# Erratum to: Divisionally free arrangements of hyperplanes

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## Mathematics Subject Classification

32S22 52C35## 1 Erratum to: Invent. math. (2016) 204:317–346 DOI 10.1007/s00222-015-0615-7

The aim of this note is to correct the statement and the proof of Theorem 6.2 in the paper published in *Inventiones Mathematicae* **204** (2016), 317–346, which is not correct as it was stated. All the other results in the paper are correct as they were stated. The correct statement of Theorem 6.2 should be as follows:

### Theorem 6.2

*V*with \(\exp ({\mathcal {A}})=(1,d_1,\ldots ,d_{\ell -1})\). Assume that distinct hyperplanes \(H_1,\ldots ,H_{\ell -1} \in {\mathcal {A}}\) satisfy that,

- (1)
\({\mathcal {A}}_i':={\mathcal {A}}{\setminus } \{H_i\}\) is free with \(\exp ({\mathcal {A}}_i')= (1,d_1,d_2,\ldots ,d_{i-1},d_i-1,d_{i+1},\ldots , d_{\ell -1})\) for \(i=1,\ldots ,\ell -1\), and

- (2)
\({\mathcal {A}}':={\mathcal {A}}{\setminus } \{H_1,\ldots ,H_{\ell -1}\}\) is free with \(\exp ({\mathcal {A}}')=(1,d_1-1,\ldots ,d_\ell -1)\).

### Proof

The original proof is correct except for the part showing the following statement in its second paragraph; “\(D({\mathcal {A}}')\) has a basis \(\theta _E, \theta _1,\ldots ,\theta _{\ell -1}\) such that the derivations \(\theta _E,\alpha _{H_1}\theta _1,\ldots ,\alpha _{H_{\ell -1}}\theta _{\ell -1}\) form a basis for \(D({\mathcal {A}})\)”. To prove this statement, we have to show that “there is a derivation \(\theta _i \in D({\mathcal {A}}_i')\ (i=1,\ldots ,\ell -1)\) of degree \(d_i-1\) such that \(\theta _i \not \in D({\mathcal {A}})\), but that \(\alpha _{H_i} \theta _i \in D({\mathcal {A}})\)”. Since both \({\mathcal {A}}\) and \({\mathcal {A}}_i'\) are free, Terao’s addition-deletion theorem implies that there is a basis \(\theta _E,\theta _1^{(i)},\ldots ,\theta _{\ell -1}^{(i)}\) for \(D({\mathcal {A}})\) such that \(\deg \theta _j^{(i)}=d_j\ (j=1,\ldots ,\ell -1)\) and \(\alpha _{H_i} \mid \theta _i^{(i)}\). If \(\theta _i:=\theta _i^{(i)}/\alpha _{H_i}\), then \(\theta _i \in D({\mathcal {A}}_i') {\setminus } D({\mathcal {A}})\). Now the rest proof is the same as the original one. \(\square \)

Except for Theorem 6.2, all the results and proofs are correct in this paper.

## Notes

### Acknowledgements

The author is grateful to Torsten Hoge for his pointing out a mistake of the original Theorem 6.2