bio | website | xs4all.nl/~wilberdk/home.html |
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location | ||
age | ||
visits | member for | 5 years, 1 month |
seen | 22 hours ago | |
stats | profile views | 1,325 |
retired
Apr 15 |
awarded | Necromancer |
Mar 21 |
awarded | Yearling |
Feb 18 |
awarded | Enlightened |
Feb 18 |
awarded | Nice Answer |
Sep 29 |
comment |
Multiprecision numerical evaluation of integral: Sage vs. PARI/GP vs. mpmath
Indeed the WorkingPrecision->50 should simply be deleted. One should leave that to Mathematica. And tau should not have been made numerical. |
Sep 24 |
awarded | Autobiographer |
Sep 18 |
comment |
F-splitting and F-purity from commutative algebra viewpoint
See my expository paper (arxiv.org/a/vanderkallen_w_1) Frobenius Splittings |
Aug 23 |
comment |
What is the abutment filtration of the second spectral sequence of hypercohomology?
In some old notes I found the comment that this is much easier to understand when using the derived couples of Massey, rather than Cartan-Eilenberg resolutions. Just apply ${\bf R}T$ to the triangle $\oplus_n \tau_{\leq n-1}K\to \oplus_n \tau_{\leq n}K\to \oplus_n H^n(K)\to$. |
Aug 21 |
comment |
What is the abutment filtration of the second spectral sequence of hypercohomology?
One should not take the spectral sequence of a bicomplex as the template. This spectral sequence is just one (too particular) case of the spectral sequence of a filtered complex. One should take the spectral sequence of a filtered complex as the template. Look for a more relevant filtration of $\text{Tot}(L)$. |
Aug 20 |
comment |
What is the abutment filtration of the second spectral sequence of hypercohomology?
Just like the naive truncation the canonical filtration can be understood as a filtration by subcomplexes. So we eventually end up with something that looks suspiciously like the spectral sequence for cohomology of a filtered complex. One must replace cohomology $H^n$ with ${\bf R}^nT$ in the theory for that spectral sequence. |
Jul 30 |
comment |
When is an exponential functor a bialgebra?
Did you read section 5 of arXiv:0902.4459 ? It may be related. |
Jul 21 |
revised |
GIT over integers
activated links |
Jul 20 |
revised |
GIT over integers
More general result |
Jul 19 |
comment |
GIT over integers
One may expect it. By inverting a nonzero element one may first kill the $R$ torsion in $A$, then hope to kill the $R$ torsion in $H^1(G,A)$. One hopes that $H^1(G,A)$ is finitely generated as an $A^G$ module. But I only seem to know that when $R$ contains a field, cf. arXiv:1109.5822 |
Jul 19 |
comment |
GIT over integers
I am not sure what the question is. If your Dedekind domain $R$ is of finite characteristic $p$, then the result just amounts to the trivial fact that you kill $H^1$ when inverting $p$. Our result that you may kill $H^1$ by inverting some integer is of interest only when $R$ itself does not have this property that you may kill it that way. |
Jul 19 |
answered | GIT over integers |
Jun 19 |
comment |
About G-modules with good filtrations
One may also observe that the two definitions are equivalent for submodules of the injective hull of a finite dimensional module. The definition of Jantzen gets unpleasant only when multiplicities of weight spaces of $U$-fixed points become infinite. |
Jun 6 |
comment |
An exact sequence which does not split
It seems that $f"$ must be injective and that $g"$ must be surjective. If there is a section of $(f',f")$ then use that $End(X)$ is local? |
Apr 22 |
comment |
Frobenius splitting of affine flag varieties
Olivier Mathieu, Asterisque 159-160 (1988) is about applying Frobenius splittings in Kac-Moody context. |
Mar 21 |
awarded | Yearling |