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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.

edited Sep 17 at 19:08

Igor Pak
8,206●2●17●57

asked Sep 17 at 18:46

Jeremy Martin
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Igor Pak · Accepted Answer · 2012-09-17 19:07:33Z UTC

5

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.

answered Sep 17 at 19:07

Igor Pak
8,206●2●17●57

	(wipes egg off face) Quite right. Thanks, Igor. – Jeremy Martin Sep 17 at 20:23
	This might be a better question: Is there a matroid that is representable over $\mathbb{R}$, but not over the algebraic closure of $\mathbb{Q}$? – Jeremy Martin Sep 18 at 18:30
	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 – Igor Pak Sep 20 at 23:20

Matroid representable over $\mathbb{R}$ but not over $\mathbb{Q}$?

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