1,609 reputation
1024
bio website math.berkeley.edu/~pablo
location Berkeley
age
visits member for 5 years, 11 months
seen Aug 4 at 16:07

Jun
24
awarded  Citizen Patrol
Jun
24
comment Properties of schemes determined by field valued points
I have edited the question to address the comments thus far. Is the phrasing still problematic? Can it be considered for re-opening?
Jun
16
comment Properties of schemes determined by field valued points
@WillSawin I have made edits to state the question more precisely. Mostly I want to consider $X(K)$ as just a topological space. Is there other structure of $X(K)$ from a functorial point of view that come to mind?
Jun
16
revised Properties of schemes determined by field valued points
added 858 characters in body
Jun
15
asked Properties of schemes determined by field valued points
Feb
9
awarded  Nice Question
Jan
9
awarded  Notable Question
Oct
15
awarded  Notable Question
Sep
28
awarded  Yearling
Sep
26
comment Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
Both of the previous claims can be boiled down to a Lie algebra calculation using that $Gr_{k,n} = GL_n/P$. Factor $P = GL_k\times GL_{n-k}\times U$. Let $U^-$ be the transpose of $U$. Then tangent space of $GL_n/P$ is naturally $Lie(U^-)$ and can check tangent map is injective because $Lie(U^-)$ acts non trivially on highest wt. vector.
Sep
26
comment Conceptual algebraic proof that Grassmannian is closed in Plucker embedding
Closed immersion? The map $GL_n\to \mathbb{P}(\wedge^k \mathbb{C}^n)$ is not even injective so cannot be an immersion. It is a submersion; since $GL_n$ acts transitively on the image so its enough to check map on tangent spaces is surjective at identity. Similarly, the map from $Gr_{k,n} \to \mathbb{P}(\wedge^k \mathbb{C})$ is a closed embedding again because map on tangent spaces is injective in this case.
Aug
16
awarded  Nice Question
Aug
14
comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$
The idea is that if $P$ is a nontrivial $G$ bundle on $U$ and $W$ is any etale neighborhood of $(u^k,v^k,u v)$ then $P$ restricted to $W \cap U$ remains nontrivial.
Aug
14
comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$
I believe this argument can be adapted to show that on $Spec\ A$ every principal $G$ bundle is trivial whenever $G$ is a split reductive group.
Aug
13
accepted vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$
Aug
13
comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$
If I understand correctly then the conclusion should be the same for higher rank: any vector bundle on $V$ is trivial so $Spec\ A$ you will see $B_{r_1} \oplus ... \oplus B_{r_n}$ and the fiber at $(u^k,v^k, u v)$ will be bigger than $n$ unless all $r_i = 0$. So $Spec\ A$ has only trivial vector bundles.
Aug
13
comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$
Great! The point I was missing is if a locally free on $U$ is to extend to a locally free on $Spec A$ it must be the reflexive extension.
Aug
13
asked vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$
Jul
2
awarded  Inquisitive
Jul
2
awarded  Curious