Hello everyone – MatroidUnion is back!<\/p>\n
Rudi and Jorn wrote\u00a0a nice post<\/a>\u00a0earlier this year about questions\u00a0in asymptotic matroid theory, and beautiful new results they’ve obtained in this area. While reading one of their papers on this topic, I saw that they restated the conjecture that almost all matroids on $n$ elements are non-representable. This was first explicitly written down by Mayhew, Newman, Welsh and Whittle [1] but earlier alluded to by Brylawski and Kelly [2] (in fact, the latter authors claim that the problem is an ‘exercise in random matroids’ but give no clue how to complete it). Indeed I would argue that most of us would independently come up with the same conjecture after thinking about these questions for a few minutes; surely representable matroids are vanishingly rare among all matroids!<\/p>\n In any case, reading this conjecture reminded me that, like many ‘obvious’ statements in asymptotic matroid theory it was still open, and seemed somewhat hard to approach with existing techniques. I’m happy to say that this is no longer the case; as it happened, I discovered a short proof that I will now give in this post. The proof is also on the arXiv [3]<\/a>; there, it is written with the bounds as tight as possible. Here, I will relax the calculations a little here to make the proof more accessible, as well as using more elementary bounds so the entire argument is self-contained. The theorem we prove is the following:<\/p>\n Theorem: <\/strong> For $n \\ge 10$, the number of representable matroids with ground set $[n] = \\{1,\\dotsc,n\\}$ is at most $2^{2n^3}$.<\/p>\n The number of matroids on $n$ elements is well-known<\/a> to be doubly exponential in $n$, so the above gives the ‘almost all’ statement we need, in fact confirming the intuition that representable matroids are extremely<\/em> rare among matroids in general. The bound in the proof can in fact be improved to something of the form $2^{n^3(1\/4 + o(1))}$, and I believe the true count has this form; see [3]<\/a> for more of a discussion.<\/p>\n Our path to the proof is indirect; we proceed by considering a more general question on `zero-patterns’ of polynomials, in the vein of [4]. Let $f_1, \\dotsc, f_N$ be integer polynomials in variables $x_1, \\dotsc, x_m$. Write $\\|f\\|$ for the absolute value of the largest coefficient of a polynomial $f$, which we call its height<\/em>; it is fairly easy to prove that $\\|f + g\\| \\le \\|f\\| + \\|g\\|$ and that $\\|fg\\| \\le \\binom{\\deg(f) + \\deg(g)}{\\deg(f)} \\|f\\|\\ \\|g\\|$ for all $f$ and $g$. We will map these polynomials to various fields; for a field $\\mathbb{K}$ and a polynomial $f$, write $f^{\\mathbb{K}}$ for the polynomial in $\\mathbb{K}[x_1, \\dotsc, x_m]$ obtained by mapping each coefficient of $f$ to an element of $\\mathbb{K}$ using the natural homomorphism $\\phi\\colon \\mathbb{Z} \\to \\mathbb{K}$.<\/p>\n Given a field $\\mathbb{K}$ and some $u_1, \\dotsc, u_m \\in \\mathbb{K}$, the polynomials $f_i^{\\mathbb{K}}(u_1, \\dotsc, u_m)$ all take values in $\\mathbb{K}$, and in general some will be zero and some nonzero in $\\mathbb{K}$. We are interested in the number of different ways this can happen, where we allow both the field $\\mathbb{K}$ and the $u_j$ to be chosen arbitrarily; to this end, we say a set $S \\subseteq [N]$ is \\realisable<\/em> with respect to the polynomials $f_1, \\dotsc, f_N$ if there is a field $\\mathbb{K}$ and there are values $u_1, \\dotsc, u_m \\in \\mathbb{K}$ such that \\[S = \\{i \\in [N]: f^{\\mathbb{K}}(u_1, \\dotsc, u_m) \\ne 0_{\\mathbb{K}}\\}.\\] In other words, $S$ is realisable if and only if, after mapping to some field and substituting some values into the arguments, $S$ is the support of the list $f_1, \\dotsc, f_N$. We will get to the matroid application in a minute; for now, we prove a lemma that bounds the number of different realisable sets:<\/p>\n Lemma:<\/strong> Let $c,d$ be integers and let $f_1, \\dotsc, f_N$ be integer polynomials in $x_1, \\dotsc, x_m$ with $\\deg(f_i) \\le d$ and $\\|f_i\\| \\le c$ for all $i$. If an integer $k$ satisfies Proof: <\/strong> If not, then there are distinct realisable sets $S_1, \\dotsc, S_k \\subseteq [N]$. For each $i \\in [k]$, define a polynomial $g_i$ by $g_i(x_1, \\dotsc, x_m) = \\prod_{j \\in S_i}f_j(x)$. Clearly $\\deg(g_i) \\le Nd$, and since each $g_i$ is the product of at most $N$ different $f_i$, we use our upper bound on the product of heights to get Let $\\ell \\in I \\cup I’$ be chosen so that $|S_{\\ell}|$ is as small as possible. We can assume that $\\ell \\in I$. Since the set $S_{\\ell}$ is realisable, there is a field $\\mathbb{K}$ and there are values $u_1, \\dotsc, u_m \\in \\mathbb{K}$ such that $S_{\\ell} = \\{i \\in [N]: f_{\\ell}^{\\mathbb{K}}(u_1, \\dotsc, u_m) \\ne 0_{\\mathbb{K}}\\}$. So $g^{\\mathbb{K}}_{\\ell}$, by its definition, is the product of nonzero elements of $\\mathbb{K}$, so is nonzero. For each $t \\in I \\cup I’ – \\{\\ell\\}$, on the other hand, since $|S_t| \\ge |S_\\ell|$ and $S_t \\ne S_\\ell$ there is some $j \\in S_t – S_\\ell$, which implies that the zero term $f^{\\mathbb{K}}_j(u_1, \\dotsc, u_m)$ shows up in the product $g^{\\mathbb{K}}_t(u_1, \\dotsc, u_m)$. It follows from these two observations that Why is this relevant to representable matroids? Because representing a rank-$r$ matroid $M$ with ground set $[n]$ is equivalent to finding an $r \\times n$ matrix $[x_{i,j}]$ over some field, for which the $r \\times r$ determinants corresponding to bases of $M$ are nonzero and the other determinants are zero. In other words, a matroid $M$ is representable if and only if the set $\\mathcal{B}$ of bases of $M$ is realisable with respect to the polynomials $(f_A \\colon A \\in \\binom{[n]}{r})$, where $f_A$ is the integer polynomial in the $rn$ variables $[x_{ij} \\colon i \\in [r],j \\in [n]]$ that is the determinant of the $r \\times r$ submatrix of $[x_{ij}]$ with column set $A$. Thus, the number of rank-$r$ representable matroids on $n$ elements is the number of realisable sets with respect to these $f_A$.<\/p>\n To bound this quantity, we apply the Lemma, for which we need to understand the parameters $N,m,a,d$. Now $m = rn \\le n^2$ is just the number of variables. $N = \\binom{n}{r} \\le 2^n$ is the number of $r \\times r$ submatrices. (We can in fact assume that $N = 2^n$ and $m = n^2$ by introducing dummy polynomials and variables). Finally, since the $f_A$ are determinants, we have $\\deg(f_A) = r \\le n$ and $\\|f_A\\| = 1$ for all $a$, so $(N,m,c,d) = (2^n,n^2,1,n)$ will do. To apply the lemma, it suffices to find a $k$ for which [1] D. Mayhew, M. Newman, D. Welsh, and G. Whittle, On the asymptotic proportion of connected matroids,<\/em> European J. Combin. 32 (2011), 882-890. Hello everyone – MatroidUnion is back! Rudi and Jorn wrote\u00a0a nice post\u00a0earlier this year about questions\u00a0in asymptotic matroid theory, and beautiful new results they’ve obtained in this area. While reading one of their papers on this topic, I saw that … Continue reading
\n\\[ 2^k > (2kc(dN)^d)^{N\\binom{Nd+m}{m}}, \\] then there are at most $k$ realisable sets for $(f_1, \\dotsc, f_N)$.<\/p>\n
\n\\[ \\|g_i\\| \\le c^N \\binom{dN}{d}^N (2kc’)^D\\], so we have a collision – there exist distinct sets $I,I’ \\subseteq [k]$ such that $g_I = g_I’$. By removing common elements we can assume that $I$ and $I’$ are disjoint.<\/p>\n
\n\\[ 0_{\\mathbb{K}} \\ne g^{\\mathbb{K}}_I(u_1, \\dotsc, u_m) = g^{\\mathbb{K}}_{I’}(u_1, \\dotsc, u_m) = 0_{\\mathbb{K}}, \\] which is a contradiction.<\/p>\n
\n\\[ 2^k > (2kc(dN)^d)^{N\\binom{Nd+m}{m}}, \\] or in other words, \\[k > N\\binom{Nd+m}{m}\\log_2(2kc(dN)^d)).\\] If you are happy to believe that $k = 2^{2n^3}\/2n$ satisfies this, then you can skip the next two estimates, but for the sticklers among us, here they are:
\nUsing $(N,m,d,c) = (2^n,n^2,n,1)$ and $n \\ge 20$ we have \\[N\\binom{Nd+m}{m} \\le 2^n(2^{n+1}n)^{n^2} = 2^{n^2(n+1 + \\log_2(n)) + n} \\le 2^{2n^3}\/6n^4.\\] (Here we need that $n^3 > n^2(1+\\log_2 n) + n + \\log_2(6n^4)$, which holds for $n \\ge 10$.) Similarly, for $k = 2^{2n^3}\/(2n)$ we have $2kc < 2^{2n^3}$, so
\n\\[\\log_2(2kc(dN)^d)) < \\log_2(2^{2n^3}(n2^n)^n) < 2n^3 + n^2 + n \\log_2 n < 3n^3.\\] Combining these estimates, we see that $k = 2^{2n^3}\/2n$ satisfies the hypotheses of the lemma, so this $k$ is an upper bound on the number of rank-$r$ representable matroids on $n$ elements. This is only a valid bound for each particular $r$, but that is what our extra factor of $2n$ was for; the rank $r$ can take at most $n+1$ values, so the number of representable matroids on $[n]$ in total is at most $(n+1)k < 2nk = 2^{2n^3}$. This completes the proof of the main theorem.<\/p>\n
\n[2] T. Brylawski and D. Kelly, Matroids and combinatorial geometries<\/em>, University of North Carolina Department of Mathematics, Chapel Hill, N.C. (1980). Carolina Lecture Series
\n[3] P. Nelson, Almost all matroids are non-representable<\/em>, arXiv:1605.04288 [math.CO]
\n[4] L. R\u00f3nyai, L. Babai and M. K. Ganapathy, On the number of zero-patterns of a sequence of polynomials<\/em>, J. Amer. Math. Soc. 14 (2001), 717\u2013735.<\/p>\n","protected":false},"excerpt":{"rendered":"