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.
Best Answer
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.
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 $n$.