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Looking into the infinitesimal view of rotations from Lie, I noticed that the vector cross product can be written in terms of the generators of the rotation group $SO(3)$. For example:

$$\vec{\mathbf{A}} \times \vec{\mathbf{B}} = (A^T \cdot J_x \cdot B) \>\> \hat{i} + (A^T \cdot J_y \cdot B) \>\> \hat{j} + (A^T \cdot J_z \cdot B) \>\> \hat{k}$$

where,

$$J_x = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix} \>\>\>;\>\>\>;J_y = \begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix} \>\>\>;\>\>\>;J_z = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix} $$

I couldn't help but think I was missing some profound connection here. Why is the vector cross product seemingly intimately related to the generators of the $SO(3)$ rotation group?

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marked as duplicate by FGSUZ, ZeroTheHero, Buzz, Qmechanic quantum-mechanics Feb 4 at 9:42

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  • $\begingroup$ The question currently has 3 migration votes. The question seems to be enough physics-inspired not to migrate to Mathematics, so I'm temporarily closing it as a duplicate to avoid migration. $\endgroup$ – Qmechanic Feb 4 at 9:45
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The mystery dissipates when we consider how this generalizes to $D$-dimensional space. The cross product only makes sense when $D=3$, but the idea of using the antisymmetric part of $A_a B_b$ to represent an oriented element of area works for any $D$, as does the concept of a rotation in a given plane. So let's see what this looks like for aribtrary $D$.

We can express the vectors $A$ and $B$ in a canonical orthonormal basis, say $$ \mathbf{E_1} =\left[\matrix{1\cr 0\cr 0\cr \vdots}\right] \hskip1cm \mathbf{E_2} =\left[\matrix{0\cr 1\cr 0\cr \vdots}\right] \hskip1cm \mathbf{E_3} =\left[\matrix{0\cr 0\cr 1\cr \vdots}\right] \hskip1cm \cdots $$ and so on (again, this is for arbitrary $D$). I'm using boldface for a vector (represented here as a single-column matrix) and using non-boldface for its components. In this basis, we have $$ \mathbf{A} = \sum_a A_a \mathbf{E}_a \hskip2cm \mathbf{B} = \sum_b B_b \mathbf{E}_b, \tag{1} $$ which gives \begin{align*} \mathbf{A}\mathbf{B}^T- \mathbf{B}\mathbf{A}^T &= ( \sum_a A_a \mathbf{E}_a)(\sum_b B_b \mathbf{E}_b^T) - (\sum_b B_b \mathbf{E}_b)( \sum_a A_a \mathbf{E}_a^T)\\ &= \sum_{a,b} A_a B_b (\mathbf{E}_a\mathbf{E}_b^T - \mathbf{E}_b\mathbf{E}_a^T) \\ &= \sum_{a,b} \frac{A_a B_b - B_a A_b}{2} (\mathbf{E}_a\mathbf{E}_b^T - \mathbf{E}_b\mathbf{E}_a^T) \tag{2} \end{align*} where $T$ means transpose. The last step uses the fact that the quantity in parentheses changes sign if the indices $a$ and $b$ are exchanged, so the value of the sum is unchanged if we replace $A_a B_b$ with $-B_a A_b$. Therefore, the value of the sum is also unchanged if we replace $A_a B_b$ with half of $A_a B_b$ plus half of $-B_a A_b$. (In other words, the sum in the second-to-last line equals half of itself plus half of itself, which gives the last line.)

Now recognize that the matrix $\mathbf{E}_a\mathbf{E}_b^T - \mathbf{E}_b\mathbf{E}_a^T$ is the generator of rotations in the $a$-$b$ plane, and recognize that $A_a B_b - B_a A_b$ are the components of the "cross product" when $D=3$. Equation (2) shows how the phenomenon noted in the OP generalizes to arbitrary $D$.

