Birch and Swinnerton-Dyer conjecture in two minutes by Ciro Santilli Updated 2024-12-15 +Created 1970-01-01
Summary:
- overview of the formula of the BSD conjecture
- definition of elliptic curve
- domain of an elliptic curve. Prerequisite: field
- elliptic curve group. Prerequisite: group
- Mordell's theorem lets us define the rank of an elliptic curve over the rational numbers, which is the . Prerequisite: generating set of a group
- reduction of an elliptic curve from to lets us define as the number of elements of the generated finite group
You select a generating set of a group, and then you name every node with them, and you specify:
- each node by a product of generators
- each edge by what happens when you apply a generator to each element
Not unique: different generating sets lead to different graphs, see e.g. two possible en.wikipedia.org/w/index.php?title=Cayley_graph&oldid=1028775401#Examples for the
How to build it: math.stackexchange.com/questions/3137319/how-in-general-does-one-construct-a-cycle-graph-for-a-group/3162746#3162746 good answer with ASCII art. You basically just pick each element, and repeatedly apply it, and remove any path that has a longer version.
Immediately gives the generating set of a group by looking at elements adjacent to the origin, and more generally the order of each element.
TODO uniqueness: can two different groups have the same cycle graph? It does not seem to tell us how every element interact with every other element, only with itself. This is in contrast with the Cayley graph, which more accurately describes group structure (but does not give the order of elements as directly), so feels like it won't be unique.
A concise to describe a specific permutation.
A permutation group can then be described in terms of the generating set of a group of specific elements given in cycle notation.
E.g. en.wikipedia.org/w/index.php?title=Mathieu_group&oldid=1034060469#Permutation_groups mentions that the Mathieu group is generated by three elements:which feels quite compact for a simple group with 95040 elements, doesn't it!
- (0123456789a)
- (0b)(1a)(25)(37)(48)(69)
- (26a7)(3945)
Elements of a Lie algebra can (should!) be seen a continuous analogue to the generating set of a group in finite groups.
For continuous groups however, we can't have a finite generating set in the strict sense, as a finite set won't ever cover every possible point.
But the generator of a Lie algebra can be finite.
And just like in finite groups, where you can specify the full group by specifying only the relationships between generating elements, in the Lie algebra you can almost specify the full group by specifying the relationships between the elements of a generator of the Lie algebra.
This "specification of a relation" is done by defining the Lie bracket.
The reason why the algebra works out well for continuous stuff is that by definition an algebra over a field is a vector space with some extra structure, and we know very well how to make infinitesimal elements in a vector space: just multiply its vectors by a constant that cana be arbitrarily small.
Minimum number of elements in a generating set of a group.
The direct product of two cyclic groups of coprime order is another cyclic group Updated 2024-12-15 +Created 1970-01-01
You just map the value (1, 1) to the value 1 of , and it works out. E.g. for , the group generated by of (1, 1) is:
0 = (0, 0)
1 = (1, 1)
2 = (0, 2)
3 = (1, 0)
4 = (0, 1)
5 = (1, 2)
6 = (0, 0) = 0