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Q1: As we know, in classical mechanics(CM), according to Noether's theorem, there is always one conserved quantity corresponding to one particular symmetry. Now consider a classical system in a $n$ dimensional general coordinates space described by the Lagrangian $L(q,\dot{q})$, where $q=(q_1,...,q_n)$ and $\dot{q}=(\dot{q_1},...,\dot{q_n})$ are general coordinates and general velocities, respectively. If the system has $SO(n)$ spatial rotation-symmetry, e.g.$\forall A\in SO(n),L(Aq^T,A\dot{q}^T)=L(q,\dot{q})$, then we can get $\frac{n(n-1)}{2}$(number of the generators of group $SO(n)$) conserved angular momentum.

My question is as follows, now note that our spatial dimension is $n$, when $n=3\rightarrow $ number of angular momentum$(\frac{n(n-1)}{2})$$=$spatial dimension$(n)$$=$3, otherwise, number of angular momentum$\neq $spatial dimension, so why spatial dimension 3 is so special? Is there any deep reason for number 3 or it's just an accidental event? Or even does this phenomena have something to do with the fact that we "live" in a 3D world?

Q2: In quantum mechanics(QM), a Hermitian operator $J=(J_x,J_y,J_z)$ is called an angular momentum iff $[J_x,J_y]=iJ_z,[J_y,J_z]=iJ_x,[J_z,J_x]=iJ_y$.

And my question is as follows, in QM, we can have 1-component momentum operator $\hat{p}$ , or 2-component momentum operator$(\hat{p_x},\hat{p_y})$, and so on. But Why we have never encountered a angular momentum with only two components $J=(J_x,J_y)$? Can we define a 2-component angular momentum? Like in the CM case, again the number 3 is special in QM case, why?

Thanks in advance.

By the way: More questions concerning the definition of rotation groups for angular momentum can be found here , who are interested in may have a look, thanks.

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Related: physics.stackexchange.com/q/9864/2451 –  Qmechanic Apr 27 '13 at 14:22
    
@ Qmechanic : Thank you very much. I think I become more clear from this related link. –  K-boy Apr 27 '13 at 18:14
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1 Answer

Angular momentum is a bivector, $J = x\wedge p$, and since the exterior/wedge product is antisymmetric, you do indeed get $n(n-1)/2$ independent components for any bivector. In general the Hodge dual provides an isomorphism between $k$-vectors and $(n-k)$-vectors. In three dimensions, $\star(x\wedge p)$ is an axial vector which and we call this Hodge-star-of-exterior-product operation 'cross product'. So three dimensions is special in the sense that bivectors are naturally isomorphic to vectors only in three dimensions.

Thus the components $(J_x,J_y,J_z)$ are stand-ins for the three independent bivector components $(J_{yz},J_{zx},J_{xy})$, and something like $(J_x,J_y) = (J_{yz},J_{zx})$ would have a missing component, since you'd necessarily have three axes.

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