Usually in Physics we define a spinor to be an element of the $\left(\frac{1}{2},0\right)$ representation space of the Lorentz group. Essentially this boils down to the 'n-tuple of numbers that transforms like a spinor' definition that physicists tend to use for vectors, covectors, and tensors.

However, vectors, covectors, and tensors also have geometric definitions that are much nicer than, and also equivalent to, the 'n-tuple of numbers' definition. For example, a vector can be thought of as an equivalence class of curves tangent at a point, or the directional derivative at a point. A covector can be thought of as a differential 1-form or as an equivalence class of functions with equal gradient at a point. Tensors are then tensor products of these spaces.

I was wondering if there is a similar definition of spinors based in differential geometry rather than just the representation theory of the Lorentz group. If so, are these specific to certain manifolds (complex, Lorentzian, etc), or are they general to all manifolds?

  • $\begingroup$ spin is related to rotation, if a thing has spin 'n' this means that this object is invariant under rotations $ 2\pi /n $ $\endgroup$ – Jose Garcia May 7 '17 at 10:59
  • $\begingroup$ @JoseGarcia Yes I know what spin is. I was asking if there was a definition of a spinor (an element of a complex vector space) that was not based in group theory. $\endgroup$ – gautampk May 7 '17 at 11:06
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    $\begingroup$ The spin groups (double covers of orientation-preserving isometry groups) are constructed in Clifford algebras, and the recursive isomorphisms for Clifford algebras are used to classify them as matrix algebras (over R, C, H) which give spin representations. I have heard some sources say these matrix representations are "non-canonical" or unnatural, in the sense of depending on coordinates and arbitrary choices, which suggests cause for pessimism in the search for direct geometric meaning. But I've never proved this statement (after satisfactorily formalizing the claim categorically first). $\endgroup$ – arctic tern May 8 '17 at 22:01
  • $\begingroup$ Not an answer but perhaps fun/interesting: cs.umb.edu/~eb/spinorSpanner.pdf $\endgroup$ – Ethan Bolker May 10 '17 at 23:58
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    $\begingroup$ Not directly what you want but I like the orientation explanation of spinors. In Misner, Thorne and Wheeler they draw a small box whose corners are connected by string to a larger box. They illustrate how rotation tangles the string but all is recorded by the spinor. $\endgroup$ – Drone Scientist May 17 '17 at 0:25

I understand that your background is in physics, but (since I have neither time nor desire to write here a crash course in the relevant mathematics), I write my answer pretending that you are a reasonably advanced graduate student in math.

Let me first try to interpret your question: You start with the standard representation of the group $G=SO(3,1)_0$ (the connected component of identity in $SO(3,1)$, equivalently, the subgroup of $SO(3,1)$ preserving the future light cone) on the Lorentzian 4-space $R^{3,1}$, the 3-dimensional real vector space equipped with a nondegenerate symmetric bilinear form $\langle, \rangle$ of signature $(3,1)$, with the associated quadratic form $$ x^2 +y^2 + z^2 -w^2. $$ The 2-fold nontrivial cover of the Lie group $G$ (the spinor group of $G$) is isomorphic to $SL(2, {\mathbb C})$ which has the natural (spinor) representation on ${\mathbb C}^2$. You would like to have an interpretation of vectors in ${\mathbb C}^2$ in terms of geometric objects associated with $R^{3,1}$.

a. First, I will do this in the case of the group $SO(2,1)_0$ since it is easier and illuminating. The natural representation of this group is on the Lorentzian 3-space $R^{2,1}$; the spinor group of $SO(2,1)_0$ is $SL(2, {\mathbb R})$ and its natural (spinor) representation is on ${\mathbb R}^2$. In order to interpret the corresponding spinors (vectors in ${\mathbb R}^2$), let me first note obstacles:

  1. The dimension mismatch: 2 versus 3.

  2. The action of $SL(2, {\mathbb R})$ on nonzero spinors is transitive (with stabilizers conjugate to the 1-parameter subgroup of strictly upper triangular matrices: Such matrices are called "unipotent"), while the action of $SO(2,1)_0$ on nonzero vectors in $R^{2,1}$ is non-transitive: There are three types of orbits, consisting of time-like, null and space-like vectors. Null-vectors will be most important for us, they are defined by the condition $\langle v, v \rangle=0$.

  3. Any "natural" class of objects which one can define in $R^{2,1}$ will be acted upon by the group $SO(2,1)_0$ while we are interested in spinors, on which $SO(2,1)_0$ does not act.

