Every Orientable 4 manifold admits Spin C Structure

Introduction

In this page, we will discuss the notion of $Spi{n}^c$–structures. Our goal is to set up the proof of Theorem \ref{maintheorem}. Recall from the previous lecture that $Spin$–structures appear as refinements of oriented vector bundles by lifting their structure group from $\text{SO}(n)$ to $\mathrm{Spin}(n)$. We will now see that $Spi{n}^c$–structures give a related refinement that also involves choosing a complex line bundle.

To recall the basic setup, a rank-$n$ real vector bundle $E\to B$ is determined by transition functions $g_{ij}$ taking values in $\mathrm{GL}(n,\mathbb{R})$. When $E$ is equipped with a continuous inner product on its fibers, the structure group may be reduced to the orthogonal group $\mathrm{O}(n)$.

If $E$ is orientable, the structure group reduces further to $\mathrm{SO}(n)$. In this case we obtain the associated principal $\mathrm{SO}(n)$—bundle of oriented orthonormal frames, denoted $P_{\mathrm{SO}}(E)$. The fiber over a point consists of all oriented orthonormal bases of the corresponding fiber of $E$, and the group $\mathrm{SO}(n)$ acts freely and transitively on each such set of frames.

Let $n\ge 3$. Recall that $\mathrm{SO}(n)$ is path—connected and that its fundamental group is

$${\pi }_1(\mathrm{SO}(n))\cong {\mathbb{Z}}_2.$$

We define $\mathrm{Spin}(n)$ to be the connected double-cover of $\mathrm{SO}(n)$, and denote the covering map by

$$p:\mathrm{Spin}(n)⟶\mathrm{SO}(n).$$

Since $\ker (p)\cong {\mathbb{Z}}_2$, this realizes $\mathrm{Spin}(n)$ as a central ${\mathbb{Z}}_2$—extension:

$$1⟶{\mathbb{Z}}_2⟶\mathrm{Spin}(n)\stackrel{p}{\to }\mathrm{SO}(n)⟶1.$$

When $n=2$, we take $\mathrm{Spin}(2)={\mathbb{S}}^1$, which double-covers $\mathrm{SO}(2)\cong {\mathbb{S}}^1$ in the usual degree $2$ way.

Since coverings of Lie groups are themselves Lie groups, it follows that $\mathrm{Spin}(n)$ is a compact Lie group. An oriented vector bundle $E\to B$ with structure group $\mathrm{SO}(n)$ is said to be spinnable if its structure group can be lifted along $p$ to a principal $\mathrm{Spin}(n)$ bundle. Equivalently, a Spin structure on $E$ consists of a principal $\mathrm{Spin}(n)$ bundle $P_{\mathrm{Spin}}(E)$ together with a bundle map

$$P_{\mathrm{Spin}}(E)⟶P_{\mathrm{SO}}(E)$$

covering the identity on $B$ and intertwining the $\mathrm{Spin}(n)$ action with the $\mathrm{SO}(n)$ action via $p$.

Remark

A word of caution regarding the above subtlety: two $\mathrm{Spin}$ structures may be equivalent as bundles, but not equivalent as $\mathrm{Spin}$ structures.

Consider $\mathrm{Spin}$ structures over the circle ${\mathbb{S}}^1$.When does a $\mathrm{SO}(n)$ bundle lift to a $\mathrm{Spin}(n)$ bundle? Is the lifting unique?

Theorem

A principal $\mathrm{SO}(n)$ bundle $E\to M$ admits a lift to a principal $\mathrm{Spin}(n)$—bundle iff its second Stiefel—Whitney class $w_2(E)\in H^2(M;{\mathbb{Z}}_2)$ vanishes.

Theorem

If $M$ is simply connected, then a principal $\mathrm{SO}(n)$—bundle admits at most one lift to a principal $\mathrm{Spin}(n)$—bundle. In general, whenever a $\mathrm{Spin}$—structure exists on $M$, the set of its isomorphism classes is a torsor for $H^1(M;{\mathbb{Z}}_2)$.

This viewpoint will lead us naturally to the notions of Spin and Spin${}^{\mathbb{C}}$ structures. A Spin${}^{\mathbb{C}}$ structure may be regarded as a complex analogue of a Spin—structure, obtained by enlarging $\mathrm{SO}(n)$ to $\mathrm{SO}(n)\times \mathrm{U}(1)$ and consider its double-cover.

