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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?

bonif
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    I'm not quite sure what you are asking. All rotations preserve orientation (they have determinant 1). Note that the bivector (a.k.a. element of $\mathfrak{so}(\mathbb{R}^4)$) is not a rotation itself but rotations are given by exponentiating it. Do you mean that $B$ preserves a plane when $B\wedge B=0$? In which case your task is to prove that $B = v\wedge w$ (i.e. a "simple wedge") exactly when that happens and the plane is then $\langle v,w \rangle$ – Callum May 16 '23 at 11:18
  • I mean what it is explained at the end of the first answer in https://math.stackexchange.com/questions/3730174/simple-bivectors-in-four-dimensions? – bonif May 16 '23 at 11:35
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    The set of simple bivectors is 5D, not 3D. And it's not a subspace (not closed under addition). – mr_e_man May 16 '23 at 11:58
  • If $B=(a\wedge b)+(c\wedge d)$ where $a,b,c,d$ are vectors, and $B$ is simple: $$0=B\wedge B=2a\wedge b\wedge c\wedge d,$$ then the vectors must be linearly dependent; they must be in a 3D subspace of $\mathbb R^4$. – mr_e_man May 16 '23 at 12:24
  • @mr_e_man Right, and every simple bivector B can be put in the form of that sum. That's what I was indicating in my question, could you maybe address now what I asked? – bonif May 16 '23 at 13:00
  • Your question is unclear to me.... – mr_e_man May 16 '23 at 13:42
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    Possibly relevant: https://math.stackexchange.com/questions/4624626/help-with-intuitively-understanding-why-the-space-of-bivectors-from-4d-euclidean – mr_e_man May 16 '23 at 17:21
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    Related: https://math.stackexchange.com/questions/1965281, https://math.stackexchange.com/questions/1887082 – bonif Nov 04 '23 at 16:54

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$ \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 $B\wedge B = 0$.

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. $$ (\gamma_0\gamma_1)(\gamma_0\gamma_2) = -\gamma_0^2\gamma_1\gamma_2 = -\gamma_1\gamma_2. $$

Nicholas Todoroff
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  • Is there a reason we're saying $W$ is 3D instead of 2D? – coiso May 16 '23 at 17:26
  • @elemelons Yes, because I'm directly addressing OP's confusion. See also my answer that OP linked to here. The fact that $B$ (not necessarily simple) lives in a 3D subspace implies that it is simple. – Nicholas Todoroff May 16 '23 at 17:31
  • Ok, thanks @NicholasTodoroff So this bivector B is a very different object from the spatial vector that I was linking it to and it does reach the corresponding six dimensions but in different 3D subspaces. This makes sense. Is there not a little abuse of notation in some physics texts when they use one index "vectors" like $J_i$ or $K_j$ in the decomposition of the 6 generators in 3+3 in the Lie algebra's commutation relations? Or that is unrelated? – bonif May 16 '23 at 20:06
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    @bonif The indices are indexing the generators themselves, not their components. So $J_1$ is a bivector, $J_2$ is a bivector, ... (probably represented as matrices). There is no abused notation here. – Nicholas Todoroff May 17 '23 at 00:36
  • Sure, but it might be misleading for newcomers to call such rearrangements of generators spatial vectors as it is reminiscent of the isomorphism between axial vectors and bivectors in $\mathbb{R^3}$ that doesn't exist in $\mathbb{R^4}$. Certainly this confused me on "the subspace versus subspaces" issue and prompted my question. – bonif May 17 '23 at 08:42
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    @bonif Do you have a reference for this terminology? Otherwise I can't directly address your question, but see the edit I made to my answer. – Nicholas Todoroff May 17 '23 at 18:58
  • If by "this terminology" you are referring to the term "spatial vector" it just means a 3D vector in the same sense that you explain in your edit, but a typical physics textbook won't be as sophisticated as going into the even subalgebra of the Clifford algebra and just calls both boosts and rotations (3D) vectors or "spatial vectors" (see for instance Maggiore's intro to QFT pg. 19 eq 2.26) – bonif May 17 '23 at 20:50
  • In classical mechanics the corresponding "3-vectors of bivectors" are the angular momentum pseudovector and the Laplace-Runge-Lenz vector and in group theory terminology the $so(3)$ subalgebras of $so(4)$ – bonif May 17 '23 at 21:05
  • @bonif As I read it, the first time Maggiore says "spatial vector" above 2.26 they use it loosely, and then they refine this by noting that 2.27 shows that the $J_i$ represent angular momentum (and hence are pseudovectors) and also by taking 2.28 as the property defining the $K_i$ as spatial vectors. – Nicholas Todoroff May 17 '23 at 21:37
  • I don't really understand how the Laplace-Runge-Lenz vector is relevant, but no one is saying there is a "3-vector of bivectors"; no one is saying that $(J_1, J_2, J_3)$ nor $(K_1, K_2, K_3)$ are "3-vectors". Instead, $K_1$ itself is a 3-vector because of how it transforms/its commutation relations, as are $K_2$ and $K_3$; and $J_1$ itself is a pseudo 3-vector, as are $J_2$ and $J_3$. Then we can form 3-vectors as linear combinations $aK_1 + bK_2 + cK_3$, and similarly for pseudo 3-vectors and the $J_i$. – Nicholas Todoroff May 17 '23 at 21:40
  • You are right, that was badly expressed. The Laplace vector is $K_i$ in the $so(4)$ algebra of the Kepler problem. – bonif May 17 '23 at 21:59