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Sep
28 |
awarded | Yearling |
Sep
23 |
awarded | Nice Question |
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 $\mathbb{C}[x,y,z]/(x y - z^k)$ |