Spin${}^{\mathbb{C}}$-structure

Definition

For each $n\ge 1$, the group ${\mathrm{Spin}}^{\mathbb{C}}(n)$ is defined as ${\mathrm{Spin}}^{\mathbb{C}}(n)=\mathrm{Spin}(n){\times }_{{\mathbb{Z}}_2}\mathrm{U}(1),$ that is, the quotient of $\mathrm{Spin}(n)\times \mathrm{U}(1)$ by the diagonal ${\mathbb{Z}}_2$ acting via $(-1,-1)$.

The group ${\mathrm{Spin}}^{\mathbb{C}}(n)$ fits into the short exact sequence

$$1⟶{\mathbb{Z}}_2⟶{\mathrm{Spin}}^{\mathbb{C}}(n)⟶\mathrm{SO}(n)\times \mathrm{U}(1)⟶1,$$

so it is a ${\mathbb{Z}}_2$—central extension of $\mathrm{SO}(n)\times \mathrm{U}(1)$.

Definition

Let $M$ be an oriented Riemannian $n$-manifold, and let $P_{\mathrm{SO}(n)}(M)$ denote its principal $\mathrm{SO}(n)$—frame bundle. A ${\mathrm{Spin}}^{\mathbb{C}}$ structure on $M$ consists of a principal ${\mathrm{Spin}}^{\mathbb{C}}(n)$-bundle $P_{{\mathrm{Spin}}^{\mathbb{C}}}(M)$ together with a bundle map $\pi :P_{{\mathrm{Spin}}^{\mathbb{C}}}(M)⟶P_{\mathrm{SO}(n)}(M),$

covering the identity on $M$ and intertwining the ${\mathrm{Spin}}^{\mathbb{C}}(n)$ action with the natural $\mathrm{SO}(n)$ action via the projection $p^{\mathbb{C}}:{\mathrm{Spin}}^{\mathbb{C}}(n)⟶\mathrm{SO}(n).$ This is encoded in the commutative diagram

In other words, for each $\tilde{p}\in P_{{\mathrm{Spin}}^{\mathbb{C}}}(M)$ and $\tilde{g}\in {\mathrm{Spin}}^{\mathbb{C}}(n)$, we have $\pi (\tilde{p},\tilde{g})=\pi (\tilde{p})p^{\mathbb{C}}(\tilde{g}).$ A manifold carrying such a structure is called a ${\mathrm{Spin}}^{\mathbb{C}}$—manifold.

Theorem

A manifold $M$ is ${\mathrm{Spin}}^{\mathbb{C}}$ iff there is a complex line bundle $L$ over $M$ such that $TM\oplus L$ has a $\mathrm{Spin}$-structure.

\begin{proof} Suppose first that $M$ is $Spi{n}^{\mathbb{C}}$, so $TM$ carries a $Spi{n}^{\mathbb{C}}$–structure. By the definition recalled above, this means that there are principal bundles $P_{U(1)}(TM)$ and $P_{Spi{n}^{\mathbb{C}}}(TM)$ over $M$ and a $Spi{n}^{\mathbb{C}}$–equivariant bundle map

$$\xi :P_{Spi{n}^{\mathbb{C}}}(TM)⟶P_{SO}(TM)\times P_{U(1)}(TM),$$

whose restriction to each fiber is the standard double cover $Spi{n}^{\mathbb{C}}(n)\to SO(n)\times U(1)$. The principal $U(1)$–bundle $P_{U(1)}(TM)$ corresponds to a complex line bundle $L$.

Since $U(1)\cong SO(2)$, there is a natural homomorphism

coming from taking Whitney sums of oriented bundles. By definition, $Spi{n}^{\mathbb{C}}(n)$ is the pullback of the covering $Spin(n+2)\to \mathrm{SO}(n+2)$ along this map, so we have

If $TM$ has a $Spi{n}^{\mathbb{C}}$–structure, then we have principal bundles $P_{\mathrm{SO}}(TM)$ and $P_{U(1)}(TM)$ together with a $Spi{n}^{\mathbb{C}}(n)$–bundle $P_{Spi{n}^{\mathbb{C}}}(TM)$ mapping equivariantly into their product. The associated $U(1)$–bundle defines a complex line bundle $L$. Applying the map $SO(n)\times U(1)\to SO(n+2)$ shows that $P_{Spi{n}^{\mathbb{C}}}(TM)$ induces a $Spin$–structure on $TM\oplus L$.