  4. While the space $C$ of future-like null-vectors is 2-dimensional (I regard the zero vector as a future null-vector for convenience), it does not have a natural structure of a vector space. Nevertheless, it does have a distinguished point (the origin) and a family of distinguished lines which are intersections of $C$ with affine 2-planes parallel to null-lines. The problem with these "lines" in $C$ is that some of them are only half-lines (the ones which pass through the origin); the rest are (connected) hyperbolas, so at least topologically they do look like lines.

Note, however, that:

  1. The null-cone is 2-dimensional, homeomorphic to the 2-plane minus the origin. (Explicitly, you can define this homeomorphism by projecting $C$ to the coordinate plane in $R^{2,1}$ via the map $(x,y,z)\mapsto (x,y)$, where the quadratic form of the Lorentzian inner product is given by $x^2 + y^2 - z^2$.) This takes care of the dimension discrepancy.

  2. The action of $SO(2,1)_0$ on the future null-cone $C$ minus the origin is transitive. The stabilizer of each nonzero null-vector is again 1-parameter (unipotent) subgroup of $SO(2,1)_0$. Under the covering map $SL(2, {\mathbb R})\to SO(2,1)_0$ the unipotent subgroups of the former map isomorphically to the unipotent subgroups of the latter. This is what we will exploit.

Now, I will let $S$ be the 2-fold cover of the punctured plane $N= C-\{{\mathbf 0}\}$. Informally, you can think of the elements of $S$ as null-vectors $v\in N$ each equipped with a "spin", a $\pm$ sign, which switches to the opposite sign as we rotate $v$ 360 degrees around the z-axis. In terms of polar coordinates we can think of the elements of $S$ as $$ (r, \theta), r\ne 0, 0\le \theta< 4\pi. $$ The reduction modulo $2\pi$ sends these to points in the punctured plane whose polar coordinates are $$ (r, \theta), 0\le \theta< 2\pi. $$ This passage to the 2-fold cover is the "unnatural step" which allows one to get spinors.

Now, the action of $SO(2,1)_0$ lifts to an action on $S$ but not of $SO(2,1)_0$ itself: The lift is the action of the spinor group $SL(2, {\mathbb R})$. One can verify (for instance, by observing that the action of $SL(2, {\mathbb R})$ on $S$ is transitive with point-stabilizers which are 1-parameter unipotent subgroups as required) that there is a diffeomorphism $S\to {\mathbb R}^2 -\{{\mathbf 0}\}$ which conjugates the action of $SL(2, {\mathbb R})$ on $S$ to the standard linear action of $SL(2, {\mathbb R})$ on ${\mathbb R}^2 -\{{\mathbf 0}\}$. Under this diffeomorphism the "lines" which I mentioned above map to affine lines in $SL(2, {\mathbb R})$; each half-line lifts to the union of two half-lines in $S$ which map to a line (minus the origin) in ${\mathbb R}^2 -\{{\mathbf 0}\}$.

This gives you a reasonably geometric "Lorentzian" interpretation of nonzero spinors in ${\mathbb R}^2$ as elements of the surface $S$: These are nonzero future null-vectors $v\in N$ "equipped with a $\pm$ sign" to indicate which sheet of the 2-fold cover they lift to. The latter description is unsatisfactory as a mathematical description but should be OK as far as your intuition goes. The rigorous definition is in terms of covering spaces as I noted above. In order to get the zero spinor as well, one can simply say that we are using a 2-fold "branched cover" of $C$, which is ramified over the origin.

b. Now, to the Lorentzian 4-space $R^{3,1}$.The difficulties are somewhat similar. Again, note nontransitivity of the action of $G$ on the set of nonzero vectors in $R^{3,1}$. Inspired by (a), one can try to use the future null-cone $C\subset {\mathbb R}^{3,1}$. However, this results in the dimensional mismatch (the cone $C$ is 3-dimensional while the spinor space is real 4-dimensional). Also, while $G$ does act transitively on $C$ (minus the origin), the stabilizers are a bit larger than the ones in ${\mathbb C}^2$: The stabilizers of nonzero vectors in ${\mathbb C}^2$ are complex 1-parameter unipotent (real 2-dimensional), conjugate to the group of strictly upper triangular complex 2-by-2 matrices $$ \left[\begin{array}{cc} 1&*\\ 0&1\end{array}\right] $$

while the stabilizers of nonzero null-vectors are 3-dimensional (in addition to 2-dimensional unipotent subgroups of $G$ which do lift to unipotent subgroups of $SL(2, {\mathbb C})$ we also have 1-parameter elliptic subgroups, isomorphic to $S^1$, which fix the null-vectors). Another problem is that $N= C- \{{\mathbf 0}\}$ is simply-connected, so taking its covering spaces would not be useful.

