From Simple Groups to Quantum Field Theory

[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 :

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

Which in polar coordinates could also be written :

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

Therefore, complex addition is just matrix addition :

$$(a+ib) + (c+id) = \left[ \begin {array}{cc} a&-b\\\noalign{\medskip}b&a\end {array} \right] + \left[ \begin {array}{cc} c&-d\\\noalign{\medskip}d&c\end {array} \right] = \left[ \begin {array}{cc} \left\{ a+c \right\} &- \left\{ b+d \right\} \\\noalign{\medskip} \left\{ b+d \right\} & \left\{ a+c \right\} \end {array} \right] = (a+c) + i (b+d)$$

And complex substraction is matrix substraction :

$$(a+ib) - (c+id) = \left[ \begin {array}{cc} a&-b\\\noalign{\medskip}b&a\end {array} \right] - \left[ \begin {array}{cc} c&-d\\\noalign{\medskip}d&c\end {array} \right] = \left[ \begin {array}{cc} \left\{ a-c \right\} &- \left\{ b-d \right\} \\\noalign{\medskip} \left\{ b-d \right\} & \left\{ a-c \right\} \end {array} \right] = (a-c) + i (b-d)$$

Complex multiplication is matrix multiplication :

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

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

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

we can then write that :

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

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

Therefore,

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

Proof:

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

 

[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 :

$$g_{ij} = ( \frac{\partial }{\partial x^i} . \frac{\partial }{\partial x^j} ) = < \frac{\partial }{\partial x^i} , \frac{\partial }{\partial x^j} >$$

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 :

$$< f , g > = \int_{b}^{a} \overline{f(t)}.g(t).dt$$

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 :

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