Spin structures

Following on from last week's post on Stiefel-Whitney classes, this post consists of the notes I made from various sources on related topics of spin structures and spinor fields. Here's one possible motivation and some definitions of what these are. Dirac wanted to take a square root of the Laplacian $\Delta$, and this in order to write down Lorentz invariant Klein-Gordon equations. I don't care about this last point, but it seems like a mathematically interesting question when operators have roots (or more generally are the result of some polynomial in some other operator). It is a useful exercise to try to solve the equation $D^2 = \Delta$ on a Euclidean space $V$ for a first order operator $D$; you will find that the coefficients have to satisfy certain relations that cannot be satisfied by ordinary real or complex numbers. The algebraic structure required to obtain these relations is provided by an algebra $A$ with $V$ as a linear subspace such that $v^2 = -||v||^2 \mathbf{1} $ in the algebra. In other words, you need to take a "square root" of your quadratic form.

A spinor bundle on a Riemannian manifold is a setting for taking a square root of the Riemannian metric. To be precise, it is a bundle $S$ on which tangent vectors act as bundle morphisms in such a way that $v^2 s = -||v||^2 s$. In Dirac's equation, the coefficients of $D$ were given by certain matrices (the "Pauli spin matrices"), and thus he was thinking of $D$ as taking values in a vector space which carries a representation of the algebra $A$. Thus the spinor bundle is a global version of that vector space.

That tells you what properties the spinor bundle is supposed to have, but it doesn't tell you what the bundle actually is. 

Here is a concrete description. Let us return to the algebra $A$ associated to a Euclidean space $V = \mathbb{R}^n$ as above. The universal example of such an algebra is the Clifford algebra $Cl(V)$, equipped with a natural left action of $V$. Choosing an orthonormal basis for $V$, one can describe $\mathbb{R}_n := Cl(V)$ as the universal algebra over $\mathbb{R}$ generated by symbols $e_1, \ldots, e_n$ subject to the relations $e_j^2 = -1$ and $e_j e_k = -e_k e_j$ for $i \neq j$. It is not hard to see that $Cl(V)$ is isomorphic as a vector space (but not as an algebra) to the exterior algebra of $V$, and thus $Cl(V)$ inherits a natural $\mathbb{Z}/2\mathbb{Z}$ grading, given by products of even / odd numbers of generators. Notice that right multiplication by the jth generator is an odd anti-involution, so a choice of orthonormal basis for $V$ gives $Cl(V)$ the structure of a $n$-multigraded super algebra.

We can define a (real) spinor bundle of a $n$-manifold to be a bundle which is locally isomorphic to the trivial bundle whose fibres are given by $\mathbb{R}_n$ equipped with a left action of the tangent bundle and a $n$-multigrading structure coming from a choice of local orthonormal frame. There is an obvious notion of complex spinor bundle as well: just use the complex Clifford algebra $\mathbb{C}_n$. Note that the fibre dimension of this bundle will be twice that of the bundle obtained via the spin representation, but the multigrading operators can be used to "reduce" this version of the spinor bundle down to the usual version. 

The existence of a real spinor bundle on a manifold M (a "Spin structure") is a rather restrictive condition. The complexification of a real spinor bundle is a complex spinor bundle, but not all complex spinor bundles ("$\mathrm{spin}^{\mathbb{C}}$ structures") arise in this way. For example, any complex manifold has a $\mathrm{spin}^{\mathbb{C}}$ structure, but even $\mathbb{CP}^2$ fails to have a spin structure. An orientation on $M$ can be recovered from a choice of $\mathrm{spin}^{\mathbb{C}}$ structure, and indeed "$\mathrm{spin}^{\mathbb{C}}$-able" is only a little bit stronger than orientable - most orientable manifolds that you can name are probably $\mathrm{spin}^{\mathbb{C}}$-able.

The reason for bringing this up is to relate spinor bundles to K-homology, the generalized homology theory dual to topological K-theory. In ordinary homology theory, a choice of orientation on an n-manifold $M$ is the same thing as a choice of fundamental class in $H_n(M)$. Similarly, a choice of real / complex spinor bundle on a $n$-manifold $M$ is the same thing as a choice of fundamental class in the nth degree real / complex K-homology of $M$ (the multigrading data are crucial here). 

Let's first see some examples of spin manifolds. We will return to $\mathrm{spin}^{\mathbb{C}}$ below.

