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A very good reference for today's material is Dan Freed's (unpublished) notes on Dirac operators, available here.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

Geometry of Curved Spacetime 5: Bianchi Identity and Einstein Equations

Background

Following last time, we are almost ready to write down the Einstein equations. Before doing any math, let's understand what we're trying to do. Minkowski realized that Einstein's special relativity was best understood by combining space and time into 4-dimensional spacetime, with Lorentzian metric
\[ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2. \]
The spacetime approach works wonderfully and even explains the Lorentz invariance of Maxwell's equations (indeed, it was Maxwell's equations that motivated Einstein to postulate his principle of relativity). However, (for reasons that I may discuss later) gravity is not a "force" but rather the geometry of spacetime itself.

By mass-energy equivalence (which is one of the most basic consequences of relativity), the gravitational field, whatever it is, must couple to the stress-energy tensor \(T_{ij}\). I won't get into details, but the stress-energy tensor is a familiar object from physics that roughly tells you what the energy-momentum density/flux is in each direction at every point in spacetime. If the matter is completely static, then it is ok to think of this as measuring the mass density, but for nonstatic matter it also takes things like pressure into account.

Now, as I said above, the gravitational field is just the geometry of spacetime, which is measured by the metric tensor \(g_{ij}\). Mass-energy equivalence says that it must couple to the stress-energy tensor \(T_{ij}\). The simplest field equation then would be
\[ G_{ij} = c T_{ij} \]
where \(G_{ij}\) is some tensor built out of \(g_{ij}\) and its derivatives, and \(c\) is some constant. The equations of Newtonian gravity are 2nd order in the gravitational field, so if we want these equations to reduce to Newton's in the appropriate limit, \(G_{ij}\) should only depend on the metric and its first two derivatives. Now there is an obvious 2nd rank tensor satisfying these constraints: \(R_{ij}\), the Ricci tensor. However, this turns out to be completely wrong (except in the vacuum).

Any reasonable matter will satisfy local energy-momentum conservation,
\[ \nabla_j T^{ij} = 0. \]
It turns out that the Ricci tensor does not satisfy this condition in general. So to look for the right tensor \(G_{ij}\), we turn to the Bianchi identity.

The Bianchi Identity

As discussed in the previous post, the curvature of a connection is the endomorphism-valued 2-form
\[ F = d\Omega - \Omega \wedge \Omega \]
where \(\Omega\) is the matrix of 1-forms telling us how to take the covariant derivative of a frame, i.e.
\[ \nabla_i e_j = \Omega_{ij} \otimes e_j. \]
Since a connection can be extended to all tensor powers in a natural way, we can consider the covariant derivative of the curvature \(F\) (thought of as a section of the appropriate bundle). Quick calcluation:
\begin{align}
\nabla F &= \nabla(d\Omega - \Omega \wedge \Omega) \\
&= d^2 \Omega - d\Omega \wedge \Omega + \Omega \wedge d\Omega \\
& \ \ + d\Omega \wedge \Omega - \Omega \wedge \Omega \wedge \Omega \\
& \ \ - \Omega \wedge d\Omega + \Omega \wedge \Omega \wedge \Omega \\
&= 0.
\end{align}
Thus the endomorphism valued 3-form \(\nabla F\) is identically 0. Writing \(F\) in components as \(R_{ijkl}\), this is equivalent to
\[ R_{ijkl|m} +  R_{ijlm|k} + R_{ijmk|l} = 0. \]
Now let's contract:
\begin{align}
0 &= g^{ik} g^{jl} R_{ijkl|m} + g^{ik} g^{jl} R_{ijlm|k} + g^{ik} g^{jl} R_{ijmk|l}\\
&= g^{ik} R_{ik|m} - g^{ik}R_{im|k} - g^{jl} R_{jm|l} \\
&= \nabla_m S - 2 \nabla^k R_{mk}
\end{align}
So we see that the tensor
\[ G_{ij} = R_{ij} - \frac{S}{2} g_{ij} \]
is divergence free. This yields the Einstein field equations:
\[ R_{ij} - \frac{S}{2} g_{ij} = c T_{ij}. \]
Actually, there is another obvious divergence free tensor: \(g_{ij}\) itself. So a more general form is
\[ G_{ij} + \Lambda g_{ij} = c T_{ij} \]
where \(\Lambda\) is a constant called the cosmological constant.

