MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 4 down vote accepted

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 $2$.

share|cite|improve this answer

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.

share|cite|improve this answer
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 $\{0\}\in F$. – Emil Jeřábek Apr 2 '12 at 13:03
@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. – Andreas Blass Apr 2 '12 at 13:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.