Tangent space basis of $S^3 \times S^3$

Question

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!

Answer 1

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.