Conversely, if there is a line bundle $L$ such that $TM\oplus L$ has a $Spin$–structure, then reducing the $SO(n+2)$–frame bundle of $TM\oplus L$ to $SO(n)\times U(1)$ yields a principal $SO(n)\times U(1)$–bundle isomorphic to $P_{SO}(TM)\times P_{U(1)}(L)$. Pulling back the $Spin$–bundle along this reduction and using the pullback square above produces a principal $Spi{n}^{\mathbb{C}}(n)$–bundle, which is precisely a $Spi{n}^{\mathbb{C}}$ structure on $TM$. Thus $TM$ is $Spi{n}^{\mathbb{C}}$ if and only if there exists a complex line bundle $L$ with $TM\oplus L$ admitting a $Spin$–structure.

Theorem

Every $Spin$-manifold has a canonical $Spi{n}^{\mathbb{C}}$-structure.

\begin{proof} Since $M$ is $Spin$, its tangent bundle has a principal $spin(n)$–bundle. To obtain a $Spi{n}^{\mathbb{C}}$–structure, we simply enlarge this by taking the fiber product with the trivial $U(1)$–bundle $U_1=M\times U(1)$ and quotienting by the diagonal ${\mathbb{Z}}_2$–action. Thus we set

$P_{Spi{n}^{\mathbb{C}}}(TM)=P_{Spin}(TM){\times }_{M,{\mathbb{Z}}_2}{U}_1,$

whose fibers are $(Spin(n)\times U(1))/\{\pm (1,1)\}=Spi{n}^{\mathbb{C}}(n)$. This gives a canonical $Spi{n}^{\mathbb{C}}$–structure on $TM$.

Theorem

An orientable manifold $M$ admits a $Spi{n}^{\mathbb{C}}$ structure if and only if its second Stiefel—Whitney class $w_2(TM)$ is the mod $2$ reduction of an integral class.

Recall that a manifold $M$ admits a $Spin$—structure exactly when the second Stiefel—Whitney class $w_2(TM)$ vanishes. Using the criterion from the previous section, $M$ has a $Spi{n}^{\mathbb{C}}$—structure if and only if there exists a complex line bundle $L$ such that $TM\oplus L$ is $Spin$. This is equivalent to $w_2(TM\oplus L)=0$.

Because Stiefel—Whitney classes are stable under Whitney sum, we have

$$=w_2(TM)+w_2(L)+w_1(TM)w_1(L).$$

Both $TM$ and $L$ are orientable (since $M$ itself is orientable and a complex line bundle always has vanishing $w_1$). Thus $w_1(TM)=w_1(L)=0$, and the formula above simplifies to

$w_2(TM\oplus L)=w_2(TM)+w_2(L).$

Hence $TM\oplus L$ is $Spin$ if and only if

$w_2(TM)+w_2(L)=0.$

Reducing mod $2$, this means $w_2(TM)=w_2(L).$

But $w_2(L)$ is always the mod $2$ reduction of the first Chern class $c_1(L)$, which is an integral class. Therefore $w_2(TM)$ must itself be the mod $2$ reduction of an integral class. This shows one direction.

Conversely, if $w_2(TM)$ is the mod $2$ reduction of some integral class $\alpha \in H^2(M;\mathbb{Z})$, we may choose a complex line bundle $L$ with $c_1(L)=\alpha$. Then $w_2(L)=\alpha \mathrm{mod}2$, and hence

$w_2(TM\oplus L)=w_2(TM)+w_2(L)=0.$

So $TM\oplus L$ is $Spin$, and by the previous criterion $M$ is $Spi{n}^{\mathbb{C}}$.

Thus an orientable manifold $M$ admits a $Spi{n}^{\mathbb{C}}$ structure if and only if $w_2(TM)$ is the mod $2$ reduction of an integral cohomology class.

Integral Lift

The main statement is the following.

