1,531 reputation
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bio website math.berkeley.edu/~pablo
location Berkeley
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visits member for 5 years
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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