2
$\begingroup$

Consider central hyperplane arrangement A with normal vectors with all combinations of -1 and 1 (is there name for it?). There is simple invariant for each chamber of A: sum of vectors, corresponding to hyperplanes, which are directed towards chamber. For example, chamber containing point (0, -1, 0) have invariant (0, -4, 0). If we get absolute values and sort it, resulting sequence (0, 0, 4) will correspond to all chambers of same shape. Is there similar invariant for joined chambers, i.e. chambers of subarrangement of A? Simple sum not working anymore.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.