Virasoro Algebra from the Free Boson without Regularization

Usual physics derivations of the Virasoro algebra from the free boson in two dimensions usually use some sort of regularization procedure to compute the central charge. Following these notes I'd like to give a purely algebraic calculation of the central charge.


Let $A = \mathbf C[x_1, x_2, \dots]$ be the polynomial algebra in countably many generators. For an integer $k > 0$, define a $k$-linear operator $a_k$ on $A$ by
$$a_k = \frac{\partial}{\partial x_k}, \ k > 0$$
Similarly, for $k < 0$ we define $a_k$ by multiplication:
$$a_{-k} = k x_k, k > 0$$
For $k = 0$, we define $a_0$ to be multiplication by some fixed complex number (which by abuse of notation we also denote by $a_0$).

Lemma. We have the commutation relation $[a_m, a_n] = m \delta_{m+n}$ as linear operators on $A$.

For any monomial in the $a_k$, we define normal ordering $::$ to be the monomial obtained by reordering the terms so that the indices are increasing. (Mathematical interpretation: it is a section of the quotient map from the tensor algebra in the $a_k$ to the symmetric algebra, defined by lexicographic order.) For example,
$$:a_j a_k:\ = \left\{ \begin{array}{rr} a_j a_k, & j \leq k \\ a_k a_j, & j > k \end{array} \right.$$

Next we formally define a set of operators $L_k$ by
$$L_k = \frac{1}{2}\sum_j :a_j a_{k-j}:$$

Proposition. The $L_k$ are well-defined as linear operators on $A$.

Proof. For sufficiently large $|j|$, at least one of $j$ or $k-j$ is positive, and hence $:a_j a_{k-j}:$ contains a differentiation (on the right). Since any element $f \in A$ is annihilated by all but finitely many of the differentiation operators $\partial_j$, the formal expression $L_k f$ contains only finitely many non-zero terms, and hence is well-defined.

Lemma. As operators on $A$, we have $[a_k, L_n] = k a_{k+n}$.

Theorem. As operators on $A$, we have
$$[L_m, L_n] = (m-n) L_{m+n} + \frac{1}{12} (m^3-m) \delta_{m+n}$$

Proof. Fix $m,n$. For the sake of simplicity we will assume $m \neq n$ and $mn \neq 0$. (The other special cases can be treated by similar arguments.) By the same argument as the proof of the preceding proposition, for any fixed element $f \in A$ there exists some $N \gg 0$ such that
$$[L_m, L_n]f = [L_m^N, L_n] f$$
where $L_m^N$ is the truncated operator
$$L_m^N = \frac{1}{2}\sum_{|j| < N} :a_j a_{m-j}:$$
Let us compute (noting that since $m \neq 0$, $:a_j a_{m-j}: = a_j a_{m-j}$)
$$\begin{align}   [L_m^N, L_n] &= \frac{1}{2} \sum_{|j| < N} [a_j a_{m-j}, L_n] \\   &= \frac{1}{2} \sum_{|j| < N} (m-j) a_j a_{m+n-j} + \frac{1}{2} \sum_{|j| < N}j a_{n+j} a_{m-j} \end{align}$$
Denote the two sums above by $S_1$ and $S_2$. It is clear that these should be related to the operator $L_{m+n}$, but to see the exact relation we will have to normal order the terms. Let's start with $S_1$. Note that $a_j a_{m+n-j}$ is already normal ordered, unless $j > m+n-j$. Hence
$$\begin{align}   S_1 &= \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{1}{2} \sum_{m+n\lt2j\lt2N} (m-j) [a_j, a_{m+n-j}] \\ &= \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{\delta_{m+n}}{2} \sum_{m+n\lt2j\lt2N} j(m-j) \\ &= \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{\delta_{m+n}}{2} \sum_{0\lt j\lt N} j(m-j) \end{align}$$

