bio | website | math.berkeley.edu/~vivek |
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location | Berkeley, CA | |
age | 31 | |
visits | member for | 4 years, 6 months |
seen | yesterday | |
stats | profile views | 3,346 |
Sep 15 |
comment |
If the direct image of f preserves coherent sheaves on notherian schemes,how to show f is proper?
why should it even be separated? the map from the affine line with the doubled origin to the affine line seems like it ought to preserve coherent sheaves |
Sep 14 |
comment |
What are the exact holomorphic Lagrangians in complex 2-space?
@JasonStarr, sure, but those are all pretty similar to the zero section. |
Sep 14 |
asked | What are the exact holomorphic Lagrangians in complex 2-space? |
Sep 10 |
accepted | curve through a point avoiding an hypersurface, II |
Sep 10 |
asked | curve through a point avoiding an hypersurface, II |
Sep 8 |
revised |
Singular points of algebraic varieties and parametrization by Puiseux series
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Sep 8 |
revised |
Singular points of algebraic varieties and parametrization by Puiseux series
added 30 characters in body |
Sep 8 |
comment |
Singular points of algebraic varieties and parametrization by Puiseux series
About the higher order coefficients: without the requirement that $\gamma$ converge and certainly without the requirement that $\gamma(1)=q$, there's a name for such spaces -- they are called arc spaces, and are heavily studied as invariants of singularities. I think they can be pretty complicated; what you're asking is for is a linear condition on the subspace of convergent arcs, which sounds at least as complicated. |
Sep 8 |
answered | Singular points of algebraic varieties and parametrization by Puiseux series |
Sep 8 |
revised |
Linearization instability and singular points of algebraic varieties
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Sep 8 |
comment |
Singular points of algebraic varieties and parametrization by Puiseux series
yeah, I just wanted to point out that you can't literally cut to a plane curve, since I at least find plane curves are easier to think about. |
Sep 8 |
comment |
Singular points of algebraic varieties and parametrization by Puiseux series
Suppose $x,y \ne 0, 1$. Then by nondegeneracy of the Vandermonde determinant, any remaining point $(w, w^2, w^3, z, z^2, z^3)$ must have $(w, z) \in \{0,1,x\} \times \{0,1,y\}$. If $x = 0, 1$, then in fact $(w,w^2,w^3)$ has to be a scalar multiple of $(1,1,1)$; this only happens $w = 0, 1$. Thus in any case your plane hits $V$ in at most 9 points and in particular contains no curves. |
Sep 8 |
comment |
Singular points of algebraic varieties and parametrization by Puiseux series
Why should such a plane exist? Strictly speaking, taking $V$ to be a non-planar curve is already a counterexample. If that's cheating, consider the product of two rational normal cubics, i.e. the variety which is the image in $\mathbb{C}^6$ of $\mathbb{C}^2$ by the map $(s,t) \mapsto (t, t^2, t^3, s, s^2, s^3)$. Say I take any plane through the points $p = (0,0,0,0,0,0)$ and $q = (1,1,1,1,1,1)$. Three points determine a plane; let $(x, x^2, x^3, y, y^2, y^3)$ be a third one. |
Aug 31 |
comment |
Different definitions of linkless graphs
yes, those are clearly different definitions. however the distinction is irrelevant for the purposes of that second paper, since flat implies [KKM]-linkless implies [RST]-linkless implies, as shown in [RST], the existence of some flat embedding. |
Aug 26 |
revised |
Twist in identification with singular cohomology
deleted 7 characters in body |
Aug 25 |
revised |
Twist in identification with singular cohomology
edited body |
Aug 25 |
answered | Twist in identification with singular cohomology |
Aug 22 |
revised |
How do I find coefficients of a product expansion
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Aug 22 |
revised |
How do I find coefficients of a product expansion
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Aug 22 |
comment |
How do I find coefficients of a product expansion
I am vaguely aware that there's an answer to this question of the form ``take a logarithm and do symmetric function combinatorics'', but I wasn't able to reconstruct the details. |