From Simple Groups to Quantum Field Theory

[Mehdi_]

Curl and Divergence

Some simple rules:

If a vector function f(x,y,z) has continuous second order partial derivatives then curl (div F) = \nabla \times (\nabla . F) = 0.

If \vec{F} is a conservative vector field then curl ( \vec{F}) = 0.

If \vec{F} is defined on all R^3 of whose components have continuous first order partial derivative and curl ( \vec{F}) = 0 then \vec{F} is a conservative vector field.

If curl ( \vec{F}) = 0 then the fluid is called irrotational.

If div ( \vec{F}) = 0 then the \vec{F} is called incompressible.

div (curl F) is always 0.

Green's theorem is a term used variously in mathematical literature to denote either the Gauss divergence theorem or the plane case (2D) of Stokes' theorem.

The first form of Green’s Theorem uses the curl of the vector field and is,

$ \oint_{ C } \vec{F}.d \vec{r} = \int \int_{D} (\nabla \times \vec{F}).\vec{k} dA $

where \vec{k} is the standard unit vector in the positive z direction.

The second form uses the divergence.
In this case we also need the outward unit normal to the curve C.

If the curve is parameterized by \vec{r} (t) = x(t) \hat{i} + y(t) \hat{j}

then the outward unit normal is given by,

\vec{n} = \frac{y'(t)} {| \vec{r'} (t)|} \hat{i} + \frac{x'(t)} {| \vec{r'} (t)|} \hat{j}

The vector form of Green’s Theorem that uses the divergence is then given by,

$ \oint_{ C } \vec{F}.\vec{n} ds = \int \int_{D} (\nabla \vec{F}) dA $

 

[Mehdi_]

The Line Element and Metric of a torus

If the major radius of this torus is c and the minor radius a ; with c>a .
The torus S(u,v) can be defined parametrically by:

x = (c + a \ cos(v) ) \ cos(u)
y = (c + a \ cos(v) ) \ sin(u)
z = a \ sin(v)

where u and v \in [0, 2 \pi ]

The coefficients E, F, and G of the first fundamental form (Line Element) are :

S_u = \frac{\partial S}{\partial u} = ( \ -(c + a \ cos(v) ) sin(u) \ , \ (c + a \ cos(v) ) \ cos(u) \ , \ 0 \ )
S_v = \frac{\partial S}{\partial v} = ( \ -(a \ cos(u) \ sin(v) ) \ , \ -(a \ sin(u) \ sin(v) ) \ , \ a \ cos(v) \ )

Therefore,

E = \frac{\partial S}{\partial u} \ . \ \frac{\partial S}{\partial u} = ( - (c + a \ cos(v) ) \ sin(u) )^2 \ + \ (( c + a \ cos(v) ) cos(u) )^2 \ + \ 0 = ( c + a \ cos(v) )^2

F = \frac{\partial S}{\partial u} \ . \ \frac{\partial S}{\partial v} = ( - (c + a \ cos(v) ) \ sin(u) ) \ -(a \ cos(u) \ sin(v) ) \ + \ (( c + a \ cos(v) ) cos(u) ) \ -(a \ sin(u) \ sin(v) ) \+ \ (0)\ a \ cos(v) \ = 0

G = \frac{\partial S}{\partial v} \ . \ \frac{\partial S}{\partial v} = ( \ -(a \ cos(u) \ sin(v) ) \ )^2 \ + \ (\ -(a \ sin(u) \ sin(v) ) \ )^2 \ + \ ( \ a \ cos(v) \ )^2 = a^2

The line element ds^2 (s here is an arc length) is :

ds^2 = E \ du^2 \ + \ 2 \ F \ du \ dv \ + G \ dv^2 \
ds^2 = ( c + a \ cos(v) )^2 \ du^2 \ + a^2 \ dv^2 \

The metric is g_{ij} is :

g_{ij} = \left [ \begin{array}{ccc} ( c + a \ cos(v) )^2 & 0 \\ 0 & a^2 \end{array}\right ]

g^{ij} = \left [ \begin{array}{ccc} \frac{1}{( c + a \ cos(v) )^2} & 0 \\ 0 & \frac{1}{a^2} \end{array}\right ]

 

[Mehdi_]

The metric g_{ij} of the torus above could also be computed by the formula :

g_{ij} = J^T \ J

where J denotes the Jacobian and J^T its transpose.

