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Apr
28
revised Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
deleted 4 characters in body
Apr
26
comment Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
Well, numbers of size about $X$ without prime factors between $X^(1/3)$ and $X^{1/2}$ are very roughly as common as the primes - but the sum of the reciprocals of the primes diverges. Or rather - the sum $\sum_n 1/n$ over all $X\leq n<2X$ without such prime factors is about a constant times $1/\log X$, i.e., much larger than $1/\sqrt{X}$. The same happens if you work over much smaller intervals. So I'm afraid this cannot work as stated.
Apr
26
awarded  Nice Question
Apr
26
comment Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
And yes, the running time is as you said and I said. I'm looking for something better.
Apr
26
comment Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
Gerhard: that's exactly what I mean by "the natural algorithm".
Apr
26
revised Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
edited body
Apr
25
asked Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
Feb
29
comment Simple lie algebras, (almost-)simple groups of Lie type
Thanks for this. This is all for a treatment in front of a group of students I haven't met yet; it's unclear to me what I will be able to assume. I'll study the source, but I suspect there has to be a simpler way - one that does not use the simplicity of the Lie algebra, but rather some slightly weaker fact that can be proven more easily and without casework.
Feb
28
comment Simple lie algebras, (almost-)simple groups of Lie type
Hi - aha. I am not trying to make progress there. My original motivation (as I thought had originally been made clear) was to give a clear, quick conceptual proof of several facts in simple groups of Lie type -- above all the fact that, if $G$ is irreducible and almost simple, and $V$, $W$ are a subvarieties of positive codimension of $G$, then, for all $g$ outside a subvariety of $G$ of positive codimension, $g V g^{-1} W$ has dimension larger than $V$. This is known, but proofs are a bit long and can be hard to motivate. If $\mathfrak{g}$ is simple, this is immediate.
Feb
27
comment Simple lie algebras, (almost-)simple groups of Lie type
Well, there is a reasonable relationship, in that, if $T_e(V)$ is not an ideal of $\mathfrak{g}$, then there is certainly a $g\in G$ such that $g V g^{-1} V$ (or $g V g^{-1} V^{-1}$) has dimension larger than $V$ (to be precise: its Zariski closure has at least one component of dimension larger than that of $V$).
Feb
26
comment Simple lie algebras, (almost-)simple groups of Lie type
Well, what is going on then, exactly? So $n_a \mathfrak{v} n_a^{-1} = \mathfrak{v}$ , yet the dimension of $n_a V n_a^{-1} V^{-1}$ is greater than the dimension of $V$?
Feb
25
comment Simple lie algebras, (almost-)simple groups of Lie type
See my newest question above.
Feb
25
revised Simple lie algebras, (almost-)simple groups of Lie type
added 491 characters in body
Feb
25
comment Simple lie algebras, (almost-)simple groups of Lie type
Interesting. Is $G$ still $\Spin_{2n+1}$ here? Just for concreteness - can you come up with a "small" enough example that you can write this in terms of matrices? I'd really like to see what is going on here. Also, see the new question at the end of the original post. Thanks!
Feb
22
comment Simple lie algebras, (almost-)simple groups of Lie type
Aha, interesting - thanks! Now I'm at a loss, however. Say you have a proper, irreducible subvariety $V$ of $G$ ($\dim V>0$) such that its tangent space $\mathfrak{v}$ at the origin is invariant under $Ad_G$. Does it follow that $g V g^{-1} = V$ for all $g$ in $G$? Is that even possible, given that $G$ is simple? It certainly isn't when the stabilizer $Stab(V) = \{h\in G: h V = V\}$ is non-trivial, since then the stabilizer would be a non-trivial, normal, proper subgroup of $G$.
Feb
20
comment Simple lie algebras, (almost-)simple groups of Lie type
A smooth subscheme - or in any event an affine (closed) subvariety that is nonsingular at the origin. Yes, I am sure Jim Humphreys can see a way out.
Feb
19
comment Simple lie algebras, (almost-)simple groups of Lie type
But is there a relatively straightforward proof that works in all characteristics? I mean a proof of the statement in the current version of the question, after "What I really need is the following".
Feb
19
revised Simple lie algebras, (almost-)simple groups of Lie type
added 617 characters in body
Feb
19
comment Simple lie algebras, (almost-)simple groups of Lie type
But are all ideals of small dimension? That is, of dimension smaller than the tangent space to any variety $V$ of positive dimension in $G$? I've edited the question; see above.
Feb
19
comment Simple lie algebras, (almost-)simple groups of Lie type
Oh, I see. You get an extra dimension!