Similarly, we normal order the terms in $S_2$:
$$\begin{align}   S_2 &= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{1}{2} \sum_{m-n\lt2j\lt2N}j [a_{m-j}, a_{n+j}] \\ &= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{\delta_{m+n}}{2} \sum_{m\lt j\lt N}j (m-j) \\ &= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{\delta_{m+n}}{2} \sum_{m\lt j\lt N}j (m-j) \\ &= \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: -\frac{\delta_{m+n}}{2} \sum_{m\lt j\lt N}j (m-j) \end{align}$$

Hence we have
$$S_1 + S_2 = \frac{1}{2} \sum_{|j| < N} (m-j) :a_j a_{m+n-j}: +  \frac{1}{2} \sum_{|j| < N}j :a_{n+j} a_{m-j}: + \frac{\delta_{m+n}}{2} \sum_{0\lt j\leq m} j(m-j)$$
Now that everything is normal ordered, we can take the limit $N \to \infty$ without fear. After a simple cancellation, and explicitly summing the last term using well-known formulas for sums of powers of integers, we obtain:
$$[L_m, L_n] = (m-n) L_{m+n} + \frac{1}{12} (m^3-m) \delta_{m+n}$$
For the usual conventions of the Virasoro algebra, this shows that this representation corresponds to central charge $c=1$.

Remark. If we tried to take the limit $N \to \infty$ in each of the terms $S_1, S_2$ separately, before taking their sum, we would obtain a formal infinite constant $\sum_{j} j(m-j)$. In any physics textbook, the author will simply zeta-regularize this sum to obtain a fininte result. However, the above calculation shows that this is not necessary. By taking care that each term in the expression $S_1+S_2$ was normal-ordered, before taking the limit, we obtain only finite constants with no need to regularize. Zeta regularization certainly has its uses (for example in rigorous definitions of functional determinants), but as the above calculation shows, it can also be an unnecessary crutch that obscures the underlying mathematical phenomena.

Remark. There is an analogous calculation which shows that one obtains a Virasoro representation from affine Lie algebras. Physically, this corresponds to the WZW model. Roughly, the generators of the affine Lie algebra behave as an infinite set of harmonic oscillators, similar to the Heisenberg algebra above. Sometime in the future I may write a sequel to this post giving the details of this calculation.

