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I have been hearing different people saying this in different contexts for quite some time but I still don't quite get it.

I know that compact operators map bounded sets to totally bounded ones, that the perturbation of a compact operator does not change the index, and that the calkin algebra is an indispensable tool in the study of operators in the sense that 'essentially something' becomes a useful notion.

But I still suspect why they are 'small'. Now Connes says they are like 'infinitesimals' in commutative function theory, which makes me even more confused. So I guess I just post this question here and hopefully I can hear some quite good explanations about the reasoning behind this intuition.

Thanks!

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    @Yemo$n$Choi When one try to explain a joke it won't be a joke anymore. Have you ever seen the elephants in the bush tomato. I think no, because they are well hidden.2012-07-21

2 Answers 2

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It may help to think of the special case of diagonal operators, that is, elements of $\ell_\infty $ acting on $\ell_2$ by multiplication. Here compact operators correspond to sequences which tend to 0, "are infinitesimally small". This is a commutative situation, in which everything reduces to multiplication of functions. So, general compact operators can be called noncommutative infinitesimals.

A shorter explanation, but with less content: every ideal in a ring can be thought of as a collection of infinitesimally small elements, because they are one step (quotient) away from being zero.

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    @HuiYu They are essentially small in the following sense, for example. A sequence in $\ell^\infty$ tending to $0$ may have a few big terms, but in the end, for every \epsilon > 0, the infinitely long tail of the sequence is smaller than $\epsilon$ whereas only a few finite terms exceed it.2012-07-20
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Finite rank operators are small (in that they squish a large space into a small one). In a Hilbert space (or more generally, a Banach space with the approximation property, which includes most familiar examples), compact operators are precisely the operator-norm limits of finite rank operators. That's what I think of when I hear that statement.

Alternatively, a compact operator from $X$ to $Y$ squishes the unit ball of $X$ (which is "big") into a compact subset of $Y$ (which, in Banach spaces, is typically "small" in that it has empty interior).

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    ...over a Hilbert space. This is false for general Banach spaces.2012-07-20