⇤Math diary
May 10, 2024: Spinors on Riemann Surfaces
Intro to spinors
we are interested in understanding the representations of the lie algebra $\mathfrak{so}(2n)$. These are defined as the group of determinant 1, orthogonal matrices, so they come with an obvious representation of dimension $2n$. We call this the standard representation $V$. From this, we can construct many new irriducible representations. For example, alternating 2-tensors $\wedge^2 V$ or symmetric 2-tensors $\text{Sym}^2(V)$. In general, we can form irriducible representations (irreps) by looking at the subspace of tensors transforming in a certain way under permutation of their coordinates. For the algebra $\mathfrak{su}_n$, this process reproduces all finite dimensional irreps, a manifestation of Schur-Weyl duality. But this only gives half the representations of $\mathfrak{so}(2n)$.
An example is worth a thousand words. Take $\mathfrak{so}(3)$. The standard representation is $V_s \cong \mathbb{C}^3$. The action comes from complexifying from the defining action of $SO(3)$ rotating of $\mathbb{R}^3$. But, as a lie algebra, $\mathfrak{so}(3) \cong \mathfrak{su}(2)$. So we get another free representation, now from the standard representation $SU(2)$ on $\Delta \cong \mathbb{C}^2$. The centerpiece of a representation theory class is the proof that all irreps of $\mathfrak{su}(2)$ are of the form $V_n = \text{Sym}^n(\Delta)$, each with dimension $n+1$. In particular, $V_s = V_2$. Starting from $V_s$, we can only get the odd dimensional representations $V_{2n}$. Exponentiating these representations from $\mathfrak{so}(3)$ to $SO(3)$, only the $V_{2n}$ are well defined. Indeed, by general principles, we can always exponentiate a representation from $\mathfrak{su}(2)$ to $SU(2)$, because $SU(2)$ is simply connected. Then, $SO(3) \cong SU(2)/{\pm I}$. The representations $V_{2n+1}$ are not well defined because the element $-I$ in $SU(2)$ gets mapped to $-1$ in $V_{2n+1}$, so it does not descend to a representation of $SO(3)$.
Moral of the story is, the lie algebra of $SO(n)$ has some represesentations that we can’t see from the group, exactly because $SO(n)$ is not simply connected. Indeed, $\pi_1(SO(n) = \mathbb{Z}_2$. To be able to find the mystery representations, we need to lift to the simply connected cover. This is called the spin group, $Spin(n)$. For $n=3$, the law of small numbers saved us. coincedentally, $\mathfrak{so}(3) = \mathfrak{su}(2)$, and we get $Spin(3)=SU(2)$.
How might we construct these secret representations, if we didn’t have this lucky $SU(2)$ showing up? How do we find $Spin(n)$? Let me take it from the top. Every element of $O(n)$ is formed as the product of reflections. A even number of reflections has determinant 1, so lives in $SO(n)$. The primary insight is to swtich the objects of study from the reflections themselves, to the vector perpendicular to the hyperplane preserved by the reflection. For a vector $v\in V \cong \mathbb{R}^n$, denote the reflection across the hyperplane orthogonal to $v$ by $R_v$. Both $v$ and $-v$ generate the same reflection, which wil eventually give us our desired double cover.