The Basic Idea of the Quantum BV Complex

$\newcommand{\h}{\hbar} \newcommand{\X}{\mathfrak{X}} \newcommand{\PV}{\text{PV}} \newcommand{\K}{{\mathbb K}} \newcommand{\R}{{\mathbb R}} \newcommand{\too}{\xrightarrow} \newcommand{\im}{\text{Im }}$ Let $M$ be an oriented $n$-dimensional manifold and $\X^\bullet(M):=\Gamma(M,\wedge^{-\bullet} TM)$, so that $\X^\bullet(M)$ is concentrated in non-positive degree. Let $\mu\in \Omega^n(M)$ a volume form on $M$. Then interior product with $\mu$ gives an isomorphism $\vee\mu: \X^k(M)\too{\cong}\Omega^{n-k}(M)$. From this, we induce a degree 1 differential $\Delta_\mu$ on $\X^\bullet(M)$ from the de Rham differential $d$ on $\Omega^\bullet(M)$, defined by $\Delta_\mu= (\vee\mu)^{-1}\circ d\circ\vee\mu$, making $\X^\bullet(M)$ into a cochain complex isomorphic to $\Omega^\bullet(M)[k]$; in particular $H^k(\X)=H^{n+k}(\Omega)$.
If $M$ is compact and connected, then $H^0(\X)=H^n(\Omega)$ is one dimensional, and upon fixing a basis to identify $H^0(\X)\too{\cong}\R$, the quotient map gives a map $\pi:C^\infty(M)\to \R$. We have the following:
Proposition Let $1\in C^\infty(M)$ be the constant function taking the value $1$. Then $[1]\in H^0(\X)$ is nontrivial and after choosing it as a basis the resulting map $\pi:C^\infty(M)\to \R$ is given by $$f\mapsto\frac{ \int_M f\mu}{\int_M \mu}$$ Proof Let $f\in C^\infty(M)$ and suppose $f=\Delta_\mu(X)$ for $X\in\X^1(M)$. Then $$f\mu = f\vee \mu = \Delta_\mu(X)\vee\mu= d(X\vee \mu)$$ so that $f\mu$ is exact and by Stokes' theorem integrates to $0$ on $M$. Thus all $f\in \im \Delta_\mu$ integrate to zero against $\mu$ on $M$, so that the above map indeed descends to the quotient.
The above arguement also implies that were $1\in\im\Delta_\mu$ then $\mu$ would integrate to zero, contradicting that $\mu$ is a volume form, so that $[1]\in H^0(\X)$ is indeed nontrivial. The map is thus well-defined, linear, and has the appropriate action on a basis so that it is correct as claimed. $\square$
This recovers the standard definition of the expectation of an observable in the path integral picture of quantum field theory. The upshot is that having formulated the integration homologically, we can hope to extend this homological definition of expectation to situations where the integral itself is not well defined.
Consider the case $M=V$ a vector space, which in particular described the situation in coordinates on $M$. We are interested in measures of the form $\mu=e^{-S/\h}\mu_0$ where $\mu_0$ is the Lesbesgue measure on $V$, given by $\mu_0=dx^1\wedge ...\wedge dx^n$. Let $X\in \X^k(M)$, we have: $$\begin{align*} d(X\vee \mu) & = d(e^{-S/\h}(X\vee\mu_0)) \\ & = de^{-S/\h}\wedge (X\vee \mu_0)+e^{-S/\h}d(X\vee \mu_0)\\ & = -\frac{1}{\h}e^{-S/\h} dS\wedge (X\vee \mu_0) + e^{-S/\h} (\Delta_{\mu_0}X)\vee \mu_0 \\ & = -\frac{1}{\h}e^{-S/\h} (dS\vee X)\vee \mu_0 + (\Delta_{\mu_0}X)\vee \mu\\ & = \left(-\frac{1}{\h}dS\vee X + \Delta_{\mu_0}X \right)\vee \mu \end{align*}$$ so that $$\Delta_\mu=\Delta_{\mu_0}-\frac{1}{\h}\iota_{dS}$$ Further, we can explicitly compute $\Delta_{\mu_0}$: for $X\in\X^k(M)$ we have that $X=\sum_I X^I \partial_I$, where the sum is over increasing $k$-tuples $I\subset\{1,...,n\}$. We have $$\begin{align*} d(X\vee \mu_0) & = d\left( \sum_I X^I \partial_I\vee(dx^1\wedge ...\wedge dx^n)\right)\\ & = d\left( \sum_I X^I (-1)^?dx^1\wedge ...\wedge \hat{dx^I} \wedge ...\wedge dx^n \right)\\ & = \sum_I \sum_{i\in I} \partial_{x^i} X^I (-1)^?dx^i\wedge dx^1\wedge ...\wedge \hat{ dx^I}\wedge ...\wedge dx^n\\ & = \left( \sum_i \partial_{x^i} (dx^i\vee X) \right) \vee \mu \end{align*}$$ so that $$\Delta_{\mu_0} = \sum_i \partial_{x^i} \iota_{dx^i}$$ To be slightly more careful about the formal variable $\h$ and allow the $\h\to 0$ limit to be more clear, we refine our complex to be: $$\X^\bullet(M)[[\h]] \quad\quad\text{equipped with}\quad\quad \h \Delta_\mu= \h\Delta_{\mu_0} -\iota_{dS}$$ Next, we restrict consideration only to polynomial observables (functions), and vector fields with polynomial coefficients, denoting the resulting complex $\PV^\bullet$. One can check that $H^0(\PV)$ is still one dimensional, so that the above proposition holds and we maintain the integral intepretation of this cohomology.
Now, we can identify $\PV^\bullet[[\h]]$ with the graded-commutative graded algebra $\K[[x^1,...,x^n,\xi_1,...,\xi_n,\h]]$ where $x^1,...,x^n,\h$ are in degree 0 and $\xi_1,...,\xi_n$ are in degree -1 as follows: $$x^i \mapsto x^i \quad\quad \partial_i\mapsto \xi_i\quad\quad \partial_i\wedge\partial_j\mapsto \xi_i\xi_j \quad\quad \h\mapsto \h$$ Under this identification, the map $\iota_{dx^i}:PV^k(M)\to \PV^{k-1}(M)$ is identified with $\partial_{\xi_i}$. Now, we require our action function $S:M\to \R$ also be polynomial, and further, that $$S(x)=\frac{1}{2}\sum_{i,j} a_{ij}x^ix^j- b(x)$$ for $a_{ij}$ symmetric and non-degenerate and for $b\in I^3$ where $I=(x^1,...,x^n)\subset \K[[x^1,...,x^n]]$, that is, $b(x)$ a polynomial with no terms of degree less than 3. This implies that $$\begin{align*} \h\Delta_\mu & = \h\Delta_{\mu_0} -\iota_{dS} \\ & = \h \sum_i \partial_{x^i} \iota_{dx^i} - \sum_i (\partial_{x^i}S) \iota_{dx^i} \\ & = \h \sum_i \partial_{x^i} \iota_{dx^i} + \sum_i (\partial_{x^i}b) \iota_{dx^i} - \sum_{i,j}a_{ij}x^i\iota_{dx^j}\\ \end{align*}$$ and under our identification this becomes $$\h\Delta_\mu = \h \sum_i \partial_{x^i}\partial_{\xi_i} + \sum_i (\partial_{x^i}b)\partial_{\xi_i} - \sum_{i,j}a_{ij}x^i\partial_{\xi_j}$$ The computation of the degree 0 cohomology of a given polynomial $f\in\K[[x_1,...,x_n]]$ under this differential is taken up in Gwilliam, Johnson-Freyd where it is shown the answer is precisely the Feynman diagram expansion for the expectation of $f$ which we expect.

