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Kasper Andersen's user avatar
Kasper Andersen's user avatar
Kasper Andersen's user avatar
Kasper Andersen
  • Member for 9 years, 11 months
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Compute corestriction map on group cohomology in Magma
I think the first error is due to wrong syntax, the command should be f(<G.1,G.1>) and not f([G.1,G.1]).
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Formula for bound on number of smooth projective toric Fano varieties of dimension n
The number of smooth projective toric Fano varieties is known up to dimension 9 by work of Mikkel Øbro, cf. sequence A140296 in OEIS. No formula or good upper bound seems to be known.
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Why do these finite group Dedekind matrices seem to have integer spectrum when specialized to the order of group elements?
I just checked that all groups of order less than 128 have integral spectrum, so this seems to hold not only for dihedral groups!
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rational homology of SO(2,1) over number fields
I don't have a reference, but the same argument as for $\mathbb{R}$ should work: Let $F$ be any field of characteristic $0$. Then $G=\text{SL}_2(F)$ acts on $V=F^2$. The action of $G$ on the symmetric square $S^2(V)$ gives a homomorphism $G\rightarrow \text{SL}_3(F)$ with kernel $\{\pm 1\}$, and choosing the right basis, the image preserves the quadratic form $x^2+y^2-z^2$. This gives the desired isomorphism $\text{PSL}_2(F) \stackrel{\cong}{\longrightarrow} \text{SO}(2,1)(F)$.
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rational homology of SO(2,1) over number fields
Isn’t $\text{SO}(2,1)(F)$ the same as $\text{PSL}_2(F)$? In this case the rational cohomology should be the same as for $\text{SL}_2(F)$.
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Why, conceptually, does the torus normalizer in $G_2$ split?
Another classical reference is Normalizers of maximal tori by Curtis, Wiederhold and Williams which handles all simple cases.
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Possible new series for $\pi$
@PeterTaylor I think one needs $\Re(\lambda)>-1$ for the series to converge.
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What is the Stirling formula for x(x+1)(x+2)...(x+n-1)?
Shouldn't this be $\sqrt{2\pi}$ in the numerator?
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Integer valued polynomials and divided power algebra
@Z. M Ok, thanks. So this is actually a slip up in Eisenbuds book I guess.
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