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On reading about matroids representable over partial fields, one learns about the 6th root of unity partial field, but other even-th root of unity partial fields seem to be absent from the standard repertoire of examples. Why is this? Is the matroid theory boring or not useful for those partial fields? I'd be particularly interested in knowing what is known about matroids over $\{1, -1, i, -i\}$ because it might be useful to me in another problem.

Thanks.

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2 Answers 2

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Consider the partial field $P:=\{p\in \mathbb{C}: |p|=1\}$. It is a theorem of Whittle that if a matroid $M$ is representable over $P$, then $M$ is sixth-root of unity.

$P$ contains each $P_k:=\{k^\text{th}\text{ roots}\}$, so all the $k$-root of unity matroids are also sixth-root-of-unity.

To represent $U_{2,4}$ over $P_k$ you need a $p\in P_k$ so that $(1-p)\in P_k$ - which happens only if $k$ is a multiple of $6$. Otherwise, the matroids over $P_k$ are both binary and sixth-root, and hence regular.

If you're looking for a really interesting class extending the sixth-root-of-unity matroids, I can recommend the 'quaternionic unimodular' matroids defined here: https://arxiv.org/abs/1106.3088 They are pretty wild, but they should not be completely unmanageable: they do not contain $U_{2,7}$ if I remember right.

It is a conjecture in that paper that quaternionic unimodular matroids have the half plane property.

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  • $\begingroup$ I think that explains very well why the 6th root of unity matroids are so important. Still, matroid classes like the {1,-1,i,-i} partial field representable ones would live inside the 6th root of unity matroids and it isn't clear to me what that subclass would look like. $\endgroup$ Commented Jan 16, 2017 at 18:59
  • $\begingroup$ $M$ is representable over $\{1,-1, i,-i\}$ if and only if $M$ is regular. $\endgroup$ Commented Jan 19, 2017 at 21:41
  • $\begingroup$ Thanks Rudi. That is very helpful. Why is it true? $\endgroup$ Commented Jan 23, 2017 at 15:39
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    $\begingroup$ The uniform matroid $U_{2,4}$ cannot be represented over $\{1,-1,i,-1\}$. Therefore, if $M$ is representable over $\{1,-1,i,-1\}$, then $M$ cannot have $U_{2,4}$ as a minor, hence $M$ is binary. But since such $M$ is also sixth-root, $M$ is ternary as well. Binary+ternary = regular. $\endgroup$ Commented Jan 24, 2017 at 15:58
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I am not an expert, but I would guess the reason you hear so much about 6th root of unity matroids is because there is the nice result that being a 6th root of unity matroid is equivalent to being representable over $GF(3)$ and $GF(4)$. I suppose if someone can prove a similarly nice theorem for another partial field we will hear more about them. One result which applies to the types of matroids in the question is Corollary 8.2 (b) of Homogeneous multivariate polynomials with the half-plane property by Choe, Oxley, Sokal, and Wagner which states:

Let $M$ be a rank-$r$ matroid on $n$ elements that can be represented over $\mathbb C$ by an $r \times n$ matrix $A$ for which every $r \times r$ subdeterminant is either zero or else of modulus 1. Then $M$ has the half-plane property.

A polynomial has the half plane property if it is nonvanishing on a half plane. A matroid has the half plane property if its basis generating function does. Of course having all minors $n$th roots of unity is stronger so perhaps more can be said, but this at least gives a result which applies to the matroids in the question. Maybe knowing this will help in your literature search.

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  • $\begingroup$ I agree with your sense that the cases studied in this area are driven by what appears in certain nice results. Very sensible, though when coming from a different direction it is easy to end up outside these studied cases and wish for a more systematic theory. Thanks re half plane property. $\endgroup$ Commented Dec 5, 2016 at 14:45

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