Ciro Santilli @cirosantilli 34

Solution $Z$ for given $X$ and $Y$ of:where $e$ is the exponential map.

If we consider just real number, $Z=X+Y$, but when X and Y are non-commutative, things are not so simple.

Furthermore, TODO confirm it is possible that a solution does not exist at all if $X$ and $Y$ aren't sufficiently small.

This formula is likely the basis for the Lie group-Lie algebra correspondence. With it, we express the actual group operation in terms of the Lie algebra operations.

Notably, remember that a algebra over a field is just a vector space with one extra product operation defined.

Vector spaces are simple because all vector spaces of the same dimension on a given field are isomorphic, so besides the dimension, once we define a Lie bracket, we also define the corresponding Lie group.

Since a group is basically defined by what the group operation does to two arbitrary elements, once we have that defined via the Baker-Campbell-Hausdorff formula, we are basically done defining the group in terms of the algebra.

The size of a set.

For finite sizes, the definition is simple, and the intuitive name "size" matches well.

But for infinity, things are messier, e.g. the size of the real numbers is strictly larger than the size of the integers as shown by Cantor's diagonal argument, which is kind of what justifies a fancier word "cardinality" to distinguish it from the more normal word "size".

The key idea is to compare set sizes with bijections.

An ordered pair of two real numbers with the complex addition and multiplication defined.

Forms both a:

Unlike the simple case of a matrix, in infinite dimensional vector spaces, the spectrum may be continuous.

The quintessential example of that is the spectrum of the position operator in quantum mechanics, in which any real number is a possible eigenvalue, since the particle may be found in any position. The associated eigenvectors are the corresponding Dirac delta functions.

Adds special relativity to the Schrödinger equation, and the following conclusions come basically as a direct consequence of this!

Experiments explained:

Experiments not explained: those that quantum electrodynamics explains like:See also: Dirac equation vs quantum electrodynamics.

The Dirac equation is a set of 4 partial differential equations on 4 complex valued wave functions. The full explicit form in Planck units is shown e.g. in Video 1. "Quantum Mechanics 12a - Dirac Equation I by ViaScience (2015)" at youtu.be/OCuaBmAzqek?t=1010:Then as done at physics.stackexchange.com/questions/32422/qm-without-complex-numbers/557600#557600 from why are complex numbers used in the Schrodinger equation?, we could further split those equations up into a system of 8 equations on 8 real-valued functions.

Every invertible matrix $M$ can be written as:where:Note therefore that this decomposition is unique up to swapping the order of eigenvectors. We could fix a canonical form by sorting eigenvectors from smallest to largest in the case of a real number.

Intuitively, Note that this is just the change of basis formula, and so:

An elliptic curve is defined by numbers $a$ and $b$. The curve is the set of all points $(x,y)$ of the real plane that satisfy the Equation 1. "Definition of the elliptic curves"

Equation 1. "Definition of the elliptic curves" definies elliptic curves over any field, it doesn't have to the real numbers. Notably, the definition also works for finite fields, leading to elliptic curve over a finite fields, which are the ones used in Elliptic-curve Diffie-Hellman cyprotgraphy.

A ring where multiplication is commutative and there is always an inverse.

A field can be seen as an Abelian group that has two group operations defined on it: addition and multiplication.

And then, besides each of the two operations obeying the group axioms individually, and they are compatible between themselves according to the distributive property.

Basically the nicest, least restrictive, 2-operation type of algebra.

Examples:

general linear group over a finite field of order $m$. Remember that due to the classification of finite fields, there is one single field for each prime power $m$.

Exactly as over the real numbers, you just put the finite field elements into a $n\times n$ matrix, and then take the invertible ones.

There are 3: real numbers, complex numbers and quaternions.

Notably, the octonions are not associative.

A function that takes input function and outputs a real number.

Invertible matrices. Or if you think a bit more generally, an invertible linear map.

When the field $F$ is not given, it defaults to the real numbers.

Non-invertible are excluded "because" otherwise it would not form a group (every element must have an inverse). This is therefore the largest possible group under matrix multiplication, other matrix multiplication groups being subgroups of it.

The term is not very clear, as it could either mean:

Given a function $f$:we want to find the points $x$ of the domain of $f$ where the value of $f$ is smaller (for minima, or larger for maxima) than all other points in some neighbourhood of $x$.

In the case of Functionals, this problem is treated under the theory of the calculus of variations.

A metric is a function that give the distance, i.e. a real number, between any two elements of a space.

A metric may be induced from a norm as shown at: Section "Metric induced by a norm".

Because a norm can be induced by an inner product, and the inner product given by the matrix representation of a positive definite symmetric bilinear form, in simple cases metrics can also be represented by a matrix.

One major difference between the elliptic curve over a finite field or the elliptic curve over the rational numbers the elliptic curve over the real numbers is that not every possible $x$ generates a member of the curve.

This is because on the Equation "Definition of the elliptic curves" we see that given an $x$, we calculate $x^3+ax+b$, which always produces an element $y^2$.

But then we are not necessarily able to find an $y$ for the $y^2$, because not all fields are not quadratically closed fields.

For example: with $a=1$ and $b=1$, taking $x=1$ gives:and therefore there is no $y\in \mathbb{Q}$ that satisfies the equation. So $x=1$ is not on the curve if we consider this elliptic curve over the rational numbers.

That $x$ would also not belong to Elliptic curve over the finite field ${\mathbb{F}}_4$, because doing everything $\mathrm{m}\mathrm{o}\mathrm{d}4$ we have:Therefore, there is no element $y\in {\mathbb{F}}_4$ such that $y\times y=2$ or $y\times y=3$, i.e. $2$ and $3$ don't have a multiplicative inverse.

For the real numbers, it would work however, because the real numbers are a quadratically closed field, and $\sqrt{3}\in \mathbb{R}$.

For this reason, it is not necessarily trivial to determine the number of elements of an elliptic curve.

This section is about functions that operate on numbers such as the integers or real numbers.

By default, we think of polynomials over the real numbers or complex numbers.

However, a polynomial can be defined over any other field just as well, the most notable example being that of a polynomial over a finite field.

For example, given the finite field of order 9, $GP(3)$ and with elements $\{0,1,2\}$, we can denote polynomials over that ring aswhere $x$ is the variable name.

For example, one such polynomial could be:and another one:Note how all the coefficients are members of the finite field we chose.

Given this, we could evaluate the polynomial for any element of the field, e.g.:and so on.

We can also add polynomials as usual over the field:and multiplication works analogously.