Mnev's universality theorem claims that any semialgebraic set is the realization space of some oriented matroid. Moreover, the rank of the or matroid can be prescribed in advance.

1.-Are there interesting corollaries to Mnev's theorem? I am aware of interesting algorithmic consequences.

Geometric consequences? Examples in which the theorem is used to prove that other moduli spaces can also be wild?

MacPherson's definition of "combinatorial differentiable manifolds" and oriented matroid bundles are based on a local system of oriented matroids over a simplicial complex. Is there some implication from Mnev's theorem to the theory of combinatorial differentiable manifolds.

What about proofs that would be easy (or statemens that would be true) if realization spaces of oriented matroids where better behaved, say connected, or contractible..

2.-Are there quantitative versions of this theorem relating (say) the number and degrees of the defining polynomial (in)equalities or the betty numbers of the semialgebraic set with the rank and number of elements in the corresponding or-mat.

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    $\begingroup$ You did not cite it, but I assume you know Vakil's paper "Murphy's law in Algebraic Geometry". He uses Mnev's theorem to prove that "every singularity of finite type over $\mathbf{Z}$" (up to smooth parameters) appears on: the Hilbert scheme of curves in projective space; and the moduli spaces of smooth projective general-type surfaces (or higher-dimensional varieties), plane curves with nodes and cusps, stable sheaves, isolated threefold singularities, and more" $\endgroup$ Aug 5, 2011 at 8:13
  • $\begingroup$ Thanks I did glance at this paper. It looks fun, and even for someone like me who knows no algebraic geometry the moral of the story is kind of clear. Are the proofs somehow like algorithmic complexity reductions? Do you know what is in Lafforgue's related paper? $\endgroup$ Aug 5, 2011 at 15:01
  • $\begingroup$ Thanks, do you know about implications of Mnev's universality to the theory of Matroid Bundles?? $\endgroup$ Aug 14, 2011 at 18:49
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    $\begingroup$ Simple-minded matroid bundles are ill defined thanks to universality. The problem showed up in attempts to use matroids for formulas for pontrjagin classes. But there are still subjects to work on. $\endgroup$ Aug 26, 2011 at 5:30
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    $\begingroup$ A comment about a quantitative bound: Mnëv’s proof provides a polynomial-time reduction from the existential theory of the reals to the realizability of an oriented matroid (see Shor 1991; available online at math.mit.edu/~shor/papers/Stretchability_NP-hard.pdf). This implies that if you have a semialgebraic set defined by polynomials whose coefficients are algebraic numbers, then the size of the oriented matroid constructed from it by Mnëv’s proof can be bounded by a polynomial in the bit-length of the description of the polynomials. $\endgroup$ Aug 26, 2011 at 18:40

3 Answers 3


Here are some references http://www.pdmi.ras.ru/~mnev/bhu.html

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    $\begingroup$ Wow. Thanks. I'll have a look. What do you mean by ill defined? The Folkman-Lawrence representation theorem doesn't come to save the day? Didn't Anderson, Babson and Gelfand and Macpherson managed to extend some results about characteristic classes?? $\endgroup$ Sep 1, 2011 at 21:38
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    $\begingroup$ All this persons made a great job, But up to now we don't know the relations between matroid and vector bundles in full details. we know some complcate formula for the first pontrjagin class of a manifold (not a bundle), we don'know formulas for whitney and even euler class of a bundle (for euler i know, may be it will be published some day). The trouble is that the matroind stratification of the Grassmanian have to be a cell complex, but by the universality it is a terrible thing actually. One have to desingularize the stratifification. It is very possible, and i hope to see good things. $\endgroup$ Sep 2, 2011 at 12:01
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    $\begingroup$ Here arxiv.org/abs/1108.4733 i have a very simple rational local formula for the chern-euler class of a triangulated $S^1$ bundle. Im shure that it can be rewritten for rank 2 matroid bundles. It can be fun $\endgroup$ Sep 2, 2011 at 12:19

For your first question, you might be interested in Ravi Vakil's paper "Murphy's law in algebraic geometry". He uses Mnëv's theorem to show that a large family of moduli spaces which are known to have singularities are in fact "as singular as possible", by which he means that every possible type of singularity defined over $\mathrm{Spec}(\mathbb{Z})$ will appear at some point of the moduli space.

Here's a different application. Kontsevich defined for every graph $G$ a hypersurface $Y_G$ in a way motivated by QFT and the theory of Feynmann integrals. Motivated by computer experiments, he suggested that period integrals on the $Y_G$ should always be multiple zeta values. I am not sure of the precise relationship here, but I believe that this is (at least morally) the same thing as stating that the cohomology of $Y_G$ contains only mixed Tate motives. This is a very strong condition to impose and would say that the cohomology of $Y_G$ is extremely special. In particular this would imply that the function $q \mapsto \#Y_G(\mathbf F_q)$ that counts the number of points on $Y_G$ over a finite field is always given by a polynomial in $q$. Belkale and Brosnan in "Matroids, motives and a conjecture of Kontsevich" disproved this conjecture in the strongest possible way: they showed that for ANY scheme $X$ of finite type over $\mathbf Z$, the function $q \mapsto \#Z(\mathbf F_q)$ is a finite linear combination of functions $q \mapsto \#Y_G(\mathbf F_q)$ for graphs $G$. Their proof uses Mnëv's theorem in a crucial way.


So the universality theorem by Mnev, shows that every semi-algebraic set is stably equivalent to the realization space of a oriented matroid or a chirotope. So stably equivalent is a slightly stronger statement than homotopy equivalent. And homotopy equivalent implies for instance that all betti numbers are the same. Which in turn implies (to give a simple example) that the number of connected components are the same.

(Thanks to Michael Dobbins, who explained this to me.)

best Till


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