# Seiberg-Witten Theory 2: Clifford Structures and Spinors

Here is the second post on setting up the Seiberg-Witten equations on a 4-manifold, based on our learning seminar at UT Austin. The first post is here.

Clifford Algebras and Structures

For a vector space V with inner product g, its Clifford Algebra is defined as the tensor algebra of V modded out by all relations generated by setting $v\otimes v=-g(v,v)1$,

$Cl(V)= \otimes V/\langle v\otimes v +g(v,v)1: v\in V\rangle.$

For a vector bundle $E\to M$, any map $c: T^*M\to End(E)$ satisfying $c(u)c(v)+c(v)c(u)=-2g(u,v)I_E$ for all $u,v\in \Gamma(T^*M)$ (equivalently satisfying $c(v)^2=-|v|I_E$ for all v) extends to a representation
$c:Cl(T^*M)\to End(E)$
Such a map is called a Clifford structure.

There are two reasons we are interested in Clifford algebras and Clifford structures for Seiberg-Witten theory. The first is their relation to Spin and Spinc structures. The second is their relation to Dirac operators. In this post we will focus on their relation to Spin and Spinc structures, and discuss Dirac operators next.

Clifford algebras and Spin

Let $Cl(n)$ denote the Clifford algebra of $\mathbb{R}^n$ with its standard inner product. Let $(e_1,\cdots, e_n)$ denote the standard orthonormal basis for $\mathbb{R}^n$. Consider the multiplicative subgroup of $Cl(n)$ generated by unit vectors of $\mathbb{R}^n$. This is called $Pin(n)$.

There is a natural $\mathbb{Z}/2$ grading on $Cl(n)$ induced by a bijection $Cl(n)\leftrightarrow \bigwedge^* \mathbb{R}^n$ identifying $e_{i_1}\cdots e_{i_k} \leftrightarrow e_{i_1}\wedge \cdots \wedge e_{i_k}$. The integer grading on the exterior power reduces to a $\mathbb{Z}/2$ grading (even/odd) on the Clifford algebra. This yields a splitting $Cl(n)= Cl^+(n)\oplus Cl^-(n)$. Define $Spin(n)$ to be the intersection of $Pin(n)$ with the even summand $Cl^+(n)$.

Before, we defined $Spin(n)$ to be the universal double cover of $SO(n)$. We can show this new definition of Spin agrees with the old definition, by explicitly constructing a double cover map from this subset of $Cl(n)$ to $SO(n)$.

There is an action of $Cl(n)$ on $\mathbb{R}^n$ given by signed conjugation (using the multiplicative structure of the Clifford algebra). If $v\in \mathbb{R}^n$ is a unit vector (i.e. a generator of $Pin(n)$) then for any $x\in \mathbb{R}^n$
$-vxv^{-1} = vxv = x-2\langle x,v \rangle v$
Here we have used the fact that for unit vectors $-vv=1$ so $v^{-1}=-v$, and the relation $vx+xv=-2g(v,x)$ in the Clifford algebra. This can be interpreted geometrically: the action $(v,x)\mapsto -vxv^{-1}$ is the reflection of x over the hyperplane orthogonal to v.

The group of orthogonal transformations is generated by reflections over hyperplanes, so we have a representation called the twisted adjoint representation:
$\rho: Pin(k)\to O(k)$
defined by $\rho(y)x = yx\varepsilon(y^{-1})$ where $\varepsilon(Cl^\pm(n))=\pm 1$ (extend linearly). Restricting this to $Spin(k)$ this is just usual conjugation, which corresponds to an even number of reflections so the image lies in $SO(k)$:
$\rho: Spin(k)\to SO(k)$
This map is a surjective group homomorphism, and by studying the elements of $Spin(k)$ which lie in the center of $Cl(k)$, we see that the kernel of $\rho: Spin(k)\to SO(k)$ is two elements $\{\pm 1\}$. Because these are nice smooth compact Lie groups, this implies that $\rho$ is a covering map. To check it is not the trivial double cover, we can find a path in $Spin(k)$ between -1 and 1 given by
$\gamma(t)=\cos(t)+e_1e_2\sin(t)=-(e_1\cos(t/2)+e_2\sin(t/2))(e_1\cos(t/2)-e_2\sin(t/2))$
for $t\in [-\pi,\pi]$ [observe this path is a product of two unit vectors at each t and is thus in $Spin(n)$].
Therefore this definition of $Spin(n)$ agrees with the previous one.

