Upper, lower, mixed indices - clarification needed

Question

Consider the following code

{A,B,C}::Indices(position=fixed); 
M^{#}::AntiSymmetric;
ex1:=M_{A B}+ M_{B A};
ex2:=M^{B}_{A} + M_{A}^{B};
ex3:=M^{A B} + M^{B A};

I wanted M to be anti-symmetric in all upper indices(however many they are)- for this I used #. But, somehow Cadabra is taking M to be antisymmetric even for lower and mixed indices. I thought if I declare

M{#}::AntiSymmetric. 

then Cadabra takes M to be antisymmetric irrespective of position and number of indices that M carries (and that's what it in fact does). Please let me how if Cadabra should behave this way in the above example or if this is a bug?

Answer 1

It behaves as expected; for historical reasons M{#} means 'M with any arguments or indices, regardless of position'. M^{#} or M_{#} is interpreted in the same way. It's used a lot to declare \delta{#}::KroneckerDelta, in which case you really need it to mean 'both upper and lower indices'.

Perhaps this historical choice was not a particularly good one, but I'm reluctant to change it now. I could potentially make M^{#} and M_{#} mean what you thought they would mean, because that should not clash with existing notebooks. Let me think this through a bit more.

For the time being, your only options are to either declare M^{A B}::AntiSymmetric explicitly (in which case you have to repeat it for any number of indices that you need) or accept that it behaves as in your example.

Comments

Thanks. Actually M{#} as a notation is a very good (for example for my problem writing \epsilon{a1? a2? a3? a4? a5? a6? a7? a8? a9? a10?} is a real pain and this feature saves the day). And it would be really nice if you could make M^{#} and M_{#} would mean what one naively would assume them to mean.