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
$$\Gamma_{ij}^k = \frac{g^{kl}}{2}\left(g_{jl,i} + g_{li,j} - g_{ij,l} \right)$$

Path Integrals 1: Feynman's Derivation

Consider the Hilbert space $\mathcal{H} = L^2(\mathbb{R})$ with Lebesgue measure and a Hamiltonian $H = T(k) + V(x)$ (a sum of kinetic and potential energy). Then the quantum hamiltonian $\hat{H}$ acts as

$$\begin{align} (\hat{H}\psi)(y) &= \frac{1}{2\pi} \int e^{ik(y-x)} T(k) \psi(x) dx dk + V(y)\psi(y) \\ &= \frac{1}{2\pi}\int e^{ik(y-x)} H(k,x) \psi(x) dx dk \\ \end{align}$$

Now consider the Schrodinger equation
$$\frac{\partial \psi}{\partial t} = i \hat{H} \psi.$$

We would like to obtain a formula for the solution operator $U_t = e^{-i \hat{H} t}$. Let us consider its Schwartz kernel $\langle{y}|U_t|{x}\rangle$. Let $N$ be a large integer so that $\Delta t = t/N$ is "small". Then we can write

$$U_t = U_{\Delta t}^N,$$

Now consider a single term:

$$\begin{align}  \langle{y}|U_{\Delta t}|x\rangle &\simeq \langle{y}|1-i\Delta{t}\hat{H}|x\rangle \\ &= \delta(y-x) -i \Delta t \langle x|\hat{H}|y\rangle \\ &= \frac{1}{2\pi}\int e^{ik(y-x)}( 1 - i \Delta t H(k,x)) dk \\ &\simeq \frac{1}{2\pi}\int e^{ik(y-x) - i \Delta t H(k,x)} dk\\ \end{align}$$
Now we have (taking $x_0 = x$ and $x_N = y$)
$$\begin{align} \langle y|U_t|x\rangle &= \int dx_1 \cdots dx_{N-1} \\ & \ \times \langle x_N|U_{\Delta t}|x_{N-1}\rangle \cdots \langle x_1|U_{\Delta t}|x_0\rangle \\ &= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\ & \ \times \ e^{ik_N(x_N-x_{N-1}) - i \Delta t H(k_{N-1},x_{N-1})} \cdots e^{-ik_N(x_1-x_0) - i \Delta t H(k_0,x_0)} \\ &= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\ & \ \times \ \exp \sum_{j=0}^{N-1} ik_j(x_{j+1} - x_j) - i \Delta t H(k_j,x_j) \end{align}$$
Note this kind of integral seems like it should have a natural interpretation in symplectic or contact geometry, since the expression $dxdk/(2\pi)^N$ is very nealy the Liouville measure. This is the most general form of the path integral.

Now assume that $H(k,x) = k^2/2m + V(x)$. Then the $k$-dependent terms have the form
$$\int e^{ik(y-x) - i\Delta t k^2/2m}.$$
Complete the square
$$\begin{align}  ik(y-x) -i\Delta t k^2/2m &= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x)) \\ &= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x) + \frac{m^2}{\Delta t^2}(y-x)^2 -\frac{m^2}{\Delta t^2}(y-x)^2) \\ &= -\frac{i \Delta t}{2m}(k - \frac{m}{\Delta t}(y-x))^2 + \frac{i m}{2 \Delta t}(y-x)^2 \end{align}$$
Now, using that
$$\int e^{-ak^2} = \sqrt{\frac{\pi}{a}}$$
We have
$$\int e^{-\frac{i \Delta t}{2m}(k-\frac{m}{\Delta t}(y-x))^2} = \sqrt{\frac{2\pi m}{i \Delta t}}$$
Putting it altogether, we get the more familiar version of the path integral,
$$\begin{align} \langle y|U_t|x\rangle &\simeq C^N \int dx_1 \cdots dx_{N-1} \\ & \ \times \ \exp i\sum_{j=0}^{N-1} \frac{m}{2\Delta t}(x_{j+1}-x_j)^2 - V(x_j) \Delta t \end{align}$$
where
$$C = \frac{1}{2\pi} \sqrt{\frac{2 \pi m}{i \Delta t}} = \sqrt{\frac{m}{2\pi i \Delta t}}$$