Theorem

Every orientable, connected 4-manifold admits $Spi{n}^{\mathbb{C}}$-structure.

To proof the main theorem, we need lemmas \ref{existence} and \ref{lift}. We will working on the short exact sequence,

and have a following sequence through applying the Universal Coefficient theorem in ${\mathbb{Z}}_2$ coefficient,

Let’s recall the Wu’s formula and Wu’s theorem.

\begin{lemma}\label{existence} $i(w_2)(\tau )=0$ for $\tau \in T$ where $T$ is the torsion subgroup of $H_2(M)$.

Let $i(w_2)\in \mathrm{Hom}(H_2(M),{\mathbb{Z}}_2)$ as a Kronecker pairing, $i(w_2)=<w_2,x>\text{for }x\in H_2(M).$ Let $\gamma \in H^2(M)$ such that $\gamma \cap [M]=\tau$. Since $\gamma$ is a torsion element and $H^4(M)=\mathbb{Z}$, $\gamma \cup \gamma =0$. So $f:H^{\ast }(M;\mathbb{Z})\to H^{\ast }(M;{\mathbb{Z}}_2)$ ring map that $f(\gamma \cup \gamma )=f(\gamma )\cup f(\gamma )=0.$ Since $M$ is orientable, the first Steifel-Whitney class $w_1=v_1$ vanishes. By the Wu’s formula and theorem $w_2=S{q}^2(1)+v_2=v_2$, $<v_2(TM)\cup f(\gamma ),[M]>=<w_2\cup f(\gamma ),[M]>=<S{q}^2(f(\gamma )),[M]>=0.$ Therefore,

$$\begin{array}{rl}i(w_2)(\tau )& =<w_2,f(\tau )>\\ & =w_2\cap f(\tau )\\ & =w_2\cap f(\gamma \cap [M])\\ & =w_2\cap (f(\gamma )\cap f([M]))\\ & =(w_2\cup f(\gamma ))\cap f([M])\\ & =0.\end{array}$$

This lemma implies we can identifying with $i(w_2)$ with the map $\varphi \in \mathrm{Hom}(H_2(M)/T),{\mathbb{Z}}_2)$. We can ask whether this $\varphi$ can be lifted to $\mathrm{Hom}(H_2(M);{\mathbb{Z}}_2)$.

Lifting

Let $M$ be a connected, orientable 4-manifold. Then we can lift the map $\varphi \in \text{Hom}(H_2(M),{\mathbb{Z}}_2)$ to $\tilde{\varphi }\in \text{Hom}(H_2(M),\mathbb{Z})$. Proof Let $T$ be the kernel of the homomorphism $H_2(M)\to {\prod }_y\mathbb{Z}$

which maps $x$ to the vector with components $x\cdot y$ for all $y\in H_2(M;\mathbb{Z})$ where $\cdot$ is a intersection number between $x$ and $y$. In the closed case Poincare duality implies that $T$ is the torsion subgroup of $H_2(M;\mathbb{Z})$. torsion submodule of $H_2(M)$, so that $H_2(M)/T$ is finitely generated free module. So $H_2(M)/T$ is the free $\mathbb{Z}$-module that can always find the lift of $\tilde{\varphi }\in Hom(H_2(M)/T,{\mathbb{Z}}_2)$ to $\varphi \in Hom(H_2(M)/T,\mathbb{Z})$.

\begin{proof}[Proof of Theorem \ref{maintheorem}] If it has a boundary, then take a oriented double to make it closed. So without loss of generality, we can consider $M$ as a closed manifold. Universal Coefficient theorem is natural on the coefficient, we have a following commute diagram with the Bockstein homomorphism $\beta$ and the definition of $\mathrm{Ext}$ functor.

