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Gadgeguy2009

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Something funky happened when you posted that- Is this the rest of it?

Hello all,

could please someone point me in the right direction (or explain)?

I really like the black steel breitling reps. However I am not exactly sure what black steel is. If it is PVD, than no thank you. I have an expensive gen PVD watch, and only after a short time there are a couple of nicks on it, where the steel shines through. That drives my crazy, as I really anal about my watches. (A nick on a SS watch is much less visible, and therefore disturbs me way less).

Is black steel PVD? Or is it the same material deeper down too?

Which model / vendor would have a guaranteed Breitling black steel ?

Thank you for your advice!

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Okay, I'll explain it again. But just this once more. Please listen carefully this time.

A topological group is locally compact if and only if the identity e of the group has a compact neighborhood. This means that there is some open set V containing e whose closure is compact in the topology of G. A locally compact group G carries an essentially unique natural measure, the Haar measure, which allows one to consistently measure the "size" of sufficiently regular subsets of G. "Sufficiently regular subset" here means a Borel set; that is, an element of the σ-algebra generated by the compact sets. More precisely, a right Haar measure on a locally compact group G is a countably additive measure μ defined on the Borel sets of G which is right invariant in the sense that μ(A x) = μ(A) for x an element of G and A is a Borel subset of G and also satisfies some regularity conditions (spelled out in detail in the Haar measure definition).

The take-home concept is, except for positive scale factors, Haar measures are unique.

Now then do we need to go over this again? :nerd:

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Okay, I'll explain it again. But just this once more. Please listen carefully this time.

A topological group is locally compact if and only if the identity e of the group has a compact neighborhood. This means that there is some open set V containing e whose closure is compact in the topology of G. A locally compact group G carries an essentially unique natural measure, the Haar measure, which allows one to consistently measure the "size" of sufficiently regular subsets of G. "Sufficiently regular subset" here means a Borel set; that is, an element of the σ-algebra generated by the compact sets. More precisely, a right Haar measure on a locally compact group G is a countably additive measure μ defined on the Borel sets of G which is right invariant in the sense that μ(A x) = μ(A) for x an element of G and A is a Borel subset of G and also satisfies some regularity conditions (spelled out in detail in the Haar measure definition).

The take-home concept is, except for positive scale factors, Haar measures are unique.

Now then do we need to go over this again? :nerd:

no no, explain that again :p

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Well I left out some of it, hoping that he'd be able to connect the dots. *dripping sarcasm*

What I meant to imply was a probabilistic proof of the Weyl integration formula on the unitary group. This relies on a suitable definition of Haar measures conditioned to the existence of a stable subspace with any given dimension. The developed method leads to the following: for this conditional measure, writing $Z_U^{(p)}$ for the first non-zero derivative of the characteristic polynomial at 1, $Z_U^{(p)}$ can be decomposed as an explicit product of $n-p$ independent random variables. This implies a central limit theorem for $\log Z_U^{(p)}$ and asymptotics for the density of $Z_U^{(p)}$ near 0. I mean, duh. Similar limit theorems apply for the orthogonal and symplectic groups. Now if you are mapping x->Ux, Let {ei, e2s ... en} be the complete orthonormal system of Rn, so that if xn = <x, en >=

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From Two and a Half Men ...

Charlie: (trying to explain to Jake, why he had to go out with a divorced woman who gave him her number, even though he promised he wouldn't) Ok, ok...let's say you're a hunter! If a deer takes your gun, shoots itself, then straps itself to the roof of your car...you have to take it home and eat it!

Jake: What?!?

Charlie: I'm sorry...i can't make it any clearer!

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