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I have been given a function $\varphi(A,B):\mathbb{H}\to\mathbb{H}$, $h\mapsto AhB^{-1}$ where $A,B\in SU(2)\times SU(2)$. I don't understand how this forms a well-defined map, nor how this would even return quaternion.

Going off the answer given here, it is straightforward enough to define map $F(C,D):h\mapsto ChD^{-1}$ given $C,D\in SU(2)$ - this clearly yields a quaternion. However, I can't then assume that $Ah$ is $ChD^{-1}$.

Thomas Pluck
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  • yes, this is an abuse of notation. $SU(2)$ is isomorphic to the group of unit quaternions. – Malkoun Oct 19 '19 at 10:16
  • The only possible interpretation that I can think of is to construct a 4x4 block matrix with 2 SU(2) matrices (top-left, bottom-right) accordingly, zeros elsewhere. This is a perfectly fine way of writing SU(2)xSU(2) and then multiplying these by 4D vector form of quaternions.

    Which agrees with some of my mathematical sensibilities but is far from a unique interpretation. I will see if this convention allows me to prove some properties of $\varphi$ and come back to the post.

     – Thomas Pluck Oct 19 '19 at 10:24
  • ok I will help you by writing a proper answer. – Malkoun Oct 19 '19 at 10:26
  • Update: foiled by simple matrix multiplication – Thomas Pluck Oct 19 '19 at 10:31

1 Answers1

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Every quaternion $q \in \mathbb{H}$ can be thought of as a special complex $2$ by $2$ matrix of the form:

$$\left( \begin{array}{cc} u & -\bar{v} \\ v & \bar{u} \end{array} \right) $$

where $q = u + j v$. One can check the algebra of such matrices is isomorphic to $\mathbb{H}$. Note that the group of unit quaternions can be shown to be isomorphic to $SU(2)$.

So essentially the map they are talking about can be understood either in a purely quaternionic way, where you multiply a quaternion from the left and right by unit quaternions, or using complex two by two matrices. I hope I made things a bit clearer.

Malkoun
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  • So am I to understand that, if A is in SU(2)xSU(2) then the action Ah is ChD for some C,D in SU(2)? – Thomas Pluck Oct 19 '19 at 10:35
  • You probably read a part that is slightly badly written. My guess is that if $(C,D) \in SU(2) \times SU(2)$, and if $h$ is a quaternion represented as a complex two by two matrix, then the action of $(C,D)$ on $h$ is $ChD^{-1}$. This gives a group homomorphism from $SU(2) \times SU(2)$ onto $SO(4)$, which is a $2$ to $1$ cover. Hence $Spin(4)$ is isomorphic to $SU(2) \times SU(2)$. Sorry for throwing too much information at once. – Malkoun Oct 19 '19 at 10:36
  • Thank you this really clarifies everything. I have no idea why you would use $SU(2)\times SU(2)$ when the same function can be defined with $SU(2)$ alone :) – Thomas Pluck Oct 19 '19 at 10:43
  • @TPluck, you cannot reduce that action of $SU(2) \times SU(2)$ to the action of a single copy of $SU(2)$ (if I understood your comment correctly). However, another action you should be aware of is the action of $B \in SU(2)$ on a quaternion $h$ given by $h$ goes to $B h B^{-1}$. Note that if $h$ is a purely imaginary quaternion, then its image is also purely imaginary. This gives a group homomorphism from $SU(2)$ onto $SO(3)$, which is also $2$ to $1$, so that $Spin(3)$ is isomorphic to $SU(2)$. – Malkoun Oct 19 '19 at 10:48
  • Well more along the lines that $\varphi(A,B)(h)=AhB^{-1}$ where $A,B\in SU(2)$ defines an action of $SU(2)\times SU(2)$ on the quaternions. However changing $A,B \in SU(2)\times SU(2)$ gives the same action through the abuse of notation using the above reasoning. Assuming that left and right actions are identical. – Thomas Pluck Oct 19 '19 at 10:52
  • I think the author probably just meant that $(A,B) \in SU(2) \times SU(2)$ or, which is the same, that $A$ and $B$ are elements of $SU(2)$. I am guessing they just forgot the parentheses... – Malkoun Oct 19 '19 at 10:56
  • The simplest explanation is often the best. Thank you again for the concise answer and clarifying in a long comment chain. – Thomas Pluck Oct 19 '19 at 10:57
  • Welcome. When people work for a while with quaternions, they make identifications often without mentioning it. You can view quaternions as special complex two by two matrices, or as special four by four real matrices. But when one writes an article, one should always keep in mind the audience so to speak. Anyway, I am glad I helped. – Malkoun Oct 19 '19 at 11:00