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Feb
6
comment Weight 12 cusp forms for $\Gamma_0(p)$
Why do you write $S_k$ and $S_k^{\rm new}$ if the weight is 12?
Feb
6
comment Unifying (& “justifying”) the various definitions for differential operators
About your “naïve” first definition, remember the theorem of Peetre: an R-linear morphism of sheaves from ${\cal O}_X$ to itself is a differential operator (locally of finite order).
Jan
31
comment A technical question about affine grassmanian
Precisely, see Theorem 1.9.
Jan
31
reviewed Approve Can Gradient be controlled by Curl and Divergence in Morrey spaces
Jan
26
comment Rings in which every non-unit is a zero divisor
@MartinBrandenburg: Sorry to revive a thread that was dead long ago, but your characterization of $\dim(A)=0$ is a particular case of the elementary (almost first order) definition of Krull dimension by Coquand, Lombardi and Roy.
Jan
25
comment Existence of prime ideals and Axiom of Choice.
@LaurentMoret-Bailly: The PhD Thesis of Hervé Perdry, hlombardi.free.fr/liens/TheseHervePerdry.pdf, took care of the question of constructive noetherianity by rewriting the definition of noetherian as “every increasing sequence pauses” (meaning “is not strictly increasing”).
Jan
24
comment What are the basic possibilities for a tensor product of two fields?
cf. mathoverflow.net/questions/82083/when-is-the-tensor-product- of-two-fields-a-field for the result of Grothendieck mentioned in the question.
Jan
24
comment Finite groups as intersection of algebraical groups
As stated, yes, for stupid reasons. Take for a 2-dimensional group the product of G with (Ga)^2, and the two subgroups G times Ga times {0} and G times {0} times Ga. As probably wanted, no, for the more interesting reason that a connected one-dimensional algebraic group is commutative.
Jan
16
comment Name for a type of weak path connectedness?
In the definition of the 1st property, do you mean "a single path component" ?
Jan
11
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Jan
4
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Dec
21
comment Does the cohomology comparison part of GAGA hold over the reals?
So the only way to make the statement non trivially false would be to change the definition of $X^{an}$. A natural definition would be the quotient of the analytic space $X(\mathbf C)$ by the action of complex conjugation (this is what Berkovich's definition gives, in any case). Then the theorem is a trivial consequence of the complex GAGA.
Dec
11
comment $\{$Affine schemes over $S$$\}$ $\cong$ $\{$$\mathcal{O}_S$ - algebras$\}$?
Your "inverse functor" is not an inverse... Try to compute its effect on the example I give.
Dec
11
comment $\{$Affine schemes over $S$$\}$ $\cong$ $\{$$\mathcal{O}_S$ - algebras$\}$?
I added one line to the answer to show that it does not hold at all.
Dec
11
revised $\{$Affine schemes over $S$$\}$ $\cong$ $\{$$\mathcal{O}_S$ - algebras$\}$?
added 114 characters in body
Dec
11
answered $\{$Affine schemes over $S$$\}$ $\cong$ $\{$$\mathcal{O}_S$ - algebras$\}$?
Oct
24
comment Demonstrating that rigour is important
@KConrad: I have been told the story of a mathematician who, while making a talk and being asked about a proof of some results he had claimed, exclaimed: “Why do you need a proof? It's a theorem!”
Oct
11
comment Why is every l-adic Galois representation conjugate to one over the l-adic integers?
@JakobD.Hüwer: Indeed, that's very funny that they summon a lattice by giving its definition. Ad the existence of a lattice: let $V=\mathbf Q_\ell^n$ endowed with its $\ell$-adic norm and, for $x\in V$, set $\nu(x) = \sup_{g\in G} \| g\cdot x\|$. Then $\nu$ is a G-invariant norm on $V$ with values in $\ell^{\mathbf Z}$. A classical lemma (see Weil, Basic Number Theory) then implies that $V$ has a basis $(e_i)$ consisting of vectors of $\nu$-norm equal to $1$. It generates the sought-for lattice.
Oct
7
comment What can we say about tropical maps $\mathbb{P}^1 \to A$ for an Abelian variety $A$?
This was exactly the example I wanted to give showing that there are morphisms from a tropical $\mathbf P^1$ to an abelian variety. Actually, some papers define $\mathbf T\mathbf P^1$ as $\mathbf R^2/(1,1)\R$. Moreover, the analogy with non-archimedean geometry makes this example very reasonable. When you say that the map $\Gamma\to\mathbf R^g$ can't obey the balancing condition at a leaf, this presumes the existence of a leaf...
Oct
6
comment Andre-Oort for conjecture
No, it's not true. Take for a Shimura variety the simplest of all, namely the $j$-line parameterizing elliptic curves. It is known (but non-trivial) that the CM points are dense for the analytic topology, but for an arbitrary set of CM-points to be Zariski dense, it is enough that it be infinite. So consider any infinite set of CM-points which are located in a given disk — it will be Zariski dense but not analytically dense.