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Define a skew-symmetric form $(\cdot,\cdot)$ on $\mathbb{R}^{2k}$ by $$(e_i,e_j) = \begin{cases} 1 &\text{if $i<j$},\\ -1 &\text{if $i>j$},\\ 0 & \text{if $i=j$.}\end{cases}$$ Given a permutation $\pi:\{1,\dotsc, 2 k\}\to \{1,\dotsc, 2 k\}$, let $V_\pi$ be the space spanned by $e_{\pi(2i)}-e_{\pi(2i-1)}$ for $1\leq i\leq k$.

For which permutations $\pi$ is the restriction $(\cdot,\cdot)|_{V_\pi\times V_\pi}$ non-degenerate?

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    $\begingroup$ Note that $k$ must be even for any such $\pi$ to exist (assuming that non- degenerate means symplectic). $\endgroup$ – BS. Jul 16 '19 at 14:25
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    $\begingroup$ In different words, you have $k$ segments on the line with distinct endpoints, and the skew-symmetric matrix with rows and columns enumerated by them is $M([a,b],[c,d])=1, a<c<b<d$; $=-1$ if $c<a<d<b$, 0 otherwise. The question is when is it non-degenerated. $\endgroup$ – Fedor Petrov Jul 16 '19 at 14:59
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    $\begingroup$ In reference jstor.org/stable/2000524 (by D. M. Jackson), the formula for $e_1^{(2)}(n)$ on page 797 gives the number of $\sigma$'s as $(2k)!/(k+1)!/2^k$ (for even $k$), so the number of $\pi$'s is indeed what you write. $\endgroup$ – BS. Jul 16 '19 at 18:06
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    $\begingroup$ The rank of the skew form on $V_\pi$ is more generally given by $k+1$ minus the number $b$ of cycles of $i\mapsto\sigma(i)+1$, because one one hand it is $2g$ where $g$ is the genus of the corresponding surface $S$ (with $b$ boundary components), and on the other hand it is $2-\chi(S)-b$, where Euler characteristic is $\chi(S)=1-k$ by construction. $\endgroup$ – BS. Jul 16 '19 at 18:55
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    $\begingroup$ @H A Helfgott : I should perhaps explain why $(\cdot,\cdot)$ on $V_\pi$ represents the (opposite of) intersection form on the first homology group $H$ of $S=S_\pi$. A basis for $H$ is given by the $k$ cycles represented by simple closed curves $h_{a,b}$ going (say) clockwise around the (disjoint) handles $H_{a,b}$ (with $(a,b)=(\pi(2i-1),\pi(2i))$ one of the transpositions in $\sigma$) and closing in the big lower rectangle. Then the intersection number $h_{a,b}\cdot h_{c,d}$ is $-M([a,b],[c,d])$ in Fedor's notation above ($-1$ if $a<c<b<d$, $1$ if $c<a<d<b$, $0$ else). $\endgroup$ – BS. Jul 16 '19 at 19:25
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I'll give it a try (clearly, comments aren't enough, and the references I found are too much).

Given the permutation $\pi$ of $\{1,2,\dots,2k\}$ one can build an orientable surface with boundary as follows.

Take a topological disk, say the rectangle $[0,2k+1]\times[0,1]$ in the complex plane, and consider its quotient by the relation which identifies for each $i=1,\dots,k$ a small interval $[a-\delta,a+\delta]$ ($0<\delta<1/2$) around point $a=\pi(2i-1)$ with interval $[b-\delta,b+\delta]$ around $b=\pi(2i)$, with orientation reversed, that is $a+x\simeq b-x$.

[added: this is reversed from my comments under the question : here we glue things on the bottom edge of the rectangle. This has the purely cosmetic effect of numbering points in the usual --counterclockwise-- orientation of the boundary].

The result is a compact, connected, orientable surface with boundary, which you can picture in $3$-space by protruding "nearly horizontal" rectangles from the identified intervals, making them meet only in their tips and the prescribed pair (I'm really sorry for not being able to draw that here, and being probably very bad at describing mental pictures too). The orientation comes from projection to the horizontal plane.

The resulting surface $S_\pi$ depends only on the partition of $\{1,2,\dots,2k\}$ into the pairs $\{\pi(2i-1),\pi(2i)\}$, $i=1,\dots,k$, that is on the $\pi$-conjugate $\sigma$ of the involution $\sigma_0=(12)(34)\dots(2k-1)(2k)$. There are $k!2^k$ different $\pi$'s for one $\sigma$ (this amounts to what @Ycor said in a comment).

Let $b$ be the number of boundary components of $S_\pi$. A book-keeping exercise tells you that $b$ is the number of cycles of the permutation $i\mapsto \sigma(i)+1$ (or its inverse $i\mapsto\sigma(i-1)$, or any of their conjugates, like $i\mapsto\sigma(i)-1$, that's the same number).

Now the all powerful surface classification theorem tells us that this (compact, connected, orientable) $S_\pi$ is (homeomorphic to) an (orientable) surface of some genus $g$ with the interiors of $b$ disjoint closed discs removed.