Geometry of Curved Spacetime 4

Today I had to try to explain connections and curvature in local frames (as opposed to coordinates), and I really feel that Wald's treatment of this is just awful (this is one of the few complaints I have with an otherwise classic textbook). It is particularly baffling since the treatment in Misner, Thorne, and Wheeler is just perfect. What follows is the modern math (as opposed to physics) point of view. This is more abstract than any introductory GR (or even Riemannian geometry) text I've seen, but in this case the abstraction absolutely clarifies and simplifies things.

Let \(M\) be a smooth manifold and suppose \(E\) is a smooth vector bundle over \(M\). A connection on \(E\) is a map \nabla taking sections of \(E\) to sections of \(T^\ast M \otimes E\), \(\mathbb{R}\)-linear and satisfying the Leibniz rule
\[ \nabla(f\sigma) = df \otimes \sigma + f \nabla \sigma. \]

Now consider the sheaf of \(E\)-valued \(p\)-forms on \(M\). Call it \(\Omega^p(E)\). Then we can extend the connection to a map
\[ \nabla: \Omega^p(E) \to \Omega^{p+1}(E) \]
via the Leibniz rule:
\[ \nabla(\eta \otimes \sigma) = d\eta \otimes \sigma + (-1)^p \eta \wedge \nabla \sigma. \]
Let us define the curvature \(F\) associated to a connection \(\nabla\) by the composition
\[ F = \nabla^2: \Omega^p(E) \to \Omega^{p+2}(E). \]

Claim \(F\) is \(C^\infty\)-linear, i.e. it is tensorial.

Proof
\begin{align}
\nabla(\nabla(f \sigma)) &= \nabla( df \otimes \sigma + f \nabla \sigma) \\\
&= d^2 f \otimes \sigma - df \wedge \nabla \sigma + df \wedge \nabla \sigma + f \nabla^2 \sigma \\\
&= f \nabla^2 \sigma.
\end{align}

So far we have not made any additional choices (beyond \(\nabla\)). In order to actually compute something locally, we have to make some choices. Let \(\hat{e}_a\) be a frame, i.e. a local basis of sections of \(E\). Then \(\nabla \hat{e}_a\) is an \(E\)-valued 1-form, hence it can be expressed as a sum
\[ \nabla \hat{e}_a = \sum_{b} \omega_a^b \otimes \hat{e}_b \]
where the coefficients \(\omega_a^b\) are 1-forms, often called the connection 1-forms. Let \(\Omega\) denote the matrix of 1-forms whose entries are exactly \(\omega_a^b\).

Claim Let \(\sigma = \sigma^a \hat{e}_a\). Then we have
\[ \nabla \sigma = d\sigma + \Omega \sigma. \]

Proof The coefficients \(\sigma^a\) are functions (i.e. scalars), so \(\nabla \sigma^a = d\sigma^a\). Using the Leibniz rule we have
\begin{align}
\nabla(\sigma^a \hat{e}_a) &= (\nabla \sigma^a) \hat{e}_a + \sigma^a \nabla \hat{e}_a \\\
&= d\sigma^a \hat{e}_a + \sigma^a \omega_a^b \hat{e}_b \\\
&= d\sigma^a \hat{e}_a + \omega_c^a \sigma^c \hat{e}_a \\\
&= (d\sigma + \Omega \sigma)^a \hat{e}_a.
\end{align}

Claim The curvature satisfies \(F = d\Omega - \Omega \wedge \Omega\).

Proof Just apply the above formula twice using Leibniz.