If the torus can be defined parametrically by :

x = (c + a \ cos(v) ) \ cos(u)
y = (c + a \ cos(v) ) \ sin(u)
z = a \ sin(v)

The Jacobian J therefore is :

J = \left[ \begin {array}{cc} - \left( c+a\cos \left( v \right) \right) \sin \left( u \right) &amp;-a\sin \left( v \right) \cos \left( u \right) <br /> \\\noalign{\medskip} \left( c+a\cos \left( v \right) \right) \cos<br /> \left( u \right) &amp;-a\sin \left( v \right) \sin \left( u \right) <br /> \\\noalign{\medskip}0&amp;a\cos \left( v \right) \end {array} \right] <br />

And its transpose is :

J^T = \left[ \begin {array}{ccc} - \left( c+a\cos \left( v \right) \right) \sin \left( u \right) &amp; \left( c+a\cos \left( v \right) <br /> \right) \cos \left( u \right) &amp;0\\\noalign{\medskip}-a\sin \left( v<br /> \right) \cos \left( u \right) &amp;-a\sin \left( v \right) \sin \left( u<br /> \right) &amp;a\cos \left( v \right) \end {array} \right]

Therefore :

g_{ij} \ = J^T \ J = \left[ \begin {array}{cc} \left( c+a\cos \left( v \right) \right) ^{2} \left( \sin \left( u \right) \right) ^{2}+ \left( c+a\cos \left( <br /> v \right) \right) ^{2} \left( \cos \left( u \right) \right) ^{2}&amp;0<br /> \\\noalign{\medskip}0&amp;{a}^{2} \left( \sin \left( v \right) \right) ^{<br /> 2} \left( \cos \left( u \right) \right) ^{2}+{a}^{2} \left( \sin<br /> \left( v \right) \right) ^{2} \left( \sin \left( u \right) \right) <br /> ^{2}+{a}^{2} \left( \cos \left( v \right) \right) ^{2}\end {array}<br /> \right] <br />

g_{ij} \ = \left[ \begin {array}{cc} \left( c+a\cos \left( v \right) \right) ^{2}&amp;0\\\noalign{\medskip}0&amp;{a}^{2} \left( \left( \sin \left( v<br /> \right) \right) ^{2} \left( \left( \cos \left( u \right) \right) ^<br /> {2}+ \left( \sin \left( u \right) \right) ^{2} \right) + \left( \cos<br /> \left( v \right) \right) ^{2} \right) \end {array} \right] <br />

g_{ij} \ = \left [ \begin{array}{ccc} ( c + a \ cos(v) )^2 &amp; 0 \\ 0 &amp; a^2 \end{array}\right ]

 

[Mehdi_]

Holonomic bases

A holonomic basis for a manifold is a set of basis vectors e_k for which all Lie derivatives vanish:[e_j, e_k] = 0

Given coordinates x^a, we define basis vectors e_a and basis one forms \omega^a in the following way:
e_a = \partial_a= \frac{\partial }{\partial x^a}
\omega^a = dx^a

Holonomic or coordinate bases is then defined in terms of derivatives with respect to coordinates.

Spherical polar coordinates (holonomic) basis vectors are :

e_r = \partial_r= \frac{\partial }{\partial r}
e_{\theta} = \partial_{\theta}= \frac{\partial }{\partial \theta}
e_{\phi} = \partial_{\phi} = \frac{\partial }{\partial \phi}

Coordinate bases need not necessary to be of unit length.

