As the title,my question is why $SU(2)$ is diffeomorphic to $S^3$ ? How can I structure a diffeomorphic map between $SU(2)$ and $S^3$?
Please help me,and give me more details!
Thank you very much!
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3$\begingroup$ Write out the elements of $SU(2)$. They satisfy the equation of the sphere $\endgroup$– Exit pathNov 2, 2018 at 4:30
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$\begingroup$ I can not understand what you mean,please give sufficient evidence. $\endgroup$– Radeha LongaNov 2, 2018 at 4:45
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2 Answers
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By definition, we have that $\displaystyle SU(2)=\left\{\begin{pmatrix}a &-\overline{b}\\ b &\overline{a}\end{pmatrix}: a, b\in\mathbb{C}, |a|^2+|b|^2=1\right\}.$
Since $\mathbb{R}^4\cong\mathbb{C}^2 $, we may think of $S^3$ as $\displaystyle S^3=\{(a,b)\in\mathbb{C}^2: |a|^2+|b|^2=1\}$.
Define a map from $S^3$to $SU(2)$ as: \begin{aligned}F: S^3&\to SU(2)\\ F(a,b)&=\begin{pmatrix}a &-\overline{b}\\ b &\overline{a}\end{pmatrix}. \end{aligned}
Can you show that $F$ is well defined, $F$ is injective, and $F$ is surjective (i.e. an isomorphism)? After that, you should show that $F$ and $F^{-1}$ are smooth (or $C^{\infty}$). (Hint: Show $\displaystyle SU(2)$ is a submanifold of of the set of 2 x 2 matrices with complex entries, or better yet $\mathbb{R}^8$. What can you say about $F$ and its inverse restricted to these submanifolds?)
answered Nov 2, 2018 at 4:43
Vegeta the Prince of SaiyansVegeta the Prince of Saiyans
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$\begingroup$ Hi. This answer is very clear but could you maybe help me understanad why $SU(2)$ is a submanifold of $\mathbb{R}^8 \cong \mathbb{C}^4 \cong M_2(\mathbb{C})$? $\endgroup$– user661541Dec 18, 2019 at 14:23
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$\begingroup$ Does the condition $|a|^2+|b|^2=1$ not reduce the degrees of freedom by 1, to 3. Afterall, the lie algebra of SU(2) has 3 degrees of freedom:$x\sigma_x+y\sigma_y+z\sigma_z$. Wouldn't SU(2) then map better to just a normal three-dimensional sphere? $\endgroup$– Anon21Jan 30, 2020 at 23:16
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Every $SU(2)$ matrix is a linear combination of Pauli matrices $\sigma_1, \sigma_2, \sigma_3 $ and the unit matrix I, which is a base of the associated Lie algebra of the SU2 group. So every $SU(2)$ matrix $X$ can be written a $X=x_1\sigma_1 +x_2\sigma_2 +x_3\sigma_3 +x_4.I$ and the condition that $det{X} = 1$ gives $x_1^2 + x_2^2 +x_3^2 +x_4^2 = 1$ which is the $S^3$ sphere. And every point on that sphere represents an $SU(2)$ matrix
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$\begingroup$ 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 $\operatorname{SU}(2)$. $\endgroup$– AlbertAug 27 at 17:52