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I am working with the group morphism $\rho: S^3 \times S^3 \rightarrow SO(4)$ where $\rho(q,r)x = qxr^{-1}$ for $q,r \in S^3$ and $x \in \mathbb{R}^4$ and trying to compute the differential of this map at the identity, namely $\rho_{*,(1,1)}$, and show it is an isomorphism. I am trying to compute the Jacobian needed, by finding a basis of the tangent space in $SO(4)$ which I already have and a basis for the tangent space of the product $S^3 \times S^3$ and filling out the Jacobian as appropriately. However, I am having trouble understandin how a vector in the tangent space of $S^3 \times S^3$ at the identity would look like and how I could find a basis of this vector space. Can anyone help me with how I should think of the vector elements in $T_{1}S^3 \times T_{1}S^3 $ in order to find the desired Jacobian? Thanks!

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  • $\begingroup$ It should $T_1S^3\times T_1S^3$ not $T_{(1,1)}S^3\times T_{(1,1)}S^3$. Think of $S^3$ as the unit quaternions, and there is a natural identification $T_1S^3$ with $\operatorname{Im}\mathbb{H}$. $\endgroup$ – user10354138 Oct 1 '18 at 9:27
  • $\begingroup$ So the way I computed the tangent basis of $S^3$ at the identity before was by considering the map $\phi(a +bi +cj+dk) = \left( \begin{matrix} a - bi & -(c+di) \\ c-di & a + bi \end{matrix} \right)$ where I ended up getting that the basis is the matrices $\left( \begin{matrix}0 & i \\ i & 0 \end{matrix}\right)$, $\left( \begin{matrix}0 & -1 \\ 1 & 0 \end{matrix}\right)$, $\left( \begin{matrix}i & 0 \\ 0 & -i \end{matrix}\right)$ $\endgroup$ – user110320 Oct 1 '18 at 9:32
  • $\begingroup$ From there, I am not sure how to use that basis of $T_1S^3$. Any further comments I would really appreciate! Plus, You are right about the notation, I fixed it. $\endgroup$ – user110320 Oct 1 '18 at 9:32
  • $\begingroup$ I should mention that my intution would tell me that we have 9 basis vectors in $T_1S^3 \times T_1S^3$, just by counting the possible combinations of the basis vectors I found above. But I know this can't be because we have an isomorphism between $T_1S^3 \times T_1S^3$ and $so(4)$ where the last vector space has dimension 6. $\endgroup$ – user110320 Oct 1 '18 at 9:37
  • $\begingroup$ Those matrices are not a basis for $T_1S^3$. They are the image of that basis under the differential $dF_1$ where $F:S^3\to\mathfrak{su}(2)$. The basis you want is $\{i, j, k\}$. $\endgroup$ – cderwin Oct 2 '18 at 3:35
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You have a vector space $V \times W$. (In your case, $V$ and $W$ are isomorphic, but I'm gonna talk about the general case for clarity).

You have a basis $v_1, v_2, v_3$ for $V$, and a similar basis for $W$.

Your conjecture, I think, is that a basis for $V \times W$ consists of all pairs $$ (v_i, w_j) $$ where $i, j = 1, 2, 3$.

The correct claim is that $$ (v_1, 0), (v_2, 0), (v_3, 0), (0, w_1), (0, w_2), (0, w_3) $$ constitute a basis; the vector $(v, w)$ can be expressed in this basis by writing each of $v$ and $w$ in the respective bases: $$ v = a_1v_1 + \ldots + a_3 v_3\\ w = b_1w_1 + \ldots + b_3 w_3 $$ Once you've done that, you have

$$ (v, w) = a_1(v_1, 0) + a_2(v_2, 0) + a_3(v_3, 0) + b_1(0, w_1) + b_2(0, w_2) + b_3 (0, w_3). $$

Perhaps the key point hiding in here is that there's a nice isomorphism between $$ T_{q,r}(S^3 \times S^3) $$ and $$ T_q(S^3) \times T_r(S^3) $$ which lets you consider the latter vector space rather than the former. The isomorphism is induced by the projections on the two factors.

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First we will describe the tangent spaces on the 3-sphere. Consider $S^3=\{p\in\mathbb{R}^4\mid \|p\|=1\}$ as the unit vectors in $\mathbb{R}^4$. We shall regard $\mathbb{R}^4$ as the quaternions with basis elements $1$, $i$, $j$ and $k$. One can prove that $\langle uv , uw \rangle = \langle u,u\rangle\langle v,w\rangle$. From this it follows that $p \perp p \alpha$ for every $p\in S^3$ and every imaginary quaternion $\alpha$. Therefore $$ T_p S^3 = \mathrm{span}\{pi, pj, pk\}. $$

Now we can look at the product $S^3 \times S^3$. As John Hughes already pointed out, at every $(p,q)\in S^3\times S^3$, we can use the isomorphism $$ T_{(p,q)}(S^3\times S^3) \cong T_p S^3 \times T_q S^3. $$ Therefore the 6 vectors $$ (pi,0), (pj,0), (pk,0), (0,qi), (0,qj), (0,qk) $$ constitute a basis of the tangent space $T_{(p,q)}(S^3\times S^3)$.

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  • $\begingroup$ Is there a mistake in my answer? Feel free to comment... $\endgroup$ – Ernie060 Oct 1 '18 at 13:06

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