Not Even Wrong

In this posting I’ll work out some examples of Lie algebra cohomology, still for finite dimensional Lie algebras and representations.

If [tex]G[/tex] is a compact, connected Lie group, it can be thought of as a compact manifold, and as such one can define its de Rham cohomology [tex]H^*_{deRham}(G)[/tex] as the cohomology of the complex

[tex]0\longrightarrow \Omega^0(G)\stackrel{d}\longrightarrow \Omega^1(G)\stackrel{d}\longrightarrow\cdots\stackrel{d}\longrightarrow\Omega^{dim\ G}(G)\longrightarrow 0[/tex]

where [tex]\Omega^i(G)[/tex] are the differential i-forms on [tex]G[/tex] (note, we’ll use complex-valued forms), and [tex]d[/tex] is the deRham differential.

For a compact group, one has a bi-invariant Haar measure [tex]\int_G[/tex], and can use this to “average” over an action of the group on a space. For a representation [tex](\pi, V)[/tex], we get a projection operator [tex]\int_g \Pi (g)[/tex] onto the invariant subspace [tex]V^G[/tex]. This projection operator gives explicitly the invariants functor on [tex]\mathcal C_{\mathfrak g}[/tex]. It is an exact functor, taking exact sequences to exact sequences.

The differential forms [tex]\Omega^*(G)[/tex] give a representation of [tex]G[/tex] in two ways, taking the induced action on forms by pullback, using either left or right translation on the group. If [tex](\Pi(g), \Omega^*(G))[/tex] is the representation by left translations, we can use this to apply our “averaging over [tex]G[/tex]” projection operator to the de Rham complex. This action commutes with the de Rham differential, so we get a sub-complex of left-invariant forms

[tex]0\longrightarrow \Omega^0(G)^G\stackrel{d}\longrightarrow \Omega^1(G)^G\stackrel{d}\longrightarrow\cdots\stackrel{d}\longrightarrow\Omega^{dim\ G}(G)^G\longrightarrow 0[/tex]

Since elements of the Lie algebra [tex]\mathfrak g[/tex] are precisely left-invariant 1-forms, it turns out that this complex is nothing but the Chevalley-Eilenberg complex considered last time to represent Lie algebra cohomology, for the case of the trivial representation. This means we have [tex]C^*(\mathfrak g, \mathbf R)= \Lambda^*(\mathfrak g^*)=\Omega^*(G)^G[/tex], and the differentials coincide. So, what we have shown is that

[tex]H^*(\mathfrak g, \mathbf C)= H^*_{de Rham}(G)[/tex]

If one knows the cohomology of [tex]G[/tex], the Lie algebra cohomology is thus known, but this identity is normally used in the other direction, to find the cohomology of [tex]G[/tex] from that of the Lie algebra. To compute the Lie-algebra cohomology, we can exploit the right-action of G on the group, averaging over the induced action on the left-invariant forms [tex]\Lambda^*(\mathfrak g)[/tex], which again commutes with the differential. We end up with a complex
[tex]0\longrightarrow (\Lambda^0(\mathfrak g^*))^G \longrightarrow (\Lambda^1(\mathfrak g^*))^G\longrightarrow\cdots\longrightarrow (\Lambda^{\dim\ \mathfrak g}(\mathfrak g^*))^G\longrightarrow 0[/tex]

where all the differentials are zero, so the cohomology is given by

[tex]H^*(\mathfrak g,\mathbf C)=(\Lambda^*(\mathfrak g^*))^G=(\Lambda^*(\mathfrak g^*))^{\mathfrak g}[/tex]

the adjoint-invariant pieces of the exterior algebra on [tex]\mathfrak g^*[/tex]. Finding the cohomology has now been turned into a purely algebraic problem in invariant theory. For [tex]G=U(1)[/tex], [tex]\mathfrak g=\mathbf R[/tex], and we have shown that [tex]H^*(\mathbf R, \mathbf C)=\Lambda^*(\mathbf C)[/tex], this is [tex]\mathbf C[/tex] in degrees 0, and 1, as expected for the de Rham cohomology of the circle [tex]U(1)=S^1[/tex]. For [tex]G=U(1)^n[/tex], we get

[tex]H^*(\mathbf R^n, \mathbf C)=\Lambda^*(\mathbf C^n)[/tex]