A composition of reflections $R_v R_w$ is rotation in the plane defined by $v \wedge w$, by twice the angle between $v$ and $w$. Let us try to characterize the behavior of rotations by finding some algebra which contains the structure of reflections. We represent a sequence of reflections by a sequence of vectors
\[R_v R_w \mapsto v\cdot w\]Let $\theta$ denote the angle between $v$ and $w$. We said that $R_v R_w$ rotates $u$ in the $v\wedge w$ plane by $2 \theta$. If we swap the order, then $R_w R_v$ rotates $u$ in the $v \wedge w$ by $-2\theta$. The sum of $R_v R_w u$ and $R_w R_v u$ is proportional to $u$. More precisely,
\[R_v R_w u + R_w R_v u = 2 \cos(2\theta) u\]Now we turn this to a bilinear map, which agrees with the reflections when $v,w$ are unit vectors. We impose the relations
\[v\cdot w + w\cdot v = \langle v,w \rangle\](This almost agrees with above, but like not quite. its the wrong angle. I need to do more work to make this precise intuition). Starting with a vector space $V$ and inner product $\langle \cdot, \cdot \rangle$, we produce its Clifford algebra $Cl(V)$. Denoting the full tensor algebra by $T^\circ V$, the Clifford algebra is the quotient of $T^\circ V$ by the ideal generated by the relation above:
\[Cl_n = T^\circ V / \langle v\otimes w + w \otimes v = \langle v,w\rangle1\rangle\]We denote multiplication in the clifford algebra by $v \cdot w$. Note that above, $v\cdot v = \frac{1}{2}\vert v\vert ^2 \cdot 1$. If we choose a basis $e_i$ of $V$, then we have basis of $Cl(V)$ consisting of products $e_{i_1}\cdot \dots \cdot e_{i_n}$, with no repeating $e_i$. As a vector space, $Cl(V)$ is isomorphic to the exterior algebra $\wedge^n V$, but the product structure is different. $Cl(V)$ carries a natural filtration by the degree of the tensor. The product on $Cl(V)$ agrees with the product on $\wedge^n V$ up to subleading order terms. So, $\wedge^n V$ is the associated graded algebra of $Cl(V)$. Because of this, we say $Cl(V)$ is a deformation, or a quantization, of the exterior algebra
We can identify some familiar subspaces of the Clifford algebra
- The ideal is generated by even degree elements, so the degree of the product of elements in $Cl(V)$ is well defined mod 2. We have a splitting into even and odd parts $Cl(V) \cong Cl^{e}(V)\oplus Cl^{o}(V)$.
- The degree 1 elements are isomorphic to $V$
- The degree 2 elements are isomorphic to $\wedge^2 V$, which are isomorphic to the lie algebra $\mathfrak{so}(V)$. (This arises if you think of elements of $\mathfrak{so}(V)$ as antisymmetric matricies.) The lie bracket on $\mathfrak{so}(V)$ agrees with the Clifford algebra commutator.
- The unit norm elements of the Clifford algebra which preserve $V$ under conjugation constitute $Spin(V)$,
- Complexify $V\cong \mathbb{R}^{2n}$ into $V^\mathbb{C} \cong \mathbb{C}^{2n}$. Then, find an isotropic splitting $V^{\mathbb{C}} \cong W \oplus W’$. The Half spin representations are $S^+ = \wedge^{odd} W$ and $S^- = \wedge^{even} W$. In fact, $Cl^{even}(V^\mathbb{C})$ is isomorphic to the matrix algebra over $S^+$
- If $V\cong \mathbb{R}^{2n+1}$
Example: Clifford algebra for 3D
Let’s check that this repoduces the discussion for $SO(3)$. Denote the standard representation by $V_s \cong \mathbb{R}^3$. The clifford algebra $Cl(V_s)$ is 8 dimensional (from the dimension of the exterior algebra). The degree 1 part is isomorphic as a vector space to $V_s$. The degree 2 part is $\wedge^2 V_s \cong V_s^*$. Indeed, the lie algebra $\mathfrak{so}_3$ is 3 dimensional, and identified with $\mathbb{R}^3$. the lie bracket is the cross product of two vectors. the clifford algebra splits into even and odd parts:
