k$, by the orbit-stabilizer theorem.
Writing $\mathcal{J}_n$ for the $\mathcal{J}$ of $\mathbb{R}^{2k}$, we can apply all of $\mathcal{J}_n$ to a particular nonzero vector to get a fibration:
$$ \mathcal{J}_{n-2}\to\mathcal{J}\to S^{n-2}. $$
We can restrict to $\mathcal{J}_L$ too, if we wanted. For $n=6$ we may be able to use the fact $\mathcal{J}_4=S^2\times S^0$, or the accidental isomorphism $\mathrm{SU}(4)/\{\pm I\}\cong\mathrm{SO}(4)$, for something of value.