# Relation Between Cross Product and Infinitesimal Rotations, Generators, Etc [duplicate]

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?

## marked as duplicate by FGSUZ, ZeroTheHero, Buzz, Qmechanic♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 4 at 9:42

• 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. – Qmechanic Feb 4 at 9:45

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.
• Isn't \vec{ } easier than the boldface command? – FGSUZ Feb 3 at 20:13
• @FGSUZ You're right, "\vec{ }$is a bit easier to type, although I was using a lot of copy-and-paste anyway. For readability, I usually default to boldface when lots of vectors are involved because it reduces the number of strokes that the eye has to parse; but one could argue that distinguishing between boldface and non-boldface causes some eyestrain, too. IMO, choosing good notation (consistent, concise, uncluttered, unambiguous, and a host of other mutually-conflicting requirements) is one of the hardest parts of physics! – Chiral Anomaly Feb 3 at 20:30 • @EthanT I checked the multi-line equation and didn't see a typo, but it's possible that I'm just not being perceptive enough. I added an extra step and some extra explanation in the text following the equation. I also numbered more of the equations to make it easier to refer to them in comments, in case that's needed. – Chiral Anomaly Feb 4 at 5:10 • @EthanT Regarding the definition of$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$is just used to absorb the proportionality factor. – Chiral Anomaly Feb 4 at 5:13 • @EthanT The significance of the identity$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. – Chiral Anomaly Feb 4 at 5:22