Note that complexifying the Lie algebra and working with [tex]\mathfrak g_{\mathbf C}=\mathfrak g\otimes \mathbf C[/tex] commutes with taking cohomology, so we get

[tex]H^*(\mathfrak g_{\mathbf C},\mathbf C)= H^*(\mathfrak g,\mathbf C)\otimes \mathbf C[/tex]

Complexifying the Lie algebra of a compact semi-simple Lie group gives a complex semi-simple Lie algebra, and we have now computed the cohomology of these as

[tex]H^*(\mathfrak g_{\mathbf C}, \mathbf C) = (\Lambda^*(\mathfrak g_{\mathbf C}))^{\mathfrak g_\mathbf C}[/tex]

Besides [tex]H^0[/tex], one always gets a non-trivial [tex]H^3[/tex], since one can use the Killing form [tex]< \cdot,\cdot>[/tex] to produce an adjoint-invariant 3-form [tex]\omega_3(X_1,X_2,X_3)=[/tex]. For [tex]G=SU(n)[/tex], [tex]\mathfrak g_{\mathbf C}=\mathfrak{sl}(n,\mathbf C})[/tex], and one gets non-trivial cohomology classes [tex]\omega_{2i+1}[/tex] for [tex]i=1,2,\cdots n[/tex], such that

[tex]H^*(\mathfrak{sl}(n,\mathbf C))=\Lambda^*(\omega_3, \omega_5,\cdots,\omega_{2n+1})[/tex]

the exterior algebra generated by the [tex]\omega_{2i+1}[/tex].

To compute Lie algebra cohomology [tex]H^*(\mathfrak g, V)[/tex] with coefficients in a representation [tex]V[/tex], we can go through the same procedure as above, starting with differential forms on [tex]G[/tex] taking values in [tex]V[/tex], or we can just use exactness of the averaging functor that takes [tex]V[/tex] to [tex]V^G[/tex]. Either way, we end up with the result

[tex]H^*(\mathfrak g, V)=H^*(\mathfrak g, \mathbf C)\otimes V^{\mathfrak g}[/tex]

The [tex]H^0[/tex] piece of this is just the [tex]V^{\mathfrak g}[/tex] that we want when we are doing BRST, but we also get quite a bit else: [tex]dim\ V^{\mathfrak g}[/tex] copies of the higher degree pieces of the Lie algebra cohomology [tex]H^*(\mathfrak g, \mathbf C)[/tex]. The Lie algebra cohomology here is quite non-trivial, but doesn’t interact in a non-trivial way with the process of identifying the invariants [tex]V^{\mathfrak g}[/tex] in [tex]V[/tex].

In the next posting I’ll turn to an example where Lie algebra cohomology interacts in a much more interesting way with the representation theory, this will be the highest-weight theory of representations, in a cohomological interpretation first studied by Bott and Kostant.

The Invariants Functor

The last posting discussed one of the simplest incarnations of BRST cohomology, in a formalism familiar to physicists. This fits into a much more abstract mathematical context, and that’s what we’ll turn to now.

Given a Lie algebra $\mathfrak g$, we’ll consider Lie algebra representations as modules over $U(\mathfrak g)$. Such modules form a category $\mathcal C_{\mathfrak g}$: what is interesting is not just the objects of the category (the equivalence classes of modules), but also the morphisms between the objects. For two representations $V_1$ and $V_2$ the set of morphisms between them is a linear space denoted $Hom_{U(\mathfrak g)}(V_1,V_2)$. This is just the set of linear maps from $V_1$ to $V_2$ that commute with the action of $\mathfrak g$:

$$Hom_{U(\mathfrak g)}(V_1,V_2)=\{\phi\in Hom_{\mathbf C}(V_1,V_2): \pi(X)\phi=\phi\pi(X)\ \forall X\in \mathfrak g\}$$

Another conventional name for this is the space of intertwining operators between the two representations.

For any representation $V$, its $\mathfrak g$-invariant subspace $V^{\mathfrak g}$ can be identified with the space $Hom_{U(\mathfrak g)}(\mathbf C, V)$, where here $\mathbf C$ is the trivial one-dimensional representation. Having a way to pick out the invariant piece of a representation also allows one to solve the more general problem of picking out the subspace that transforms like a specific irreducible $W$: just find the invariant subspace of $V\otimes W^*$.