The quantity $\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T$ is a matrix representation of a bivector, the oriented element of area defined by the two vectors $\mathbf{A}$ and $\mathbf{B}$. The antisymmetry isolates their "mutually orthogonal part," which is the part we need for specifying an element of area. In fact, the bivector $\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T$ is the (unnormalized) generator of a rotation in the $\mathbf{A}$-$\mathbf{B}$ plane. The corresponding rotation matrix is $$ R(\theta)=\exp\left(\theta\frac{\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T}{N}\right) \tag{3} $$ where $N$ is defined by the condition $$ \left(\frac{\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T}{N}\right)^3 =-\frac{\mathbf{A}\mathbf{B}^T-\mathbf{B}\mathbf{A}^T}{N}. \tag{4} $$ Notice that the generator $\mathbf{E}_a\mathbf{E}_b^T - \mathbf{E}_b\mathbf{E}_a^T$ satisfies this condition with $N=1$. The minus sign is a consequence of antisymmetry, and this is also what ensures that $R$ is an orthogonal matrix: $R^TR =R R^T=I$.

When $D=3$, given any plane through the origin, there is a unique line through the origin that is orthogonal to the given plane. For this reason, we can get away with pretending that a bivector is a vector when $D=3$, and this is why we can define the "cross product" when $D=3$. (Vectors and bivectors still behave differently under reflections, though.) But for general $D$, such as $D=2$ or $D\geq 4$, that trick doesn't work, and we need to use a two-index quantity to represent a bivector.

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  • $\begingroup$ Isn't \vec{ } easier than the boldface command? $\endgroup$ – FGSUZ Feb 3 at 20:13
  • $\begingroup$ @FGSUZ You're right, "\vec{ }$ is a bit easier to type, although I was using a lot of copy-and-paste anyway. For readability, I usually default to boldface when lots of vectors are involved because it reduces the number of strokes that the eye has to parse; but one could argue that distinguishing between boldface and non-boldface causes some eyestrain, too. IMO, choosing good notation (consistent, concise, uncluttered, unambiguous, and a host of other mutually-conflicting requirements) is one of the hardest parts of physics! $\endgroup$ – Chiral Anomaly Feb 3 at 20:30
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    $\begingroup$ @EthanT I checked the multi-line equation and didn't see a typo, but it's possible that I'm just not being perceptive enough. I added an extra step and some extra explanation in the text following the equation. I also numbered more of the equations to make it easier to refer to them in comments, in case that's needed. $\endgroup$ – Chiral Anomaly Feb 4 at 5:10
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    $\begingroup$ @EthanT Regarding the definition of $N$, another way to say it is that if $A$ and $B$ happen to be orthogonal unit vectors like $(1,0,0)$ and $(0,1,0)$, then $AB^T-BA^T$ is just one of the generators $J$, which satisfies $J^3=-J$ (with $N=1$). The way I wrote it in the answer works even if $A$ and $B$ aren't orthogonal to each other and even if they aren't unit vectors. If we define $J=AB^T-BA^T$ without the factor of $N$, then we'd still get $J^3\propto -J$ (this is not obvious at all, by the way; but it can be checked). The factor of $N$ is just used to absorb the proportionality factor. $\endgroup$ – Chiral Anomaly Feb 4 at 5:13
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    $\begingroup$ @EthanT The significance of the identity $J^3=-J$ is similar to the significance of the identity $i^3=-i$ for the imaginary unit $i$. This ensures that $\exp(i\theta)$ is a periodic function of $\theta$ with period $2\pi$, and $J^3=-J$ ensures that $\exp(\theta J)$ is a periodic function of $\theta$ with period $2\pi$, like a rotation matrix. To prove this, expand the exponential using $\exp(M)=\sum_{n\geq 0} M^n/n!$, which is valid for any matrix $M$. Then use $J^3=-J$ to write this as a linear combo of $1$, $J$, and $J^2$, with $\sin(\theta)$ and $\cos(\theta)$ in the coefficients. $\endgroup$ – Chiral Anomaly Feb 4 at 5:22

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