Nevertheless, what we can do is to take a future nonzero null-vector $v\in C$ and equip it with a half-plane $P$ which is tangent to the cone $C$ along the line spanned by $v$. Now, the stabilizer of each "flag" $(v,P)$ in $G$ is real 2-dimensional (a unipotent subgroup which lifts to a complex 1-dimensional unipotent subgroup of $SL(2, {\mathbb C})$ as required). I let $F$ denote the space of such "flags" $(v,P)$. It is not hard to check that this space is connected with the fundamental group isomorphic to ${\mathbb Z}_2$, which means that $F$ does have a connected 2-fold cover $S\to F$. One can also describe $F$ as the total space of the tangent bundle of the 2-sphere with the image of the zero section deleted. Now, we can play the same game as in (a): The elements of $S$ can be though of as flags $(v,F)$ equipped with a "spin", a $\pm$ sign which changes after we "spin" the half-plane $F$ around $v$ 360 degrees. One then verifies that the action of $G=SO(3,1)_0$ lifts to $S$ to an action of the spinor group $SL(2, {\mathbb C})$ on ${\mathbb C}^2$ minus the origin (again, by comparing the structure of point-stabilizers).

This is again a reasonably geometric description of (nonzero) spinors as elements of the 4-dimensional manifold $S$. The drawback of this description is that we do not see directly a complex structure on $S$ and the fact that the spinor group acts holomorphically; the linear structure is also very nontransparent. If this does not bother you, stop reading here; if it does bother you, proceed to the item (c).

c. I will now give a description of spinors which is derived from the Lorentzian geometry of ${\mathbb R}^{3,1}$, where the complex linear structure is transparent, but mathematics required to understand it gets harder.

Let's go back to the 3-dimensional null-cone $C\subset {\mathbb R}^{3,1}$. The space $\Sigma$ of future null-rays in $C$ is naturally diffeomorphic to the 2-sphere $S^2$; the action of $G$ on $C$ under the map of $\Sigma$ to $S^2$ becomes the conformal action of $PSL(2, {\mathbb C})$ on the Riemann sphere. The conformal structure on $\Sigma$ can be described as follows. For each nonzero null-vector $v\in N$, the restriction of the Lorentzian inner product to $T_vC$ (the tangent space of the cone $C$ at $v$, which is 3-dimensional) is degenerate positive semidefinite: The vector $v$ pairs to zero with each vector $w\in T_vC$. However, the projection $N\to \Sigma$ (sending the positive ray through $v\in C$ top a single point) divides out the line ${\mathbb R}v$ and, hence, $\langle, \rangle$ projects to a positive-definite inner product on the tangent plane to $\Sigma$ at the equivalence class $[v]$ of $v$. The action of $G$ on $\Sigma$ preserves the conformal class of the resulting Riemannian metric on $\Sigma$; the orientation is also preserved, hence, the action is conformal. (One can also describe the almost complex structure on $T\Sigma$ more directly but I will skip this.) The tangent bundle $T\Sigma$ is a complex one-dimensional vector bundle on $\Sigma$; algebraic geometers would call it the anticanonical bundle. It has degree $+2$, hence, there exists a "half-anticanonical line bundle" $L$ on $\Sigma$ so that the tensor square of $L$ is isomorphic to $T\Sigma$. The line bundle $L$ has degree $+1$; algebraic geometers call it the "hyperplane bundle" of ${\mathbb C}P^1\cong S^2$.

Remark. the total space $E$ of $L$ can be described as the 2-fold branched cover over $T\Sigma$ which is ramified over the image of the zero section of $T\Sigma$. Now, you may start to see a connection to Part (b). Fiberwise - this branched cover is nothing by the 2-fold branched cover over the complex plane ramifield at the origin. Now, you see a connection to Part (a).

The group $PSL(2, {\mathbb C})$ acts on $T\Sigma$ via (holomorphic) bundle automorphisms and this action lifts to an action of $SL(2, {\mathbb C})$ on $L$. It is easy to check that the space of holomorphic sections $\Gamma(L)$ of $L$ is complex-2-dimensional. The action of $SL(2, {\mathbb C})$ on the vector space $\Gamma(L)$ is manifestly complex linear, nontrivial. Hence, we obtain the spinor representation of $SL(2, {\mathbb C})$ on $\Gamma(L)\cong {\mathbb C}^2$.

