How do simple bivectors in $\mathbb{R^4}$ generate rotations?

Question

Simple 6-dimensional bivectors in $\mathbb{R^4}$ are restricted to a 3D subspace by $B\wedge B=0$. They generate simple rotations in 4D. Intuitively I'm failing to completely understand how a bivector restricted to 3D (with the information of a spatial vector axis? ) can generate a rotation that preserves the orientation of a plane on a simple rotation in 4D, so there must be something that I'm missing or misconstruing here.Wouldn't it be required that such bivectors had access to the 6 planes of rotation in 4D? I mean, how a bivector with the info of a spatial vector is enough to characterize a rotation in 4D when it usually does so just in 3D?

Answer 1

$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\MVects[1]{\mathop{\textstyle\bigwedge^{\mkern-1mu#1}}} \newcommand\Cl[1]{\mathrm{Cl_{#1}}(\mathbb R)} \newcommand\ClE[1]{\mathrm{Cl}^+_{#1}(\mathbb R)} $For a fixed bivector $B$, the condition $B\wedge B = 0$ means that $B$ lives in some 3D subspace. It does not mean that all bivectors live in the same 3D subspace.

To be precise, let $V$ be the vector space in question. For every subspace $W \subseteq V$ we can naturally identify $\Ext W$ as a subspace of $\Ext V$. The statement is:

• Suppose $V$ has dimension 4 and that $B \in \MVects2V$. If $B\wedge B = 0$ then there exists a subspace $W \subseteq V$ of dimension 3 such that $B \in \Ext W$.

This is not the same as the following statement (which is false):

• Suppose $V$ has dimension 4. Then there is a subspace $W \subseteq V$ of dimension 3 such that $B \in\Ext W$ for every $B \in\MVects2V$ satisfying $B\wedge B = 0$.

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Some of these bivectors are "spatial vectors", just not all of them and not for all observers.

Space-time can be studied through the Clifford algebra $\Cl{1,3}$ which is naturally isomorphic to the exterior algebra as a vector space. It's even subalgebra $\ClE{1,3}$ (the subalgebra of all sums of products of an even number of 4-vectors) is isomorphic to $\Cl3$, the algebra of 3D Euclidean space. A choice of such an isomorphism is equivalent to a choice of non-null 4-vector and is called a space-time split.

With the standard basis $\gamma_0, \gamma_1, \gamma_2, \gamma_3$, if we do a space-time split with $\gamma_0$ then the following three bivectors $$ \gamma_0\gamma_1,\quad \gamma_0\gamma_2,\quad \gamma_0\gamma_3 $$ are exactly an orthonormal basis for "spatial vectors" in $\Cl3 \cong \ClE{1,3}$. This space-time split corresponds to an observer whose 4-velocity is $\gamma_0$. They also generate boosts for such an observer. We are left with three other basis bivectors $$ \gamma_2\gamma_3,\quad \gamma_3\gamma_1,\quad \gamma_1\gamma_2. $$ These generate rotations for a $\gamma_0$-observer; you can see that these are in fact "spatial bivectors" with respect to the $\gamma_0$-split since e.g. $$ (\gamma_0\gamma_1)(\gamma_0\gamma_2) = -\gamma_0^2\gamma_1\gamma_2 = -\gamma_1\gamma_2. $$