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8:01 PM
So I'm looking at $X=V(xv-yu)\subseteq\Bbb A^2_k\times\Bbb P^1_k$, where $x,y$ are coordinates on $\Bbb A^2_k$ and $u,v$ on $\Bbb P^1_k$ (the blowup of the affine plane at the origin) and at the projection $X\to\Bbb A^2_k$
I want to compute $\Omega_{X/\Bbb A^2_k}$ on an affine cover, so I want to use the usual two patches on $\Bbb P^1_k$ given by $U_1=\{u\neq 0\}$ with coordinate $a=v/u$ and $U_2$ defined similarly with coordinate $b=u/v$
Let $X_1=\Bbb A^2_k\times U_1\simeq \Bbb A^3_k$, here $X$ is cut out by $xa-y$, so $X\cap X_1$ is $\mathrm{Spec}A$ where $A= k[x,y,a]/(xa-y)$
And $\Omega_{X_1/\Bbb A^2_k}$ should be the $A$-module generated by $da$, with relation $xda$. Makes sense so far?
 
yes
 
why do people make the same observations independently at the same time?
Leibniz and Newton both came up with calculus at the same time
and then Tate and Iwasawa came up with "Tate's thesis" at the same time
 
With a completely symmetric argument $\Omega_{X_2/\Bbb A^2_k}$ is the $k[x,y,b]/(x-yb)$ module generated by $db$ with relation $ydb$
 
i think you meant $b$ but yes
 
I did, thanks
 
8:13 PM
@Leaky probably timing? Like the theory reaches an impasse at a certain time so you have several people working on the same thing
Also hi all
 
gott kvöld
 
So now I want to find the support of $\Omega_{X/\Bbb A^2}$. I think that $\Omega_{X_1/\Bbb A^2}$ is supported at the single point $(x)$ and the other one at $(y)$, but this feels suspicious
 
Gott kvøld :)
 
hi , does anyone here know where I could find a script with useful properties for union and interesection of indexed families of sets ( such like distributivity and stuff ) ?
 
Well if it's supported at the single point $(x)$ (which is not closure), it's also supported along the closure of this point - which is exactly the exceptional divisor in your chart

So that seems fine to me
 
8:28 PM
Ah wait I know nothing about divisors
 
oh I just mean the $\mathbb{P}^1$
i.e. the pre-image of (0,0)
 
And why is it not closed?
 
the point $(x)$ is not closed in $k[x,y,a]/(xa-y) = k[x,a]$!
 
Ohh, of course, it's not maximal there, derp
 
here's maybe how you can think about it: the blow up is an isomorphism away from the origin, and you should expect $\Omega_{X/\mathbb{A}^2}$ to be supported away from the locus where your map is an isomorphism
 
8:31 PM
Hmm so the module $(Ada)/(xda)$ should be isomorphic to $A/(x)$ and its support should be $\mathrm{Spec}(A/(x))$ which is isomorphic to $\mathrm{Spec}(k[a])$ as you pointed out so it's supported on a line
 
yeah
 
@loch That makes sense intuitively, there should be no "relative tangent vectors" where I have an isomorphism
 
exactly!
 
So the other half is supported on $\mathrm{Spec}(k[b])$ by symmetry and now it's a question of deciding how those two supports fit together and that should be as a copy of $\Bbb P^1$ (which makes sense with the origin heuristic you were talking about!)
 
Is proofwiki.org reliable website ?
 
8:36 PM
mostly
 
@AlessandroCodenotti yep
 
Also I can conclude immediately that $\Omega_{X/\Bbb A^2}$ is not locally free on $X$ since it's not supported everywhere, right?
 
yep
 
Awesome, thanks!
 
is this chat reliable ?
:)
 
8:44 PM
We could say yes, but that's exactly what we'd say it the chat were unreliable
 
@JamesGroon no it isn't
including me in particular
 
That situation reminds me of how you cannot trust theories talking about their own consistency. A theory can swear to be consistent while it is in fact inconsistent, but even worse a theory can prove its own inconsistency while being consistent
 
@AlessandroCodenotti might be interesting to note that in your charts, your \P^1 is cut out by a single element. This is in fact a general phenomenon -- it will turn out to define the universal property of blow up -- i.e. the pre-image of the locus you're blowing up will be something which is locally cut out by a single equation
 
Oh, I see
We only talked about blowing up $\Bbb A^n$ at the origin this term, we should talk about blow ups in general in the next semester
 
 
1 hour later…
10:09 PM
@AlessandroCodenotti what, a theory can prove its own inconsistency?
 
10:27 PM
Problem: Give an example of a bounded positive function on [0,1] that is not Lebesgue integrable....Since the measure space is finite and the function of interest is bounded, isn't this equivalent to finding a non-measurable function? If so, by Vitali's theorem there is a nonmeasurable subset $A$ of $[0,1]$, so take $f = 1_{A} + \frac{1}{2}$. Would this work?
 
@user193319 LGTM
 
Sweet! Thanks!
 
11:20 PM
Feels like when a video game gives you a map of all the places to explore
 
@AkivaWeinberger That orientation though.
 
Yeah I'm not quite sure what exactly the projection is
There's an absolutely massive selection in Asia, to no one's surprise
 
i'm kind of surprised there are so little in the US...
 
HCM = Ho Chi Minh, I assume
 
the projection looks like they tried flattening the globe into a thin sheet of paper and then tried twisting the paper...
 
11:25 PM
I feel like Beijing and Tianjin are basically the same place
so that area is like a double-megacity
Same thing with Guangzhou and Shenzhen (and Hong Kong which is by there)
See also: Kolkata and Dhaka
though they're in different countries so I guess that doesn't really count
 
a formal definition is needed.
 
Well, Guangzhou, Shenzhen, and Hong Kong are all part of the Pearl River delta
Tianjin borders Beijing
 
@LeakyNun yes
 
11:50 PM
@MatheinBoulomenos why?
 
because $F^\times/U$ is discrete
the dual of a discrete group is compact
oh wait is the codomain of your characters $\Bbb C^\times$ or $S^1$?
if it's $\Bbb C^\times$, then it's not compact, but for $S^1$ it is
 
oh
@MatheinBoulomenos thanks
the codomain is $\Bbb C^\times$ but it's irrelevant right
 
no, that's not irrelevant
 
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