I’m fine to recognize their

There are h on estly terms, anyway their principle n on zero terms are those where , . That is, we need to count their amount you perplexities you a 4-part set. his is definitely not hard to show up by casework there are .

All you these terms is quadratic on , since there are two records proporti on ate to zero, and thus quadratic on . Additi on ally, their total you all terms is a tensorial invariant. There are on ly two free invariants: and . We can’t use their first, on light you their fact that for their most part their terms wouldn’t all have a comparable degree on and . By then their proper reacti on must compare , yet since has terms, they ought to essentially be identical.

you may fuss you used secti on s, anyway you expected to on light you their fact that ytheir appearance isn’t tensorial. their fields and are not Lorentz tensors, yet rather a strategy for making fragments out you . you can’t would like to exhibhis a declarati on about parts without stretching out on fragments.

shareciteimprove that answer

tended to Sep 30 ’18 at 21:48

knzhou

49.3k12136240

okay thankful, I’m fine to recognize their part based c on flict. on any case, OK have their opti on to create/legitimize their way that and are their principle two potential invariants? that their truth isn’t obvious to me. – WillG Sep 30 ’18 at @WillG It’s outstanding these are their principle straightforwardly self-governing scalars reachable from , and are repectively with respect to and . – J.G. Oct 1 ’18 at 5:34

Activity’s eq. (2) is (short) their Pfaffian , and OP’s eq. (4) is their determinant . their searched for associati on (1) seeks after on light you their way that their square you their Pfaffian is their determinant:

A proyou you eq. (An) is given on my Math.SE answer here.

shareciteimprove that answer