Schemes 1: Introduction

  Рет қаралды 44,944

Richard E Borcherds

Richard E Borcherds

Күн бұрын

Пікірлер: 28
@bobclarke5543
@bobclarke5543 3 жыл бұрын
Absolutely the clearest, most comprehensible introduction to the subject I've ever seen. Clearer, more business-like, and better motivated than Vakil's attempt (which, I must say, is also very good but has an entirely different character). I highly recommend this series to anyone who wants to learn algebraic geometry.
@georgediderrich43
@georgediderrich43 2 жыл бұрын
What a pleasure to view these lectures. Professor Borcherds rocks!
@Ibakecookiess
@Ibakecookiess 4 жыл бұрын
Watching this from Colombia. Thank you!
@ipudisciple
@ipudisciple 4 жыл бұрын
18:25 I think we have two options. We can choose a dimension d and consider the sheaf of vector fields of dimension d. Or we can lump these all together and work with vectors in the direct sum ⨁ ℝ^n, which is just the countably dimensional vector space. Technically, Richard says the second, but I think he might be thinking of the first.
@juliavilageliu6797
@juliavilageliu6797 11 ай бұрын
You really helped me through my final degree project :) Thank you so much
@vinayakgupta2368
@vinayakgupta2368 4 жыл бұрын
Hey, i am from india. Thank you for you these free lectures.
@domingogomez6999
@domingogomez6999 3 жыл бұрын
Sorry, it might be trivial, I don't see why the "constant" presheaf is not a sheaf. I mean that, if the empty set, U_1, U_2 and their union U = U_1\union U_2 are the only open sets, why does F(U) = AxA hold?
@ruinenlust_
@ruinenlust_ 3 жыл бұрын
Constant pre-sheaf assigns A to every U. Clearly, if F were to be a sheaf then F(U) must be AxA. But, since F assigns A to every open set, F cannot be a sheaf in general
@domingogomez6999
@domingogomez6999 3 жыл бұрын
@@ruinenlust_ the clearly part is not evident for me. Could you develop more, please?
@harshavr
@harshavr 3 жыл бұрын
@@domingogomez6999 If U is partitioned into U1 and U2 with no intersection, then the sheaf condition says that section on U1 and section on U2 uniquely determines a section on U ( the compatibility condition is trivial because U1 and U2 dont intersect), So F(U)=F(U1)xF(U2)=AxA. The key here is that space is disconnected, For a connected space, the constant sheaf is a sheaf. (U1 and U2 will intersect above and the constant function should restrict to the same constant on the intersection) For any space, the locally constant functions form a sheaf.
@yiyangjia1967
@yiyangjia1967 2 жыл бұрын
look at the injectivity of the exact sequence
@premkumar-so3ff
@premkumar-so3ff 4 жыл бұрын
Please make videos explaining on cohomology of sheaves unit 3 of hartshrone. Thank you
@mathgeeks3598
@mathgeeks3598 4 жыл бұрын
nice, video, i am in uSa. thanks for good videos
@禿田与平
@禿田与平 3 жыл бұрын
Thank you so much for your online couse !
@markcanedo3
@markcanedo3 2 жыл бұрын
Thank you for this magistral lessons, professor
@nostalgia_1439
@nostalgia_1439 Жыл бұрын
2:35 Isn't k[x] a k-algebra? I mean I think it's even a finite k-algebra? What is his definition of k-algebra? I know what an algebra \phi: R\to S is (e.g. in the sense of Altman Kleinman's commeutative algebra text, and "Algebra" by Hungerford)
@ac-dp3jk
@ac-dp3jk 3 жыл бұрын
Thank you very much for your videos !
@donatoebbasta2797
@donatoebbasta2797 2 жыл бұрын
Does anyone have a clear cut pdfLaTeX-compiled version of these lectures?
@okoyoso
@okoyoso Жыл бұрын
I believe his name is pronounced "Hart's horn" (hart being an old word for a stag, type of buck)
@sdfdsf4162
@sdfdsf4162 Жыл бұрын
haha me too
@okoyoso
@okoyoso 10 ай бұрын
@@sdfdsf4162 You too as in you pronounce it that way too?
@sdfdsf4162
@sdfdsf4162 10 ай бұрын
@@okoyoso yeeep. dude im an eastern europe dog, of course i suck speaking eng =)
@okoyoso
@okoyoso 10 ай бұрын
@@sdfdsf4162 I mean a lot of people call him "hart shorne"
@tapioms
@tapioms 6 ай бұрын
🎉
@HERŞEYKÖTÜ
@HERŞEYKÖTÜ 2 жыл бұрын
I dont understand how people love this series, I for myself, find it very confusing and not clear
@guiwenluo1774
@guiwenluo1774 2 жыл бұрын
Second. I found it not newcomer friendly. Probably those who love this series are already learnt Algebraic geometry for at least one time.
@mbgdemon
@mbgdemon 2 жыл бұрын
@@guiwenluo1774 That is why the series is called Algebraic Geometry II.
@bennoarchimboldi6245
@bennoarchimboldi6245 2 ай бұрын
People love him because he’s a fields medalist not because he’s a good teacher.
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