7
$\begingroup$

Does there exist a matroid that is representable over $\mathbb{R}$ but not over $\mathbb{Q}$?

In particular, can one give a positive answer using a nonrational polytope, i.e., a combinatorial polytope that cannot be realized as the convex hull of rational vertices? (Such things do exist; see, e.g., p.94 of Grünbaum's Convex Polytopes.) The vertex sets of the faces of a convex polytope certainly form the flats of a matroid, but it's not clear to me why the same matroid could not be realized by affine dependences of a set of points not in convex position.

$\endgroup$

1 Answer 1

11
$\begingroup$

Jeremy, on the very same page 94 you will find a "point and line configuration" called Perles configuration which when viewed as set ov vectors in $\Bbb R^3$ is a matroid that is realizable over $\Bbb Q[\sqrt{5}]$ but not over $\Bbb Q$. In my book I even prove it (Ex 12.3) - sorry to make a plug, this is the only place with a proof I know.

$\endgroup$
3
  • $\begingroup$ (wipes egg off face) Quite right. Thanks, Igor. $\endgroup$ Sep 17, 2012 at 20:23
  • $\begingroup$ This might be a better question: Is there a matroid that is representable over $\mathbb{R}$, but not over the algebraic closure of $\mathbb{Q}$? $\endgroup$ Sep 18, 2012 at 18:30
  • 4
    $\begingroup$ Um, Jeremy, I think you should continue reading p.94, starting with "As a matter of fact..." To clarify for those without a book, the answer is NO. BTW, this is the reason Grünbaum won the Steele Prize: math.washington.edu/newsletter/2005/grunbaum.html $\endgroup$
    – Igor Pak
    Sep 20, 2012 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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