Where is the error in my argument that SO(4) contains a 7-dimensional subspace? I have read online that SO(4) is of dimension 6 - an element in SO(4) can be described as a continuous matrix function of 6 parameters. So therefore it shouldn't contain a 7-dimensional figure. But when I try to count dimensions, I obtain the opposite - where is my error? (Forgive my perhaps poor notation, I'm new to group theory!)
First, I consider a rotation $T$ which is an element of $SO(4)$ - a rotation of $\mathbb{R}^4$. I consider the case where $T$ has a 2-dimensional invariant plane $V^2$, which means that the complement $\mathbb{R}^4 / V^2 = W^2$ (another 2-dimensional plane) is also an invariant plane for $T$. So we have $\mathbb{R}^4 = V^2 \otimes W^2$. $T$ when restricted to these planes is just a 2-dimensional rotation). So to describe the dimension of this subset of $SO(4)$, I need to count the dimensions of: the number of ways to decompose $\mathbb{R}^4$ into two 2-dimensional planes $V^2 \otimes W^2$; and the number of 2-dimensional rotations $T|_{V^2}$, $T|_{W^2}$.
For the first part, we can describe a 2-dimensional plane $V^2$ as being spanned by two unit vectors in $\mathbb{R}^4$ which are orthogonal. To pick the first unit vector is to pick an element of $S^3$ (the 3-dimensional sphere embedded in $\mathbb{R}^4$), and to pick the second unit vector is to do the same, except the orthogonality condition means we have to pick from a slice of $S^3$ which is orthogonal to the first - ie. we pick from $S^2$. And then to find $W^2$, we must set it equal to $\mathbb{R}^4/V^2$, the orthogonal complement - no choice there. Therefore, the number of dimensions here is the sum of the dimensions of $S^3, S^2$, which is 5.
For the second part, we just have to pick two elements of $SO(2)$, which I know to be 1. Therefore, we get 2 dimensions there.
Therefore, my argument says that $SO(4)$ contains a 5+2=7-dimensional subspace. This can't be right - where is the flaw in my counting? I suspect that it's in the first counting step (5 dimensions), but I can't see why. There is redundancy in picking, say, the plane spanned by vectors $v_1, v_2$ versus $-v_1, v_2$ and so on with all the sign changes - but that would only cut the number of distinct choices by a constant factor of 4, which wouldn't reduce the dimension of the resulting set (only its size).
 A: The mistake is at counting the dimension of "2 dimensional subspaces" in $\mathbb R^4$. This is called the Grassmannian of 2 planes in $\mathbb R^4$ and has dimension $4$.
In your argument you first start with $v_1\in \mathbb S^3$ and then another $v_2$ in the two sphere in the plane orthogonal to $v_1$. But you need to cut down the dimension by one: there's a rotation in the $v_1-v_2$ plane so that if $w_1, w_2$ are formed by rotations along the $v_1-v_2$ plane, then both $\mathrm{span}\{v_1, v_2\}$ and $\mathrm{span}\{ w_1, w_2\}$ correspond to the same two plane $V^2$ in $\mathbb R^4$.
So the dimension of all $2$ planes in $\mathbb R^4$ is $4$ instead of $5$.
A: I have just figured out the flaw in my logic after trying out my construction in $R^3$ instead. After picking two orthogonal unit vectors (such as $(0, 0, 1)$ and $(0, 1, 0))$ to define a plane, there is an extra factor of $SO(2)$ because we can rotate the plane around, getting a new set of spanning vectors (such as $(0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ and $(0, -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$), but still defining the same plane. Therefore, we lose $SO(2)$ dimensions - ie. 1 dimension. Therefore, the set of orthogonal 2-dimensional planes is of dimension 4, not 5.
