combinatorial lemma (is it well-known?) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:52:27Z http://mathoverflow.net/feeds/question/119830 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119830/combinatorial-lemma-is-it-well-known combinatorial lemma (is it well-known?) vaevictis 2013-01-25T13:16:30Z 2013-01-27T00:31:20Z <p>The following should be something well?-known, but i haven't seen it anywhere, neither have i met any references about.</p> <p>Let $M^{n}$ be a $n$-dimensional oriented closed manifold with a (sufficiently small) triangulation $\tau$. We "colour" the vertices of $\tau$ with $n+2$ colors: $v^{o}\rightarrow w(v^{o})\in$ {$1,2,...,n+2$ } and we shall say that the correspondence $w$ is a "coloring" of $\tau$. Take an arbitrary color $i\in$ {$1,2,...,n+2$ } and consider the $n$-simplices whose vertices are colored with exactly the colors {$1,2,...,n+2$ }$\backslash{i}$. Let $\Delta^{n}$ be such a simplex and $v_{1},...,v_{n+1}$ be its vertices ordered according to the positive orientation of $\Delta^{n}$ induced by the orientation of $M^{n}$. Then we write $\sigma_{i}(\Delta^{n})=1$, if the permutation $(w(v_{1}),...,w(v_{n+1}))$ is even, and $\sigma_{i}(\Delta ^{n})=-1$ otherwise. Set $\sigma_{i}(\Delta^{n})=0$ if some vertex of $\Delta^{n}$ is colored $i$, or there are two identically colored vertices. Let finally</p> <p>$\sigma_{i}(w)=\sum\sigma_{i}(\Delta^{n})$,</p> <p>where the sum is over all $n$-simplices.</p> <p><strong>The Claim:</strong> The number $\sigma_{i}(w)$ does not depend on $i$: $\sigma_{1}(w)=\sigma_{2}(w)=...=\sigma_{n+2}(w)$. So we have a global invariant $\sigma(w)$ of the coloring $w$.</p> <p>This invariant has a <strong>geometrical meaning</strong>: Consider the dual cell complex of the triangulation $\tau$, then since each cell corresponds to a vertex $v^{o}$ of $\tau$, we may color this cell by the color $w(v^{o})$. Let $F_{i}$ be the union of all cells colored $i$, then we get a covering $\lambda=${${F_{1},...,F_{n+2}}$} of $M^{n}$. It is easy to see that the intersection of all $F_{i}$ is empty, so the canonical map of $M^{n}$ into the nerve of $\lambda$ may be considered as a map of $M^{n}$ into the $n$-sphere $\mathbb{S}^{n}$: $\varphi:M^{n}\rightarrow\mathbb{S}^{n}$. Then the degree of $\varphi$ equals $\sigma(w)$:</p> <p>$\deg\varphi=\sigma(w)$.</p> <p>As the proofs are not sophisticated at all and the construction seems conceptual, maybe it is worth including this material in an elementary topology textbook. Note also that it gives a method for calculating the degree without smooth approximation.</p> <p>Of course, i don't want to repeat well-known things without citation, so any references are welcome.</p> http://mathoverflow.net/questions/119830/combinatorial-lemma-is-it-well-known/119976#119976 Answer by Igor Pak for combinatorial lemma (is it well-known?) Igor Pak 2013-01-27T00:31:20Z 2013-01-27T00:31:20Z <p>This is closely related to the Generalized Sperner's Lemma, which holds for all for simplicial manifolds with or without boundary. See <a href="http://www.math.ucla.edu/~pak/papers/tilesurvey.pdf" rel="nofollow">my old survey</a> for a quick introduction (Section 8.1). Classical references include A.B. Brown and S.S. Cairns, <a href="http://www.pnas.org/content/47/1/113.full.pdf+html" rel="nofollow">Strengthening of Sperner's lemma applied to homology theory</a>, PNAS, 1960, and D.I.A Cohen, <a href="http://www.sciencedirect.com/science/article/pii/S0021980067800620" rel="nofollow">On the Sperner lemma</a>, JCT (1967). I don't immediately see how your result follows from the lemma, but recall that many extensions and generalizations are known. I would start with these references and search forward to find your particular version. </p>