Clifford Algebras and Spinors III: Bochner identity

Let $M$ be a Riemannian manifold, and let $Cl(M)$ be its Clifford bundle. Let $E \to M$ be any vector bundle with connection, and assume that $C^\infty(M, E)$ is a $Cl(M)$-module. We can define a Dirac operator $\mathcal{D}$ acting on sections of $E$ via the formula
$$\mathcal{D} \sigma = \sum_{i=1}^n e_i \cdot \nabla_i \sigma$$
for any orthonormal frame $\{e_1, \dots, e_n\}$ on $M$, and where $\cdot$ denotes the Clifford module action. We demand that the connection on $E$ is compatible with Clifford multiplication in the following sense:
$$\nabla_j (e_i \cdot \sigma) = (\nabla_j e_i) \cdot \sigma + e_i \cdot \nabla_j \sigma.$$

Let $R$ denote the curvature of $E$, i.e. we have
$$[\nabla_i, \nabla_j] \sigma =  R(e_i, e_j) \sigma+ \nabla_{[e_i, e_j]} \sigma$$

We can define an endomorphism $\mathcal{R}$ on $E$ by
$$\mathcal{R} = \frac{1}{2} \sum_{ij} R(e_i, e_j).$$

Theorem. We have the identity $\mathcal{D}^2 = -\Delta + \mathcal{R}$.