The map $V\rightarrow V^{\mathfrak g}$ that takes a representation to its $\mathfrak g$-invariant subspace is a functor: it takes the category $\mathcal C_{\mathfrak g}$ to $\mathcal C_{\mathbf C}$, the category of vector spaces and linear maps ($\mathbf C$ – modules and $\mathbf C$ – homomorphisms). If, instead of taking

$$V\rightarrow V^{\mathfrak g}$$

one takes

$$V\rightarrow V^{\mathfrak h}$$

where [tex]\mathfrak h[/tex] is a Lie subalgebra of [tex]\mathfrak g[/tex], one again gets a functor. If [tex]\mathfrak h[/tex] is an ideal in [tex]\mathfrak g[/tex] (so that [tex]\mathfrak g/\mathfrak h[/tex] is a Lie algebra), then this functor takes [tex]\mathcal C_{\mathfrak g}[/tex] to [tex]\mathcal C_{\mathfrak g/\mathfrak h}[/tex]. This is a simple version of the situation of interest in the case of gauge theory: if [tex]V[/tex] is a state space with [tex]\mathfrak h[/tex] acting as a gauge symmetry, then [tex]V^{\mathfrak h}[/tex] will be the physical subspace, carrying an action of the algebra of operators [tex]U(\mathfrak g/\mathfrak h)[/tex].

Some Homological Algebra

It turns out that when one has a category of modules like [tex]\mathcal C_{\mathfrak g}[/tex], these can usefully be studied by considering complexes of modules, and this is the subject of homological algebra. A complex of modules is a sequence of modules and homomorphisms

$$\cdots\stackrel{\partial}\longrightarrow U\stackrel{\partial}\longrightarrow V \stackrel{\partial}\longrightarrow W\stackrel{\partial}\longrightarrow\cdots$$

such that [tex]\partial\circ\partial =0[/tex]. If the complex satisfies [tex]im\ \partial=ker\ \partial[/tex] at each module, the complex is said to be an “exact complex”.

To motivate the notion of exact complex, note that

$$0\longrightarrow V_0\longrightarrow V \longrightarrow 0$$

is exact iff [tex]V_0[/tex] is isomorphic to [tex]V[/tex], and an exact sequence

$$0\longrightarrow V_1 \longrightarrow V_0 \longrightarrow V \longrightarrow 0$$

represents the module [tex]V[/tex] as the quotient [tex]V_0/V_1[/tex]. Using longer complexes, one gets the notion of a resolution of a module [tex]V[/tex] by a sequence of n modules [tex]V_i[/tex]. This is an exact complex

$$0\longrightarrow V_n \longrightarrow\cdots\longrightarrow V_1 \longrightarrow V_0\longrightarrow V\longrightarrow 0$$

The deviation of a sequence from being exact is measured by its homology $H^*=\frac{ker\ \partial}{im\ \partial}$. Note that if one deletes [tex]V[/tex] from its resolution, the sequence

$$0\longrightarrow V_n \longrightarrow\cdots\longrightarrow V_1 \longrightarrow V_0\longrightarrow 0$$

is exact except at [tex]V_0[/tex]. Indexing the homology in the obvious way, one has [tex]H^i =0[/tex] for [tex]i>0[/tex], and [tex]H^0=V[/tex]. A sequence like this whose only homology is [tex]V[/tex] at [tex]H^0[/tex] is another manifestation of a resolution of [tex]V[/tex].

The reason this construction is useful is that, for many purposes, it allows us to replace a module whose structure we may not understand by a sequence of modules whose structure we do understand. In particular, we can replace a [tex]U(\mathfrak g)[/tex] module [tex]V[/tex] by a sequence of free modules, i.e. modules that are just sums of copies of [tex]U(\mathfrak g)[/tex] itself. This is called a free resolution, and more generally one can work with projective modules (direct summands of free modules).

A functor that takes exact complexes to exact complexes is called an exact functor. Homological invariants of modules come about in cases where one has a functor on a category of modules that is not exact. Applying such a functor to a free or projective resolution gives the homological invariants.