Examples and non-examples: 

• If \( H^2(M, \mathbb{Z}_2) \) vanishes, \( M \) is spin. For example, \( S^n \) is spin for all \( n \neq 2 \). (Note that \( S^2 \) is also spin, but for different reasons; see below.)
• All even-dimensional complex projective spaces \( \mathbb{CP}^{2n} \) are not spin.
• All odd-dimensional complex projective spaces \( \mathbb{CP}^{2n+1} \) are spin.
• All compact, orientable manifolds of dimension 3 or less are spin.
• All Calabi--Yau manifolds are spin.

A genus $g$ Riemann surface admits \( 2^{2g} \) inequivalent spin structures; When spin structures exist, the inequivalent spin structures on a manifold have a one-to-one correspondence (not canonical) with the elements of \( H^1(M, \mathbb{Z}_2) \), which by the universal coefficient theorem is isomorphic to \( H_1(M, \mathbb{Z}_2) \). More precisely, the space of the isomorphism classes of spin structures is an affine space over \( H^1(M, \mathbb{Z}_2) \).

Intuitively, for each nontrivial cycle on \( M \) a spin structure corresponds to a binary choice of whether a section of the \( \mathrm{SO}(N) \) bundle switches sheets when one encircles the loop. If \( w_2 \) vanishes then these choices may be extended over the two-skeleton, then (by obstruction theory) they may automatically be extended over all of \( M \). See also this discussion on theta characteristics.

However, the definition given above for spin isn't the usual one, so let's see how they are equivalent.

Let \( M \) be a paracompact topological manifold and \( E \) an oriented vector bundle on \( M \) of dimension \( n \) equipped with a fibre metric. This means that at each point of \( M \), the fibre of \( E \) is an inner product space. A spinor bundle of \( E \) is a prescription for consistently associating a spin representation, i.e. a choice of spinor as discussed above, to every point of \( M \). When this is possible, $E$ is said to be spin. 

This may be made rigorous through the language of principal bundles. The collection of oriented orthonormal frames of a vector bundle form a frame bundle \( P_{\mathrm{SO}}(E) \), which is a principal bundle under the action of the special orthogonal group \( \mathrm{SO}(n) \). A spin structure for \( P_{\mathrm{SO}}(E) \) is a lift of \( P_{\mathrm{SO}}(E) \) to a principal bundle \( P_{\mathrm{Spin}}(E) \) under the action of the spin group \( \mathrm{Spin}(n) \), by which we mean that there exists a bundle map \( \phi : P_{\mathrm{Spin}}(E) \rightarrow P_{\mathrm{SO}}(E) \) such that
\[
\phi(pg) = \phi(p)\rho(g),
\]
for all \( p \in P_{\mathrm{Spin}}(E) \) and \( g \in \mathrm{Spin}(n) \),
where \( \rho : \mathrm{Spin}(n) \rightarrow \mathrm{SO}(n) \) is the mapping of groups presenting the spin group as a double cover of \( \mathrm{SO}(n) \).

In the special case in which \( E \) is the tangent bundle \( TM \) over the base manifold \( M \), if a spin structure exists then one says that \( M \) is a spin manifold. Equivalently \( M \) is spin if the \( \mathrm{SO}(n) \) principal bundle of orthonormal bases of the tangent fibres of \( M \) is a \( \mathbb{Z}_2 \) quotient of a principal spin bundle.

Obstruction theory and classification

As is common in algebraic topology, there is a cohomology class, which we interpret as an obstruction class, that determines whether or not one can lift the map from $SO(n)$ to $\rm{Spin(n)}$, and this turns out to be a familiar friend.

Theorem (Borel--Hirzebruch): For an orientable vector bundle \( \pi_E : E \rightarrow M \) a spin structure exists on \( E \) if and only if the second Stiefel--Whitney class \( w_2(E) \) vanishes.  
Furthermore, in the case \( E \rightarrow M \) is spin, the number of spin structures are in bijection with \( H^1(M, \mathbb{Z}/2) \). 

