The Mystery of Spin

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
$$\sigma:V\rightarrow V$$

  • 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.

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20 Responses to The Mystery of Spin

  1. john says:

    I have been following this series called “Spinors for Beginners” and they are up to video 20 in the series which covers the Lorentz Group. It’s good background information to help me understand the spinor topics in your blog posts.

  2. Peter – You open your remarks with
    “…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.”

    In geometric representation of Clifford algebra, wavefunction interactions can be modeled by the Clifford product, sum of dimension-reducing dot and dimension-increasing wedge products. So product of two vectors is scalar plus bivector, where bivector is what i think you are calling a spinor. This seems to be the inverse of what you say.

    In geometric representation of Clifford algebra, product of two bivector spinors is scalar plus quadvector. No one-dimensional vector there.

    Please can you explain to me how our perspectives seem inverted in this?

  3. Garrett Lisi says:

    The dance between spinors and biquaternions makes an appearance in a recent short paper I wrote, “C, P, T, and Triality.” I would send you the hep-th arXiv link but the paper has been “on hold” for five weeks now. This is especially annoying since I don’t think there’s anything technically wrong with the paper. I strongly suspect one or more arXiv moderators just doesn’t like me or my work so I’m getting special treatment.

  4. @Garrett

    why not also host it elsewhere until then so people can see it?

  5. Garrett Lisi says:

    The arXiv is where it belongs, and the most reasonable place people would see it and possibly cite it. If they reject it, which I see no good reason for, maybe I’ll retract all my papers in disgust.

  6. Peter Woit says:


    Sorry to hear that. I hope anyone reading this who is in the good graces of the arXiv moderators will contact them to ask them to fix this.

    I’ll also add to the posting what I should have originally had there, that “biquaternions” is a standard name for the complexification of the quaternions.

  7. Peter Woit says:

    Peter Cameron,
    A bivector is not a spinor.
    What I’m writing about in this posting is specific to 3d, but there’s a very general story explaining what you bring up, simplest in even dimensions.

    In dimension d=2n, the Clifford algebra (which includes vectors, your bivectors, etc), when complexified (allow complex coefficients) is isomorphic to the matrix algebra $M(2^n,\mathbf C)$. Spinors are column vectors $\mathbf C^n$ that these matrices act on. More invariantly, the Clifford algebra is the algebra of linear transformations of the spinor space S.

    What you can build out of vectors is not spinors, but linear transformations of spinors. Since the linear maps on $S$ are isomorphic to the tensor product $S\otimes S^*$, the general story is that (in even dimensions) the spinors are not something you can build out of vectors, but a “square root” of what you can build out of vectors (the complexified Clifford algebra).

  8. @Garrett

    I’m not saying don’t keep pushing to get it on the arXiv, I’m saying pop it on your own website for now (or somewhere, anywhere, public), so that people can judge for themselves the content of the paper, now that you’ve just advertised it on a widely-read blog…. By doing nothing you’re giving the arXiv moderation process the power to keep the paper out of sight.

  9. Garrett Lisi says:

    Very well, I’ve made it available here:

  10. flippiefanus says:

    “I still think spinor geometry is more fundamental than geometry based on vectors.”
    There is a subtlety with this notion that has frustrated me for a long time. It is the fact that the formulation of geometry in terms of vectors allows shift invariance (because there is nothing physically special about the choice of the origin), while the formulation in terms of spinors does not seem to make shift invariance so apparent. The choice of origin is somehow encoded into the associated spinors that cannot be so easily accessed. Perhaps you know of a way around this problem.

  11. Peter Woit says:


    Yes, vectors are essentially the same as shifts, they generate translations (momentum is a vector). If you define vectors in terms of spinors, nothing changes about translations. But if you express your vectors in terms of spinors (e.g. as matrices), you have some new things. In particular you have the Dirac operator, which is a vector, but acts on spinors.

    In the usual discussion, the Dirac operator is written $D=\gamma^\mu \frac{\partial}{\partial x_\mu}$ which looks like it is a scalar. It really is a vector though…

  12. To a mathematician, a spinor is not an element of C^2, but an element of H. Then you have absolutely no problem getting a real vector out of the square of the spinor, because you take the quaternionic tensor product of two 1-dimensional quaternion representations, which is automatically a 4-dimensional real representation.

  13. The main problem, I think, with regarding a spinor as living in C^2 rather than H, is that it picks out a particular direction of spin as being somehow special. The question then is, which direction is it? This is not an easy question to answer. But it is an important question to ask. So I don’t understand why physicists try to insist that it isn’t a question.

  14. Peter Woit says:

    Robert A. Wilson,
    If you’re doing QM with spin, you need to have a complex state space. In particular, if you want to couple to EM, you need a distinguished choice of how the complex numbers act. I do suspect there’s a mystery there.

  15. Yes, I agree there is a bit of a mystery there – why does the mathematical symmetry of the quaternions need to be broken to C^2 to describe observed physics? There is no doubt that it does need to be broken, but why? And how?

  16. GS says:

    I think there are typos in your definition of “antilinear”. I think it should be
    \sigma (\lambda v) = \bar{\lambda} \sigma (v)

  17. Peter Woit says:

    Thanks! fixed.

  18. akhmeteli says:

    Peter Woit,

    “A bivector is not a spinor.”

    For what it’s worth, according to E. Cartan, The Theory of Spinors (Dover, 1981), Section 154, in the space of special relativity, “the semi-spinor is, then, the polarised bivector”. See also E.T. Whittaker, On the relations of the tensor-calculus to the spinor-calculus. Proc. Roy. Soc. A 158, 38–46 (1937).

  19. Peter Woit says:

    The Whittaker reference makes clear that this is about the Klein correspondence, which is at the basis of the idea of twistors.
    What’s going on is you can identify the Grassmannian of complex two planes (the half-spinor space at a point) in complex four dimensions (twistor space) with the space of two-forms satisfying a particular equation. So, yes, you can relate spinors to certain two form (“bivectors”). But, spinors are not bivectors, this is about special complex bivectors, and is very special to four dimensions.

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