Proof. We compute
$$\begin{align} \mathcal{D}^2 \sigma &= \sum_{ij} e_i \nabla_i \left( e_j \nabla_j \sigma \right) \\ &= \sum_{ij} e_i e_j \nabla_i \nabla_j \sigma + e_i ( \nabla_i e_j ) \nabla_j \sigma \\ &= -\Delta \sigma + \frac{1}{2}\sum_{ij}e_i e_j [\nabla_i, \nabla_j] \sigma + \sum_{ij} e_i ( \nabla_i e_j) \nabla_j \sigma \\ &= -\Delta \sigma + \frac{1}{2}\sum_{ij}e_i e_j R(e_i, e_j) \sigma + \frac{1}{2}\sum_{ij} e_i e_j \nabla_{[e_i, e_j]} \sigma+ \sum_{ij} e_i ( \nabla_i e_j) \nabla_j \sigma \\ &= (-\Delta + \mathcal{R})\sigma + \frac{1}{2} \sum_{ij} \left( e_i e_j \nabla_{[e_i, e_j]}\sigma +  e_i (\nabla_i e_j) \nabla_j + e_j (\nabla_j e_i) \nabla_i \right)\sigma \end{align}$$
We will be done provided we can show that the last term vanishes. Notice that it is fully tensorial, since it can be expressed as $\mathcal{D}^2 + \Delta - \mathcal{R}$. On the other hand, the terms $[e_i, e_j]$ and $\nabla_j e_i$ are (by definition!) proportional to Christoffel symbols. Since we can always choose a frame so that these vanish at a point, these terms must vanish identically. Hence we have $0 = \mathcal{D}^2 + \Delta - \mathcal{R}$, as desired.

Clifford Algebras and Spinors, Part II: Spin Structures and Dirac Operators

A very good reference for today's material is Dan Freed's (unpublished) notes on Dirac operators, available here.

Spin(n)

Consider the Clifford algebra $Cl(\mathbb E^n)$ as constructed in yesterday's post. Define maps $t, \beta: Cl(\mathbb E^n) \to Cl(\mathbb E^n)$ via
$$(e_1 \cdots e_k)^t = e_k \cdots e_1, \ \beta(e_1 \dots e_k) = (-1)^k e_k \dots e_2 e_1$$
 There is a natural inclusion $\mathbb E^n \hookrightarrow Cl(\mathbb E^n)$. Given $x \in Cl(\mathbb E^n)$ and $v \in \mathbb E^n$, we can consider the product $x v x^t$. In general, this might not be contained in $\mathbb E^n \subset Cl(\mathbb E^n)$.

Definition. We define the group $Pin(n)$ to consist of all those $g \in Cl(\mathbb E^n)$ such that
$$g \beta(g) = 1, \ \ g v \beta(g) \subset \mathbb E^n \ \forall\ v \in \mathbb E^n.$$
Similarly, we define the group $Spin(n)$ to be the subgroup of $Pin(n)$ such that $gg^t = 1$.

Theorem. The natural action of $Pin(n)$ on $\mathbb E^n$ is by othogonal transformations, giving a natural map $Pin(n) \to O(n)$. This map is a double cover. Similarly, $Spin(n)$ is a double cover of $SO(n)$. If $n \geq 2$, $Spin(n)$ is simply connected.

The importance of the spin groups is due to the following basic fact. Suppose that $G$ is a Lie group with Lie algebra $\mathfrak{g}$. Any representation of $G$ induces a representation of $\mathfrak{g}$. However,  given a representation of $\mathfrak{g}$, it is not always possible to integrate it to a representation of $G$. But it is always possible to integrate a representation of $\mathfrak{g}$ to produce a representation of the universal cover of $G$. For $n \geq 2$, $Spin(n)$ is the universal cover of $SO(n)$.

Spin Structures

Let $(M^n, g)$ be a Riemannian manifold. Recall that the frame bundle $O(M)$ is the manifold consisting of pairs $(x, \mathbb{e})$ where $x \in M$ and $\mathbb{e} = \{e_1, \dots, e_n\}$ is an orthonormal frame in $T_x M$. Since the orthogonal group $O(n)$ acts on the set of orthonormal frames, this makes $F(M)$ into a principal $O(n)$ bundle over $M$. Let us assume that $M$ is oriented, so that we may reduce its structure group to $SO(n)$.

Suppose that $V$ is a representation of $SO(n)$. Then we may form the associated bundle $SO(M) \times_{SO(n)} V$, which is a vector bundle over $M$ with structure group $SO(n)$. If we take the defining representation then we obtain the tangent bundle, but of course there are many others. Unfortunately, since $SO(n)$ is not simply connected, not every representation of $\mathfrak{so}_n$ can be integrated to a representation of $SO(n)$. At the level of geometry, this means that in a certain sense there are certain vector bundles over $M$ that are "missing"! Even more disturbing, is that these "missing" bundles appear to be necessary to describe many of the fundamental particles that appear in the standard model--so this has real world consequences. The solution is to equip $M$ with a spin structure.

