Ciro Santilli @cirosantilli 40

The set of all algebraic numbers forms a field.

This field contains all of the rational numbers, but it is a quadratically closed field.

Like the rationals, this field also has the same cardinality as the natural numbers, because we can specify and enumerate each of its members by a fixed number of integers from the polynomial equation that defines them. So it is a bit like the rationals, but we use potentially arbitrary numbers of integers to specify each number (polynomial coefficients + index of which root we are talking about) instead of just always two as for the rationals.

Each algebraic number also has a degree associated to it, i.e. the degree of the polynomial used to define it.

A set $S$ plus any number of functions $f_i:S\times S\to S$, such that each $f_i$ satisfies some properties of choice.

Key examples:

A vector field with a bilinear map into itself, which we can also call a "vector product".

Note that the vector product does not have to be neither associative nor commutative.

Examples: en.wikipedia.org/w/index.php?title=Algebra_over_a_field&oldid=1035146107#Motivating_examples

A solution to a computational problem!

Draft by Ciro Santilli with cross language input/output test cases: github.com/cirosantilli/algorithm-cheat

By others:

Metric created by IonQ.

Materials paywalled: www.cs.ox.ac.uk/teaching/courses/2022-2023/algorithms/

Funding rounds:

About their qubit:

A linear map $A$ can be seen as a (1,1) tensor because:is a number, $v\ast$. is a dual vector, and W is a vector. Furthermoe, $T$ is linear in both $v\ast$ and $w$. All of this makes $T$ fullfill the definition of a (1,1) tensor.