The tensor of the inner product on an infinite vector space Let $V$ be an infinite dimensional complex vector space. What is the general form of an inner product as a (0,2) tensor on this space:
$$
\langle\cdot,\cdot\rangle \equiv \sum \alpha_{ij}\, \overline{\mathbf{e}^i}\otimes\mathbf{e}^j(\cdot,\cdot):= \sum \alpha_{ij}\, \overline{\mathbf{e}^i}(\,\cdot\,)\mathbf{e}^j(\,\cdot\,)
$$
Where {$\mathbf{e}^j$} is dual basis of the (topological or algebraic, I don't know) dual of $V$ (usually denoted $V^\prime$ or $V^*$ resp.) and {$\overline{\mathbf{e}^i}$} is the corresponding basis in the complex conjugate dual space $\overline{V^*}$. For complex conjugate vector space see Wiki. For sesquilinear forms as tensor product see Wiki
How can an inner product be represented as a tensor. What are the conditions on $\alpha_{ij}$, if that is possible at all, so that the sum above is an inner product?
I know this may sound weird but at the end an inner product is a symetric positive definite sesquilinear form which may be represented as some tensor. I know also that some requirements on the vector space are missing, like topology or norm but I don't know very much about this. In quantum mechanics we have a Hilbert space that comes with an inner product. But we know that there are many other inner products on the same Hilbert space.
 A: Some comments.

*

*Very strictly speaking, complex inner products aren't tensors; tensors have to be multilinear, and complex inner products are sesquilinear.


*If $V$ is an infinite-dimensional vector space, with no further structure, then we have a map from $V^{\ast} \otimes V^{\ast}$ to the vector space $\text{Bilin}(V \times V, \mathbb{C})$ of bilinear forms $V \times V \to \mathbb{C}$, but this map is no longer an isomorphism; its image is the subspace of bilinear forms of "finite rank," by which I mean bilinear forms $B$ such that the induced map $V \ni v \mapsto B(v, -) \in V^{\ast}$ has finite rank. In particular all such bilinear forms are highly degenerate ("nondegenerate" here means that this induced map is injective) so none of them can be inner products.


*So 1 and 2 are two ways in which inner products on $V$ can't be represented as tensors, strictly speaking. Fixing 1 is not so bad, we just allow some conjugate-linearity into the definitions. Fixing 2 is harder because naively we'll want to express an inner product as an infinite sum so we need extra structure on $V$ that lets us do that.
We can ignore all of these problems by working in the following very simple setting. Take $V$ to be the free vector space on an infinite set $I = \{ e_i \}$ and let $A : V \times V \to \mathbb{C}$ be a sesquilinear form on $V$. $A$ is uniquely and freely determined by its "matrix entries"
$$A_{ij} = A(e_i, e_j)$$
even though we have not written these entries as components of a tensor. Now we can ask for conditions on the $A_{ij}$ making $A$ an inner product. We already have sesquilinearity; symmetry is equivalent to $\overline{A_{ij}} = A_{ji}$ just as in the finite-dimensional case; so the only remaining thing to understand is positive-definiteness.
In this case what saves us is that, by the definition of $V$, its elements consist of only finite linear combinations of the basis $e_i$ (no need to consider a topology for infinite sums), so $A$ is positive-definite iff it's positive-definite when restricted to each finite-dimensional subspace. This means that $A$ is positive-definite iff each of the finite submatrices defined by
$$A_{ij}^F = A(e_i, e_j), i, j \in F \subset I, F \text{ finite}$$
is positive-definite. And in fact a slightly stronger statement is true.

Claim: $A$ is positive-definite iff the determinant of each of the finite submatrices $\det A^F$ above is positive.

This is a mild variant of Sylvester's criterion. If $I = \mathbb{N}$ is countable it suffices to check the determinants corresponding to $F = \{ 1, 2, \dots n \}$ for all $n$.
