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How is Lie algebra named? Why is it $\mathfrak{su}(2)$ for group $SU(2)$, but $\mathfrak{o}(3)$ for group $SO(3)$? What does the "$\mathfrak{s}$" in algebra $\mathfrak{su}(2)$ mean?

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The "s" stands for special. Lie algebras are usually named after Lie groups. $\text{U}(2)$ and $\text{SU}(2)$ have different Lie algebras so they're named $\mathfrak{u}(2)$ and $\mathfrak{su}(2)$, but as it turns out, $\text{O}(3)$ and $\text{SO}(3)$ have the same Lie algebra, so $\mathfrak{o}(3) \cong \mathfrak{so}(3)$.

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Can you elaborate a bit on how $\mathfrak{o}(3) \cong \mathfrak{so}(3)$? –  Siyuan Ren May 24 '12 at 13:23
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@Karsus: well, as you presumably know, the Lie algebra of a Lie group is its tangent space at the identity. $\text{O}(3)$ and $\text{SO}(3)$ have the same tangent space at the identity because the latter is just the connected component containing the identity of the former (recall that elements of $\text{O}(3)$ have determinant $\pm 1$). –  Qiaochu Yuan May 24 '12 at 13:41
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@KarsusRen Qiaochu's explanation is the simplest but if you are not familiar with Lie groups (or if you want a proof using only the definitions of the relevant Lie algebras), then note that $\mathfrak{o}(3)$ consists of all skew-symmetric real $3\times 3$ matrices and $\mathfrak{so}(3)$ is the Lie subalgebra of $\mathfrak{o}(3)$ consisting of all traceless $3\times 3$ matrices. Of course, skew-symmetric matrices are traceless, i.e., the extra condition is redundant and $\mathfrak{o}(3)=\mathfrak{so}(3)$ ... –  Amitesh Datta Jun 26 '12 at 9:55
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@KarsusRen ... Of course, the proof above is tautological in a sense because I have defined $\mathfrak{o}(3)$ and $\mathfrak{so}(3)$ in such a way that the equality $\mathfrak{o}(3)=\mathfrak{so}(3)$ is trivial. However, a generalization of this idea is useful in the corresponding case of the inclusion of Lie groups $SU(2)\subseteq U(2)$. Note that $\mathfrak{u}(2)$ consists of all skew-hermitian complex $2\times 2$ matrices and $\mathfrak{su}(2)$ is the Lie subalgebra of $\mathfrak{u}(2)$ consisting of all traceless $2\times 2$ matrices. Note that $\mathfrak{u}(2)\neq \mathfrak{su}(2)$ ... –  Amitesh Datta Jun 26 '12 at 9:59
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@KarsusRen ... A skew-hermitian complex $2\times 2$ matric is not necessarily traceless, e.g., consider the diagonal complex $2\times 2$ matrix whose diagonal entries are all equal to $i\in\mathbb{C}$. –  Amitesh Datta Jun 26 '12 at 9:59

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