Through the lemma \ref{existence} and \ref{lift}, $i(w_2)=f(\tilde{\varphi })$ for some $\tilde{\varphi }\in Hom(H_2(M),\mathbb{Z})$. Since $i^{\prime }$ is surjective, $\exists b\in H^2(M;\mathbb{Z})$ such that $i^{\prime }(b)=\tilde{\varphi }$. Since the diagram computes, $if(b)=f{i}^{\prime }(b)=f(\varphi )=i(w_2)$.So that $w_2-f(b)\in Ker(i)=Im(j)$. Hence, there is a unique $a\in \mathrm{Ext}(H_1(M;\mathbb{Z}),{\mathbb{Z}}_2)$ such that $j(a)=w_2-f(b)$. Similarly, $\exists a^{\prime }\in Ext(H_1(M;\mathbb{Z}),\mathbb{Z})$ such that $f(a^{\prime })=a$ where $jf(a^{\prime })=f{j}^{\prime }(a^{\prime })=j(a)=w_2-f(b)$. Therefore $f(j^{\prime }(a^{\prime })+b)=w_2$ which means $w_2$ can be lifted to the integral class. Then by the theorem \ref{equivalent condition}, every compact, connected, orientable 4-manifold admits $Spi{n}^{\mathbb{C}}$ structure.

Examples

Even the manifold $M^4$ could not have a $Spin$ structure, but still can have a $Spi{n}^{\mathbb{C}}$ structure. For example, ${\mathbb{CP}}^2$. By the Wu’s formula, $<v_i(TM)\cup -,[{\mathbb{CP}}^2]>=<S{q}^i(-),[{\mathbb{CP}}^2]>.$

Since ${\mathbb{CP}}^2$ is orientable, $w_1=0$, so that $v_2(TM)=w_2(TM)$. From the intersection theory, $<S{q}^2(w_2(TM)),[M]>=<w_2(TM)\cup w_2(TM),[M]>=X\cdot X$ where $X$ is the corresponding embedded surface in ${\mathbb{CP}}^2$ which is the sphere. It turns out $X\cdot X=1$, so that $w_2^2(TM)\ne 0$ which implies $w_2(TM)\ne 0$. So by the theorem \ref{Spin structure conditions}, it does not admit $Spin$ structure.

Application and Appendix

Equivariant Vector Bundle

$Spi{n}^{\mathbb{C}}$ is a compact Lie group.

Suppose we have: \begin{itemize} \item A \textbf{principal $G$-bundle} $P\to Y$ over a smooth manifold $Y$, \item A \textbf{unitary representation} $\rho :G\to U(V)$, \item Equivariance of $P$ with respect to the $G$-action. \end{itemize}

Then we can construct the associated vector bundle:

$$E_{\rho }=P{\times }_{\rho }V\to Y,$$

where $G$ acts on $V$ via $\rho$, and the bundle is glued using the equivariant $G$-action on $P$.

1. Define spinor bundles Using the spinor representation $\Delta$, we obtain $S=P{\times }_{\Delta }{\mathbb{C}}^{2^k}\to Y.$

   This is the space of spinors for Seiberg—Witten or Dirac operators.

2. Define determinant line bundles Using the top exterior power, $L=\det (S)={\Lambda }^{2^k}S,$

   which plays a key role in gauge theory and Chern classes.

3. Twist bundles Tensor powers of the determinant line bundle $S\otimes L^{\otimes m}$ allow construction of twisted spinor bundles, e.g., for different ${\mathrm{Spin}}^c$ connections.

4. Define Clifford multiplication Equivariance ensures that for $v\in TY$,

   is well-defined and compatible with the $G$-action.

5. Construct Dirac operators and gauge-theoretic equations $D_A:\Gamma (S)\to \Gamma (S),A\text{ a connection on }L,$ which is fundamental in Seiberg—Witten theory and Floer homologies.

6. Use representation theory for classification Every associated vector bundle is classified by a representation of $G$; this also connects naturally to the classifying space $BG$.

Representation of $Spi{n}^{\mathbb{C}}(n)$	Associated bundle
Spinor representation $\Delta$	Spinor bundle $S=P{\times }_{\Delta }{\mathbb{C}}^{2^k}$
Determinant representation	Line bundle $L=\det (S)$
Trivial representation	Trivial bundle $Y\times \mathbb{C}$
Tensor powers of $\det$	$L^{\otimes m}$

A $Spi{n}^{\mathbb{C}}$ structure is actually equivalent to the bundle $S=S_{-}\oplus S_{+}$ together with the Clifford action. From this point of view it easy to describe the $H^2(M;Z)$ action on $S_M$: an element $\alpha \in H^2(M;\mathbb{Z})$ sends $S$ to $S\otimes E$, where $E$ is the line bundle with $c_1(E)=\alpha$.