Line element of spherical coordinates is ds^2 = dr^2 \ + \ r^2 \ d \theta^2 + \ r^2 \sin^2(\theta) \d \phi^2


In a coordinate basis, basis vector satisfy : e_1 \ . \ e_2 = g_{12}


e_r \ . \ e_r = 1 \ \ \ \ \ |e_r| = 1

e_{ \theta } \ . \ e_{ \theta } = r^2 \ \ \ \ \ |e_{ \theta }| = r

e_{\phi} \ . \ e_{\phi} = \ r^2 \sin^2(\theta) \ \ \ \ \ |e_{\phi}| = \ r \sin(\theta)

Since two of these vectors do not have unit length, these coordinate basis are therefore not orthonormal.

To choose spherical polar noncoordinates that are orthonormal, we need to define nonoholonomic (noncoordinate) bases given by the following :

e_R = \partial_r
e_{\Theta} = \frac{1}{r} \partial_{\theta}
e_{\Phi} = \frac{1}{r \ sin(\theta)} \partial_{\phi}

 

[Mehdi_]

The metric g_{ij} = J^T \ J of the sphere :

x = cos(\theta)sin(\phi)
y = sin(\theta)sin(\phi)
z = cos(\phi)

The Jacobian J is :

J = \left[ \begin {array}{ccc} \cos \left( \theta \right) \sin \left( \phi \right) &amp;-r\sin \left( \theta \right) \sin \left( \phi \right) &amp;r<br /> \cos \left( \theta \right) \cos \left( \phi \right) <br /> \\\noalign{\medskip}\sin \left( \theta \right) \sin \left( \phi<br /> \right) &amp;r\cos \left( \theta \right) \sin \left( \phi \right) &amp;r\sin<br /> \left( \theta \right) \cos \left( \phi \right) \\\noalign{\medskip}<br /> \cos \left( \phi \right) &amp;0&amp;-r\sin \left( \phi \right) \end {array}<br /> \right]<br />

And its transpose is :

J^T = \left[ \begin {array}{ccc} \cos \left( \theta \right) \sin \left( \phi \right) &amp;\sin \left( \theta \right) \sin \left( \phi \right) &amp;<br /> \cos \left( \phi \right) \\\noalign{\medskip}-r\sin \left( \theta<br /> \right) \sin \left( \phi \right) &amp;r\cos \left( \theta \right) \sin<br /> \left( \phi \right) &amp;0\\\noalign{\medskip}r\cos \left( \theta<br /> \right) \cos \left( \phi \right) &amp;r\sin \left( \theta \right) \cos<br /> \left( \phi \right) &amp;-r\sin \left( \phi \right) \end {array} \right] <br />

\left[ \begin {array}{ccc} \cos( \theta)^{2} \sin(\phi)^{2}+ <br /> \sin(\theta)^{2} \sin( \phi)^{2} + \cos( \phi)^{2}&amp;0<br /> &amp; \cos( \theta)^{2} \sin( \phi) r\cos( \phi) + \sin( \theta)^{2}\sin( \phi) r\cos( \phi) -\sin( \phi) r\cos( \phi) <br /> \\0&amp;{r}^{2} ( \sin( \theta)^{2} \sin( \phi)^{2}+{r}^{2} \cos( \theta)^{2} \sin(\phi)^{2}&amp;0\\ \cos( <br /> \theta)^{2}\sin( \phi) r\cos( \phi) + \sin( \theta)^{2}\sin( <br /> \phi) r\cos( \phi) -\sin( \phi) r\cos( \phi) &amp;0&amp;{r}^{2} \cos( \theta)^{2} \cos( \phi)^{2}+{r}^{2}<br /> \sin( \theta)^{2} \cos(\phi)^{2}+{r}^{2} \sin( \phi)^{2}\end {array} \right] <br />


g_{ij} \ = \left[ \begin {array}{ccc} 1&amp;0&amp;0\\\noalign{\medskip}0&amp;{r}^{2} \left( \sin \left( \phi \right) \right) ^{2}&amp;0\\\noalign{\medskip}0&amp;0&amp;{r}^{2<br /> }\end {array} \right] <br />

Therefore the line element ds^2 is :


ds^2 = dr^2 \ + \ r^2 \sin^2(\phi) \ d \theta^2 + \ r^2 \ d \phi^2

Here \theta is the polar angle in the xy-plane from the x-axis with 0 \leq \theta \leq 2 \phi while \phi is the azimuthal angle from the z-axis with 0 \leq \phi \leq 2 \phi

However, because we are dealing with a sphere (symmetry!), the symbols \theta and \phi could be reversed.

