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

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!

• 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 $\operatorname{Im}\mathbb{H}$. – user10354138 Oct 1 '18 at 9:27
• 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 $\left( \begin{matrix}0 & i \\ i & 0 \end{matrix}\right)$, $\left( \begin{matrix}0 & -1 \\ 1 & 0 \end{matrix}\right)$, $\left( \begin{matrix}i & 0 \\ 0 & -i \end{matrix}\right)$ – user110320 Oct 1 '18 at 9:32
• 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. – user110320 Oct 1 '18 at 9:32
• 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. – user110320 Oct 1 '18 at 9:37
• 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 $\{i, j, k\}$. – cderwin Oct 2 '18 at 3:35

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.

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 $$T_{(p,q)}(S^3\times S^3)$$.

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