bio | website | math.berkeley.edu/~pablo |
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location | Berkeley | |
age | ||
visits | member for | 5 years, 3 months |
seen | 2 days ago | |
stats | profile views | 1,261 |
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 |
Apr 15 |
comment |
Holomorphic convergence conditions on $\mathbb C((z))$-valued points of a group $G$
Oh and $T \subset G$ is a maximal torus. |
Apr 15 |
accepted | Holomorphic trivialization of $(x,y) \subset \mathbb{C}[x,y]/(y^2 - x^3 + x)$ |
Apr 14 |
awarded | Critic |
Apr 14 |
comment |
Holomorphic convergence conditions on $\mathbb C((z))$-valued points of a group $G$
There is an obvious thing that breaks in passing from $\mathbb{C}[z^\pm]$ to $\mathbb{C}((z))$: the former has ring homomorphisms to $\mathbb{C}$ and the latter doesn't. On the other hand $G(\mathbb{C}((z))) = G(\mathbb{C}[z^{-1}])T(\mathbb{C}[z^\pm])G(\mathbb{C}[[z]])$. You can talk about convergence for $G(\mathbb{C}[[z]])$ in the sense that you can ask if $g \in GL_n(\mathbb{C}[[z]])$ lands in $G(\mathbb{C}[z]/z^n) \subset GL_n(\mathbb{C}[z]/z^n)$ for all $n$. |
Apr 14 |
asked | Holomorphic trivialization of $(x,y) \subset \mathbb{C}[x,y]/(y^2 - x^3 + x)$ |
Mar 5 |
comment |
The localization of a regular local ring is regular
It seems to me $\sum_i x^n/y^{n^2}$ is an element of $k((y))[[x]]$ that will not be in $k[[x]]((y))$. In the latter ring any element can be brought into $k[[x,y]]$ by multiplying by a finite power of $y$. See mathoverflow.net/questions/34010/… for related differences. |
Feb 20 |
awarded | Popular Question |