 

[Mehdi_]

Complex numbers & orthogonal matrices

A complex number is written in the form z=a+ib where a and b are real numbers while i is a symbole which satisfy i^2=-1.

In polar coordinates, z=r(cos(\theta)+isin(\theta)) where r is the magnitude and \theta is the angle.

However complex numbers could also be viewed as linear transformation and therefore matrices which obey linear algebra.

A complex number could then be represented by the orthogonal matrice :

z = \left[ \begin {array}{cc} a&amp;-b\\\noalign{\medskip}b&amp;a\end {array} \right] = \left[ \begin {array}{cc} a&amp;0\\\noalign{\medskip}0&amp;a\end {array} \right] + \left[ \begin {array}{cc} 0&amp;-1\\\noalign{\medskip}1&amp;0\end {array} \right] \left[ \begin {array}{cc} b&amp;0\\\noalign{\medskip}0&amp;b\end {array} \right] = a+ib

Which in polar coordinates could also be written :

z = \left[ \begin {array}{cc} a&amp;-b\\\noalign{\medskip}b&amp;a\end {array} \right] = \left[ \begin {array}{cc} r&amp;0\\\noalign{\medskip}0&amp;r\end {array} \right] \left[ \begin {array}{cc} cos(\theta)&amp;-sin(\theta)\\\noalign{\medskip}sin(\theta)&amp;cos(\theta)\end {array} \right] = r(cos(\theta)+isin(\theta))

Therefore, complex addition is just matrix addition :

(a+ib) + (c+id) = \left[ \begin {array}{cc} a&amp;-b\\\noalign{\medskip}b&amp;a\end {array} \right] + \left[ \begin {array}{cc} c&amp;-d\\\noalign{\medskip}d&amp;c\end {array} \right] = \left[ \begin {array}{cc} \left\{ a+c \right\} &amp;- \left\{ b+d \right\} \\\noalign{\medskip} \left\{ b+d \right\} &amp; \left\{ a+c<br /> \right\} \end {array} \right] = (a+c) + i (b+d)

And complex substraction is matrix substraction :

(a+ib) - (c+id) = \left[ \begin {array}{cc} a&amp;-b\\\noalign{\medskip}b&amp;a\end {array} \right] - \left[ \begin {array}{cc} c&amp;-d\\\noalign{\medskip}d&amp;c\end {array} \right] = \left[ \begin {array}{cc} \left\{ a-c \right\} &amp;- \left\{ b-d \right\} \\\noalign{\medskip} \left\{ b-d \right\} &amp; \left\{ a-c<br /> \right\} \end {array} \right] = (a-c) + i (b-d)

Complex multiplication is matrix multiplication :

(a+ib) (c+id) = \left[ \begin {array}{cc} a&amp;-b\\\noalign{\medskip}b&amp;a\end {array} \right] \left[ \begin {array}{cc} c&amp;-d\\\noalign{\medskip}d&amp;c\end {array} \right] = \left[ \begin {array}{cc} \left\{ ac-bd \right\} &amp;- \left\{ bc+ad \right\} \\\noalign{\medskip} \left\{ bc+ad \right\} &amp; \left\{ ac-bd<br /> \right\} \end {array} \right] = (ac-bd) + i (bc + ad)

The complex conjugate \overline{z} of z is :

\overline{z} = a -ib = \left[ \begin {array}{cc} a&amp;b\\\noalign{\medskip}-b&amp;a\end {array} \right]

we can then write that :

|z|^2 = z \overline{z} = (a + i b) (a -i b) = \left[ \begin {array}{cc} a&amp;-b\\\noalign{\medskip}b&amp;a\end {array} \right] \left[ \begin {array}{cc} a&amp;b\\\noalign{\medskip}-b&amp;a\end {array} \right] = \left[ \begin {array}{cc} {a}^{2}+{b}^{2}&amp;0\\\noalign{\medskip}0&amp;{a}^{2}+{b}^{2}\end {array} \right] =a^2 + b^2

|z| = \sqrt{a^2 + b^2} is then the modulus.