The Koszul Resolution and Lie Algebra Cohomology

There are many possible choices of a free resolution of a module. For the case of [tex]U(\mathfrak g)[/tex] modules, one convenient choice is known as the Koszul (or Chevalley-Eilenberg) resolution. To construct a resolution of the trivial module [tex]\mathbf C[/tex], one uses the exterior algebra on [tex]\mathfrak g[/tex] to make free modules

$$Y_k=U(\mathfrak g)\otimes_{\mathbf C}\Lambda^k(\mathfrak g)$$

and get a resolution of [tex]\mathbf C[/tex]

$$0\longrightarrow Y_{dim\ \mathfrak g}\stackrel{\partial_{dim\ \mathfrak g -1}}\longrightarrow\cdots\stackrel{\partial_1}\longrightarrow Y_1\stackrel{\partial_0}\longrightarrow Y_0\stackrel{\epsilon}\longrightarrow \mathbf C\longrightarrow 0$$

The maps are given by
$$\epsilon : u\in Y_0=U(\mathfrak g) \rightarrow \epsilon (u) = const.\ term\ of\ u$$

and
$$\partial_{k-1} (u\otimes X_1\wedge\cdots\wedge X_k)=$$
$$\sum_{i=1}^k(-1)^{i+1}(uX_i\otimes X_1\wedge\cdots\wedge\hat X_i\wedge\cdots \wedge X_k)$$
$$+\sum_{i<j} (-1)^{i+j}(u\otimes[X_i,X_j]\wedge X_1\wedge\cdots\wedge \hat X_i\wedge\cdots\wedge \hat X_j\wedge\cdots\wedge X_k)$$

To get Lie algebra cohomology, we apply the invariants functor

$$V\longrightarrow V^{\mathfrak g}=Hom_{U(\mathfrak g)}(\mathbf C, V)$$

replacing the trivial representation by its Koszul resolution. This gives us a complex with terms

$$C^k(\mathfrak g, V)=Hom_{U(\mathfrak g)}(Y_k,V)= Hom_{U(\mathfrak g)}(U(\mathfrak g)\otimes \Lambda^k(\mathfrak g),V)$$
$$=Hom_{U(\mathfrak g)}(U(\mathfrak g),Hom_{\mathbf C}(\Lambda^k(\mathfrak g),V))$$
$$=Hom_{\mathbf C}(\Lambda^k(\mathfrak g),V) =V\otimes\Lambda^k(\mathfrak g^*)$$

and induced maps $d_i$

$$0\longrightarrow C^0(\mathfrak g, V)\stackrel{d_0}\longrightarrow C^1(\mathfrak g, V)\cdots\stackrel{d_{dim\ \mathfrak g -1}}\longrightarrow C^{dim\ \mathfrak g}(\mathfrak g, V)\longrightarrow 0$$

The Lie algebra cohomology $H^*(\mathfrak g, V)$ is just the cohomology of this complex, i.e.

$$H^i(\mathfrak g, V)=\frac{ker\ d_i}{im\ d_{i-1}}|_{C^i(\mathfrak g, V)}$$

This is exactly the same definition as that of the BRST cohomology defined in physicist’s formalism in the last posting with $\mathcal H =C^*(\mathfrak g, V)$.

One has $H^0(\mathfrak g, V)=V^{\mathfrak g}$ and so gets the $\mathfrak g$-invariants as expected, but in general the cohomology will be non-zero also in other degrees.

This is all rather abstract, so in the next posting some examples will be worked out, as well as the relationship of all this to the de Rham cohomology of the group. Anthony Knapp’s book Lie Groups, Lie Algebras, and Cohomology is an excellent reference for details on Lie algebra cohomology.

My initial plan was to have the second part of these notes be about gauge symmetry and the problems physicists have encountered in handling it, but as I started writing it quickly became apparent that explaining this in any detail would take me into various issues that are quite interesting, but far afield from what I want to get to. So, I hope to get back to this at some point, but for now will just assume that most of my readers know what gauge symmetry is, and that the rest just need to know that:

The actual situation is quite a bit more complicated than this, but for now we’ll focus on the simplest version of the mathematical problem that comes up here, and see how the BRST formalism deals with it. This posting will begin explaining one part of this story, starting with the simplest version of BRST cohomology, in a language familiar to physicists. The next posting will deal with Lie algebra cohomology in a more general mathematical context and work out some examples. For more about the material in this posting, see, for instance, Green, Schwarz and Witten, volume I, section 3.2.1, where they go on to apply this to the Virasoro algebra, or these lecture notes from Jose O’Figueroa-Farrill .