Proof: Consider the associated principal \( \mathrm{SO}(n) \)-bundle \( P_E \rightarrow M \). Notice this gives a fibration
\[
\mathrm{SO}(n) \rightarrow P_E \rightarrow M
\]
hence the Serre spectral sequence can be applied. From general theory of spectral sequences, there is an exact sequence
\[
0 \rightarrow E_3^{0,1} \rightarrow E_2^{0,1} \xrightarrow{d_2} E_2^{2,0} \rightarrow E_3^{2,0} \rightarrow 0
\]
where
\[
E_2^{0,1} = H^0(M, H^1(\mathrm{SO}(n), \mathbb{Z}/2)) = H^1(\mathrm{SO}(n), \mathbb{Z}/2)
\]
\[
E_2^{2,0} = H^2(M, H^0(\mathrm{SO}(n), \mathbb{Z}/2)) = H^2(M, \mathbb{Z}/2)
\]
In addition, \( E_\infty^{0,1} = E_3^{0,1} \) and \( E_\infty^{0,1} = H^1(P_E, \mathbb{Z}/2)/H^1(H^1(P_E, \mathbb{Z}/2)) \) for some filtration on \( H^1(P_E, \mathbb{Z}/2) \), hence we get a map
\[
H^1(P_E, \mathbb{Z}/2) \rightarrow E_3^{0,1}
\]
giving an exact sequence
\[
H^1(P_E, \mathbb{Z}/2) \rightarrow H^1(\mathrm{SO}(n), \mathbb{Z}/2) \rightarrow H^2(M, \mathbb{Z}/2)
\]

Now, a spin structure is exactly a double covering of \( P_E \) fitting into a commutative diagram
\[
\begin{array}{ccccc}
\mathrm{Spin}(n) & \rightarrow & \tilde{P}_E & \rightarrow & M \\
\downarrow & & \downarrow & & \downarrow \\
\mathrm{SO}(n) & \rightarrow & P_E & \rightarrow & M
\end{array}
\]
where the two left vertical maps are the double covering maps. Now, double coverings of \( P_E \) are in bijection with index 2 subgroups of \( \pi_1(P_E) \), which is in bijection with the set of group morphisms \( \mathrm{Hom}(\pi_1(E), \mathbb{Z}/2) \). But, from Hurewicz theorem and change of coefficients, this is exactly the cohomology group \( H^1(P_E, \mathbb{Z}/2) \). Applying the same argument to \( \mathrm{SO}(n) \), the non-trivial covering \( \mathrm{Spin}(n) \rightarrow \mathrm{SO}(n) \) corresponds to \( 1 \in H^1(\mathrm{SO}(n), \mathbb{Z}/2) = \mathbb{Z}/2 \), and the map to \( H^2(M, \mathbb{Z}/2) \) is precisely the \( w_2 \) of the second Stiefel--Whitney class, hence \( w_2(1) = w_2(E) \). If it vanishes, then the inverse image of 1 under the map
\[
H^1(P_E, \mathbb{Z}/2) \rightarrow H^1(\mathrm{SO}(n), \mathbb{Z}/2)
\]
is the set of double coverings giving spin structures. Now, this subset of \( H^1(P_E, \mathbb{Z}/2) \) can be identified with \( H^1(M, \mathbb{Z}/2) \), showing this latter cohomology group classifies the various spin structures on the vector bundle \( E \rightarrow M \). This can be done by looking at the long exact sequence of homotopy groups of the fibration
\[
\pi_1(\mathrm{SO}(n)) \rightarrow \pi_1(P_E) \rightarrow \pi_1(M) \rightarrow 1
\]
and applying \( \mathrm{Hom}(-, \mathbb{Z}/2) \), giving the sequence of cohomology groups
\[
0 \rightarrow H^1(M, \mathbb{Z}/2) \rightarrow H^1(P_E, \mathbb{Z}/2) \rightarrow H^1(\mathrm{SO}(n), \mathbb{Z}/2)
\]
Because \( H^1(M, \mathbb{Z}/2) \) is the kernel, and the inverse image of \( 1 \in H^1(\mathrm{SO}(n), \mathbb{Z}/2) \) is in bijection with the kernel, we have the desired result. $\blacksquare$

But what does this mean? Orientability means the tangent bundle trivializes over a 1-skeleton. Dually you could think of that as saying the complement of a co-dimension 2 subcomplex has a trivial tangent bundle. A surface is orientable if and only if it contains no Moebius bands -- a regular neighbourhood of any simple closed curve must be a cylinder. In higher dimensions this translates into a manifold being orientable if and only if it contains no twisted bundles $D^{n-1} \rtimes S^1$, i.e. regular neighbourhoods of simple closed curves are diffeomorphic to $D^{n-1} \times S^1$.

