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Rotations in $\mathbb{R}^3$ are generated by differential operators $$ O_{yz}=y\frac{d}{dz}-z\frac{d}{dy}, O_{zx}=z\frac{d}{dx}-x\frac{d}{dz}, O_{yx}=y\frac{d}{dx}-x\frac{d}{dy},$$ which satisfy the commutation relations $$[O_{yz},O_{zx}]=O_{yx}$$ and $[O_{ab},O_{bc}]=O_{ac}$ in general, with $O_{ab}=-O_{ba}$.

I suppose rotations in $\mathbb{R}^4$ must be generated by those three generators plus three others $O_{xw}$, $O_{yw}$, $O_{zw}$, defined analogously. This is consistent with the dimension of algebra $so(4)$ being 6.

Now I have seen that $so(4)\sim so(3)\oplus so(3)$, so there should be possible to construct linear combinations of the six $O_{ab}$ generators in $\mathbb{R}^4$ to produce two separate $so(3)$. How is this done?

thedude
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  • @TorstenSchoeneberg I can see that it is related, but I cannot read off the answer to my question from there. I know it must be there somehow, but am not able to translate – thedude Nov 29 '20 at 19:40
  • (To be clear, one has to do a little more work than there. But if you read the parameters $a,...,f$ in that answer as real numbers, you have $so(4) \simeq su(2) \times su(2)$, and then use an iso from su(2) to so(3).) – Torsten Schoeneberg Nov 29 '20 at 19:51
  • I thought I would read $(a,b,c,d,e,f)$ as $(O_{xy},O_{xz},O_{xw},O_{yz},O_{yw},O_{zw})$, but that doesn't work – thedude Nov 29 '20 at 19:55
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    Does this help? (Specifically pages 1-3) – RyanK Nov 29 '20 at 20:03
  • @RyanK Yes, this is exactly what I wanted. – thedude Nov 29 '20 at 20:13

1 Answers1

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$$L_{xy}=\tfrac12(O_{xy}+O_{zw}),\quad L_{yz}=\tfrac12(O_{yz}+O_{xw}),\quad L_{zx}=\tfrac12(O_{zx}+O_{yw})$$

$$R_{xy}=\tfrac12(O_{xy}-O_{zw}),\quad R_{yz}=\tfrac12(O_{yz}-O_{xw}),\quad R_{zx}=\tfrac12(O_{zx}-O_{yw})$$

mr_e_man
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