*The following makes no claims to originality or any physical significance on its own. For a better explanation of some of the math and the physical significance of the use of quaternions here, see this lecture by John Baez. *

*
**I’ve been spending a lot of time thinking about spinors and vectors in four dimensions, where I do think there is some important physical significance to the kind of issue discussed here. See chapter 10 here for something about four dimensions. A project for the rest of the semester is to write a lot more about this four-dimensional story.*

Until recently I was very fond of the following argument: in three dimensions the relation between spinors and vectors is very simple, with spinors the more fundamental objects. If one uses the double cover $SU(2)=Spin(3)$ of the rotation group $SO(3)$, the spinor (S) and vector (V) representations satisfy

$$ S\otimes S = \mathbf 1 \oplus V$$

which is just the fact well-known to physicists that if you take the tensor product of two spinor representations, you get a scalar and a vector. The spinors are more fundamental, since you can construct $V$ using $S$, but not the other way around.

I still think spinor geometry is more fundamental than geometry based on vectors. But it’s become increasingly clear to me that there is something quite subtle going on here. The spinor representation is on $S=\mathbf C^2$, but one wants the vector representation to be on $V_{\mathbf R}=\mathbf R^3$, not on its complexification $V=\mathbf C^3$, which is what one gets by taking the tensor product of spinors.

To get a $V_{\mathbf R}$ from $V$, one needs an extra piece of structure: a real conjugation on $V$. This is a map

$$\sigma:V\rightarrow V$$

which

- commutes with the $SU(2)$ action
- is antilinear

$$ \sigma(\lambda v)=\overline{\lambda}\sigma(v)$$
- satisfies $\sigma^2=\mathbf 1$

$V_{\mathbf R}$ is then the conjugation-invariant subset of $V$.

If we were interested not in usual 3d Euclidean geometry and $Spin(3)$, but in the geometry of $\mathbf R^3$ with an inner product of $(2,1)$ signature, then the rotation group would be the time-orientation preserving subgroup $SO^+(2,1)\subset SO(2,1)$, with double cover $SL(2,\mathbf R)$. In this case the usual complex conjugations on $\mathbf C^2$ and $\mathbf C^3$ provide real conjugation maps that pick out real spinor ($S_{\mathbf R}=\mathbf R^2\subset S$) and vector

$(V_{\mathbf R}=\mathbf R^3\subset V=S\otimes S)$ representations.

For the case of Euclidean geometry and $Spin(3)$, there is no possible real conjugation map $\sigma$ on $S$, and while there is a real conjugation map on $V$, it is not complex conjugation. To better understand what is going on, one can introduce the quaternions $\mathbf H$, and understand the spin representation in terms of them. The spin group $Spin(3)=SU(2)$ is the group $Sp(1)$ of unit-length quaternions and the spin representation on $S=\mathbf H$ is just the action on $s\in S$ of a unit quaternion $q$ by left multiplication

$$s\rightarrow qs$$

(we could instead define things using right multiplication).

There is an action of $\mathbf H$ on $S$ commuting with the spin representation, the right action on $S$ by elements $x\in \mathbf H$ according to

$$s\rightarrow s\overline{q}$$

(this is a right action since $\overline {q_1q_2}=\overline q_2\ \overline q_1$).

This quaternionic version of the spin representation is a complex representation of the spin group, since the right action by the quaternion $\mathbf i$ provides a complex structure on $S=\mathbf H$. While there are no real conjugation maps $\sigma$ on the spin representation $S$, there is instead a quaternionic conjugation map, meaning an anti-linear map $\tau$ commuting with the spin representation and satisfying $\tau^2=-\mathbf 1$. An example is given by right multiplication by $\mathbf j$

$$\tau (q)=q\mathbf j$$

Note that in the above we could have replaced $\mathbf i$ by any unit-length purely imaginary quaternion and $\mathbf j$ by any other unit-length purely imaginary quaternion anticommuting with the first.

In general, a representation of a group $G$ on a complex vector space $V$ is called

- A real representation if there is a real conjugation $\sigma$. In this case the group acts on the $\sigma$-invariant subspace $V_\mathbf R\subset V$ and $V$ is the complexification of $V_\mathbf R$.
- A quaternionic representation if there is a quaternionic conjugation $\tau$. In this case $\tau$ makes $V$ a quaternionic vector space, in a way that commutes with the group action.