The spinor representation

We have already seen that $Spin(4)\cong SU(2)\times SU(2)$ so $Spin(4)$ and $Spin^c(4)$ naturally admit two complex rank two representations coming from the projections onto the two factors of $SU(2)$. However, it is useful to understand these representations from the Clifford algebra perspective so that the representations carry the additional information of a Clifford structure. In fact, there is a complex representation of the entire (complexified) Clifford algebra $Cl(4)$ which splits into a direct sum of two complex rank two representations, which behave nicely with respect to the $\mathbb{Z}/2$ grading on the Clifford algebra. More specifically:

Theorem: There is a complex vector space $\mathbb{S}=\mathbb{S}^+\oplus \mathbb{S}^-$ with $\dim_{\mathbb{C}}\mathbb{S}^+=\dim_{\mathbb{C}}\mathbb{S}^-=2$, and an $\mathbb{C}$-linear isomorphism
$c: Cl(4)\otimes \mathbb{C}\to End(\mathbb{S})$
such that $c(Cl^+(4))\cong End(\mathbb{S}^+)\oplus End(\mathbb{S}^-)$ and $c(Cl^-(4)\cong Hom(\mathbb{S}^+,\mathbb{S}^-)\oplus Hom(\mathbb{S}^-,\mathbb{S}^+)$.

To prove this, we have to define $\mathbb{S}$, $\mathbb{S}^\pm$, and the map c, and then verify that c is an algebra isomorphism satisfying the specified properties. There are a lot of things to check so I will define everything, and say a few things about how the map c works which hopefully make it more believable that c is an algebra isomorphism.

Let $V=\mathbb{R}^4$ with standard coordinates and standard almost complex structure J. This almost complex structure gives rise to a splitting of $V\otimes \mathbb{C} = V^{1,0}\oplus V^{0,1}$, where $V^{1,0}$ is the i-eigenspace of J and $V^{0,1}$ is the -i-eigenspace of J. We have orthonormal bases for these pieces given by:
$V^{1,0}=span\left(\varepsilon_1 := \frac{1}{\sqrt{2}}(e_1-if_1), \varepsilon_2 := \frac{1}{\sqrt{2}}(e_2-if_2)\right)$
$V^{0,1}=span\left(\overline{\varepsilon}_1 := \frac{1}{\sqrt{2}}(e_1+if_1), \overline{\varepsilon}_2 := \frac{1}{\sqrt{2}}(e_2+if_2)\right)$

Define $\mathbb{S}:= \bigwedge^* V^{1,0}$, and its splitting by $\mathbb{S}^+ := \bigwedge^{even}V^{1,0}$ and $\mathbb{S}^- := \bigwedge^{odd}V^{1,0}$.

Now we need to define $c: Cl(V)\otimes \mathbb{C} \to End (\mathbb{S})$ with the properties specified in the theorem. We will define c on elements of $V\otimes \mathbb{C}$ and then extend this to a map on the Clifford algebra by setting $c(e_{i_1}\cdots e_{i_k})=c(e_{i_1})\cdot \cdots \cdot c(e_{i_k})$ and extending complex linearly. To specify c on $V\otimes \mathbb{C}$, it suffices to say what c does to vectors in $V^{1,0}$ and $V^{0,1}$.

For $v\in V^{1,0}$, $c(v)$ is the endomorphism of $\mathbb{S}$ obtained by wedging with v:
$c(v)(u_1\wedge \cdots u_k)=\sqrt{2}v\wedge u_1\wedge \cdots \wedge u_k$

For $\overline{v}\in V^{0,1}$ $c(\overline{v})$ is contraction with $\overline{v}$:
$c(\overline{v})(u_1\wedge \cdots u_k) = \sqrt{2}\sum_{j=1}^k (-1)^j g(v,u_j)u_1\wedge \cdots \wedge \widehat{u_j} \wedge \cdots u_k$

One needs to check that this respects the Clifford algebra structure, and is an isomorphism. Initially, this may look wrong because for example when $v\in V^{1,0}$
$c(v)^2(u_1\wedge \cdots \wedge u_k) = v\wedge v\wedge u_1\wedge \cdots \wedge u_k=0$
and it seems like we should have $c(v)^2=-|v|^2I$. However, the algebra structure we want to preserve is complex linear on $Cl(V)\otimes \mathbb{C}$ and has the Clifford structure only on the $Cl(V)$ piece. Therefore, for example when $v=e_j-if_j\in V^{1,0}$,
$0=c(e_j-if_j)^2 = (c(e_j)-ic(f_j))^2 = (c(e_j))^2-ic(e_j)c(f_j)-ic(f_j)c(e_j)-(c(f_j))^2 = |e_j|^2-i2g(e_j,f_j)-|f_j|^2$