Definition. A spin structure on $M$ is a principal $Spin(n)$-bundle $Spin(M)$ over $M$ together with a bundle morphism $Spin(M) \to SO(M)$ which is a reduction of structure (i.e., satisfies the obvious axioms).

As you might expect, not every manifold admits a spin structure, and spin structures may not be unique. Loosely speaking, a spin structure is a slightly stronger notion of orientability. Spin structures may always be chosen locally, and the obstruction to consistent gluing is not too difficult to characters as a certain $\mathbb Z_2$ cohomology class, called the second Stiefel-Whitney class.

Spin Connection and Dirac Operators

The reduction of structure $Spin(M) \to SO(M)$ allows us to pull back the Levi-Civita connection on $SO(M)$ to obtain a connection on $Spin(M)$, called the spin connection. Let $S_0$ be the spinor module described in the previous post. Then we may construct the associated bundle
$$S = Spin(M) \times_{Spin(n)} S_0$$
which is called the spinor bundle. Moreover, since $S_0$ is a Clifford module, there is well-defined notion of Clifford multiplication on sections of $S$. We may then define the Dirac operator $\mathcal{D}$ by
$$\mathcal{D} = \sum_{a=1}^n c(e_a) \nabla_{e_a}$$
where $\{e_a\}$ is any orthonormal frame, $\nabla$ is the spin connection, and $c$ denotes Clifford multiplication.

Next time: the Weitzenböck formula, and maybe a vanishing theorem.

Clifford Algebras and Spinors

Clifford Algebras

Today I'd like to write some brief notes about Clifford algebras and spinors. A classic reference is the paper "Clifford Modules" by Atiyah-Bott-Shapiro. Clifford algebras not only useful in algebra and geometry, but are essential for the construction of theories with fermions. Let $V$ be a vector space with a non-degenerate symmetric bilinear form $B$. We define the Clifford algebra $Cl(V, B)$ to be the unital associative algebra generated by $v \in V$ subject to the relation
$$vw + wv = -2B(v,w)$$
Equivalently, the definiting relation is $v^2 = -B(v,v)$.

The Clifford algebra inherits a $\mathbb Z$-filtration as well as a $\mathbb Z_2$-grading from the tensor algebra. In fact, we have an analogue of the Poincare-Birkhoff-Witt theorem for Lie algebras:

Theorem The associated graded algebra of $Cl(V,B)$ is naturally isomorphic to the exterior algebra on $V$.

In this way, we may view the Clifford algebra as a quantization of the exterior algebra, much in the same way that $U(\mathfrak g)$ is a quantization of the Poisson algebra of functions on $\mathfrak g^\ast$ for a Lie algebra $\mathfrak g$.

Example. Take (V,B) to be the Euclidean space $\mathbb E^1$. Then we have a single generator $e$ satisfying the relation $e^2 =  -1$. Hence
$$Cl(\mathbb R) \cong \mathbb R \cdot 1 \oplus \mathbb R \cdot e \cong \mathbb C$$
Where the isomorpism is given by $e \mapsto i = \sqrt{-1}$.

Example. Take $\mathbb E^2$. We have generators $e_1, e_2$ both squaring to -1, and additionally we have $e_1 e_2 = e_2 e_1$. We can define an isomorphism from $Cl(\mathbb E^2)$ to the quaternions $\mathbb H$ by $e_1 \mapsto i, e_2 \mapsto j$.

Spinors

Now consider the complexified Clifford algebra, denoted $\mathbb{C}l(V)$. Since we can now take square roots of negative numbers, the complex Clifford algebra is insensitive to the signature (as long as our bilinear form is non-degenerate). Denote by $C_n$ the complex Clifford algebra $Cl(\mathbb C^n)$,
where $\mathbb C^n$ is equipped with the standard bilinear form $(x,y) = \sum_{i=1}^n x_i y_i$.