Returning to our original situation of the relation $S\otimes S= 1 \oplus V$ between complex representations, $S$ is a quaternionic representation, with a quaternionic conjugation $\tau$. Applying $\tau$ to both terms of the tensor product the minus signs cancel and one gets a real conjugation $\sigma$ on $V$.

What’s a bit mysterious is not the above, but the fact that when we do quantum mechanics, we have to work with complex numbers, not quaternions. We then have to find a consistent way to replace quaternions by complex two by two matrices when they are rotations and and complex column vectors when they are spinors (so $S=\mathbf C^2$ rather than $\mathbf H$).

In my book on QM and representation theory I use a standard sort of choice that identifies $\mathbf i,\mathbf j,\mathbf k$ with corresponding Pauli matrices (up to a factor of $i$):

$$1\leftrightarrow \mathbf 1=\begin{pmatrix}1&0\\0&1\end{pmatrix},\ \ \mathbf i\leftrightarrow -i\sigma_1=\begin{pmatrix}0&-i\\ -i&0\end{pmatrix},\ \ \mathbf j\leftrightarrow -i\sigma_2=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$$

$$\mathbf k\leftrightarrow -i\sigma_3=\begin{pmatrix}-i&0\\ 0&i\end{pmatrix}$$

or equivalently identifies

$$q=q_0 +q_1\mathbf i +q_2\mathbf j + q_3\mathbf k \leftrightarrow \begin{pmatrix}q_0-iq_3&-q_2-iq_1\\q_2-iq_1 &q_0 +iq_3\end{pmatrix}$$

Note that this particular choice incorporates the physicist’s traditional convention distinguishing the $3$-direction as the one for which the spin matrix is diagonalized.

The subtle problem here is the same one discussed above. Just as the vector representation is complex with a non-obvious real conjugation, here complex matrices give not $\mathbf H$ but its complexification

$$M(2,\mathbf C)=\mathbf H\otimes_{\mathbf R}\mathbf C$$

*Note added: complexified quaternions are often called “biquaternions”*

The real conjugation is not complex conjugation, but the non-obvious map

$$\sigma (\begin{pmatrix}\alpha&\beta \\ \gamma & \delta\end{pmatrix})= \begin{pmatrix}\overline\delta &-\overline\gamma \\ -\overline\beta & \overline\alpha \end{pmatrix}$$

Among mathematicians (see for example Keith Conrad’s Quaternion Algebras), a standard way to consistently identify $\mathbf H$ with a subset of complex matrices as well as with $\mathbf C^2$, (giving the spinor representation) is the following:

- Identify $\mathbf C\subset \mathbf H$ as

$$z=x+iy\in \mathbf C \leftrightarrow x+\mathbf i y \in \mathbf H$$
- Identify $\mathbf H$ as a complex vector space with $\mathbf C^2$ by

$$q=z +\mathbf j w \leftrightarrow \begin{pmatrix}z\\ w\end{pmatrix}$$

Note that one needs to be careful about the order of multiplication when writing quaternions this way (where multiplication by a complex number is on the right), since

$$z+w\mathbf j= z+\mathbf j\overline w$$
- Identify $\mathbf H$ as a subset of $M(2,\mathbf C)$ by

$$q=z +\mathbf jw \leftrightarrow \begin{pmatrix}z&-\overline{w}\\ w& \overline z\end{pmatrix}$$
This is determined by requiring that multiplication of quaternions in the spinor story correspond correctly to multiplication of an element of $\mathbf C^2$ by a matrix.

With this identification

$$\mathbf i\leftrightarrow \begin{pmatrix}i&0\\ 0&-i\end{pmatrix},\ \ \mathbf j\leftrightarrow \begin{pmatrix}0&-1\\ 1&0\end{pmatrix},\ \ \mathbf k\leftrightarrow \begin{pmatrix} 0&-i\\ -i&0\end{pmatrix}$$

This is a bit different than the Pauli matrix version above, but shares the same real conjugation map identifying $\mathbf H$ as a subset of $M(2,\mathbf C)$.

**Update**: There’s a very new video here, where Keith Conrad discusses quaternions, especially the case of quaternion algebras over $\mathbf Q$ and their relation to quadratic reciprocity.