\[Cl^{even}(V) = 1 \oplus \wedge ^2 V \qquad Cl^{odd}(V) = V \oplus \wedge ^3 V\]Notice that $Cl^{even} \cong {Cl^{odd}}^*$, with the pairing induced by the wedge product. As an algebra, $Cl(\mathbb{R}^3) \cong \mathbb{H} \oplus \mathbb{H}$, where $\mathbb{H}$ is the quaternions. In particular, $Cl^0(V) \cong \mathbb{H}$. To see this, choose an orthonoromal basis $x,y,z$ of $\mathbb{R}^3$. Then, the clifford algebra has relations
\[\begin{align} \{x,x\} &= \{y,y\} = \{z,z\} = -2 \\ \{x,y\} &= \{y,z\} = \{x,z\} = 0 \end{align}\]$Cl^0(\mathbb{R}^3)$ is spanned by $1,xy,yz,zx$. For convience, we define a basis $e_z = xy, e_y = zx, e_x = yz$. The commutator relation is
\[[e_z,e_x] = \frac{1}{2}(xyyz - yzxy )= \frac{1}{2}(-xz - zyyx) = \frac{1}{2}(+zx +zx = 2zx) = e_y\]and likewise, we have the relations $[e_x,e_y]=e_z$ and $[e_y,e_z]=e_x$. So, the Clifford algebra commutator agrees with the lie bracket of $\mathfrak{so}_3$. Using the identification of $Cl^0(V)$ with $\mathbb{H}$, 1 is the unit in $\mathbb{H}$ and $\wedge^2 V$ is the imaginary part of $\mathbb{H}$, spanned by $i,j,k$. The exponential of $\text{span}(i,j,k)$ starting at the unit $1$ gives the unit sphere $S^3\subset \mathbb{H}$. We see $S^3 \cong SU(2)$ is a simply connected lie group with lie algebra $\wedge^2 V \cong \mathfrak{so}_3$. We can characterize the unit sphere more abstractly. These are all the unit norm elements of $Cl^e(V)$ which preserve $V\subset Cl(V)$ under conjugation. Indeed, the infinitesimal action of $x \in \wedge^2 V$ on $V$ via commutator generates rotation at velocity $|x|$ in the plane perpendicular to $x$. This exponentiates to the conjugation action. In terms of quaternions, conjugation by the unit sphere preserves the 3D space of imaginary quaternions.
Finally, we need to check the spin representation agrees with the standard representation. To build the spin representtion, we first complexify $V = \mathbb{R}^3$ into $V^{\mathbb{C}} = \mathbb{C}^3$. Then, we split into isotropic subspaces $W,W’$, and an additional 1-dimensional space $U$, such that $V^{\mathbb{C}} = W \oplus W’ \oplus U$. For $V^{\mathbb{C}}\cong \mathbb{C}^3$, we have $W \cong W’ \cong U\cong \mathbb{C}$. The clifford algebra is a matrix algebra over the exterior algebras of these isotropic subspaces:
\[Cl(V) = \text{Mat}(\wedge^* W)\oplus \text{Mat}(\wedge^* W')\]And the even part of the clifford algebra is
\[Cl^{even}(V) = \text{Mat}(\wedge^* W) \cong \wedge^0W \oplus \wedge^1W \cong \mathbb{C}^2\]This is an irreducible representation of $\mathfrak{so}_3$, so must agree with the representation described above.
Example: Clifford algebra for 2D
To understand riemann surfaces, which have 2 real dimensions, we need to understand clifford modules for $\mathbb{R}^2$. First, we have a vector space structure for the clifford algebra:
\(Cl(\mathbb{R}^2) = 1 \oplus \mathbb{R}^2 \oplus \wedge^2 \mathbb{R}^2\) Choosing a basis $x,y$ of $\mathbb{R}^2$, the Clifford algebra has basis
\[1,x,y,xy\]with relations
\[\{x,x\}= \{y,y\} = -1 \qquad \{x,y\}=0\]This algebra is isomorphic to the quaternions $\mathbb{H}$. Under the splitting $Cl(\mathbb{R}^2) \cong Cl^e(\mathbb{R}^2) \oplus Cl^o(\mathbb{R}^2)$, the even part is a subalgebra and the odd part is a module over that subalgebra. We observe
\[Cl^e(\mathbb{R}^2) = 1 \oplus \wedge^2 \mathbb{R}^2= \text{span}(1,xy)\]Notice that $(xy)^2 = xyxy = -xxyy = -1$. The element $xy$ is a square root of minus 1, so as an algebra $Cl^e(\mathbb{R}^2) \cong \mathbb{C}$. The imaginary part of $\mathbb{C}$ is $\wedge^2 \mathbb{R}$. We can think of this as the tangent space of the identity of $U(1)$, the group of norm 1 units of $\mathbb{C}$. Therefore, $\wedge^2 \mathbb{R} \cong \mathfrak{u}(1)$. Indeed, we can explicitly compute the exponential map:
\[\begin{align} e^{t \cdot xy} &= 1 + t xy + \frac{t^2}{2}(xy)^2 + \frac{t^3}{3!}(xy)^3 + \dots \\ &= (1 - \frac{t^2}{2} + \frac{t^4}{4!} - \dots ) + xy (t - \frac{t^3}{3!} + \dots )\\ &= \cos(t) + xy \sin(t) \end{align}\]Following the general principles of Clifford algebras, this $Spin(2)\cong U(1)$ subgroup of $Cl(\mathbb{R}^2)$ should be a double cover of $SO(2)$. Indeed, $SO(2)$ is also isomorphic to a circle, so topologically its double cover is a circle. Let’s verify this explicitly. $e^{t \cdot xy}$ acts on $\mathbb{R}^2 \subset Cl(\mathbb{R}^2)$ by conjugation:
\[\begin{align} e^{-t \cdot xy}(ax + by)e^{t \cdot xy} &= (\cos(t) - xy \sin(t)) (ax+by)(\cos(t)+xy\sin(t))\\ &= (\cos(t)^2-\sin(t)^2)(ax + by) - 2\cos(t)\sin(t) (-ay+bx) \\ &= \begin{pmatrix} a & b\end{pmatrix} \begin{pmatrix}\cos(2t) & \sin(2t) \\ -\sin(2t) & \cos(2t) \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} \end{align}\]This shows the action of $Spin(2)\cong U(1)$ on $\mathbb{C}$ factors through the standard action of $U(1)$ on $\mathbb{C}$ via the doubling map $\theta \mapsto 2 \theta$.
Next we build the spin representation. We follow the standard procedure for even-dimensional vector spaces. We start with $V = \mathbb{R}^2$, then we complexity to get $V^\mathbb{C} = \mathbb{C}^2$. Then, we split into isotropic subspaces. There is a standard choice of splitting, the holomorphic and antiholomorphic parts:
\[V^{\mathbb{C}} = V^{(1,0)} \oplus V^{(0,1)}\]Where $V^{(1,0)}\cong V^{(0,1)}$ is a complex vector space isomorphic as a real vector space to $V$. The spin representation is $\wedge^* V^{(1,0)}$. This splits into irreducible representations
\[S = \wedge ^* V^{(1,0)} \qquad S^+ = \wedge^{even} V^{(1,0)} \qquad S^- = \wedge^{odd} V^{(1,0)}\]called the half spin representations. The even part of the Clifford algebra is a matrix algebra over the spin representation
\[Cl^{even}(V) = \text{Mat}(S)\]Applied to the two-dimensional case, we get that $V^{(1,0)}\cong \mathbb{C}$, so $S^+ = \wedge^0 V^{(1,0)} \cong \mathbb{C}$ and $S^- = \wedge^1 V^{(1,0)}\cong \mathbb{C}$. as a representation of $U(1)$, $S^+$ has weight 0 and $S^-$ has weight 2.
Spinors
Finally we are out of the algebra mire. The next goal is to turn the algebra into geometry. For any riemannian manifold $M$, the tangent spaces have an inner product structure. There is a canonically defined bundle of Clifford algebras, each based at a tangent space. Symbolically, the fiber over the point $x$ is the Clifford algebra $Cl(T_xM)$.
The riemannian structure is equivalent to a principle $SO(n)$ bundle over $M$. Can we lift this $SO(n)$ bundle to a $Spin(n)$ bundle? A choice of this lifting is called a spin structure on the manifold. A natural question is, is there even a spin structure? We can certainly find a spin structure if the tangent bundle is trivial, but what if it’s nontrivial? Denote the principle $SO(n)$ bundle by $P$. We have a fibration
\[SO(n) \to P \to M\]passing to the long exact sequence of the fibration, we have
\[H^1(X,\mathbb{Z}_2) \to H^1(P,\mathbb{Z}_2) \to H^1(SO(n),\mathbb{Z}_2) \to H^2(M,\mathbb{Z}_2)\to \dots\]since $\pi_1(SO(n))=\mathbb{Z}_2$,, there is a canonical representative $H^1(SO(n),\mathbb{Z}/$
To undertand, we can think of the double cover $Spin(n) \to SO(n)$ by passing to the determinant. On the level of representations, the standard rep $V$ of $SO(n)$ induces a one dimensional representation $\wedge^n V$. Lifting to $Spin(n)$, we get another one dimensional representaiton.
One way to see the double cover $Spin(n) \to SO(n)$ is to pass to the determinant line bundle