Physicists always begin by choosing a basis, in this case a basis [tex]X_i[/tex] of [tex]\mathfrak g[/tex] satisfying [tex][X_i,X_j]=f_{ij}^kX_k[/tex], where [tex]f_{ij}^k[/tex] are called the structure constants of [tex]\mathfrak g[/tex]. A representation [tex](\pi,V)[/tex] is then a set of linear operators [tex]K_i=\pi (X_i)[/tex] on [tex]V[/tex] satisfying [tex][K_i,K_j]=f_{ij}^kK_k[/tex]. Let [tex]\alpha^i[/tex] be a basis of the dual space [tex]\mathfrak g^*[/tex], dual to the basis [tex]X_i[/tex].

Now, extend [tex]V[/tex] to [tex]\mathcal =V\otimes \Lambda^* (\mathfrak g^*)[/tex], where [tex]\Lambda^* (\mathfrak g^*)[/tex] is the exterior algebra on [tex]\mathfrak g^*[/tex]. On this space, define the “ghost” operator [tex]c^i[/tex] to be wedge-product with [tex]\alpha^i[/tex], and “anti-ghost” operator [tex]b_i[/tex] to be contraction (interior product) with [tex]X_i[/tex]. These operators satisfy “fermionic” anti-commutation relations

[tex]\{c^i,c^j\}=\{b_i,b_j\}=0,\ \ \{c^i,b_j\}=\delta^i_j[/tex]

and one can get all vectors in [tex]\mathcal H[/tex] from linear combinations of decomposable elements of [tex]\mathcal H[/tex] (those given by repeated application of the [tex]c^i[/tex] to the “vacuum vector” [tex]V\otimes \mathbf 1[/tex]).

The ghost number operator [tex]N=c^ib_i[/tex] on [tex]\mathcal H[/tex] has eigenvectors the decomposable elements, with integer eigenvalues from 0 to dim [tex]\mathfrak g[/tex], given by the number of ghost operators needed to produce the eigenvector from a vacuum vector.

The BRST operator is given by

[tex]Q=c^iK_i -\frac{1}{2}f_{ij}^kc^ic^jb_k[/tex]

which increases the ghost number by one, and has the crucial property of [tex]Q^2=0[/tex] (this comes from the fact that the [tex]f_{ij}^k[/tex] satisfy the Jacobi identity). The BRST cohomology is given by considering the space [tex]ker\ Q[/tex] of elements [tex]\chi[/tex] of [tex]\mathcal H[/tex] that are “BRST-closed”, i.e. satisfy [tex]Q\chi=0[/tex], and identifying two such elements if they are “BRST-exact”, i.e. differ by [tex]Q\lambda[/tex] for some [tex]\lambda[/tex]. So BRST cohomology is defined by

[tex]H^*_Q(V)=\frac{ker\ Q}{im\ Q}|_{V\otimes\Lambda^*(\mathfrak g^*)[/tex]

with [tex]H^j_Q(V)[/tex] the component of the BRST cohomology of ghost number [tex]j[/tex].

A vector [tex]\chi=v\otimes\mathbf 1[/tex] of ghost number zero satisfies [tex]Q\chi =0[/tex] iff and only if [tex]K_iv=0[/tex] for all i, so we can identify [tex]H^0_Q(V)[/tex] with the space [tex]V^\mathfrak g[/tex] of [tex]\mathfrak g[/tex] – invariant vectors in [tex]V[/tex].

The essence of the BRST method is to replace the problem of finding the invariant subspace [tex]V^{\mathfrak g}[/tex] of a representation [tex]V[/tex] by the problem of finding the degree zero BRST cohomology [tex]H^0_Q(V)[/tex].

There are two different ways of putting an inner product on [tex]\Lambda^*(\mathfrak g*)[/tex] and thus getting an inner product on [tex]\mathcal H[/tex] ([tex](\pi,V)[/tex] is assumed to be unitary, so preserves a given inner product on [tex]V[/tex]).

This is the first posting of a planned series that will discuss the BRST method for handling gauge symmetries and related mathematical topics. I’ve been writing a more formal paper about this, but given the substantial amount of not-well-known background material involved, it seems like a good idea to first put together a few expository accounts of some of these topics. And what better place for this than a blog?

Many readers who are used to my usual attempts to be newsworthy and entertaining, often by scandal-mongering or stirring up trouble of one kind or another, may be very disappointed in these posts. They’re quite technical, hard to follow, and of low-to-negative entertainment value. You probably would do best to skip them and wait for more of the usual fare, which should continue to appear from time to time.