Connection 1-forms from Christoffel symbols. Suppose now that we are in the Riemannian setting and we already know the Christoffel symbols in some coordinates. Then we can express our frame \(\hat{e}_a\) in terms of coordinate vector fields, i.e.
\[ \hat{e}_a = \hat{e}_a^i \frac{\partial}{\partial x^i} \]
Then we have that
\[ \nabla_j \hat{e}_a^i = \frac{\partial \hat{e}_a^i}{\partial x^j} + \Gamma^i_{jk} \hat{e}_a^k \]
So, as a vector-valued 1-form, we have
\[ \nabla \hat{e} = \frac{\partial \hat{e}_a^i}{\partial x^j} dx^j \otimes \frac{\partial}{\partial x^i}
+ \Gamma^i_{jk} \hat{e}_a^k dx^j \otimes \frac{\partial}{\partial x^i}. \]
Juggling things a bit using the metric, we find
\[ \nabla \hat{e}_a = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j \otimes \hat{e}_b
 + \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j \otimes \hat{e}_b. \]
So the connection 1-forms are given by
\[ \omega_a^b = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j
 + \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j. \]

To come later (if I ever get around to it): some explicit computations.

Geometry of Curved Spacetime 3

Today, some numerology. The Riemann curvature tensor is a tensor \(R_{abcd}\) satisfying the identities:

1. \(R_{abcd} = -R_{bacd}.\)

2. \(R_{abcd} = R_{cdba}. \)

3. \(R_{abcd} + R_{acdb} + R_{adbc} = 0. \) (First Bianchi)

4. \(R_{abcd|e} + R_{acec|d} + R_{abde|c} = 0. \) (Second Bianchi)

By 1, the number of independent \(ab\) indices is \(N = n(n-1)/2\), and similarly for \(cd\). By 2, the number of independent pairs of indices is \(N(N+1)/2\). Now the cyclic constraint 3 can be written as
\[ R_{[abcd]} = 0, \]
and thus constitutes \({n \choose 4}\) equations. So the number of independent components is
\begin{align}
 N(N+1)/2 - {n \choose 4} &= \frac{n(n-1)((n(n-1)/2+1)}{4} - \frac{n(n-1)(n-2)(n-3)}{24} \\
&= \frac{(n^2-n)(n^2-n+2)}{8} - \frac{(n^2-n)(n^2-5n+6}{24} \\
&= \frac{n^4-2n^3+3n^2+2n}{8} - \frac{n^4-6n^3+11n^2-6n}{24} \\
&= \frac{2n^4-2n^2}{24} \\
&= \frac{n^4-n^2}{12} \\
&= \frac{n^2(n^2-1)}{12}
\end{align}

Now consider the Weyl tensor \(C_{abcd}\) which is defined as the completely trace-free part of the Rienmann tensor. The trace is determined by the Ricci tensor \(R_{ab}\) which as \(n(n+1)/2\) indepdendent components, so the Weyl tensor has
\[ \frac{n^2(n^2-1)}{12} - \frac{n^2-n}{2} = \frac{n^4-7n^2+6n}{12} \]
independent components. Now, for \(n = 1\) we see that \(R_{abcd}\) has no independent components, i.e. it vanishes identically. In \(n=2\), it has only 1 independent component, and so the scalar curvature determines everything. In \(n=3\), it has 6 independent components. Note that in this case, the Weyl tensor has no independent components, i.e. it is identically 0. So we see that in \(n = 2, 3\) every Riemannian manifold is conformally flat. So things only start to get really interesting in \(n=4\), where the Riemann tensor has 20 independent components, and the Weyl tensor has 10.

Geometry of Curved Space, Part 2

Disclaimer: as before, these are (incredibly) rough notes intended for a tutorial. I may clean them up a bit later but for now it will seem like a lot of unmotivated equations (with typos!!).


