Showing the two sides solve the same differential equation with the same initial conditions seems like a satisfactory answer to my question. But I don't understand a lot of the steps you're showing here. I think I follow $d_{\theta}\sigma_{\theta} \propto [n\cdot \sigma, \sigma]_{\theta}$. That this is proportional to $(n\times \sigma)_{\theta}$ is not obvious to me. And then neither of the two steps for the right-hand-side are clear to me. Could you spell these out in more detail? I'm curious if they are proven algebraically or if they are specific to these specific Lie groups/algebras.

Ok, actually I've worked out the expressions you have for the left hand side. But I think you have an extra factor of $i$ in the latter two lines. The $i$ in the first line gets canceled by an $i$ that arises in converting the commutator to a cross product. I'm now trying to understand why $(n\times J)A = n\times A$ then I think I will follow this answer.

There were also some errors, which didn't help for clarity... I've added some details and there shouldn't be any mistakes now

Ok, so it seems like one essential element of this proof is that $(J_i)_{jk} = -\epsilon_{ijk}$. That is, this proof requires using the explicit forms for $J$. But these forms only look this way for $SO(3)$, not for $SO(n)$ I don't think. I think this gets at the uniqueness of $SO(3)$ for this particular expression that is spelled out in physics.stackexchange.com/a/679357/128186

Well, that's usually how you define $J$, not from writing out the matrices as you did. It comes from the adjoint representation, the components are the structure coefficients of the Lie algebra.

right, but that characterization of $J$ only holds for $\mathfrak{so}(3)$, I think the structure coefficients are a lot more complicated for $\mathfrak{so}(n)$. Right?

It's just that you need to be more careful with your book keeping since you cannot use the Levi-Civita symbol. But the relations are the same:$$[J_{ij},J_{kl}] = \delta_{jk}J_{li}-\delta_{ik}J_{lj}+\delta_{il}J_{kj}-\delta_{jl}J_{ki}$$

Thanks for all your responses. Yes, that is exactly what I had in mind for the structure factors for $\mathfrak{so}(n)$. I guess my point is there are some peculiarities coming from that fact that we're thinking about $SO(3)$ instead of $SO(n)$ here. I guess a further question, and I think it's maybe the same question as the linked question, is does the exponential expression hold for larger $n>3$ or $n\neq 3$. But there I don't even know if it's possible to parametrize rotations

simply with an axis and an angle.

So it's probably true from some kind of abstract Adjoint map stuff that $U^{\dagger}(R)XU(R) = RX$ in higher dimensions, But the matrix exponential representation maybe looks different..

The pattern that generalises in arbitrary dimensions is that a rotation is decomposed in terms of elementary rotations along planes that are orthogonal to each other

The $J_{ij}$ are the generators of rotations in the $ij$ plane. Using the conversion in 3D $J_{ij} = \epsilon_{ijk}J_k$ you can check the equivalence

Similarly, you would not have a Pauli vector anymore, you would rather have a Pauli bivector.

But yes, it all boils down to the adjoint formula $Ad\circ\exp = \exp\circ ad$

Right... so it seems like the equation $(*)$ doesn't immediately generalize to higher dimensions. The structure of the symbols are just all different.

Yes, exactly, I think $Ad\circ\exp = \exp\circ ad$ is a bit what I'm trying to understand.

And this does generalize for any Lie groups, or at least nice enough Lie groups

Yes, the proof that I provided generalises and this is why the more abstract answers refer to

Abstractly, I think $(*)$ can be understood as $Ad \circ \exp = \exp \circ ad$ together with the isomorphism between $\mathfrak{su}(2) \sim \mathfrak{so(3)}$?

In your case, it is applied to $\mathfrak{so}(3)$. $\mathfrak{so}(2)$ just enters the game if you want to implement the representation

More formally, what only matters is the action of $SO(3)$ on the components of $\sigma$ which is precisely the adjoint representation of $\mathfrak{so}(3)$

Well, there are already many answers to this on this site. You can check out a previous answer of mine physics.stackexchange.com/q/805505/333454, and in the linked answers of QMechanics as well. Is there something ore specific you are interested in?

This is where I feel the isomorphism between \mathfrak{su}(2) and \mathfrak{so}(3) is needed to convert the \sigma that appears on the RHS into a J

Just copying your message with the dollars

So we have $Ad \circ \exp = \exp \circ ad$. I'm trying to understand exactly how that applies in this case. We have $U^{\dagger}\sigma U = R\sigma$. The LHS can be written as $Ad_{U} \sigma$ and $U$ is $\exp(i(\theta/2)(\hat{n}\cdot \sigma))$ so that already covers the LHS. The RHS is tricker.

In the ad/Ad formula it seems we would calculate $\exp(ad(i\theta/2 \hat{n}\cdot \sigma)$, but in the expression in the question we have $J$ appear on the RHS, not $\sigma$

This is where I feel the isomorphism between $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ is needed to convert the $\sigma$ that appears on the RHS into a $J$

No, for the adjoint, it is only a matter of $\mathfrak{so}(3)$. The 3D space spanned by $\sigma_x,\sigma_y,\sigma_z$ is a representation of $\mathfrak{so}(3)$.

this is seen explicitly by the Lie bracket. Let $x,y\in\mathbb R^3$, then $$[x\cdot s,y\cdot s] = (x\times y)\cdot s$$ with $s=\sigma/2$, i.e. the Lie bracket structure of $\mathfrak{so}(3)$

Follow up question: physics.stackexchange.com/questions/819663/…

You can render mathjax in chat! math.ucla.edu/~robjohn/math/mathjax.html

Ok I'll write up a proper answer when I find the time