Digression: Recall that if $E$ is a line bundle on $M$, the first Chern class $c_1(E)\in H^2(M;\mathbb{Z})$ is (at least if $M$ is a closed oriented manifold) the Poincar´e dual of the zero set of a generic section of $E$. On any $M$, $c_1$ gives a bijection between the set of isomorphism classes of line bundles on $M$ and $H^2(M;Z)$.¹

Floer Theory

A $Spi{n}^{\mathbb{C}}$–structure $\mathfrak{s}$ on a 3– or 4–manifold determines:

1. a spinor bundle (S)
2. a determinant line bundle $L=\det S$ with $c_1(\mathfrak{s})=c_1(L)\in H^2(\cdot ;\mathbb{Z})$.

Because the Seiberg–Witten (monopole) equations use the Dirac operator associated to $\mathfrak{s}$, each $Spi{n}^{\mathbb{C}}$–structure yields its own configuration space and hence its own Floer chain complex. Therefore monopole Floer homology splits as

Heegaard Floer homology has an analogous decomposition. Every generator $\mathbf{x}\in {\mathbb{T}}_{\alpha }\cap {\mathbb{T}}_{\beta }$ determines a $Spi{n}^{\mathbb{C}}$–structure ${\mathfrak{s}}_z(\mathbf{x})$, and the differential preserves this label, so

$${\mathrm{HF}}^{\circ }(Y)\cong \underset{\mathfrak{s}\in Spi{n}^{\mathbb{C}}(Y)}{⨁}{\mathrm{HF}}^{\circ }(Y,\mathfrak{s}).$$

Gradings in both theories depend on $Spi{n}^{\mathbb{C}}$.
For a fixed $Spi{n}^{\mathbb{C}}$–structure $\mathfrak{s}$, relative gradings are given by spectral flow / Maslov index, and in the torsion case there is an absolute $\mathbb{Q}$–grading.

Cobordism maps are also indexed by $Spi{n}^{\mathbb{C}}$–structures.
If $W:Y_0\to Y_1$ carries a $Spi{n}^{\mathbb{C}}$–structure $\mathfrak{t}$ restricting to ${\mathfrak{s}}_0,{\mathfrak{s}}_1$, then there is a map

$$F_{W,\mathfrak{t}}:\mathcal{F}(Y_0,{\mathfrak{s}}_0)\to \mathcal{F}(Y_1,{\mathfrak{s}}_1),$$

and its grading shift is determined by the index formula

$$\mathrm{deg}(F_{W,\mathfrak{t}})=\frac{c_1(\mathfrak{t}{)}^2-2\chi (W)-3\sigma (W)}{4}.$$

Classifying Space Obstruction Theory

We can rephrase the main theorem using obstruction language.

We have a classifying space $BSO(n)$ such that principal $\text{SO}(n)$-bundle has a correspondence between $[M,BSO(n)]$. So the question for having a $Spi{n}^{\mathbb{C}}$ structure is, for $f\in [M,BSO(n)]$, can we lift to $\tilde{f}\in [M,BSpi{n}^{\mathbb{C}}(n)]$.

, so the obstruction cocycle $[\theta (f)]$ lies in $H^3(M;{\pi }_2(BU(1))=H^3(M;\mathbb{Z})$.

Therefore, having a $Spi{n}^{\mathbb{C}}$ structure is equivalent to the obstruction cocyle in $H^3(M;\mathbb{Z})$ vanishes. The exact sequence

where $\beta$ is the Bockstein homomorphism. So $ker\beta =Im(mod2)$ that $W_3=\beta (w_2)=0$ means $w_2$ can be lifted to the integral class.

Therefore, having a $Spi{n}^{\mathbb{C}}$ structures is equivalent to $w_2$ can be lifted to the integral class. Thus $Spi{n}^{\mathbb{C}}$–structures serve as:

1. the geometric background needed to define the Dirac operator in gauge–theoretic Floer homology,
2. the canonical way of decomposing Floer groups into topological summands,
3. the indexing system for cobordism maps and their grading shifts,
4. the common language in which equivalences between Floer theories (e.g.\ Heegaard $\leftrightarrow$ monopole) respect and preserve structure.

Footnotes

1. Copied from Taubes and Hutchings Notes. ↩