The Energy Functional
\[ S = \int_0^T |\dot{\gamma}|^2 dt \]
Letting \(V^i = \dot{\gamma}^i\), this is 
\[ S = \int_0^T g_{ij}(\gamma(t)) V^i V^j dt = \int_0^T L dt \]
where the Lagrangian \(L\) is
\[ L = g_{ij} V^i V^j \]
Now, 
\[ \frac{\partial L}{\partial x^k} = (\partial_k g_{ij}) V^i V^j \]
and 
\[ \frac{\partial L}{\partial V^k} = g_{ij} \delta^i_k V^j + g_{ij} V^i \delta^i_k = 2 g_{jk} V^j \]
Now, 
\[ \frac{d}{dt} \frac{\partial L}{\partial V^k} = 2 (\partial_i g_{jk}) V^i V^j + 2 g_{jk} \dot{V}^j \]
Plugging these expressions into the Euler-Lagrange equations, we have
\[ 2 g_{jk} \dot{V}^j + \left(\partial_i g_{jk} + \partial_j g_{ik}- \partial_k g_{ij}\right) V^i V^j = 0 \]
Multiplying by the inverse metric, we have
\[ \dot{V}^k + \frac{g^{kl}}{2} \left( \partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij} \right) V^i V^j = 0 \]
Which is the geodesic equation (recall the formula for the Christoffel symbols).


Orthonormal Frames (Lorentzian and Riemannian) (tetrads, vielbeins, vierbeins, ...)
Locally, we can find an orthonormal basis of vector fields \(e^\mu_i\). Greek indicates coordinates, whereas Latin indicates label in the basis. These necessarily satisfy
\[ g_{\mu\nu} e^\mu_i e^\nu_j = \eta_{ij} \]
where \(\eta_{ij}\) is the flat/constant metric (of whatever signature we are working in).


Methods for Computing Curvature (from Wald)
0. Getting the Christoffel symbols from the geodesic equation.
See e.g. sphere or spherical coordinates.


1. Coordinates. By definition,
\[ \nabla_a \nabla_b \omega_c = \nabla_b \nabla_a \omega_c + {R_{abc}}^d \omega_d \]
Writing things explicitly, this gives
\[ R_{abc}^d = \partial_b \Gamma^d_{ac} - \partial_a \Gamma^d_{bc}\]
\[+\Gamma^e_{ac}\Gamma^d_{be} - \Gamma^e_{bc}\Gamma^d_{ae}\]
(todo: fix typesetting.)


Do this for eg unit sphere in \(\mathbb{R}^3\).


2. Curvature in Frames (equivalent to coordinates but totally different flavor)
(note: Misner-Thorne-Wheeler seems much better than Wald for this stuff).
 Using MTW notation. Fix a frame \(\mathbf{e_\mu}\) and a dual frame \(\omega^\mu\). The connection 1-forms are defined by
\[ 0 = d\omega^\mu + \alpha^\mu_\nu \wedge \omega^\nu \]
We also have
\[ dg_{\mu\nu} = \omega_{\mu\nu} + \omega_{\nu\mu}\]
So metric compatibility yields
\[ \omega_{\mu\nu} = -\omega_{\nu\mu}\]
Antisymmetry means fewer independent components. In this language, the curvature 2-form is given by
\[ R^\mu_\nu = d\alpha^\mu_\nu + \alpha^\mu_\sigma \wedge \alpha^\sigma_\nu \]




Gaussian Coordinates
Via Wald. Suppose \(S \subset M\) is a codimension 1 submanifold. If \(S\) is not null, we can find a normal vector field \(n^a\) which is everywhere orthogonal to \(S\) and has unit length. (Probably also need orientation to make it unique!) We can pick any coordinates \(x^1, \cdots, x^{n-1}\) on \(S\), and we pick the last coordinate to be the distance to \(S\), measured along a geodesic with initial tangent vector \(n^a\) (i.e. we use exponential coordinates in the normal direction). 


Once we pick these coordinates, we obtain a family of hypersurfaces \(S_t\) given by
\(x^n = t\). These have the property that they are orthogonal to the normal geodesics through \(S\). Proof: (X are vector fields which are tangent to \(S_t\))
\[ n^b \nabla_b (n_a X^a) = n_a n^b \nabla_b X^a \] 
\[= n_a X^b \nabla_b n^a \] 
\[= \frac{1}{2}X^b \nabla_b (n^a n_a) = 0 \]
(first: geodesic, second: they lie-commute since they are coordinate vector fields).


