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Determine the Lie algebra of the unit-quaternions $S^3 \subset \mathbb{H}^*$ and their left-invariant vector fields.

Unfortunately I am struggling with quaternions. I computed the differential of left-multiplication in $\mathbb{H}^*$ but that didn't get me very far.

Thanks in advance for any help !

Paul Joh
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    There are several posts here with an answer, e.g. this one. I don't know which duplicate is the best. See also this post. You should say what you can use. – Dietrich Burde Mar 21 '22 at 19:24
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    See also the answer here. – Dietrich Burde Mar 21 '22 at 19:41
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    I would like to also advertise my answer https://math.stackexchange.com/a/4020854/96384. It is much more general, but includes (see "Note 1") your case as the basic case $SL_1(\mathbb H)$ and writes the Lie algebra explicitly as $\mathfrak{su}_2$ and/or the space of pure quaternions with commutator bracket (which by the way is further $\simeq (\mathbb R^3, \times)$, the cross-product). – Torsten Schoeneberg Mar 21 '22 at 20:28

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Thanks for all the help guys !
I decided to use the Lie group-Lie algebra correspondence https://en.wikipedia.org/wiki/Lie_group%E2%80%93Lie_algebra_correspondence.\ We know that $\mathbb{H} \cong \mathbb{R}\cdot SU(2)$, then it is easy to see that $S^3 \cong SU(2)$. By the above correspondence, it suffices to compute $T_{id}SU(2)$. For that, say we have a smooth curve $c:(-\epsilon,\epsilon) \to SU(2)$ with $c(0) = id$ and $c'(0) = X \in T_{id}SU(2)$. Since $c(t)$ lies in $SU(2) \: \forall t \in (-\epsilon,\epsilon)$ we have that $c^{*}(t)\cdot c(t) = id$. Differentiating both sides for t and evaluating at $t = 0$ yields:
$X+X^{*} = 0$. Therefore $T_{id}SU(2) \subset V:=\{X \in SU(2) \:| X^{*}+X = 0\}$. And since both spaces have $\mathbb{R}-$dimension 3 they must be equal. If we write $X^{*}+X$ explicitly we get: \begin{equation} \begin{bmatrix} u & -\overline{v} \\ v & \overline{u} \end{bmatrix} + \begin{bmatrix} \overline{u} & \overline{v}\\ -v & u\\ \end{bmatrix} = \begin{bmatrix} 0 & 0\\ 0 & 0\\ \end{bmatrix} \end{equation} and therefore that $u + \overline{u} = 0$ which means u must be purely imaginary. But with the standard identification $\mathbb{H} \cong SU(2)$ $u = a+i\cdot b$. So $T_{id}S^3 = Im(\mathbb{H})$.

Now for the left-invariant vector-fields:
Let $z \in S^3$. To compute $(l_z)_{*,id}$, take $h \in T_{id}S^3 = Im(\mathbb{H})$, and a curve $c:(-\epsilon,\epsilon) \to S^3$ with $c(0) = id$ and $c'(0) = h$. Then $(l_z)_{*,id}(h) = \frac{d}{dt}|_{t=0}(l_z \circ c(t)) = \frac{d}{dt}|_{t=0}(z \cdot c(t)) = z \cdot h$. So the differential is just left-multiplication. I now choose the basis $\{i,j,k\}$ of $T_{id}S^3$. This is then a basis of the left-invariant vector-fields on S^3. To compute them explicitly:
$X_i(a,b,c,d) = (a+b\cdot i + c\cdot j + d \cdot k)\cdot i = -b + a\cdot i + d\cdot j - c\cdot k$.
$X_j(a,b,c,d) = (a+b\cdot i + c\cdot j + d \cdot k)\cdot j = -c - d\cdot i + a\cdot j + b \cdot k$.
$X_k(a,b,c,d) = (a+b\cdot i + c\cdot j + d \cdot k)\cdot k = -d + c\cdot i-b \cdot j + a\cdot k$.

I hope everything is correct.

Paul Joh
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