Naïve question about the general form of elements of $SU(2)$ and a homomorphism from $S^3$ to $SU(2)$ .

Question

I have a follow-up question to this post.

Is there a difference if we set $$SU(2)=\{\begin{bmatrix} \alpha & \beta\\ -\overline{\beta} & \overline{\alpha} \end{bmatrix}:\alpha,\beta\in\mathbb{C}\}\text{ or }SU(2)=\{\begin{bmatrix} \alpha & -\overline{\beta}\\ \beta & \overline{\alpha} \end{bmatrix}:\alpha,\beta\in\mathbb{C}\}?$$

For example, I want to establish a homomorphism $\varphi_1$ between a group of unit quaternions $S^3$ and $SU(2)$ i.e. $$\varphi_1:S^3\to SU(2)$$ where I send $(x+jy=(a+bi)+j(c+di))\mapsto\begin{bmatrix} x & y\\ -\overline{y} & \overline{x} \end{bmatrix}$. We can see that $$\varphi_1(i)=\begin{bmatrix} i & 0\\ 0 & -i \end{bmatrix},\varphi_1(j)=\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix},\text{ and }\varphi_1(k)=\begin{bmatrix} 0 & i\\ i & 0 \end{bmatrix}.$$ So, $\varphi_1(ij)=\varphi_1(i)\varphi_1(j)$.

But, if I define a different group homomorphism $\varphi_2(x+jy)=\begin{bmatrix} x & -\overline{y}\\ y & \overline{x} \end{bmatrix}$, then I don't have anymore that $\varphi_2(ij)=\varphi_2(i)\varphi_2(j)$. But, I have that $$\varphi_2(ij)=\varphi_2(j)\varphi_2(i)$$ where $$\varphi_2(i)=\begin{bmatrix} i & 0\\ 0 & -i \end{bmatrix},\varphi_2(j)=\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix},\text{ and }\varphi_2(k)=\begin{bmatrix} 0 & i\\ i & 0 \end{bmatrix}.$$

Does it happen because $\varphi_2(x)=\varphi_1(x)^T$? And, how is it connected to our choice of general form of elements?

Answer 1

There's no difference between using the two forms you described; you pass from one to the other by sending $\beta \mapsto - \overline{\beta}$.

There is a difference between $\varphi_1$ and $\varphi_2$, because the multiplication on the quaternions is fixed. As you've computed, $\varphi_1$ is a homomorphism and $\varphi_2$ isn't. You could make $\varphi_2$ into a homomorphism by redefining the multiplication on the quaternions if you wanted to.

Answer 2

Yes, taking the transpose reverts the order of matrix multiplication, $(AB)^T=B^TA^T$, and you end up with an "antihomomorphism" instead. That is $\phi_2(q_1q_2)=\phi_2(q_2)\phi_2(q_1)$ for all $q_1,q_2$. Observe that whenever $q_1$ and $q_2$ commute the "anti"-feature is nowhere to be seen. This happens for example when $q_1$ and $q_2$ are both complex numbers.

If you are unhappy with the usual homomorphism from the unit quaternions, you can use conjugation to replace it with a different one. So fix a unit quaternion $r$ and look at $\phi_2(q)=\phi_1(rqr^{-1})$. Alternatively you can do the conjugation on the matrix side, and conjugate the values of $\phi_1$ by a fixed matrix $R\in SU(2)$.

One more thing. If you want $\phi_2(i)=\phi_1(i)$, then you need to select $r$ from the set of unit complex numbers also, essentially leaving you with the choice of an angle $\alpha$ such that $\phi_2(j)=\cos\alpha\phi_1(j)+\sin\alpha\phi_1(k)$. In the case of quaternions it is probably straightforward to prove that claim. It is an instance of a more powerful result known as Skolem-Noether theorem.