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This is a question from Jeevanjee's Introduction to Tensors and Group Theory for Physicists:

Show that for a metric $g$ on $V$,

$$g_i^{~~j} = \delta_i^{~~j}$$

so the $(1,1)$ tensor associated with $g$ (via $g$!) is just the identity operator.

The only way I can think of associating $g$ with a $(1,1)$ tensor is to define $\tilde{g}(v,f)=g(v,L^{-1}(f))$, where $L:V\to V^*$ such that given $v \in V$,

$$(L(v))(w) = g(v,w)~~\forall w \in V.$$

The components of the $(1,1)$ tensor can be expressed as

$$\tilde{g}_i^{~~j} = \tilde{g}(e_i, e^j) = g(e_i, L^{-1}(e^j))$$

Now $L^{-1}(e^j) = \sum_k g^{jk}e_k$, where $[g^{ij}]$ is the inverse matrix of $[g_{ij}]$.

$$\tilde{g}_i^{~~j} = g(e_i, \sum_k g^{jk}e_k) = \sum_k g^{jk}g(e_i,e_k) = \sum_k g^{jk} g_{ik}$$

There's a discrepancy with the order of the indices: it should either be $\sum_k g^{jk} g_{ki}$ or $\sum_kg^{kj}g_{ik} = \sum_kg_{ik}g^{kj}$. How to resolve this? I guess I'm missing some fact about the notation.

Edit: As pointed out in an answer, this is fine as long as $g$ is symmetric. What if it's not?

Shirish Kulhari
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1 Answers1

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The statement you're trying to prove is generally true for any nondegenerate rank $2$ tensor, essentially because in components, a matrix times its inverse is the identity. At the end, note that the metric is symmetric, so the index ordering you're worrying about doesn't matter.

For a non-symmetric tensor, you need to be more careful with the conventions, especially the order of contraction and order of arguments. In relativity we are not consistent because the metric is symmetric, so it doesn't make a difference; for example we might write $v^j = g^{jk} \omega_k$ one day and $v^j = g^{kj} \omega_k$ the next. However, when we do the representation theory of $SL(2, \mathbb{C})$ we have the antisymmetric "metric" $\epsilon_{ij}$ and it's essential to set consistent conventions. You can play around with finding a set of such conventions if you want, but personally I would just be thankful that at this stage, you don't need them most of the time!

knzhou
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  • Thanks! If possible, could you give a concrete explanation on how we'd handle the notation convention for a non-symmetric tensor in the question? What if $g$ were non-symmetric? (I've edited the question too) – Shirish Kulhari Jun 21 '18 at 11:46
  • @ShirishKulhari A supergravity text should cover it, with the galore of dotted spinor indices, non-symmetric indices and even anti-commuting quantities. – JamalS Jun 21 '18 at 16:32