Introduction to the work of Mirzakhani
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- čas přidán 5. 11. 2014
- Maryam Mirzakhani wins 2014 Fields medal - first woman to do so; AWARD CEREMONY FIELDS MEDAL SEOUL 2014 PART2; Maryam Mirzakhani Is First Woman Fields Medalist; 2014 Fields medal - Iranian Maryam Mirzakhani is first female winner; Maryam Mirzakhani (مریم میرزاخانی), an Iranian-born professor is first woman to win Fields Medal; ICM2014 Curtis McMullen, Laudation for Fields Medalist: Maryam Mirzakhani; Maryam Mirzakhani on "Dynamics on the Moduli Spaces of Curves", I; Who Is Maryam Mirzakhani?; Maryam Mirzakhani on "Dynamics on the Moduli Spaces of Curves", III
- Hudba
What a great lost for the humanity: The first woman of the Fields medal. I had invited her to come to France as a honorary speaker but she gently answered me that she cannot travel too much.
bless you for inviting her.. so very sad that we have lost her..
Thanks a lot for the clear explanation! As a physicist who knows something about the Poincare's disk, it's thrilling to learn the connection between hyperbolic surfaces and moduli space. RIP Prof. Mirzakhani ...
Rip maryam 😔❤💐
Great
We can imagine what she could have achieved if she had more time!😔😔
Unbelievable sad for all people who loved her, condolences to her Father& Mother, husband Jan Vondrák , her little daughter Anahita and all her families.
She did not die in vain or at all, she lives in the mathematics. There, we are able to find her, and that widely spoken beauty. There, we are able to find her. Thank you so much for posting this introduction!
great thank you
Excellent!
Thank you very much Professor for a wonderful explanation.
when I learned about her death my eyes were full of tears, its a great loss for humanity we lost a one of the great minds. RIP
I just stumbled on this. My soul became immediately sad. I am not a mathemation, if there is such a thing, at all. I am a painter, but I am painting on 2 cycles that have from my point of view a certain connection. One is the development of the term field and that other I developed out of the "new Geometry" from Raimondo Lullo, the perfect fluid pink square and cube of aesthetics. Thanks for your unknown inspiration Maryam! My condolences to her family.
55:45 Any quotient of H by a finite index subgroup of SL_2(Z) has finite area. Since the measure on H shrinks rapidly as you approach vertical infinity it's easy to see that Y(0) has finite area, and any such quotient can be covered by finitely many SL_2(Z) translations of Y(0).
Thank you for making this, I really appreciate it.
Clear and concise.
UM Thank you for watching!
schang1971 Your channel seems like a gold mine! Very interesting topics, at least to me.
I'll be watching them all :-)
Rest in peace Prof. Mirzakhani.
We will celebrate your life and achievements, here is to Dr. Maryam
is there any video available that Maryam explains her work in Farsi ?
49:40 Please, EICHLER-Shimura congruence relation (theorem).
RIP Maryam
Line space in R^2 is projective dim. 2 real = cross cap. Embedding into projective dim. 2 complex makes the cross cap into a torus, since genus 1 was obtained? The projective plane is not orientable, like the klein bottle, which is by lifting in 4 dim real(C^2) then homeomorphic to the torus?
+Brian Elsen Hi there. They were two separate examples: the collection of lines and the collection of elliptic curves. Thanks for watching :)
RIP Maryam!
Sad loss, may RIP
Rest in peace young guineas lady
can we approach the sphere the other way around thus rendering it with negative curvature?
what happens to around the handle with the metrics and curvature?
Given a space with two handles(as shown in the example you gave) can we isolate the handles or we need to consider them analytically as an equation that can not be solved in the general case(no 0) but can be mapped in n-D space in way that allows us to control the capacity of the spaces that instantiate it?
Can we say that in a cases with more than one handles there are paired handles and un-paired handles(for example in the previous question a pair will be an equation and the "free" one will be a second equation)?
Is it correct that the number of handles relates to the number of unique parameters used to construct the space(s)?
czcams.com/video/Q-H0jEsJUWk/video.html
why is it the same line? It is effectively a new instance of the same type, but then given the concept of a module space it is defined by the arch-type of a line, so any "instance" must be considered unique(unless we are looking for intersecting lines only, which however must be specified somehow)
czcams.com/video/Q-H0jEsJUWk/video.html
I lost it! We have constructed an arch-type(a module space), then for some reason we reduced the information it holds by defining what an unique instance is(no parallel lines in the example), and then we somehow rendered it to a sphere by dividing each line on itself thus kind of fabricating a positive curvature. Why?
czcams.com/video/Q-H0jEsJUWk/video.html
?
czcams.com/video/Q-H0jEsJUWk/video.html
but what if the handles(a=0 and b=0) are not symmetrical on the sphere? You will end up with two different spaces that require projection the way you showed it at the beginning(metrics) since |a->b| != |b->a|. There is still symmetry since the original surface has a curvature of 1, but by introducing more than one handle you bend the space in the general case effectively filtering the contained instances in two sub-categories and by doing so taking advantage of the different metrics(basically compressing the available instances in regards of absolute position value).
czcams.com/video/Q-H0jEsJUWk/video.html, is this matrix a type of movement/coordinate system matrix? Does it mean that we can reduce the instance count in the global space(all projections/planes) by constructing clusters of similar(isomorph) instances defined by variations of the members of the matrix, where all instances that can not be fit in newly constructed domain of clusters are considered invalid for that domain and thus render as a handle(an invalid cluster- sort of the error/constant remaining after the differentiation)?
Is the determinant of the matrix definitive for the curvature index of domain it is used to construct or the relation is coincidental? If the latter is not a coincidence, given the classification of spaces based on their curvature index, how do we handle cluster domains with higher indexes- by decomposing those to fit the given classification manipulating the metrics or there is some other way?
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Hi ! I am a first year undergraduate student but i want to understand this. Can you please tell me the prerequisites ?
Probably Topology and Complex Analysis for this video; for further learning, knowing a bit about Algebraic Topology and Manifolds will also help!
listening to this I realise how big of an idiot I really am and wat kind of mathimatical "gangster rebel" superstar she actually was. R.I.P.
what a same exact feel i have
Big loss, RIP Professor Mirzakhani...RIP..
Rip Maryam
Today she passed away because of cancer
RIP Maryam