The Global Invariant of Signed Graphic hyperplane Arrangements

On the Falk invariant of Shi and Linial arrangements

“…Then an easy computation shows that in this case dim(span(F 3 )) = 19. Using Example 5.9, the same exact argument used in this case will prove that dim(span(F 4 3 )) = 19s 3 . ⊓ ⊔ Notice that the argument of the previous lemma cannot be utilized to compute dim(span(F 2 3 )).…”

“…For a pair of parallel edges (i, j) that form an unbalanced circle, we will consider the 3-circuit (0, i, j). Hence, e 0 will also appear in F 3 3 and F 4 3 . Lemma 5.6.…”

“…Since elements in F 3 3 and F 4 3 share at most three indices, and if it is exactly three, one of them is 0, this implies that no elements of F 3 3 can be written as a linear combination of elements of F 4 3 . Vice versa, no elements of F 4 3 can be written as a linear combination of elements of F 3 3 , and hence, we have the second part of the statement. ⊓ ⊔ Differently from the situation discussed in [6], in this setting it might happen that span(F 2…”

“…In [13], Schenck and Suciu studied the lower central series of arrangements and described a formula for the Falk invariant in the case of graphic arrangements. In [4], the authors gave a formula for φ 3 in the case of simple signed graphic arrangements. In [5], the authors extended the previous result for signed graphic arrangements coming from graphs without loops.…”

On the Falk Invariant of Signed Graphic Arrangements

“…Similarly to theprevious examples, we can directly compute dim(span(F 3 )) for several sign graph. In particular, if we consider D 3 , K 4 and K3 3 , then dim(span(F 3 )) = 10. Assume that in the sign graph G there are exactly g • = p distinct subgraphs isomorphic to a G • , G 1 , .…”