Where is the slot for the metric so that the summation doesn't blindly treat my tensor's covector slots like vector slots? Maybe I am being a bit too cynical, but it is a shame that people have come to know about "Einstein summation convention" without ever learning that it is a special case of Ricci calculus and was not invented by Einstein. Every time I see someone do u_i v_i as an implied sum I know I am going to have a hard time. "The sum over the 0 index has a minus sign," yes because you need to contract the metric there to convert between a covector and a vector to get a sum.

I made a C++ operator overloading implementation of Ricci calculus once upon a time. That way you could define your tensors and then write statements like:

  T = (rho + p)*u("^a")u("^b") + p * g("^a^b");
  traceT = T("^a_a");

Anyways, I find the domain specific language of this particular numpy function to be a bit abhorrent. Why not write "ik,kj->ij" as "M_ij = A_ik * B_kj" and make it much more obvious what is happening instead of requiring this understanding. Heck, if you used kwargs, you could even avoid passing the same argument multiple times like in the quadratic form example:

  x = np.ricci_calculus("M_ij v^i v^j", M, v)

Here I am leaving out that "x = ..." prefix in the DSL since you can take an ordering convention and a scalar has no slots anyways.