By Joel Friedman

During this paper the writer establishes a few foundations relating to sheaves of vector areas on graphs and their invariants, akin to homology teams and their limits. He then makes use of those rules to end up the Hanna Neumann Conjecture of the Nineteen Fifties; in truth, he proves a bolstered kind of the conjecture

Show description

Read Online or Download Sheaves on graphs, their homological invariants, and a proof of the Hanna Neumann conjecture PDF

Best combinatorics books

Proofs from THE BOOK

This revised and enlarged 5th version gains 4 new chapters, which include hugely unique and pleasant proofs for classics corresponding to the spectral theorem from linear algebra, a few more moderen jewels just like the non-existence of the Borromean jewelry and different surprises. From the Reviews". .. within PFTB (Proofs from The publication) is certainly a glimpse of mathematical heaven, the place smart insights and gorgeous rules mix in brilliant and excellent methods.

Combinatorial Algebraic Geometry: Levico Terme, Italy 2013, Editors: Sandra Di Rocco, Bernd Sturmfels

Combinatorics and Algebraic Geometry have loved a fruitful interaction because the 19th century. Classical interactions comprise invariant conception, theta features and enumerative geometry. the purpose of this quantity is to introduce fresh advancements in combinatorial algebraic geometry and to procedure algebraic geometry with a view in the direction of functions, equivalent to tensor calculus and algebraic statistics.

Finite Geometry and Combinatorial Applications

The projective and polar geometries that come up from a vector house over a finite box are really worthwhile within the development of combinatorial gadgets, equivalent to latin squares, designs, codes and graphs. This booklet presents an creation to those geometries and their many purposes to different components of combinatorics.

Extra resources for Sheaves on graphs, their homological invariants, and a proof of the Hanna Neumann conjecture

Example text

Wm are compartmentally distinct, then w1 , . . 6. MAXIMUM EXCESS AND SUPERMODULARITY 33 if) they are each non-zero. Second, W ⊂ W is compartmentalized only if (and if) there exist quotients, Qs , of Ws for s ∈ S such that π induces an isomorphism Qs → W/W . 10. 24. Let W be a finite dimensional vector space with a decomposition. Let w1 , . . , wm be compartmentally distinct, and let W ⊂ W be a compartmentalized subspace of W . Then the images of w1 , . . , wm in W/W are linearly independent (in W/W ) iff they are nonzero (in W/W ).

All higher Tor groups vanish in sheaves of vector spaces over graphs), so we get a new short exact sequence 0 → F1 ⊗ FG → F2 ⊗ FG → F3 ⊗ FG → 0; now note that for any sheaf, F, on G we have F ⊗ FG = FG . 4. (F1 ) = 0. ((F3 )G ) for all open subsets, G , of G, which could be a much stronger inequality (and is much stronger for the setting of the SHNC). Let us state a slightly stronger “contagious vanishing” theorem that we shall apply to the maximum excess. 15. By a scaling first quasi-Betti number, α1 , we mean a rule that, for some field, F, and any digraph, G, assigns a non-negative real number to each sheaf of F-vector spaces over G, such that (i) α1 is a first quasi-Betti number when restricted to sheaves on G for any digraph, G; (ii) for any covering map ϕ : K → G of digraphs and any sheaf, F, on G we have α1 (ϕ∗ F) = α1 (F) deg(ϕ); and (iii) for an ´etale ϕ : K → G and any sheaf, F, on K we have α1 (ϕ!

In this case we have h1 (G ) = 1 − χ(G ) = 1 − pχ(G) = pρ(G) + 1. 18 with F = F (so that μ∗ F = F on G ) we have htwist (F) 1 h1 (G , F)/p = h1 (G )/p = ρ(G) + (1/p). Letting p → ∞ we conclude htwist (F) 1 htwist (F) 1 ρ(G). But the “trivial lower bound” gives −χ(F) = ρ(G). If G is not connected then we apply the above to each of its connected components and conclude the following theorem. (F). 19. For any digraph, G, we have ρ(G) = htwist 1 30 1. 3. The Maximum Excess Bound. Let F be a sheaf of F-vector spaces on a digraph, G, and let U ⊂ F(V ).

Download PDF sample

Rated 4.18 of 5 – based on 22 votes