Universal metric of a bivariant tensor $t^i_j$.

Question

I'm reading Elements for Physics by Albert Tarantola, and get stuck on page 16.

Before this page, the author defined a metric on a linear space $S$ as a map to its dual $\mathbf{G} : S \to S^*$ that is invertible, linear and symmetric.

So if there is a metric defined over $S$, then we could define the norm of a vector, and of a linear form, and of a second-order tensor.

Then the author suggested a way to define a norm of a 'bivariant' tensor, when there is no metric defined over $S$, as following:

Definition 1.12 (Universal metric) For any two nonvanishing real numbers $\chi$ and $\psi$, the operator with components $$ g_j{}^j{}_k{}^l = \chi \delta_i^l \delta_k^j + \frac{\psi-\chi}{n} \delta_i^j \delta_k^l $$ maps the space of bivariant('contravariant-covariant') tensors into its dual (i.e., the space $S^* \otimes S$), is symmetric, and invertible. Therefore, it defines a metric over $S\otimes S^*$, that we shall call the universal metric.

I have two question:

1. What does the author mean by the coponents notations of the operator? That is, what is the image of, say, tensor $\mathbf{t} = t^i{}_j e_i \otimes e^j$, under the operator he defined? (Here we take a basis $e_1,...,e_n$ of $S$ and the dual basis $e^1,...,e^n$ of $S^*$.)

2. I'm also confused by the definition of metric here, and this is the first time I've seen metric is defined as a map from a vector space to its dual. Does this often happen?

Thanks in advance!

Answer 1

For the first question, the image of a tensor $\mathbf{t} = t^i\phantom{}_j \mathbf{e}_i \otimes \mathbf{e}^j \in S \otimes S^*$ will be the tensor $\mathbf{p} = p_i\phantom{}^j \mathbf{e}^i \otimes \mathbf{e}_j \in S^*\otimes S$ given by $$p_i\phantom{}^j = g_i\phantom{}^j\phantom{}_k\phantom{}^l \, t^k\phantom{}_l.$$ This is indicated in a footnote in the edition I have, but in a sloppy way.

For the second, the definition of metric here is not standard, but the idea that inner product is the same as isomorphism between $S$ and $S^*$ is common. The usual way is to define an inner product as a symmetric positive definite bilinear form $S \times S \to \mathbb{R}$ and then say that it can be used to define a map $\mathbf{G} \colon S \to S^*$ as $\mathbf{Gx} \colon \mathbf{y} \mapsto \left(\mathbf{x}, \mathbf{y}\right)$. In terms of tensors the inner product corresponds to the metric tensor $g_{ij}$ and the map $\mathbf{G}$ is the "index lowering" map which maps a vector $x^i \mathbf{e}_i$ to a covector $x_i \mathbf{e}^i$ where $x_i = g_{ij} x^j$.