Definition. A subspace $W \subset \mathbb C^n$ is isotropic if the restriction of the standard bilinear form to $W$ is identically 0. A maximal isotropic subspace is an isotropic subspace that is not properly contained in any other isotropic subspace.

Theorem. Let $W$ be a maximal isotropic subspace, and let$\{w_1, \dots, w_k\}$ be a basis of $W$. Let $\omega = w_1 \cdots w_k \in C_n$, and let $S = C_n \cdot \omega$. If n is even, then $S$ is an irreducible Clifford module. If n is odd, then $S=S^+ \oplus S^-$ is a direct sum irreducible Clifford modules, and $S^+ \cong S^-$.

Irreducible Clifford modules are called spinor modules. This description of spinor modules allows one to prove straightforwardly the following complete classification of complex Clifford algebras.

Corollary. We have $C_{2m} \cong \mathrm{End}(\mathbb C^m)$ and $C_{2m+1} \cong \mathrm{End}(\mathbb C^m) \oplus \mathrm{End}(\mathbb C^m)$.

Note that this classification depends on n mod 2, which is closely related to Bott periodicity. There is a similar classification of real Clifford algebras.

Dirac Operators

Now we come to the real importance of Clifford algebras. Consider Euclidean space $\mathbb{E}^n$ and let $S$ be a spinor module for its Clifford algebra. We define the Dirac operator acting on $S$-valued functions as
$$D f = \sum_{i=1}^n e_i \cdot \partial_i f$$
Now the amazing property of $D$ is the following:
$$D^2 = \sum_{i,j} e_i e_j \partial_i \partial_j = \sum_i e_i^2 \partial_i^2 + \sum_{i,j} e_i e_j [\partial_i, \partial_j] = -\Delta$$
hence the Dirac operator provides an algebraic (as opposed to pseudodifferential) square root of the Laplacian.

To Be Added in an Update...

Supersymmetric point particle, Dirac operators on spin manifolds, Weitzenböck formula, spinor reps of Lorentz algebra, N=1 susy.

Virasoro Algebra

Conformal Invariance in 2D

To begin, recall that in two dimensions, the conformal transformations are generated by holomorphic and anti-holomorphic transformations. At the infinitesimal level, let $\ell_n := -z^{n+1} \partial_z$ be a basis of holomorphic vector fields. These satisfy the Witt algebra
$$[\ell_m, \ell_n] = (m-n)\ell_{m+n}.$$
Similarly, we can define $\bar{\ell}_m = -\bar{z}^{n+1} \partial_{\bar{z}}$, and in addition to the Witt algebra these new generators satisfy $[\bar{\ell}_m, \ell_n]=0$.

Now, we could try to define a 2D conformal quantum field theory to be a unitary representation of the Witt algebra (or rather, of two copies of the Witt algebra, since we have both holomorphic and anti-holomorphic vector fields--but nevermind that). But this is too naive.

Central Extensions

Recall that in quantum mechanics, states are represented by vectors in some Hilbert space $\mathcal{H}$. However, the state $|\phi\rangle$ and $\alpha|\phi\rangle$ are physically equivalent for any non-zero complex number $\alpha$. The reason, of course, is that the expectation value of an operator $\mathcal{O}$ is defined to be $\langle \phi|\mathcal{O}|\phi\rangle / \langle \phi|\phi\rangle$, and such expressions are invariant under rescaling in $\mathcal{H}$.