For basis elements, the map c is a sum of the exterior and interior products. To compute for example, $c(e_j)$ we split this into the $V^{1,0}$ and $V^{0,1}$ parts, so
$c(e_j)=c\left(\frac{1}{2}(e_j-if_j)+\frac{1}{2}(e_j+if_j)\right)=\frac{\sqrt{2}}{2}\left((e_j-if_j)\wedge\cdot +\iota_{e_j-if_j} \right)$
If you want to be slightly more convinced without completing the proof that $c(v)^2=-|v|^2I$ for real elements of $Cl(V)$ it is fairly easy at this point to check that $c(e_j)^2=-I$ at least on the $\bigwedge^0V^{1,0}$ part of $\mathbb{S}=\bigwedge V^{1,0}$ (since any map that starts with contraction vanishes and $\iota_x(y\wedge f)=-fg(x,y)$ for $f\in \bigwedge^0V^{1,0}$ a complex number, and $x,y\in V\subset Cl(V)\otimes 1$).

We get the last property in the theorem easily from the definition of c. For $v\in V\otimes \mathbb{C}$, $c(v)$ either raises or lowers by 1, wedge power of an element of $\mathbb{S}=\bigwedge V^{1,0}$. Therefore $c(v)$ sends $\mathbb{S}^+$ to $\mathbb{S}^-$ and vice versa. Extending this over the entire Clifford algebra, we see that the endomorphisms in $c(Cl^+(4))$ preserve $\mathbb{S}^+$ and $\mathbb{S}^-$ (since they switch between $\mathbb{S}^\pm$ an even number of times) and $c(Cl^-(4))$ sends $\mathbb{S}^\pm$ to $\mathbb{S}^\mp$.

Note: We can rewrite the isomorphism $c: Cl(4)\otimes \mathbb{C}\to End(\mathbb{S})$ as a map
$c: Cl(4)\otimes \mathbb{C}\otimes \mathbb{S}\to \mathbb{S}$.
This will be useful when we use this representation to form associated bundles and consider sections of those bundles and maps between the spaces of sections.

This theorem generalizes for $Cl(2n)$, producing a complex vector space $\bigwedge V^{1,0}$ which splits where $dim(V)=2n$, whose endomorphisms are isomorphic to $Cl(2n)\otimes \mathbb{C}$, where $Cl^+$ preserves the splitting and $Cl^-$ switches the components. In the odd dimensional case, the situation is slightly different, but reduces to the even case by showing that $Cl(2n-1)\cong Cl^+(2n)$. For the purposes of Seiberg-Witten Floer homology, it will be useful to know $Cl(3)\cong Cl^+(4)$ which implies $Cl(3)\otimes \mathbb{C}\cong End(\mathbb{S}^+)\oplus End(\mathbb{S}^-)$.

Spinor bundles

Now that we have this representation of the complexification of the Clifford algebra, we can restrict to get a representation of Spin. Because $Spin(4)\subset Cl^+(4)$, and $c(Cl^+(4))$ preserves the splitting $\mathbb{S}=\mathbb{S}^+\oplus \mathbb{S}^-$, we get two representations
$\rho_{\pm}: Spin(4)\to Aut(\mathbb{S}^\pm)$
Note the image of $Spin(4)$ lands in automorphisms instead of only endomorphisms because elements of $Spin(4)$ are invertible in $Cl(4)$. These two representations correspond to the same ones we obtain by identifying $Spin(4)\cong SU(2)\times SU(2)$ and projecting onto one component.

We can extend these maps to $Spin^c$ by defining
$\rho^c_{\pm}: Spin^c(4)\to Aut(\mathbb{S}^\pm)$
by $\rho^c_\pm((g,z))=z\rho_{\pm}(g)$ for $g\in Spin(4)$, $z\in U(1)$.

Note this is well defined since $\rho^c_\pm((-g,-z))=\rho^c_\pm((g,z))$.

Given a Spin or Spinc structure on a manifold, these representations give rise to associated bundles $S^\pm \to M$. These bundles show up in the set-up for the Seiberg-Witten configuration space, which I will get to in another post.