Jacobi Fields, Focusing and Growth, Conjugate Points
Geodesic deviation. Suppose we have a 1-parameter family of geodesics \(\gamma_s\) with tangent \(T^a\) and deviation \(X^a\). (draw pictures!) By the geodesic equation, we have
\[ T^a \nabla_a T^b = 0 \]
What can we say about \(X^a\)? By change of affine parameter if necessary, we can assume that \(T^a\) and \(X^a\) are coordinate vector fields, and in particular they commute. So
\[ X^a \nabla_a T^b = T^a \nabla_a X^b \]
Then it is easy to see that \(X^a T_a\) is constant, and so (again by change of parameter if necessary) we can assume that it is 0. Now set \(v^a = T^b \nabla_b X^a\). We interpret this as the relative velocity of nearby geodesics. Similarly, we have the acceleration
\[ a^a = T^c \nabla_c v^a = T^b \nabla_b (T^c \nabla_c X^a) \]
Some manipulation shows that
\[ a^a = -R_{cbd}^a X^b T^c T^d \]
This is the geodesic deviation equation. (Positive curvature -> focus, negative curvature ->growth.)


Now we can work this in reverse. Suppose I have a single geodesic with tangent \(T^a\). If I have some vector field \(X^a\) on the geodesic, under what conditions will it integrate to give me a family of geodesics? The above shows that we must have
\[ T^a \nabla_a (T^b \nabla_b X^c) = -R_{abd}^c X^b T^a T^d \]
Solutions to this equation are called Jacobi vector fields.


Definition Points p, q on a geodesic are said to be conjugate if there exists a Jacobi field on the geodesics which vanishes at p and q. (Picture time!)


Definition (Cut Locus in Riemannian Signature) For \(p \in M\), we define the cut locus in \(T_p M\) to be those vectors \(v \in T_p M\) for which \(\exp(tv)\) is length minimizing on \([0,1]\) but fails to be length-minimizing on \([0,1+\epsilon]\) for and \(\epsilon\). The cut locus in M is the image of the cut locus in \(T_p M\) under the exponential map.


eg. Sphere, antipodes.

Geometry of Curved Space, Part 1: Prerequisites for General Relativity

I'm TAing a course on general relativity this semester, and I'm covering some of the geometry background in tutorials. Since I need to prepare some material for these, I thought there was no harm in putting it up on this blog. So here we go.

Throughout, we'll let \(M\) be a smooth manifold equipped with a metric \(g\). Whenever it makes life easier, I'll assume that \(g\) is positive definite (rather than Lorentzian). For a point \(p \in M\), denote its tangent space by \(T_p M\).

Theorem 1 For any \(p \in M\), there exist a neighborhood \(U\) of 0 in \(T_p M\) and a neighborhood \(V\) of \(p\) in \(M\), and a diffeomorphism \(\exp_p: U \to V\) called exponential map. This map takes lines through the origin in \(T_p M\) to geodesics in \(M\) passing through \(p\).

Theorem 2 In exponential coordinates, the components of the metric are
\[ g_{ij} = \delta_{ij} + O(|x|^2) \]

Corollary 3 In exponential coordinates, the Christoffel symbols vanish at \(p\).

Corollary 4 The Christoffel symbols are not the components of a tensor.

Corollary 5 Not all metrics are equivalent: there is a local invariant, called the curvature.

Construction of exponential map. The metric on \(M\) induces a (constant) metric \(g_{ij}\) on \(T_p M\). By a linear change of coordinates on \(T_p M\), we can assume that this induced metric is just \(g_{ij} = \delta_{ij}\). Now the geodesic equation on \(M\) is a 2nd order ODE which has a unique solution once we specify an initial condition. An initial condition is just a pair \((p,v)\) consisting of a point \(p \in M\) and tangent vector \(v \in T_p M\). Since we have fixed \(p\), each \(v \in T_p M\) gives a unique geodesic through \(p\). Call it \(\gamma_v(t)\). Then define the exponential map as follows:
\[ \exp_p(v) := \gamma_v(1) \]
The fact that this map is well-defined, smooth, and 1-1 (at least locally) follows from the standard existence and uniqueness theorem for ODEs. So to see that it is a diffeomorphism near \(0\), we can just compute its differential and apply the inverse function theorem.

