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1d
revised For $P$ $\mathbb{Z}G$-projective, $\mathbb{Q}\otimes P$ is $\mathbb{Q}G$-free
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2d
comment Are rays in Carnot groups straight?
OK, thanks, I was mislead by another (wrong) interpretation of the question
2d
comment Are rays in Carnot groups straight?
No: if $V\subset\mathfrak{g}$ is the subspace defining the CC-metric, then the only geodesics that would have the form $\exp(tX)$ should satisfy $X\in V$, and hence can only reach points of $\exp(V)$.
2d
comment Polynomial roots in the ring extension
I know, but this was addressed to other readers.
2d
comment Polynomial roots in the ring extension
Note: "unitary polynomial" means "monic polynomial".
2d
revised Polynomial roots in the ring extension
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2d
revised Are rays in Carnot groups straight?
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2d
revised Rank of a locally free $\mathbb Z[G]$-module
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2d
revised References for metrics in matrix groups
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Apr
29
comment Why do we study symplectic geometry?
One motivation is just differential geometry: to each smooth manifold we can canonically associate a symplectic manifold, its cotangent bundle. So symplectic invariants (of the cotangent bundle) yield differential invariants (of the original manifolds).
Apr
28
comment Is this group given presentation isomorphic to $\mathbb{Z}_2$, and why?
Possibly they didn't realize (no GAP!), but anyway this is a reasonable exercise to compute a presentation for a subgroup of index 2 (in the same fashion, proving that there infinitely many primes is trivial if one assumes the Riemann hypothesis...).
Apr
28
revised Is this group given presentation isomorphic to $\mathbb{Z}_2$, and why?
Added an edit to reflect the OP's hope and also grant the original answer
Apr
28
comment Is this group given presentation isomorphic to $\mathbb{Z}_2$, and why?
I think that it is reasonable to ask about a proof which is not just running an algorithm (e.g., as I mentioned elsewhere, writing explicitly, say, $ab$ as a product of relators, since GAP doesn't provide this); however I agree it was awkward to just discard the answer to the original question because you didn't like it. Recall the original question was "is this group cyclic on 2 elements", and not "I know that this group has 2 elements, can you prove it?" Well, I'll edit, and hope the question will not be migrated since it's not so obvious.
Apr
28
comment A finiteness property for semi-simple algebraic groups
I was aware of the cohomological argument, but I don't see how it works. This argument (at least, the one I have in mind, using $H^1(H,\mathfrak{g})=0$) shows that two close semisimple subalgebras are conjugated under a linear automorphism of $\mathfrak{g}$. But you need to have them conjugated by some element of $G$ (or, essentially equivalently, by some Lie algebra automorphism of $\mathfrak{g}$).
Apr
28
comment Invertibility of all left multiplication maps in non-unital rings
Ah OK. Indeed bilinear laws can be freely twisted on the right by automorphisms, this affects associativity, left/right units, but not commutativity or invertibility of left/right translations.
Apr
28
revised Weil's Haar measure construction from below
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Apr
28
revised A finiteness property for semi-simple algebraic groups
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Apr
28
comment A finiteness property for semi-simple algebraic groups
The algebraically closed case seems to be well-known, and actually Dynkin provides a recipe (see books.google.fr/…). On Google I also found algorithms about the real case. But a direct proof a the finiteness result would be interesting (without going into this combinatorial cuisine)
Apr
28
comment A finiteness property for semi-simple algebraic groups
A positive answer in the case when $K$ is algebraically closed of characteristic zero implies a positive answer for real and $p$-adic fields, by the finiteness results of Borel-Serre. For the algebraically closed case, I guess it's well-known (I guess it's known that $G$ has finitely many maximal connected subgroups up to conjugacy - there are finitely many parabolic ones, and the others are reductive; if the latters are finitely many, then the finiteness of semisimple subgroups up to conjugacy follows by induction)
Apr
28
comment Invertibility of all left multiplication maps in non-unital rings
well, it's a variant of mine. More generally, you can take any linear map $f:K^n\to M_n(K)$ such that $f(x)$ is invertible for all $x\neq 0$, any linear automorphism $g$ of $K^n$ not in the range of $f$, and define the multiplication on $K^n$ $(x,y)\mapsto f(x)g^{-1}(y)$.