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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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    $\begingroup$ Have you ever seen in real a compact operator? I think no, because they are small. $\endgroup$
    – Norbert
    Jul 20, 2012 at 21:51
  • $\begingroup$ @Norbert: what on earth does that comment even mean? Look up "integral operators" for plenty of compact operators that arise naturally from applied mathematics or physics. Or just think for a moment to come up with loads of examples. $\endgroup$
    – user16299
    Jul 20, 2012 at 23:54
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    $\begingroup$ If it is not too heretical to say this: occasionally one should take the pronouncements of brilliant mathematicians with a pinch of salt, perhaps even a whole compact operator's worth... $\endgroup$
    – user16299
    Jul 20, 2012 at 23:55
  • $\begingroup$ More generally, compactness is sometimes seen, in analysis, as "the next best thing after finiteness" -- this imprecise but useful POV is expounded on slightly in e.g. Sutherland's Introduction to Metric and Topological Spaces, and probably also somewhere on Terence Tao's blog IIRC. $\endgroup$
    – user16299
    Jul 20, 2012 at 23:57
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    $\begingroup$ @YemonChoi 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. $\endgroup$
    – Norbert
    Jul 21, 2012 at 8:17

3 Answers 3

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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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  • $\begingroup$ I like your second explanation better actually. As for the first one, it is hard for me to see why sequences tending to $0$ are infinitesimally small. For instance, we do not usually say a function vanishing at infinity is infinitesimally small, right? $\endgroup$
    – Hui Yu
    Jul 20, 2012 at 4:14
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    $\begingroup$ @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. $\endgroup$
    – user12014
    Jul 20, 2012 at 5:24
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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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  • $\begingroup$ Yes. This is related to the fact that bounded sets are mapped to totally bounded ones by compact operators. But I guess there is something more intrinsic to the algebra of operators, not much to do with the underlying space. $\endgroup$
    – Hui Yu
    Jul 20, 2012 at 2:10
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    $\begingroup$ ...over a Hilbert space. This is false for general Banach spaces. $\endgroup$
    – Qiaochu Yuan
    Jul 20, 2012 at 2:18
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Connes wrote a paper It is in french. I translate a sentence at the beginning with google: Compact operators in Hilbert space H plays the same role among bounded operators in H as infinitesimals among complex numbers.
In his 1994 book Noncommutative geometry he wrote:

K = {T ∈ L(H) ; T compact} is a two-sided ideal in the algebra L(H) of bounded operators in H, and it is the largest nontrivial ideal. An operator T in H is compact iff for any ε > 0 the size of T is smaller than ε except for a finite-dimensional subspace of H. More precisely, one lets, for n ∈ N, $µ_n(T)$ = Inf $||T − R||$ ; R operator of rank ≤ n} where the rank of an operator is the dimension of its range. Then T is compact iff $µ_n(T)$ →0 when n→∞. Moreover, the $µ_n(T)$ are the eigenvalues, ordered by decreasing size, of the absolute value |T| of T. The rate of decay of the µn(T) as n→∞ is a precise measure of the size of the infinitesimal T.

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