The interpretation of admitting a spin structure is the same, but it will be that the tangent bundle trivializes over a 2-skeleton, or dually the complement of a co-dimension three subcomplex admits a trivial tangent bundle. This matches the interpretation of $n^{th}$ Stiefel-Whitney classes as obstructions to extending maps over $n$-skeleta. A surface admits a spin structure if and only if it is orientable. It's a more interesting notion in higher dimensions. The statement there is the manifold is orientable, and if you take a regular neighbourhood of any surface in the manifold, then it has a trivial tangent bundle. So manifolds like $\mathbb RP^3$ are perfectly valid spin manifolds -- $\mathbb RP^3$ contains $\mathbb RP^2$ but the total space of its normal bundle has a perfectly trivializable tangent bundle. Technically, the condition is a little stronger than that -- you can trivialize the tangent bundle of the complement of a co-dimension 3 subset. So not only can you trivialize the total spaces of normal bundles of surfaces, but even the regular neighbourhoods of unions of surfaces.

4-manifolds

After the above discussion, we can prove:

Theorem: A simply connected 4-manifold is spin iff all embedded oriented surfaces have even self-intersection number or, equivalently, if the quadratic form $H_2 (M;\mathbb{Z}) \to \mathbb{Z}$ induced by the intersection form takes even values. 

Proof:  $M$ is spin iff $w_2 (TM)=0$. By the Wu formula, the latter happens iff the linear form $H_2 (M; \mathbb{Z}/2) \to \mathbb{Z}/2$, $a \mapsto \langle w_2 (TM);a\rangle$ is null.

By the Thom conjecture (see a brief discussion in this previous post), any class $a \in H_2 (M;\mathbb{Z})$ can be represented as the fundamental class of an embedded oriented surface $F \subset M$.

$w_2 (TM)|_F = w_2 (\nu_F)$ by the product formula for Stiefel-Whitney classes and because $F$ is spin. But $w_2 (\nu_F)$ is the mod 2 reduction of the Euler class of the normal bundle of $F$, and $\langle [F]; \chi(\nu_F) \rangle$ is the self-intersection number of $F$, or equivalently, the value of the quadratic form at $[F]$. $\blacksquare$

Spin structures are of great interest to topologists due to the connections with various operators, the question of admitting metrics of positive scalar curvature, and the signature and other genera. This last point has been discussed previously on this blog. We now turn to

$\mathrm{spin}^{\mathbb{C}}$ Structures

The more common definition of a $\mathrm{spin}^{\mathbb{C}}$ structure is analogous to the one given above for a spin structure, but uses the $\mathrm{spin}^{\mathbb{C}}$ group, which is defined instead by the exact sequence

\[1\to \mathbb {Z}/{2}\mathbb {Z} \to \mathrm{spin}^{\mathbb{C}}(n)\to \operatorname {SO} (n)\times \operatorname {U} (1)\to 1.\] 

To motivate this, suppose that $\kappa : \mathrm{spin}(n) \to U(N)$ is a complex spinor representation. The centre of $U(N)$ consists of the diagonal elements coming from the inclusion $i : U(1) \to U(N)$, i.e., the scalar multiples of the identity. Thus there is a homomorphism
\[\kappa \times i\colon {\mathrm {Spin} }(n)\times {\mathrm {U} }(1)\to {\mathrm {U} }(N).\]

This will always have the element (−1,−1) in the kernel. Taking the quotient modulo this element gives the group ${\mathrm {Spin} }^{\mathbb {C} }(n)$. This is the twisted product

    \[ {\mathrm {Spin} }^{\mathbb {C} }(n)={\mathrm {Spin} }(n)\times _{\mathbb {Z}/{2}\mathbb {Z} }{\mathrm {U} }(1)\,\]

where $U(1) = SO(2) = S^1$. In other words, the group $\mathrm{spin}^{\mathbb{C}}(n)$ is a central extension of $SO(n)$ by $S^1$

Examples: 

• All oriented smooth manifolds of dimension 4 or less are $\mathrm{spin}^{\mathbb{C}}$
• All almost complex manifolds are $\mathrm{spin}^{\mathbb{C}}$
• All spin manifolds are $\mathrm{spin}^{\mathbb{C}}$.

When a manifold carries a $\mathrm{spin}^{\mathbb{C}}$ structure at all, the set of $\mathrm{spin}^{\mathbb{C}}$ structures forms an affine space. Moreover, the set of $\mathrm{spin}^{\mathbb{C}}$ structures has a free transitive action of \( H^2(M, \mathbb{Z}) \). Thus, $\mathrm{spin}^{\mathbb{C}}$-structures correspond to elements of \( H^2(M, \mathbb{Z}) \) although not in a natural way.

