2020年5月13日 · Thus, it's true that the basis elements are the same (so of course they follow the same commutation relations) but it's also true that the vector space of the SU(2) Lie algebra is much bigger since is the complex linear combination.

To prove surjectivity, you can use that SO(3) is connected and that the quotient map sends a small neighborhood of the identity in SU(2) to a small neighborhood of the identity of SO(3), so the image is an open subgroup of the latter, hence everything by connectedness. $\endgroup$

2023年12月29日 · With all due respect to Manton, if that is indeed what he does, defining the map from $\mathrm{SU}_2$ to $\mathrm{SO}_3$ by this formula is a god-awful way to describe the map!

In principle it is enough to take the exponential of the Lie algebra isomorphism and a surjective Lie group homomorphism arises this way $\phi : SU(2)\to SO(3)$: $$\phi\left(\exp\left\{-\sum_k t^k i\sigma_k/2\right\}\right) =\exp\left\{-\sum_k t^k iL_k\right\}\:.$$ The point is that one should be sure that the argument in the left-hand side ...

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In fact $\mathrm{SU}(2)$ is the double cover of $\mathrm{SO}(3)$; there is a 2-1 homomorphism from the former to the latter. How is this possible when every so(3) irrep can get raised to one of SO(3) as described above?

$\begingroup$ You seem to be confusing $\operatorname{SU}(2)$ with its algebra $\mathfrak{su}(2)$; besides, you can very easily check that any Pauli matrix $\sigma_j$ has a determinant $\det \sigma_j = -1$, so they do not even belong to …

2018年3月31日 · I read that SU(2) is the double covering of SO(3), so to each matrix in SO(3) corresponds one in SU(2). I am not sure I understand this. So if we have a 3D representation of SU(2), the 3D object it acts on are complex vectors, while the …

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