Little silly thing, on the first slide you forgot a square on the R for the circle equation. 9:11 Funny story, just after having my geometry 1 exam (basically linear algebra) I ended up chatting a bit with my professor since I was the last person that day and while packing things up he just started to semi-jokingly list all the reasons why the complex numbers are so much better than the reals.
2:24 > _"algebraic geometry studies zero sets of polynomials"_ i guess then, that it's different than Clifford's "geometric algebra" right?? 8:22 yeah, it's definitely different
Exactly 👍. AG is quite different in spirit from geometric algebra. The latter wants to include “geometry” (mostly vectors and friends) into algebra, while AG tries to make geometry as algebraic as possible.
@@VisualMath awesome note. thanks for sharing ur thoughts (: > _"latter wants to include “geometry” (mostly vectors and friends) into algebra, while AG tries to make geometry as algebraic as possible."_
The way you've defined SL_n(K), as a set of particular matrices whose determinant is 1 led me to think that SL_n it's not in fact the variety, but actually the set of loci of zeroes of the determinants of these matrices is in fact the set of varieties , not SL_N(K) . Am I correct, or is there something I'm not seeing ? I'm not a mathematician, so it's important for me to get the precise idea....
Ok, let me give it a shot; say for n=2 to have easier notation. Then SL2=all vectors [a,b,c,d] in C^4 with ac-bd-1=0. All that is done is to flatten the matrix [[a,b],[c,d]] into a vector and we think of the four possible entries as variables. The determinant = 1 condition is then a polynomial equation in four variables. I hope that helps!
@@VisualMath Thanks for the answer, although it didn't addresed my question, maybe I didn't state it clearly. I understood that the determinant is a polynomial in the case, and that its zeroes constitute a variety. The point is that the zeroes of the determinant make a variety, not the SL(K). Unless the SL(K) is constitute by the zeroes of the determinant, but it's clearly the case. So, why to call SL(K) a variety ? I understood that a variety is the locus of the zeroes of a polynomial curve or surface.
@@JoaoKogler Hmm, I think I answered your question. But maybe I misunderstood your question, so let me try again, say for SL2 😀 SL2 = “matrices [[a,b],[c,d]] with ad-bc-1=0” = “points in K^4 vanishing for f(a,b,c,d)=ad-bc-1” = an affine variety That is why I would call SL2 an affine variety. Does this answer you question?
@@VisualMath Thanks, again. After watching your 4th lesson, on Idelas & Varieties, it occured to me now that maybe the locus of zeroes of the determinant of a particular SLK matrix form a variety, a geometric object, while SLK would be its algebraic counterpart, an ideal related to that variety. Well, I'm just guessing by now, I hope I'm not bothering with this.... Anyway, let me watch a bit more of your next lessons, maybe I should have not worry about the question for a while.
You are wrong !! Modern v2 of Algebraic Geometry doesnt constitute Grobner Basis , Homotopy Continuation . It is Derived Algebraic Geometry or Higher Algebraic Geometry,
@@VisualMath very much so... Perfect presentation... You're at 23k now... I'm guessing before you finish this series you will have surpassed 30k...possibly even 35k...Hopefully the RUclips algorithm does its thing 🙏🏼
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
Little silly thing, on the first slide you forgot a square on the R for the circle equation.
9:11 Funny story, just after having my geometry 1 exam (basically linear algebra) I ended up chatting a bit with my professor since I was the last person that day and while packing things up he just started to semi-jokingly list all the reasons why the complex numbers are so much better than the reals.
Ah, didn't I say the R=1? No? Too bad 😨
Anyway, thanks for spotting 😀 I have put a warning into the description.
2:24 > _"algebraic geometry studies zero sets of polynomials"_
i guess then, that it's different than Clifford's "geometric algebra" right??
8:22 yeah, it's definitely different
Exactly 👍. AG is quite different in spirit from geometric algebra. The latter wants to include “geometry” (mostly vectors and friends) into algebra, while AG tries to make geometry as algebraic as possible.
@@VisualMath awesome note. thanks for sharing ur thoughts (:
> _"latter wants to include “geometry” (mostly vectors and friends) into algebra, while AG tries to make geometry as algebraic as possible."_
@@yash1152 Welcome 👍
Nicely presented!
Thanks for the feedback, that is very much appreciated ☺
I hope you will enjoy the series.
The way you've defined SL_n(K), as a set of particular matrices whose determinant is 1 led me to think that SL_n it's not in fact the variety, but actually the set of loci of zeroes of the determinants of these matrices is in fact the set of varieties , not SL_N(K) . Am I correct, or is there something I'm not seeing ? I'm not a mathematician, so it's important for me to get the precise idea....
Ok, let me give it a shot; say for n=2 to have easier notation.
Then SL2=all vectors [a,b,c,d] in C^4 with ac-bd-1=0. All that is done is to flatten the matrix [[a,b],[c,d]] into a vector and we think of the four possible entries as variables. The determinant = 1 condition is then a polynomial equation in four variables.
I hope that helps!
@@VisualMath Thanks for the answer, although it didn't addresed my question, maybe I didn't state it clearly. I understood that the determinant is a polynomial in the case, and that its zeroes constitute a variety. The point is that the zeroes of the determinant make a variety, not the SL(K). Unless the SL(K) is constitute by the zeroes of the determinant, but it's clearly the case. So, why to call SL(K) a variety ? I understood that a variety is the locus of the zeroes of a polynomial curve or surface.
@@JoaoKogler Hmm, I think I answered your question. But maybe I misunderstood your question, so let me try again, say for SL2 😀
SL2 = “matrices [[a,b],[c,d]] with ad-bc-1=0” = “points in K^4 vanishing for f(a,b,c,d)=ad-bc-1” = an affine variety
That is why I would call SL2 an affine variety. Does this answer you question?
@@VisualMath Thanks, again. After watching your 4th lesson, on Idelas & Varieties, it occured to me now that maybe the locus of zeroes of the determinant of a particular SLK matrix form a variety, a geometric object, while SLK would be its algebraic counterpart, an ideal related to that variety. Well, I'm just guessing by now, I hope I'm not bothering with this.... Anyway, let me watch a bit more of your next lessons, maybe I should have not worry about the question for a while.
@@JoaoKogler If I can help you with anything, do not hesitate to ask!
such a nice video! Thank you so much
You are very welcome. I hope you will enjoy your journey through AG 😚
Let's goooo brooo 🍾🍾🍾🤘🏼🤘🏼🤘🏼🙌🏼🙌🏼🙌🏼💪🏼💪🏼💪🏼🙏🏼🙏🏼🙏🏼🙏🏼
On it, friend!
You are wrong !! Modern v2 of Algebraic Geometry doesnt constitute Grobner Basis , Homotopy Continuation . It is Derived Algebraic Geometry or Higher Algebraic Geometry,
Haha, or anything people like to call "modern" 🤣
Those 14 minutes went by in a breeze.... Felt like 90 seconds.... 😢
I hope that means the video was entertaining 😂
@@VisualMath very much so... Perfect presentation... You're at 23k now... I'm guessing before you finish this series you will have surpassed 30k...possibly even 35k...Hopefully the RUclips algorithm does its thing 🙏🏼