A complex-analyst would describe sections of $L$ as "holomorphic half-vector fields" (or degree $-1/2$ holomorphic differentials) $$ \omega=f(z)dz^{-1/2}, $$ where $f(z)$ as a holomorphic function. The strange degree $-1/2$ refers to the transformation law for such differentials: If $z=g(w)$ a conformal mapping then $g_*\omega$ is given by $$ f(w) (g')^{-1/2} dw^{-1/2}. $$ If $w=\frac{az+b}{cz+d}$ then $$ f(w) (g')^{-1/2} dw^{-1/2}= f(w) (\frac{1}{(cw+d)^2})^{-1/2} dw^{-1/2} = f(w) (cw+d) dw^{-1/2}. $$ Note that this expression is meaningless unless we specify the 2-by-2 matrix $$ \left[\begin{array}{cc} a&b\\ c&d\end{array} \right]\in SL(2, {\mathbb C}). $$ This is your spinor representation.

These holomorphic differentials of order $-1/2$ are (like it or not) your spinors. The linear structure is very transparent: In order to add these fellows, you just add the functional parts:

$$ f_1(z)dz^{-1/2} + f_2(z)dz^{-1/2}= (f_1(z)+ f_2(z))dz^{-1/2}. $$ Linearity of the action of the spinor group is also clear (this action is just a change of variables in the differentials). The fact that the space of spinors is complex 2-dimensional might not be immediate but becomes clear once you think a bit about it. (The space is isomorphic to the space of holomorphic functions on the complex plane which have at worst simple pole at infinity, i.e. have the form $\alpha z + \beta$, $\alpha, \beta\in {\mathbb C}$.)

I am not sure of analytical importance of "holomorphic half-vector fields", but holomorphic half-order differentials $$ \omega=f(z)dz^{1/2}, $$ do appear naturally in complex analysis when considering 2nd order linear holomorphic ODEs, that's how I first learned about them; see for instance (a bit dated but very clearly written):

N. S. Hawley and M. Schiffer, Half-order differentials on Riemann surfaces, Acta Math. Volume 115 (1966), 199--236.

Edit. See also Chapter 1 (The geometry of world-vectors and spin-vectors) in

Roger Penrose, Wolfgang Rindler, Spinors and space-time, Volume I, 1984.


Spinors can be represented in geometric algebra as even multivectors, that is, even-graded elements of a geometric algebra.

A geometric algebra is like an exterior algebra (indeed, in a geometric algebra you can and usually do define an exterior product as well) but with a different (still associative) product between its elements: the geometric product. On vectors specifically, the geometric product is defined so that $v^2=\langle v,v\rangle$, where $\langle\cdot ,\cdot\rangle$ is the scalar product between vectors. So a geometric algebra is a Clifford algebra from another perspective.

Even-graded elements of a geometric algebra are those that are formed from products of an even number of vectors and the sums of such products. These even multivectors can act on vectors by two-sided multiplication to rotate and scale the vector (in the same way that quaternions are used to rotate vectors):

$$v'=\overline\psi v \psi$$

($\overline\psi$ is the reverse of the multivector $\psi$: just reversing the order of all factors in the geometric products that make up $\psi$.)

This is why spinors are sometimes called the "square roots" of vectors: because by two-sided multiplication, a spinor can transform a given reference vector into another vector.

So a spinor can essentially be thought of as a transformation that rotates and scales a vector. (In spacetime contexts this is a Lorentz rotation, so a spatial rotation plus a boost.) This is the most intuitive way of thinking about spinors that I have been able to find.

In differential geometry, this means that spinors are even multivectors in the geometric algebra formed from the tangent space of the manifold, and the bilinear covariants of the spinors are the results of the spinors acting by two-sided multiplication on vectors of a reference orthonormal frame on the manifold.

A little more info can be found here and here.

  • $\begingroup$ I am not sure this answers the question which was to define spinors on $R^{3,1}$ in terms of differential-geometric objects on the space-time. Clifford algebra of course can be used to define spinors (this is how it is usually done), but their geometric meaning becomes unclear with this formalism. $\endgroup$ – Moishe Kohan Jan 17 '18 at 20:46
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    $\begingroup$ See the last couple of paragraphs of my questions. I gave a geometric interpretation (spinor represents a rotation or lorentz transformation of vectors) and how it relates to differential geometry (defined from geometric algebra of the tangent space of the manifold). $\endgroup$ – Matt Dickau Jan 18 '18 at 20:59

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