Therefore,

1/z = \overline{z} / |z|^2.

Proof:

1/z = \left[ \begin {array}{cc} a&amp;-b\\\noalign{\medskip}b&amp;a\end {array} \right]^{-1} = \left[ \begin {array}{cc} {\frac {a}{{a}^{2}+{b}^{2}}}&amp;{\frac {b}{{a}^{2}+{b}^{2}}}\\\noalign{\medskip}-{\frac {b}{{a}^{2}+{b}^{2}}}&amp;{\frac {a}{{a}^{2}+{b}^{2}}}\end {array} \right] = \left[ \begin {array}{cc} a&amp;b\\\noalign{\medskip}-b&amp;a\end {array} \right] \left[ \begin {array}{cc} \left( {a}^{2}+{b}^{2} \right) ^{-1}&amp;0\\\noalign{\medskip}0&amp; \left( {a}^{2}+{b}^{2} \right) ^{-1} \end {array} \right] = \left[ \begin {array}{cc} a&amp;b\\\noalign{\medskip}-b&amp;a\end {array} \right] / \left[ \begin {array}{cc} {a}^{2}+{b}^{2}&amp;0\\\noalign{\medskip}0&amp;{a}^{2}+{b}^{2}\end {array} \right] = \overline{z} / |z|^2

 

[Mehdi_]

Inner product and metric tensor

Recall that the scalar product of two vector is v.w :

v.w = (v^{\alpha} e_{\alpha}) (w^{\beta} e_ {\beta}) = v^{\alpha} w^{\beta} (e_{\alpha} . e_{\beta}) = v^{\alpha} w^{\beta} \eta _{\alpha \beta}


(e_{\alpha} . e_{\beta}) = \eta_{\alpha \beta}

so v.w = \eta_{\alpha \beta} v^{\alpha} w^{\beta}

Where \eta_{\alpha \beta} are the component of the metric tensor.

The notation g_{ij} is however conventionally used for the components of the metric tensor.

In the tangent space, coordinate basis are \frac{\partial }{\partial x^i}, the components of the metric tensor become :

g_{ij} = ( \frac{\partial }{\partial x^i} . \frac{\partial }{\partial x^j} ) = &lt; \frac{\partial }{\partial x^i} , \frac{\partial }{\partial x^j} &gt;

The metric g_{ij} could also be computed by the formula :

g_{ij} = J^T \ J

where J denotes the Jacobian and J^T its transpose.

An example of an inner product which induces a metric we have the space of continuous complex valued functions on the interval [a,b]; the inner product is :

&lt; f , g &gt; = \int_{b}^{a} \overline{f(t)}.g(t).dt

where \overline{f(t)} denotes the conjugate of f(t)

 

[Mehdi_]

Tensors & metric

contravariant components : A = A^1 e_1 + A^2 e_2 + A^3 e_3 = A^i e_i
covariant components : A = A_1 e^1 + A_2 e^2 + A_3 e^3 = A_i e^i

Therefore :

A = A_i e^i = A^i e_i

and then

(A^i e_i) \bullet e^k = (A_i e^i) \bullet e^k
A^i (e_i \bullet e^k) = A_i (e^i \bullet e^k)
A^i (e_i \bullet e^i) = A_i (e^i \bullet e^i)
A^i = A_i g^{ii} where g^{ik} is the metric tensor
A^i = A_i because |e^i| = |e_i| = 1

The equation A^i = g^{ik} A_k is the operation of raising an index.

The equation A_i = g_{ik} A^k is the operation of lowering an index.

