Andrew Stout

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Name Andrew Stout
Member for 3 years
Seen May 13 at 19:25
Website
Location Paris
Age 27
I am a Ph.D. Candidate at the Graduate School and University Center of the City University of New York. My research focus is Motivic Integration. I'm also around UPMC this year.
Mar
13
 comment PhD in operator algebras and non-commutative geometry
il est à Vanderbilt
Mar
11
 revised What are the enforceable models of local artinian rings?
added tag for logic just in case someone searchs for questions on logic as this will probably go unanswered for some time.
Mar
10
 comment What are the enforceable models of local artinian rings?
ah, I should mention the following fact: enforceable models are existentially closed. The existentially closed models of local aritinian rings of length at most $l$ containing the field $k$ are Gorenstein aritinian rings of length $l$ containing the algebraic closure of $k$. This is per H. Schoutens' work on existentially closed models of local aritinian rings. Does requiring the model to be enforceable and not just existential closed, give us even more properties?
Mar
10
 asked What are the enforceable models of local artinian rings?
Mar
10
 awarded ● Peer Pressure
Mar
10
 awarded ● Organizer
Mar
7
 comment the perimeter of a non-convex set
*...prove the result for star-shaped domains.
Mar
7
 comment the perimeter of a non-convex set
Yes, it is true. I wonder if it could be true when the boundary is piece-wise Lipschitz. I think the easiest proof is to cover $\Omega$ by star-shaped domains and then prove the result for star-domains. Note that the convex hull of elements of your cover will cover the convex hull of $\Omega$. However, what worries me about this ad-hoc argument is I can't detect where I use the notion of $C^1$ boundary.
Mar
7
 comment finite global dimension vs integral Domain
Yes, technically, $k[x,y]/(y^2-x^3)$ is your counter example then. See this: mathoverflow.net/questions/59981/…
Mar
7
 revised Grothendieck ring of “varieties carrying a function”
edited body
Mar
7
 answered Grothendieck ring of “varieties carrying a function”
Mar
7
 comment Which topology for compactness and continuity?
Take the weakest topology on $A$ that makes $H$ continuous, then prove that $A$ must be compact. You can't do any better than that. Note that H(A) is compact. The rest follows.
Mar
6
 comment Crepant Birational Map on the Blow-up
of course, yes, segre embedding is a blow up of one point of $\mathbb{P}^2$. The exception divisors below should line up in the formula I presented.
Mar
6
 awarded ● Disciplined
Mar
6
 comment Crepant Birational Map on the Blow-up
yes, this is true, $f$ is birational, but it is not universally closed. I made an error.
Mar
6
 comment Crepant Birational Map on the Blow-up
Even if you believe $f$ is a birational morphism, since blow-ups are projective, we have that $X$ has the same hodge polynomial of $\mathbb{P}^n$ and $X'$ has the same hodge polynomial of $(\mathbb{P}^1)^n$, but I have already computed these hodge polynomials in the last question. They are not equal. Therefore, there is no crepant resolution of $X \rightarrow X'$ by Batyrev's or Kontsevich's theorem because $H(X,u,v) \neq H(X',u,v)$.
Mar
6
 comment Crepant Birational Map on the Blow-up
Maybe you don't believe me, but I still contend that $f$ cannot be a birational map. It is projective: factor as the graph of $f$ followed by the projection onto the second factor. $f$ being separated, means that the graph of $f$ is a closed immersion. Thus, $f$ is projective and therefore proper. Kontsevich proved that if $X$ and $Y$ admit a proper birational map, then the hodge numbers must be the same. In the previous question, I show that the hodge numbers are not the same: mathoverflow.net/questions/123700/…
Mar
6
 comment Crepant Birational Map
okay. good luck.
Mar
6
 accepted Crepant Birational Map
Mar
6
 awarded ● Critic
Mar
6
 comment Crepant Birational Map
anyway, this argument I posted still works. $f$ cannot be a birational map or else the hodge polynomials would be the same (by the same work of Kontsevich mentioned above). next time you should either add to your question or start a new question instead of asking a completely different question after someone has answered you.
Mar
6
 comment Crepant Birational Map
That is annoying that you completely changed the question.
Mar
6
 revised Crepant Birational Map
added 2 characters in body
Mar
6
 revised Crepant Birational Map
corrected some minor mistakes; edited body; edited body
Mar
6
 answered Crepant Birational Map
Mar
5
 answered finite global dimension vs integral Domain
Feb
25
 comment group actions on blow-ups
edit: all $z \in Z$
Feb
25
 comment group actions on blow-ups
isn't a blow-up a proper birational morphism? meaning that on $\bar{X} - \pi^{-1}(Z)$ the group action must be induced canonically via the isomorphism $\bar{X} - \pi^{-1}(Z) \cong X $. Thus, you could extend by acting on fibers in $Z$ $g\pi^{-1}(z) = \pi^{-1}(gz)$ for all $g\in G$ and all $Z\in Z$. well, this is a start anyway.
Feb
23
 comment How would you call a subscheme of a smooth $S$-scheme?
I've seen the following definition of quasi-smooth floating around: locally free kähler differentials $\Omega_{X/S}$ and $X/S$ flat (where X/S has no finiteness condition -- e.g., $X/S$ may not be of finite type). Perhaps a better name for this is pro-smooth as I could imagine the central question here is when is $X$ pro-representable by smooth schemes of finite type over $S$. Nevertheless, embeddable is better.