I'm interested in the intuition for the following calculation:

$|Ric_g|_g^2 = g^{ia}g^{jb}R_{ij}R_{ab}$,

where $R_{ij}$ are the components of the Ricci curvature tensor.

Here's my thought process: Locally, we have

$|Ric_g|_g^2 = g(Ric_g,Ric_g) = R_{ij}R_{ab} \hspace{2pt} g(dx^i\otimes dx^j, dx^a\otimes dx^b),$

which suggests we define $(g(dx^i,dx^a)g(dx^j,dx^b) =) g^{ia}g^{jb} = g(dx^i\otimes dx^j, dx^a\otimes dx^b)$.

I do not understand why, if this is correct, it is a natural way to define $g$ on two-tensors.

As a follow up, how can this be used to extend $g$ to be evaluated on other types of tensors?



1 Answer 1


Once we've fixed a metric $g$, we can write this as a total contraction $$(A, B) \mapsto \operatorname{contr}(g^{-1} \otimes g^{-1} \otimes A \otimes B) ,$$ (here, $\operatorname{contr}$ just denotes the appropriate composition of trace operators) or in abstract index notation, $$(A_{ab}, B_{cd}) \mapsto g^{ac} g^{bd} A_{ab} A_{cd} .$$ In both notations this map is manifestly coordinate-free and hence natural. (It's worth doing the exercise of showing that this really defines a fiber metric on the bundle $T^*M \otimes T^*M$. In particular, to show that is positive definitive, it is useful to expand in a local orthogonal frame.)

For a general tensor bundle, we can form a fiber metric in the same way, by using $g$ and $g^{-1}$ to pair corresponding indices. For example, on the bundle $\operatorname{End} TM = TM \otimes T^*M$ of endomorphisms of $TM$ we can define the fiber metric $$(C^a{}_b, D^c{}_d) \mapsto g_{ac} g^{bd} C^a{}_b D^c{}_d .$$ Obviously, this prescription specializes to the usual metric $g$ on $TM$ and the dual metric $g^{-1}$ on $T^*M$.

  • $\begingroup$ Thanks for your answer. In each of your examples, both tensors had the same type, and I understand how this generalizes to other tensor types. Could I extend this to define $g$ to take in two different tensor types. For example, if I have $g(Ric_g, X)$ where $X = X^i \partial_i$, would it be reasonable to define $g(Ric_g, X) = g_{ac}g^{bd}R_{ab}X^c\partial_d$? $\endgroup$
    – dannum
    Jan 6, 2018 at 21:20
  • $\begingroup$ That object isn't well-formed: $a$ appears twice as a lower index, and even without this term, the quantity $\partial_d$ doesn't make sense in isolation (i.e., without a matching index). There is no natural scalar quantity bilinear in a $(0, 2)$-tensor and a vector, even in the presence of an inner product. $\endgroup$ Jan 7, 2018 at 22:57

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