# 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.

• Conjugate symmetry of the inner product means $\alpha_{ij} = \overline{\alpha_{ji}}$. But how can one right down some condtion for the positive definiteness in terms the of $\alpha_{ij}$ ? – Physor Sep 6 '20 at 16:18

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

2. 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.

3. 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 $$n$$.

• Thanks a lot for your answer, but I still have some remarks, better to say questions on it much more important than this one: in the last paragraph before the claim, you meant : (no need to consider a topology for infinite sums), I think you meant for finite, right ? – Physor Sep 6 '20 at 19:06
• I mean "no need to consider a (topology for infinite sums)," as in "no need to consider a topology because we aren't taking any infinite sums." – Qiaochu Yuan Sep 6 '20 at 19:08
• which definition of tensors are you using here, because I,ve seen a lot of these definitions. For example: 1. Quotient space of the free vector space. 2. the set of all multilinear maps $V\times \dots \times V \times V^* \times \dots \times V^* \rightarrow \mathbb{R}$, and I don't which is appropriate for infinte dimensional spaces – Physor Sep 6 '20 at 19:13
• @Physor: I'm using the definition that a tensor is an element of a tensor product $(V^{\ast})^{\otimes n} \otimes V^{\otimes m}$. The equivalence between this definition and the definition in terms of multilinear maps only works when $V$ is finite-dimensional and in the infinite-dimensional case they define two genuinely different concepts; I'm not aware of terminology that distinguishes them. With topologies around there are multiple choices of "completed" tensor product also, which further complicates this issue: see en.wikipedia.org/wiki/Topological_tensor_product for example. – Qiaochu Yuan Sep 6 '20 at 19:17
• What is the definition of $(V^{\ast})^{\otimes n} \otimes V^{\otimes m}$ please?, I hope my questions are not anoying to you – Physor Sep 6 '20 at 19:21