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accepted Compact operators: why is the image of the unit ball only assumed to be relatively compact?
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comment Compact operators: why is the image of the unit ball only assumed to be relatively compact?
Thanks a lot! So the qualification "relative" is apparently meant for the case of Banach spaces that are not reflexive. My argument would fail there as the (closed) unit ball would no longer be weakly compact.
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accepted How to draw a complex line bundle?
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comment Compact operators: why is the image of the unit ball only assumed to be relatively compact?
According to Wikipedia, the weak topology is metrizable at least for separable Hilbert spaces. But it appears that you have a good point: in the non-separable case, weak sequential compactness need not follow from weak compactness, but I'm not entirely sure whether this is the case or not.
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asked Compact operators: why is the image of the unit ball only assumed to be relatively compact?
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comment Direct proof that the wedge product preserves integral cohomology classes?
Thanks for your answer! I'll need a while to digest it, though. :)
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answered Is it possible to store an integer in sub-logarithmic space?
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revised Find the maximum possible value of an expression, subject to a linear constraint
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Feb
22
comment How do you find the Lie algebra of a Lie group (in practice)?
@HaukeStrasdat: Yes, you set $t=0$ after differentiation. In practice, one often keeps the $t$ around before differentiation and says that the expression is "to first order in $t$", which means that one silently drops all terms $t^n$ where $n>1$. This has the same effect as differentiating at $t=0$, but this way of thinking is very useful if you are interested in the higher-order terms.
Feb
17
comment How do you find the Lie algebra of a Lie group (in practice)?
@HaukeStrasdat: Sure. Basically you just differentiate with respect to $t$ at the point $t=0$. In other words, you calculate $\frac{d}{dt}|_{t=0} A(t)^TA(t) = H^T+H$ where $H=\frac{d}{dt}_{t=0}A(t)$. The expression "first order" refers to "first order in the Taylor expansion". In other words, the formula $A(t) = I + tH + \mathcal O(t^2)$ is just the Taylor expansion of $A(t)$ "up to first order". Admittedly, this kind of language takes a while to get used to, but it's very intuitive once you get the hang of it.
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