 

[Mehdi_]

Metric & line element

The coordinates (contravariant components) of a point in a coordinate system are written x^i.
The radius vector of the point is then :

r = x^k e_k

Let ds be the arc length between two close points x^i and x^i + dx^i.
And let the vector dr joining the two points have covariant components dx_i and controvariant components dx^i

r = x^k e_k
dr = dx^k e_k + x^k de_k
dr = dx^k e_k
\frac{\partial r}{\partial x^i} dx^i = dx^k e_k
\frac{\partial r}{\partial x^i} dx^i = dx^i e_i
\frac{\partial r}{\partial x^i} dx^i = e_i dx^i
\frac{\partial r}{\partial x^i} = e_i


e_i = \frac{\partial r}{\partial x^i}

e^i = \frac{\partial r}{\partial x_i} = w^i

The line element is ds^2 :

ds^2 = |dr|^2 = dr \bullet dr = (e_i dx^i) \bullet (e_j dx^j)
Then ds^2 = (e_i e_j) dx^i dx^j = g_{ij} dx^i dx^j where g_{ij} is the metric tensor.

therefore :

ds^2 = g_{ij} dx^i dx^j = \frac{\partial r}{\partial x^i} \frac{\partial r}{\partial x^i} dx^i dx^j

 

[Pietjuh]

This is all very nice, but what is your point in writing all these things down?
This is all very basic stuff, which you can find in all the textbooks, and they do a good job in explaining it.

 

[Mehdi_]

Covariant derivatives & vectors

The covariant derivative of a scalar field or a function is :

\nabla_i \varphi = \frac{\partial \varphi}{\partial x^i}

And the covariant derivative of a vector \vec{v} = v^i e_i is just :

\nabla_i \vec{v} = \frac{\partial \vec{v}}{\partial x^j} =\frac{\partial v^i}{\partial x^j} e_ i

The basis e_ i does vary if the coordinate system is not rectangular or oblique.

If basis vectors e_i vary from point to point :

\nabla_i \vec{v} = \frac{\partial \vec{v}}{\partial x^j} = \frac{\partial v^i}{\partial x^j} e_ i + v^i \frac{\partial e_i}{\partial x^j}

e_i being a vector, we have :

\frac{\partial e_i}{\partial x^j} = \Gamma^{k}_{ij} e_k

where \Gamma^{k}_{ij} the christoffel symbols of the second kind, could be viewed as the components of this tensor.

Therefore

\nabla_i \vec{v} = \frac{\partial v^i}{\partial x^j} e_ i + v^i \Gamma^{k}_{ij} e_k

\nabla_i \vec{v} = ( \frac{\partial v^i}{\partial x^j} + v^k \Gamma^{i}_{kj} ) e_i

If however the colon and semicolon derivative notation are used, the covariant derivative of the vector is then :

\nabla_i \vec{v} = v^{i}_{;j} e_i = ( v^{i}_{,j} + v^k \Gamma^{i}_{kj} ) e_i

 

[Mehdi_]

Covariant derivatives & differential

Let e_i be local basis and x^i be generalized coordinates.

The differential of a vector \vec{v} = v^i e_i is :

d \vec{v} = d(v^i e_i) = e_i dv^i + v^i de_i

But

d \vec{v} = \frac{\partial \vec{v}}{\partial x^j} dx^j

therefore

\frac{\partial \vec{v}}{\partial x^j} dx^j = e_i \frac{\partial v^i}{\partial x^j} dx^j + v^i \frac{\partial e_i}{\partial x^j} dx^j

\frac{\partial \vec{v}}{\partial x^j} = e_i \frac{\partial v^i}{\partial x^j} + v^i \frac{\partial e_i}{\partial x^j}

Which is the covariant derivative of the vector \vec{v} = v^i e_i

\nabla_i \vec{v} = \frac{\partial v^i}{\partial x^j} e_ i + v^i \frac{\partial e_i}{\partial x^j}

\nabla_i \vec{v} = \frac{\partial v^i}{\partial x^j} e_ i + v^i \Gamma^{k}_{ij} e_k

 

[Mehdi_]

Christoffel Symbols

An affine connection in the case of a Riemannian manifold is a Levi-Civita connection if it preserves the metric and is Torsion-free.

The components of this connection with respect to a system of local coordinates are then called Christoffel symbols.