Why that ? Well, glue $b$ closed discs to the boundary components of $S_\pi$, and you have a (compact, connected, orientable) surface without boundary, connected sum of $g$ tori, by classification. Then undo.

Hence the amazing fact : there is only one model for each pair $(g,b)$ ! [this should be a footnote] That this is amazing may easily be overlooked nowadays, but can perhaps better be sensed by the fact that mapping class groups of surfaces, that is the discrete groups of components of their autoequivalences (homeo/diffeo doesn't matter here) aren't yet understood. For instance they share many properties with subgroups of linear groups, but they are not known to be linear aside from small cases. Yet they are the means to recognize a surface when we meet one.[end footnote]

Now in an oriented manifold $M^n$, one can count algebraically intersections of submanifolds or cycles of complementary dimensions, as Poincaré tolds us. Either triangulate and refine or use Thom's transversality, but in the end it's linear algebra, counting intersections by comparing the orientation of the direct sum of tangent spaces with the ambient one. This gives rise to the integer valued intersection product on cycles of complementary dimensions $p,q=n-p$, which depends only on their homology classes and is $(-1)^{pq}$-symmetric.

Crucially, it is non-degenerate when the manifold is closed (and oriented), meaning that $H_p(M,\mathbb{Z})$ mod torsion is isomorphic to $Hom(H_{n-p}(M,\mathbb{Z}),\mathbb{Z})$ via the intersection product. This is part of so-called Poincaré duality, expressed more neatly nowadays by adding cohomology to the picture (but we don't need it).

In our context of $1$-cycles on oriented surfaces, intersection is skew-symmetric.

If $S_g$ is a closed (=without boundary), connected oriented surface of genus $g$, $H_1(S_g,\mathbb{Z})$ is a free abelian group with a non-degenerate skew-symmetric form, with standard "symplectic" basis $(a_1,b_1,\dots,a_g,b_g)$, meaning

$$a_i\cdot b_j=\delta_{ij},$$ $$a_i\cdot a_j=b_i\cdot b_j = 0.$$

But when there are $b\geq 1$ boundary components, $H_1(S_{g,b},\mathbb{Z})\simeq \mathbb{Z}^{2g+b-1}$, mapping to $H_1(S_g,\mathbb{Z})\simeq \mathbb{Z}^{2g}$ with kernel of rank $b-1$, preserving (clearly) the intersection form. [added comment] This could be proved using the homology exact sequence of the pair $(S_g \supset S_{g,b})$, but here it is simpler to use homotopy invariance of homology plus the fact that for $b>0$, $S_{g,b}$ is homotopy equivalent to the graph with one vertex and $2g+b-1$ loops. This is also relevant below.

Hence the intersection form is non-degenerate on $H_1(S_{g,b},\mathbb{Z})$ iff $b\leq 1$.

Concerning the surface $S_\pi$, it should be clear that it is homotopy equivalent to a graph with one vertex and $k$ loops, and that letting $h_i$ denote the class of a closed curve going counterclockwise [NOTE: this is reversed from my comments under the question] through the handle resulting from gluing the intervals $\pi(2i-1)\pm\delta$ to $\pi(2i)\mp\delta$ (in reverse) that

  • the $h_i$, $i=1,\dots, k$ form a basis of $H_1(S_\pi,\mathbb{Z})\simeq\mathbb{Z}^k$
  • the matrix of their intersection products is the same as your skew-symmetric form on $V_\pi$

(maybe up to sign, I switched to the upper half plane hoping to correct for this).

Hope this helps.

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  • $\begingroup$ Thanks! A couple of comments. First, you are identifying segments of the top edge. This fact was more clearly stated in your original comments - reading your answer I was a bit confused at first. $\endgroup$ – H A Helfgott Jul 19 '19 at 4:05
  • $\begingroup$ Second, by "closed curve" you mean "closed curve traversed clockwise" (this seems obvious but is unstated). $\endgroup$ – H A Helfgott Jul 19 '19 at 4:26
  • $\begingroup$ Also used is the fact that the h_i's form a basis of the homology. That much seems evident - I take it is not customary to say more than that? $\endgroup$ – H A Helfgott Jul 19 '19 at 4:32
  • $\begingroup$ Also, quick double-check. All of the above looks as if it would still be valid in homology over $\mathbb{Z}/2\mathbb{Z}$. In particular, $H_1(S_g,\mathbb{Z}/2\mathbb{Z})\sim (\mathbb{Z}/2\mathbb{Z})^{2g}$ (I imagine). Thus even the rank over $\mathbb{Z}/2\mathbb{Z}$ of the matrix $M$ is as stated. $\endgroup$ – H A Helfgott Jul 19 '19 at 4:46
  • $\begingroup$ Oh, and the Euler characteristic of the surface without boundary is $1+b-k$ by triangulation, so its genus $g$ is $(1+k-b)/2$. (Just spelling out what you spelled out before.) $\endgroup$ – H A Helfgott Jul 19 '19 at 5:17

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