This has the following geometric interpretation, which is due to Edward Witten. When the $\mathrm{spin}^{\mathbb{C}}$ structure is nonzero this square root bundle has a non-integral Chern class, which means that it fails the triple overlap condition. In particular, the product of transition functions on a three-way intersection is not always equal to one, as is required for a principal bundle. Instead it is sometimes \(-1\).

This failure occurs at precisely the same intersections as an identical failure in the triple products of transition functions of the obstructed spin bundle. Therefore, the triple products of transition functions of the full spin\(^c\) bundle, which are the products of the triple product of the spin and U(1) component bundles, are either \(1^2 = 1\) or \((-1)^2 = 1\) and so the $\mathrm{spin}^{\mathbb{C}}$

 

 bundle satisfies the triple overlap condition and is therefore a legitimate bundle.

This may be made precise as follows. Consider the short exact sequence \( 0 \rightarrow \mathbb{Z} \xrightarrow{2} \mathbb{Z} \rightarrow \mathbb{Z}_2 \rightarrow 0 \), where the second arrow is multiplication by 2 and the third is reduction modulo 2. This induces a long exact sequence on cohomology, which contains
\[
\cdots \longrightarrow H^2(M; \mathbb{Z}) \overset{2}{\longrightarrow} H^2(M; \mathbb{Z}) \longrightarrow H^2(M; \mathbb{Z}_2) \overset{\beta}{\longrightarrow} H^3(M; \mathbb{Z}) \longrightarrow \cdots,
\]
where the second arrow is induced by multiplication by 2, the third is induced by restriction modulo 2 and the fourth is the associated Bockstein homomorphism \( \beta \).

The obstruction to the existence of a spin bundle is an element \( w_2 \) of \( H^2(M, \mathbb{Z}_2) \). It reflects the fact that one may always locally lift an \( \mathrm{SO}(n) \) bundle to a spin bundle, but one needs to choose a \( \mathbb{Z}_2 \) lift of each transition function, which is a choice of sign. The lift does not exist when the product of these three signs on a triple overlap is \(-1\), which yields the Čech cohomology picture of \( w_2 \).

To cancel this obstruction, one tensors this spin bundle with a U(1) bundle with the same obstruction \( w_2 \). Notice that this is an abuse of the word bundle, as neither the spin bundle nor the U(1) bundle satisfies the triple overlap condition and so neither is actually a bundle.

A legitimate $U(1)$ bundle is classified by its Chern class, which is an element of \( H^2(M, \mathbb{Z}) \). Identify this class with the first element in the above exact sequence. The next arrow doubles this Chern class, and so legitimate bundles will correspond to even elements in the second \( H^2(M, \mathbb{Z}) \), while odd elements will correspond to bundles that fail the triple overlap condition. The obstruction then is classified by the failure of an element in the second \( H^2(M, \mathbb{Z}) \) to be in the image of the arrow, which, by exactness, is classified by its image in \( H^2(M, \mathbb{Z}_2) \) under the next arrow.

To cancel the corresponding obstruction in the spin bundle, this image needs to be \( w_2 \). In particular, if \( w_2 \) is not in the image of the arrow, then there does not exist any $U(1)$ bundle with obstruction equal to \( w_2 \) and so the obstruction cannot be cancelled. By exactness, \( w_2 \) is in the image of the preceding arrow only if it is in the kernel of the next arrow, which we recall is the Bockstein homomorphism \( \beta \). That is, the condition for the cancellation of the obstruction is
\[
W_3 = \beta w_2 = 0
\]
where we have used the fact that the third integral Stiefel--Whitney class \( W_3 \) is the Bockstein of the second Stiefel--Whitney class \( w_2 \) (this can be taken as a definition of \( W_3 \)).

Integral lifts of Stiefel--Whitney classes

This argument also demonstrates that second Stiefel--Whitney class defines elements not only of \( \mathbb{Z}_2 \) cohomology but also of integral cohomology in one higher degree. In fact this is the case for all even Stiefel--Whitney classes. It is traditional to use an uppercase \( W \) for the resulting classes in odd degree, which are called the integral Stiefel--Whitney classes, and are labeled by their degree (which is always odd).