I am trying the calculate the Kretschmann scalar $K$ of a metric $$K=R^{abcd}R_{abcd}$$ where $R$ is the Riemann tensor. If I do


I get a scalar function, but is this code equivalent to the above formula? Should I raise the indices of the first Riemann tensor with the inverse of the metric, or does Mathematica do this for me? Even if the metric is pseudo-Riemannian and with complicated coefficients?


1 Answer 1


Mathematica contracts indices assuming flat Euclidean metric. You have to lower/raise the indices yourself.

g = DiagonalMatrix[{1, -1, -1, -1}];

RaiseIndex[T_, g_, i_] := 
  Module[{indices = Append[Delete[Range[Length@Dimensions@T], i], i]},
     TensorContract[TensorProduct[T, g], {i, Length[indices] + 1}], 

Rcontra = Fold[RaiseIndex[#1, g, #2] &, R, {1, 2, 3, 4}];

TensorContract[TensorProduct[R, Rcontra], {{1, 5}, {2, 6}, {3, 7}, {4, 8}}]

Edit: Explanaion of RaiseIndex

Suppose, we want to raise second index $j$ in $R_{ijkl}$: $$ R^{\ j'}_{i\ kl} = R_{ijkl}g^{jj'} $$

However, if we just do R1=TensorContract[R⊗g, {2,5}], then the order of the indices will be $i,k,l,j'$. In other words, we need to transpose the tensor to restore the order $i,j',k,l$. The order of transposition is calculated in variable indices. It says remove index i from sequence {1,2,...n} and put it to the end. In the case where $i=2$ and $n=4$: {1,2,3,4} → {1,3,4,2}. Now when we do Transpose[R1, {1,3,4,2}], with R1 having the order of dimensions $i,k,l,j'$, we basically say put dimension $i$ to the 1st place, put dimension $k$ go to the 3rd place, put dimension $l$ to the 4th place and put dimension $j'$ to the 2nd place. Thus, the order of our dimensions (indices) will be $i,j',k,l$.

Since we need to raise all of 4 indices, Fold is just a smart way to write this:

Rcontra = RaiseIndex[RaiseIndex[RaiseIndex[RaiseIndex[R, g, 1], g, 2], g, 3], g, 4];

Please, read help on Fold to get further insights.

  • $\begingroup$ Could you explain the function RaiseIndex? It is a bit too difficult for me. And why do you use Fold? $\endgroup$
    – mattiav27
    Nov 18, 2019 at 14:17
  • $\begingroup$ Added explanations $\endgroup$ Nov 18, 2019 at 15:00
  • $\begingroup$ In addition, we could probably write this: TensorContract[TensorProduct[R,R,g,g,g,g], {{1,9},{5,10},{2,11},{6,12},{3,13},{7,14},{4,15},{8,16}}]. However, it has $4^8=65k$ of components as intermediate tensor product. So, it is easier to raise indices one by one if possible. $\endgroup$ Nov 18, 2019 at 15:06
  • $\begingroup$ Thanks now I understand. $\endgroup$
    – mattiav27
    Nov 18, 2019 at 15:17

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.