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Perhaps my question is naive, but let me try.

Take a (real or complex) vector space $V$ and consider an ideal $\mathcal{I}$ of subsets of $V$ with the following property (call it (*)): for each linear map $A\colon V\to V$ and each $S\in \mathcal{I}$ we have $A(S)\in \mathcal{I}$.

The ideal of finite sets enjoys this property. Is there a maximal ideal (that is, such that the set $\{E\setminus S\colon S\in \mathcal{I}\}$ is an ultrafilter) with (*)?

What if we replace vector spaces and linear maps by Banach spaces and bounded linear operators?

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Every ideal satisfying $(\*)$ is included in an ideal maximal with respect to the property of satisfying $(\*)$, by Zorn's lemma. However, such an ideal need not be maximal as an ideal.

No maximal ideal can satisfy $(*)$, apart from trivial cases ($0$-dimensional vector spaces). Let $X$ be a subset of the base field $F$ such that $F$ is the disjoint union of $X$, $-X$, and $\{0\}$ (e.g., $X=(0,\infty)$ if $F=\mathbb R$). We can choose a basis of the vector space, hence we may assume that $V$ is the space of almost everywhere $0$ functions $I\to F$. Fix $i_0\in I$, let $A$ be the linear function which negates the $i_0$th coordinate, $B$ the function which sets the $i_0$th coordinate to $0$, and $S=\{v\in V:v(i_0)\in X\}$. If $I$ is a maximal ideal, either $S$ or its complement belongs to $I$. However, $V=S\cup A(S)\cup B(S)$ and similarly for $V\smallsetminus S$, hence $V\in I$, a contradiction. The same argument works for vector spaces over any field of characteristic other than $2$.

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Considering that the constant-zero map is linear, your ultrafilter would have to be the principal ultrafilter concentrated at 0. Even if you exclude the constant-zero map, you get the same conclusion if there is, among the linear maps you allow, one whose only fixed-point is 0 (i.e., one that doesn't have 1 as an eigenvalue). This is because, if an ultrafilter $U$ on a set $X$ is sent to itself by a function $f:X\to X$, then $U$ must contain the set of fixed-points of $f$. You still get the same result if, among the linear maps that you allow, there are finitely many whose only common fixed-point is 0.

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    $\begingroup$ Note that the required condition on ultrafilters is dual to that on ideals: if $S\in F$ and $A$ is linear, then $V-A(V-S)\in F$. For the constant $0$ map, this gives $V-\{0\}\in F$, rather than $\{0\}\in F$. $\endgroup$
    – Emil Jeřábek
    Commented Apr 2, 2012 at 13:03
  • $\begingroup$ @Emil: You're right. The "constant-zero" part of my answer should be ignored. The later parts seem to be OK if, in addition to the assumptions I made, you require the maps to be isomorphisms. Then $V-A(V-S)=A(S)$ and $A$ will map the ultrafilter to itself, as I had wrongly assumed. $\endgroup$
    – Andreas Blass
    Commented Apr 2, 2012 at 13:13

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