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Jun
17
awarded  Yearling
Jun
1
comment Is this a known formula? $\det (id, \partial_x, \partial_y)^t (f, g, h)$
@BCLC I found it while calculating automorphisms of rank-two vector bundles on a non-commutative deformation of a surface. This expression comes from rewriting three expressions containing Poisson structures as this determinant. The expressions with Poisson structures don't simplify to anything nice, so I am wondering if maybe this determinant expression is something nice I just don't know.
May
30
asked Is this a known formula? $\det (id, \partial_x, \partial_y)^t (f, g, h)$
May
29
awarded  Inquisitive
May
28
asked Computing the cotangent complex: what's the ring?
May
17
revised A proof for Atiyah-Macdonald Exercise I.21.iii
added 217 characters in body
May
17
comment A proof for Atiyah-Macdonald Exercise I.21.iii
@TakumiMurayama That is where my confusion lies. Maybe it is a shorthand for the set of all the elements in the contractions of the prime ideals containing $\mathfrak b$ (or a shorthand for the ideal generated by all these elements). The first equality in the last chain of equalities says $r (E^c) = r (E)^c$, which I only know to hold when $E$ is an ideal. I'm asking if there is a way to make sense of this "proof", because if the answer is yes, it is shorter than the other proof I found. But if the answer is no, then I will simply ignore it.
May
17
asked A proof for Atiyah-Macdonald Exercise I.21.iii
Apr
19
awarded  Taxonomist
Mar
3
accepted Solvable Lie algebra with codimension 1 ideal
Feb
5
comment First examples for topology of non-Hausdorff spaces
@EricWofsey What is an "inductive Mayer-Vietoris argument"?
Feb
5
comment First examples for topology of non-Hausdorff spaces
@PedroTamaroff The only broad thing about the question is the last paragraph, which is optional (should I say "bonus question"?).
Feb
5
asked First examples for topology of non-Hausdorff spaces
Dec
20
awarded  Constituent
Dec
8
awarded  Caucus
Sep
28
answered Calculus on Manifolds (Spivak), problem 2-41(a)
Sep
24
awarded  Autobiographer
Jul
22
comment Connection in fibre bundle from discontinuous group action
@Bombyxmori I think it is reasonable, considering that I "act in every direction". The formal argument I know invokes Poincaré duality to conclude: if a group $\Gamma$ acts (faithfully) on a contractible manifold $X$ and $\mathrm{cd} \, \Gamma = \dim X$, then $X / \Gamma$ is compact. ($\mathrm{cd}$ being the "cohomological dimension".)
Jul
22
revised Connection in fibre bundle from discontinuous group action
deleted 89 characters in body
Jul
2
awarded  Curious