The easy way out. By construction, every geodesic through 0 is of the form \(\gamma_v(t) = tv\). Plugging this into the geodesic equation,
\[ \dot{v} + \Gamma(x)^i_{jk} v^j v^k = 0 \]
we see that at 0, \(\Gamma^i_{jk}\) vanishes, and in particular, the first partial derivatives of the metric vanish.

The hard way: the differential of \(exp\) at 0. First, taylor expand the velocity of a geodesic, and evaluate at time \(t = 1\):
\[ v(t) = v + \dot{v} + \frac{1}{2} \ddot{v} + \cdots \]
Now, by the geodesic equation, \(\dot{v}\) is \(O(v^2)\). Similarly, by differentiating the geodesic equation, we find that all of the higher time derivatives are also \(O(v^2)\). So we find that the exponential map is just the identity + \(O(v^2)\), and hence its differential at 0 is just the identity.

Now, we have argued that in exponential coordinates, the Christoffel symbols vanish at 0. Recall that for any tensor \(T\), if the components of \(T\) vanish at some point \(p\) in one coordinate system, then \(T\) is identically 0 at that point (i.e. its components vanish at that point in all coordinate systems). If the Christoffel symbols were a tensor, then, the above shows that they must be identically zero at all points in \(M\), in all coordinate systems. But this is absolutely not the case--even in flat \(\mathbb{R}^n\), we can pick coordinates so that the Christoffel symbols do not vanish. Hence they are not a tensor.

Aside Though the Christoffel symbols are not a tensor, they are the components of something which does have a coordinate indepdendent definition: a connection 1-form. A connection 1-form is not a tensor but rather a section of a certain associated bundle. More on this in future posts.

Claim Suppose \(\gamma\) is a curve in \(M\) with tangent vector \(T\), and suppose \(V\) is a vector field defined on \(\gamma\). Then \(\nabla_T V\) is well-defined, independent of the smooth extension of \(V\).

Proof Suppose \(V\) and \(W\) are two smooth vector field that agree on \(\gamma\). We would like to show that
\[ \nabla_T V = \nabla_T W \]
i.e., this directional derivative depends only on their restriction to \(\gamma\). It suffices to prove this pointwise. In coordinates, we have
\[ \nabla_T V = T^k \frac{\partial V^i}{\partial x^k} + \Gamma^i_{jk} V^j T^k \]
and similarly for \(W\). The terms involving Christoffel symbols do not depend on derivatives of \(V\) or \(W\), so they agree by assumption. If \(T\) is zero at a point, there is nothing to show. So assume that \(T\) is nonzero at some point. Then near this point, we can choose coordinates \(x^i\) such that
\[ \gamma(t) = (t, 0) \]
so that
\[ T = (1, 0) \]
Then \(V\) and \(W\) agree when \(x^i = 0\) for \(i \geq 2\), and hence their partials agree. We have
\[ T^k \frac{\partial V^i}{\partial x^k} = \frac{V^i}{\partial x^1} = T^k \frac{\partial W^i}{\partial x^k} \]

Explicit Formulas for Christoffel Symbols. Using properties of covariant derivatives, we find
\[ 0 = \nabla_k g_{ij} = \frac{\partial g_{ij}}{\partial x^k} - \Gamma^l_{ki} g_{lj} - \Gamma^l_{kj} g_{il} \]
So
\[ \Gamma_{kij} + \Gamma_{kji} = g_{ij,k} \]
\[ \Gamma_{ijk} + \Gamma_{ikj} = g_{jk,i} \]
\[ \Gamma_{jki} + \Gamma_{jik} = g_{ki,j} \]
Taking (2) + (3) - (1) gives
\[ 2\Gamma_{ijk} = g_{jk,i} + g_{ki,j} - g_{ij,k} \]
Hence
\[ \Gamma_{ij}^k = \frac{g^{kl}}{2}\left(g_{jl,i} + g_{li,j} - g_{ij,l} \right) \]