Quantum Mechanics and Representation Theory

A quantum mechanical physical system is given by the following mathematical structure:

If a physical system has a symmetry group [tex]G[/tex], there is a unitary representation [tex](\Pi, \mathcal H)[/tex] of [tex]G[/tex] on [tex]\mathcal H[/tex]. This means that for each [tex]g\in G[/tex] we get a unitary operator [tex]\Pi(g)[/tex] satisfying

[tex]\begin{displaymath}\Pi(g_3)=\Pi(g_2)\Pi(g_1)\ \text{if}\ g_3=g_1g_2\end{displaymath}[/tex]

i.e. the map [tex]\Pi[/tex] from group elements to unitary operators is a homomorphism. The [tex]\Pi(g)[/tex] act on [tex]\mathcal O[/tex] by taking an operator [tex]O[/tex] to its conjugate [tex]\Pi(g)O(\Pi(g))^{-1}[/tex].

When [tex]G[/tex] is a Lie group with Lie algebra [tex](\mathfrak g, [\cdot,\cdot])[/tex], differentiating [tex]\Pi[/tex] gives a unitary representation [tex](\pi, \mathcal H)[/tex] of [tex]\mathfrak g[/tex] on [tex]\mathcal H[/tex]. This means that for each [tex]X\in \mathfrak g[/tex] we get a skew-Hermitian operator [tex]\pi(X)[/tex] on [tex]\mathcal H[/tex], satisfying

[tex]\pi(X_3)=[\pi(X_1),\pi(X_2)]\ \text{if}\ X_3=[X_1,X_2][/tex]

i.e. the map [tex]\pi[/tex] taking Lie algebra elements [tex]X[/tex] (with the Lie bracket in [tex]\mathfrak g[/tex]) to skew-Hermitian operators (with commutator of operators) is a homomorphism. On [tex]\mathcal O[/tex], [tex]\mathfrak g[/tex] acts by the differential of the conjugation action of [tex]G[/tex], this action is just that of taking the commutator with [tex]\pi(X)[/tex].

The Lie bracket is not associative, but to any Lie algebra [tex]\mathfrak g[/tex], one can construct an associative algebra [tex]U(\mathfrak g)[/tex] called the universal enveloping algebra for [tex]\mathfrak g[/tex]. If one identifies [tex]X\in \mathfrak g[/tex] with left-invariant vector fields on [tex]G[/tex], which are first-order differential operators on functions on [tex]G[/tex], then [tex]U(\mathfrak g)[/tex] is the algebra of left-invariant differential operators on [tex]G[/tex] of all orders, with product the composition of differential operators. A Lie algebra representation is precisely a module over [tex]U(\mathfrak g)[/tex], i.e. a vector space with an action of [tex]U(\mathfrak g)[/tex].

So, the state space [tex]\mathcal H[/tex] of a quantum system with symmetry group [tex]G[/tex] carries not only a unitary representation of [tex]G[/tex], but also a unitary representation of [tex]\mathfrak g[/tex], or equivalently, an action of the algebra [tex]U(\mathfrak g)[/tex]. [tex]X\in \mathfrak g[/tex] acts by the operator [tex]\pi(X)[/tex]. In this way a representation [tex]\pi[/tex] gives a sub-algebra of the algebra [tex]\mathcal O[/tex] of observables. Most of the important observables that show up in practice come from a symmetry in this way. An interesting philosophical question is whether the quantum system that governs the real world is purely determined by symmetry, i.e. such that ALL its observables come from symmetries in this manner.

Some Examples

Much of the structure of common quantum mechanical systems is governed by the fact that they carry space-time symmetries. In our 3-space, 1-time dimensional world, these include:

Another example is the symmetry of phase transformations of the state space [tex]\mathcal H[/tex]. Here [tex]G=U(1), \mathfrak g=R[/tex], and one gets an operator [tex]Q_e[/tex] that can be normalized to have integral eigenvalues.

This last example also comes in a local version, where we make independent phase transformations at different points in space-time. This is an example of a “gauge symmetry”, and the question of how it gets represented on the space of states is what will lead us into the BRST story. Next posting in the series will be about gauge symmetry, then on to BRST.

If you want some idea of where this is headed, you can take a look at slides from a colloquium talk I gave recently at the Dartmouth math department. They’re very sketchy, the postings in this series should add some detail.