Thus,  a symmetry group $G$ for a theory does not necessarily act via a map $G \to U(\mathcal{H})$. It suffices to have a projective representation $G \to PU(\mathcal{H})$. Let $\mathfrak{g}, \mathfrak{pu}$ be the Lie algebras of $G$ and $PU$, respectively. A projective representation gives a map
$$\mathfrak{g} \to \mathfrak{pu}.$$
Since $PU$ is a quotient of $U$, we have a short exact sequence
$$0 \to \mathbb{C} \to \mathfrak{u} \to \mathfrak{pu} \to 0.$$
Now let $\hat{\mathfrak{g}}$ be defined as
$$\hat{\mathfrak{g}} = \{ (\xi, \eta) \in \mathfrak{u}\oplus\mathfrak{g} \ | \ \pi(\xi) = \rho(\eta) \}$$
This comes with a natural projection $\hat{\mathfrak{g}} \to \mathfrak{g}$. If we suppose that the projective representation $\rho$ is faithful, then the kernel of this map is exactly $\mathbb{C}$. Hence, a faithful projective representation of $\mathfrak{g}$ yields a short exact sequence of Lie algebras
$$0 \to \mathbb{C} \to \hat{\mathfrak{g}} \to \mathfrak{g} \to 0.$$
We have obtained a central extension of $\mathfrak{g}$.

Virasoro Algebra

Finally, we can define the Virasoro algebra. It has generators $L_n$ and $c$, with defining relations
$$[L_m, L_n] = (m-n) L_{m+n} + \frac{c}{12}(m^3-m) \delta_{m+n,0}, [c, L_n] = 0.$$
The generator $c$ acts as a scalar in any irreducible representation, and its value is called the central charge. The factor of $1/12$ is entirely conventional. Now, the amazing fact is the following.

Theorem. Up to equivalence, the Virasoro algebra is the unique non-trivial central extension of the Witt algebra.

Proof sketch. This is essentially just a calculation. Any central extension has to be of the form
$$[L_m, L_n] = (m-n) L_{m+n} + A(m,n) c$$
for some function $A(m,n)$. If we make the replacement $L_m \mapsto L_m + a_m c$, then we have
$$[L_m, L_n] = (m-n) L_{m+n} + \left( A(m,n) + (m-n) a_{m+n} \right) c$$
Taking $n = 0$, we have
$$[L_m, L_0] = m L_{m} + \left( A(m,0) + m a_{m} \right) c$$
Hence for $m\neq0$ we can take $a_m = m^{-1} A(m,0)$. Having done this, we are now free to assume that $A(m,0) = 0$ for all $m$. Then we may apply the Jacobi identity to deduce that $A(m,n)=0$ except possibly for $m=-n$, so that $A(m,n)$ can be written in the form $A(m,n) = A_m \delta_{m+n, 0}$. Finally, another application of the Jacobi identity yields a simple recurrence relation for the coefficients $A_m$, and it is easily seen that every solution of this recurrence is proportional to $m^3-m$.

Now we can take our (preliminary, and still too naive) definition of a quantum conformal field theory to be a unitary representation of the Virasoro algebra.

Stress-Energy Tensor and OPE

The operator $L_0$ behaves like the Hamiltonian of the theory, and the Virasoro relations show that $L_n$ for $n>0$ act as lowering operators. Hence, in a physically sensible representation, the vacuum vector $|\Omega\rangle$ will be annihilated by $L_n$ for all $n > 0$. Unitary requires $L_n^\dagger = L_{-n}$, so additionally we have $\langle \Omega|L_n = 0$ for $n < 0$. Hence

$$\langle \Omega | L_m L_n | \Omega \rangle = 0 \ \textrm{unless}\ n \leq 0, m \geq 0$$

Now define the stress-energy tensor to be the operator-valued formal power series

$$T(z) = \sum_n \frac{L_n}{z^{n+2}}$$
We can consider the vacuum expectation of the product $T(z) T(w)$. By the above remarks, many terms in the expansion will vanish. In fact, it is a straightforward (but tedious!) exercise to check the following.

Theorem. The stress-energy tensor satisfies the operator product expansion
$$T(z) T(w) \sim \frac{c/2}{(z-w)^4} + \frac{2 T(w)}{(z-w)^2} + \frac{\partial_w T(w)}{z-w}$$
where $\sim$ denotes that the left- and right-hand sides are equal up to the addition of terms with vanishing vev and/or regular as $z \to w$.