$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ # Inspiration We want to be able to explicitly realize the spin representations of $\so_{2n}$ and $\so_{2n+1}$. This is tricky, because they don't occur in tensor powers of $V$ due to having half-integer spin. In particular these representations do not lift to $\SO(V)$, and instead lift to a ==**spin group**== $\Spin(n)$ which double-covers. However, the tensor product of a spin representation with its dual, $S\otimes S^*$, has all integer weights, and can be expressed in terms of $V$. Thus, we wish to take $V$ and embed it as a space which looks like $\End S$; this would yield the desired ==**spinors**==. > [!idea] > You have seen all of this play out in QM and QFT already, so you know what's coming. As a reminder of the QM picture, a spinor in a representation of $\so_{3}$ was an element of $\CC^2$. $\Spin(3)$ happens to be the same as $\su(2)$; this is a low-dimensional collision; this acts naturally on the space $\End \CC^2$. # $\so(V) \equiv \wedge^2V \implies \Cl(V)$ $\so(V)\cong \wedge^2 V$ as vector spaces via the "plane of rotation" embedding. We want to extend this to an algebra map from the universal envelope of $\so(V)$ to some "envelope" of $\wedge^2V$, which turns out to be isomorphic as a vector space to $\wedge V$. The induced algebra is called $\Cl(V)$. > [!proof]- Explain it on a basis > Give $V$ the naive basis $e_1,\dots e_N$; then $\so(V)$ consists of skew-symmetric matrices spanned by $e_{ij} - e_{ji}$, $i < j$ (no factor of $\frac12$ here; its just a choice). This can be identified in the vector space sense with $e_i\wedge e_j$. The induced lie bracket looks like $[e_a\wedge e_b, e_c\wedge e_d] = \left( \delta_{bc}e_{ad} - \delta_{ac}e_{bd} - \delta_{bd}e_{ac} + \delta_{ad}e_{bc}\right) - \text{symmetric};$the basis-free bilinear extension is $[a\wedge b, c\wedge d] = (b,c)a\wedge d - (a,c)b\wedge d - (b,d)a\wedge c + (a,d)b\wedge c.$Alright cool. Under the $\Cl(V)$ map we want to send $a\wedge b\to \frac12\left(ab - ba\right)$. One can check by explicit computation that this is a Lie algebra homomorphism if $\{a,b\} = (a,b)$. >[!proof]- Explain it naturally >$V$ is endowed with a metric $\eta$. $\so(V)$ consists of elements $\omega^\alpha_\beta$ such that $\omega^{\alpha\beta} + \omega^{\beta\alpha} = 0$. It's obvious what's happening now; the map $\so(V)\to \wedge^2V$ is contraction against $\eta$. > [!definition] Clifford Algebra > Let $V$ be a finite-dimensional vector space over algebraically closed $k$ of characteristic $\neq 2$ with a nondegenerate symmetric inner product $(\bullet,\bullet)$. The ==**Clifford Algebra**== $\Cl(V)$ is the algebra generated by $v\in V$ with defining relations $v^2 = \frac12 (v,v)$ for $v\in V$. In particular, $\{a,b\} = (a,b)$. Thus we have an embedding $\xi:\so(V)\hookrightarrow \Cl(V)$. Here's [[Clifford Algebras intuition]]. > [!idea] The explicit action of $\so(V)$. > From now on, $a\wedge b$ just means $\frac12(ab - ba)$ on $\Cl(V)$; there's no $\wedge$ operation on $\Cl(V)$, but this is a natural notation (its compatible with the grading later). > > $\so(V)$ acts by $\rho(a\wedge b)(x) = (a\wedge b)x - x(a\wedge b)$. Not deep. >[!idea] >You have seen all of this before. The $Vs inside $\Cl(V)$ are the $\gamma^\mu$ matrices, and under QFT conventions $\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}$. The full algebra is spanned by the ==**Dirac Bilinears**==, of which there are $2^4 = 16$ in $1+3$ dimensions. Then, the generators map (in QFT) as $S^{\mu\nu} = \frac{i}{4}[\gamma^\mu, \gamma^\nu]$. # Alternative for Experts: $\so(V)\curvearrowright V\hookrightarrow \Cl(V)$ is a deformation of $\bigwedge V$ The extension $\so(V)\curvearrowright V\hookrightarrow \bigwedge V$ works in a really stupid way: $0 = \rho(g)\{a,b\} = \{g(a), b\} + \{a,g(b)\} = 0$. Somehow, too many things are killed in this relation. This motivates us to consider the deformation $\{a,b\} = (a,b)$. It is compatible with the defining relation: $0 = \rho(g)(a,b) = \rho(g)\{a,b\} = g(a)b + ag(b) + bg(a) + g(b)a = (a,g(b)) + (b,g(a))$ It is clear that this is completely equivalent to the original definition: if $x = x_1\dots x_k$, the action of $a\wedge b$ via $[a\wedge b, x]$ can be expanded as$ \sum x_1\dots \widehat{x_j} [a\wedge b, x_j]\dots x_k $and one can check the action of $\so(V)\curvearrowright V$ works. I guess the mantra is **inner representations extend to inner representations on powers.** > [!proof]- Computation on basis vectors (boring) > $[e_i\wedge e_j, e_k] = e_ie_je_k - e_je_ie_k - e_ke_ie_j + e_ke_je_i = 2\delta_{jk}e_i - 2\delta_{ik}e_j$ as desired. In the same way that the [[Universal Enveloping Algebra]] is a [[deformation]] of $S(V)$ (roughly speaking, a valid first-order perturbation to the algebra operator $V^*\otimes V^*\otimes V$), the Clifford algebra is a deformation of $\wedge^n V$. The associated Poincare-Birkhoff-Witt-like theorem states that given an ordered basis of $V$, $\Cl(V)$ has basis given by ordered subsequences of this basis. We will prove this immediately. > [!idea] PBW > According to Etingof, this is actually a special case of a generalized PBW theorem for Lie super-algebras. # $\Cl(V)$ is the square of $S$ The algebra $\Cl(V)$ is isomorphic to $\Mat_{2^n}(k)$ if $\dim V = 2n$, and $\Mat_{2^n}(k) \oplus \Mat_{2^n}(k)$ if $\dim V = 2n + 1$. This is insanely good, because we just embedded $V$ into $\End M$ like we desired. I think the following construction is important to see to understand the whole picture, so let's be more explicit: > [!theorem] When $\dim V = 2n$, $\Cl(V)\cong \Mat_{2^n}(k)$ > Pick [[Dn|the usual basis]], and consider the $\Cl(V)$ module $M = \bigwedge(a_1,\dots, a_n)$ with the action of $\Cl(V)$ defined by$\rho(a_i)w = a_iw,\qquad \rho(b_i)w = \frac{\pa w}{\pa a_i}$where$\pa_{a_i} a_{k_1}\dots a_{k_r} = (-1)^{j-1} a_{k_1}\dots\widehat{a_{k_j}}\dots a_{k_r}.$ Then $\rho:\Cl(V)\to \End M$ is an isomorphism. >[!proof]- Bash > Up to sign, we can anticommute any $c_{IJ}$ to be of the form > > $ > c = a_1^{r_1}b_1^{s_1}\cdots a_n^{r_n}b_n^{s_n}, > $ > > where $r_i,s_i\in\{0,1\}$. We now proceed by induction on $n$. When $n = 0$, this is trivial. As the induction step, biject basis vectors with bitstrings and order them little-endian lexicographically. For example, when $n = 2$, I want the ordering to be $1, a_1, a_2, a_1a_2$. > > Then, if $a_1^{r_1}b_1^{s_1}\cdots a_n^{r_n}b_n^{s_n}$ is nonzero, then $r_i = s_i$ for all $i$. Then, if $c$ acts with a matrix $M$ on $M_n\cong \CC^{2^n}$, we can manually compute the actions of $ca_{n+1}^{r_{n+1}}b_{n+1}^{s_{n+1}}$: > > - If $r_{n+1}, s_{n+1} = 0$, $c$ acts as $M$ diagonally: > $ > \begin{pmatrix} > M & 0\\ > 0 & M > \end{pmatrix} > $ > - If $r_{n+1} = 1$ and $s_{n+1} = 0$, $c$ acts as > $ > \begin{pmatrix} > 0 & 0\\ > M(-1)^\sigma & 0 > \end{pmatrix} > $ > where $(-1)^\sigma$ is a digonal matrix sending a bitstring with $k$ elements to $(-1)^k$. > - If $r_{n+1} = 0$ and $s_{n+1} = 1$, $c$ acts as > $ > \begin{pmatrix} > 0 & M(-1)^\sigma\\ > 0 & 0 > \end{pmatrix} > $ > - Finally, if $r_{n+1} = s_{n+1} = 1$, $c$ acts as > $ > \begin{pmatrix} > 0 & 0\\ > 0 & M > \end{pmatrix} > $ > But it's completely obvious that these matrices are linearly independent (and thus form a basis); muliplication by $(-1)^\sigma$ is an isomorphism on the space of operators over $\CC^{2^n}$. > [!theorem] When $\dim V = 2n+1$, $\Cl(V)\cong \Mat_{2^n}(k)\oplus \Mat_{2^n}(k)$ > The usual basis now has an additional element $z$. We construct two $\Cl(V)$ modules $M_\pm = \bigwedge(a_1,\dots, a_n)$ with the action of $\Cl(V)$ defined by$\rho(a_i)w = a_iw,\qquad \rho(b_i)w = \frac{\pa w}{\pa a_i},\qquad zw = \pm\left(-1\right)^w w$where $\deg w$ is the number of terms in $w$. Then $\rho:\Cl(V)\to \End M_+\oplus \End M_-$ are isomorphisms. > [!idea] Grassmann Picture and QFT Waffles > This reminds me of the related Grassmann picture from QFT. $M$ consists of the Grassman algebra generated by $a_i$, and $\Cl(V)$ is exactly the algebra generated by $a_i$ and $\pa_{a_i}$, and $\{a_i, b_j\} = \delta_{ij}$. These $a_i$ and $b_i$ can also be thought of as raising and lowering operators. $M$ shouldn't be thought of as containing objects of the same type as $\Cl(V)$ I think. > > I think these objects correspond exactly, but I'm not completely sure. Spinors fields are functions from $\RR^{1+3}$ to $M\cong \CC^{2^2} = \CC^4$; they're spinor fields. Raising and lowering operators form independent algebras on each momentum mode. $SO(1,3)$ acts globally on $\CC^4$ and the spinor representation instructs the spinor fields how to rotate with the global rotation. I don't actually see the picture... > > In QFT, they did discuss the $\gamma$ matrices, which are elements of $\End M$, and the Dirac billinears, of which there are $\dim \End M = 16$. # Deep Connection: $\Cl(V)$ is a *filtration-preserving* deformation of $\bigwedge(V)$ The natural [[filtered algebra|filtration]] generated by $V\in F_1$ induces a surjective map $\phi: \bigwedge V\to \gr\Cl(V)$ (in particular, $\phi: \bigwedge^k V\hookrightarrow \Cl_k(V)$). By dimensions, $\phi$ is an isomorphism of vector spaces. Here's the really deep part: $\bigwedge V$ and $\Cl(V)$ as isomorphic as $\so(V)$ lie algebra representations! >[!idea]- Thought Process > Multiplication by $a\wedge b$ sends $\Cl_{k}\to \Cl_{(k+2)}$, thus naively, the adjoint action of $\so(V)$ does the same. However, we have already computed three cases in which it does not: $\so(V)$ sends scalars to scalars, vectors to vectors, and $\so(V)$ preserves $\so(V)$ (bivectors). The action of $\so(V)$ literally just rotates the basis. That's not going to change your degree! This also shows that the mapping $\bigwedge V\to \gr\Cl(V)$ has the power of picking the correct representatives for your $\Cl_{k}(V) / \Cl_{k-1}(V)$. >[!idea]- There are a ton of ways to see this: > - I mean, conjugation by $a\wedge b$ already preserves $V$. Thus $[a\wedge b, x_1x_2\dots] = [a\wedge b, x_1]x_2\dots + \dots$. This can *also* be realized explicitly, by porting the $\bigwedge V$-argument via symmetrization. > - We know that parity is conserved, thus if your degree does not increase by $2$, then it does not increase at all. Conjugation by $a\wedge b$ inside of $\bigwedge V$ annihilates you, because $a\wedge b$ is degree 2! Thus the same is true in $\phi: \bigwedge^kV\hookrightarrow \Cl_k(V)$. > Thus, the action of $\so(V)$ on $\Cl(V)$ has sub-representations on $\Cl_k(V)$. But $\so(V)$ is decomposable, thus $\so(V)\curvearrowright \Cl(V)$ as $\bigoplus_k \bigwedge^k V\equiv \bigwedge V$! # Conclusion Thus, $\xi^*M$ is a representation of $\so_{2n}$ on a space $\cong \CC^{2^n}$ via $\rho_{\xi^*M}(a) = \rho_M(\xi(a))$. This representation reduces into $(\xi^*M)_0\oplus (\xi^*M)_1$ consisting of the even and odd-degree parts, which are isomorphic to $S_+$ and $S_-$. $\xi^*M_\pm$ are representations of $\so_{2n+1}$ on spaces $\cong \CC^{2^n}$, and they turn out to both be isomorphic to $S$. > [!proof]- Setup: Porting over $\fg_\alpha$ and $\hh$. > The element of $\so_{2n}$ spanning root space $e_i - e_{i+1}$ is the following matrix in $\so_{2n}$, which becomes the following under contraction against the metric, which becomes the following element of $\wedge^2 V$ which gets mapped to the following element of $\Cl(V)$.$e_i - e_{i+1}\implies \left(\begin{array}{c c | c c} > 0 & 1 & 0 & 0 \\ > 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & 0 & 0 \\ > 0 & 0 & -1 & 0 > \end{array}\right) \implies \left(\begin{array}{c c | c c} > 0 & 0 & 0 & 0 \\ > 0 & 0 & -1 & 0\\ > \hline > 0 & 1 & 0 & 0 \\ > 0 & 0 & 0 & 0 > \end{array}\right) \implies b_i\wedge a_{i+1} \implies b_ia_{i+1}\in \Cl(V).$The last raising operator is $e_{n-1} + e_n \implies \left(\begin{array}{c c | c c} > 0 & 0 & 0 & 1 \\ > 0 & 0 & -1 & 0 \\ > \hline > 0 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0 > \end{array}\right) \implies \left(\begin{array}{c c | c c} > 0 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0\\ > \hline > 0 & 0 & 0 & 1 \\ > 0 & 0 & -1 & 0 > \end{array}\right) \implies b_{n-1}\wedge b_n\implies b_{n-1}b_n\in \Cl(V).$ > The CSA of $\so_{2n}$ corresponding to $e_i$ looks like$\left(\begin{array}{c c | c c} > 1 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & -1 & 0 \\ > 0 & 0 & 0 & 0 > \end{array}\right) \implies \left(\begin{array}{c c | c c} > 0 & 0 & -1 & 0 \\ > 0 & 0 & 0 & 0\\ > \hline > 1 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0 > \end{array}\right) \implies b_i\wedge a_i \implies b_ia_i - \frac12 \in \Cl(V).$The story is basically the same for $\so_{2n+1}$. > $e_i - e_{i+1} \implies \left(\begin{array}{c | c c | c c} > 0 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & 1 & 0 & 0 \\ > 0 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & 0 & 0 & 0 \\ > 0 & 0 & 0 & -1 & 0 > \end{array}\right)\implies b_ia_{i+1},\qquad > e_n \implies \left(\begin{array}{c | c c | c c} > 0 & 0 & 0 & 0 & 1 \\ > \hline > 0 & 0 & 0 & 0 & 0 \\ > -1 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0 & 0 > \end{array}\right)\implies \left(\begin{array}{c | c c | c c} > 0 & 0 & 0 & 0 & 1 \\ > \hline > 0 & 0 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & 0 & 0 & 0 \\ > -1 & 0 & 0 & 0 & 0 > \end{array}\right) \implies z\wedge b_n\implies zb_n\in \Cl(V)$ > The CSA of $\so_{2n+1}$ corresponding to $e_i$ looks like$\left(\begin{array}{c | c c | c c} > 0 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 1 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & 0 & -1 & 0 \\ > 0 & 0 & 0 & 0 & 0 > \end{array}\right)\implies\left(\begin{array}{c | c c | c c} > 0 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 0 & 0 & -1 & 0 \\ > 0 & 0 & 0 & 0 & 0 \\ > \hline > 0 & 1 & 0 & 0 & 0 \\ > 0 & 0 & 0 & 0 & 0 > \end{array}\right)\implies b_i\wedge a_i\implies b_ia_i - \frac12$ > [!proof]- Compute the highest weight vectors then stare at dimensions. > After all the setup, this isn't hard anymore. In $\so_{2n}$, we care about the vectors in the joint kernel of $\{b_ia_{i+1}\}$, which are all of the form $a_k\wedge\dots \wedge a_n$. But this is also supposed to be in the kernel of $b_{n-1}b_n$, so our options are $a_n$ and $1$. Observe $a_n$ is a $1$-eigenvector of all $b_ia_i$ except for $b_na_n$; this is a highest-weight vector with weight $\left(\frac12,\dots,\frac12,-\frac12\right)$. Meanwhile, $1$ is a highest-weight vector with weight $\left(\frac12,\dots,\frac12\right)$. But we know these irreps have dimension $2^{n-1} + 2^{n-1} = 2^n$, which is all of $\wedge\left(a_1,\dots, a_n\right)$, so we conclude. > > In $\so_{2n+1}$, we care about vectors of the form $a_k\wedge\dots\wedge a_n$ again. This time, we want to be in the kernel of $zb_n$, which leaves us no options; $1$ is a highest-weight vector with weight $\left(\frac12,\dots, \frac12\right)$ in both $\xi^*M_+$ and $\xi^*M_-$. It has dimension $2^n$, which fills $\wedge(a_1,\dots, a_n)$, thus we conclude. > [!idea] Weyl Fermions > In $2n = 4$, the two sectors of $\xi^*M$ form the left-handed and right-handed Weyl fermions. A Dirac or Majorana spinor is composed from Weyl fermions in ways that will be discussed in the homework.