Tensors, Differential Geometry, Manifolds
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Most of this content is based on a 2002 Caltech course taught by Kip Thorn (PH237).


On a manifold, only "short" vectors exist. Longer vectors are in a space tangent to the manifold.

There are points ($P$), separation vectors ($\Delta \vector P$),
curves ($Q(\zeta)$), tangent vectors ($\delta P / \delta \zeta \equiv
\lim_{\Delta \zeta \rightarrow 0} \frac{ \vector{ Q(\zeta+\delta \zeta) -
Q(\zeta) } }{\delta \zeta}$)

Coordinates: $\Chi^\alpha (P)$, where $\alpha = 0,1,2,3$;
$Q(\Chi_0, \Chi_1, ...)$
    there is an isomorphism between points and coordinates

Coordinate basis: $\vector{e_\alpha} \equiv \left( \frac{\partial
Q}{\partial \Chi^\alpha} \right$)
    
    for instance, on a sphere with angles $\omega, \phi$: 
    
    $\vector{e_\phi} = \left( \frac{\partial Q(\phi, \theta)}{\partial \phi}\right)_\theta$

Components of a vector:

    $\vector{A} = \frac{\partial P}{\partial \Chi^\alpha }$

Directional Derivatives: consider a scalar function defined on a manifold \Psi(P)
    $\partial_\vector{A} \Psi = A^\alpha \frac{\partial \Psi}{\partial \Chi^\alpha}$

Mathematicians like to say that the coordinate bases are actually directional derivatives

Tensors
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A **tensor** $\bold{T}$ has a number of slots (called it's **rank**), takes a vector in each slot, and returns a real number. It is linear in vectors; 
as an example for a rank-3 tensor:

$$\bold{T} ( \alpha \vector{A} + \beta \vector{B}, \vector{C}, \vector{D}) =
\alpha \bold{T} (\vector{A}, \vector{C}, \vector{D}) + \beta \bold{T}
(\vector{B}, \vector{C}, \vector{D}) $$

Even a regular vector is a tensor: pass it a second vector and take the 
inner product (aka dot product) to get a real.

Define the **metric tensor** 
$\bold{g}(\vector{A}, \vector{B}) = \vector{A} \cdot \vector{B}$. The 
metric tensor is rank two and symetric (the vectors A and B could be swapped 
without changing the scalar output value) and is the same as the inner product. 

$$\Delta P \cdot \Delta P \equiv \Delta P^2 \equiv (length of \Delta P)^2 A \cdot B = 1/4[ (A+B)^2 - (A-B)^2 ]$$

Starting with individual vectors, we can construct tensors by taking the 
product of their inner products with empty slots; for example

$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\_ ,\_ ,\_)$$ 
$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\vector{E}, \vector{F}, \vector{G}) = ( \vector{A} \cdot \vector{E})(\vector{B} \cdot \vector{F})(\vecotr{C} \cdot \vector{G}) $$ 

Spacetime
--------------

Two types of vectors.

Timelike: $\vector{\Delta P}$
    $(\vector{\Delta P})^2 = -(\Delta \Tau)^2$

Spacelike: $\vector{\Delta Q}$
    $(\vector{\Delta Q})^2 = +(\Delta S)^2$

Because product of "up" and "down" basis vectors must be a positive Kronecker 
delta, and timelikes squared come out negative, the time "up" basis must be  negative of the time "down" basis vector.