There are two kinds of Christoffel symbols :

The Christoffel symbols of the first kind (connections coefficients) which are denoted \Gamma_{ijk} or [i,jk]

[i,jk] = e_i \frac{\partial e_j}{\partial x^k} = \frac{1}{2} ( \frac{\partial g_{ij}}{\partial x^k} + \frac{\partial g_{ik}}{\partial x^j} - \frac{\partial g_{kj}}{\partial x^i} ) = [i,kj]

The Christoffel symbols of the second kind (affine connections) which are denoted \Gamma^{i}_{jk} or \left\{\begin{array}{cc}i \\jk \end{array} \right \}

\left\{\begin{array}{cc}h \\jk \end{array} \right \} = e^h \frac{\partial e_j}{\partial x^k} = \frac{1}{2} g^{ih} ( \frac{\partial g_{ij}}{\partial x^k} + \frac{\partial g_{ik}}{\partial x^j} - \frac{\partial g_{kj}}{\partial x^i} ) = \left\{\begin{array}{cc}h \\kj \end{array} \right \}

Therefore

[i,jk] = g_{hi} \left\{\begin{array}{cc}h \\jk \end{array} \right \}

and

\left\{\begin{array}{cc}h \\jk \end{array} \right \} = g^{ih} [i,jk]

where g_{hi} is the metric tensor.

And as seen above for vectors, the covariant derivative of contravariant A^{ij} and covariant A_{ij} tensors involves Christoffel symbols :

A^{ij}_{;k} = \frac{\partial A^{ij}}{\partial x^k} + \left\{\begin{array}{cc} i \\mk \end{array} \right \} A^{mj} + \left\{\begin{array}{cc} j \\mk \end{array} \right \} A^{im}

A_{ij}_{;k} = \frac{\partial A_{ij}}{\partial x^k} - \left\{\begin{array}{cc} m \\ik \end{array} \right \} A_{mj} - \left\{\begin{array}{cc} m \\jk \end{array} \right \} A_{im}

 

[Mehdi_]

Christoffel Symbols from the metric tensor

By definition the metric tensor g_{ij} is :

g_{ij} = \frac{\partial z^n}{\partial x^i} \frac{\partial z^n}{\partial x^j}

Therefore the derivative of g_{ij} is :

g_{ij} = \partial_i z^n \partial_j z^n

\partial_k g_{ij} = \partial_i \partial_k z^n \partial_j z^n + \partial_i z^n \partial_j \partial_k z^n

\partial_k g_{ij} = \partial_{ik} z^n \partial_j z^n + \partial_i z^n \partial_{jk} z^n

Using the same method, we have then for g_{ik} and g_{jk}

\partial_j g_{ik} = \partial_{ij} z^n \partial_k z^n + \partial_i z^n \partial_{jk} z^n

\partial_i g_{jk} = \partial_{ij} z^n \partial_k z^n + \partial_j z^n \partial_{ik} z^n

But by definition, the type of quantity \partial_{ij} z^n \partial_k z^n = [ij,k], is called the Christoffel symbols of the first kind

Lets add then \partial_k g_{ij} and \partial_j g_{ik}

\partial_k g_{ij} + \partial_j g_{ik} = \partial_{ik} z^n \partial_j z^n + \partial_i z^n \partial_{jk} z^n + \partial_{ij} z^n \partial_k z^n + \partial_i z^n \partial_{jk} z^n = 2 \ \partial_i z^n \partial_{jk} + \partial_{ik} z^n \partial_j z^n + \partial_{ij} z^n \partial_k z^n

Let's remove now unwanted terms by substracting \partial_i g_{jk} from the sum :

\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk} = 2 \ \partial_i z^n \partial_{jk} + \partial_{ik} z^n \partial_j z^n + \partial_{ij} z^n \partial_k z^n - \partial_{ij} z^n \partial_k z^n - \partial_j z^n \partial_{ik} z^n


\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk} = 2 \ \partial_i z^n \partial_{jk}

\frac{1}{2} (\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk}) = \partial_i z^n \partial_{jk} = [jk,i]

Therefore Christoffel symbols of the first kind could be written :

[jk,i] = \frac{1}{2} (\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk})