# Hilbert representation of a bilinear form

Let $$\sigma:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$$ be a bilinear symmetric form which is non-degenerate in the sense that for every $$0\neq u\in \mathbb{R}^n$$ there is $$v\in \mathbb{R}^n$$ with $$\sigma\left(u,v\right)\ne 0$$.

It is well-known and easy to see that there is a basis $$e_1,...,e_n$$ and $$m \in \overline{0,n+1}$$ such that $$\sigma\left(u,v\right)= \sum_{k=1}^{m} u_kv_k-\sum_{k=m}^{n} u_kv_k$$, where $$v=(v_1,...,v_n)$$ and $$u=(u_1,...,u_n)$$ with respect to $$e_1,...,e_n$$.

Endow $$\mathbb{R}^n$$ with an inner product $$\langle u,v\rangle=\sum_{k=1}^{m} u_kv_k$$, where $$v=(v_1,...,v_n)$$ and $$u=(u_1,...,u_n)$$ with respect to $$e_1,...,e_n$$. Then $$T:\mathbb{R}^n\to \mathbb{R}^n$$ defined by $$Te_k=e_k$$, for $$1\le k\le m$$ and $$Te_k=-e_k$$, for $$m\le k\le n$$ is an isometry, such that $$\sigma\left(u,v\right)=\langle Tu,v\rangle$$.

I am wondering if the following infinite-dimensional version holds:

Let $$V$$ be a real vector space, and let $$\sigma$$ be a symmetric and non-degenerate bilinear form on $$V$$. Can we find a Hilbert space $$H$$, a linear injection $$S:V\to H$$ and a surjective isometry $$T:H\to H$$ such that $$\sigma\left(u,v\right)=\langle TSu,Sv\rangle_{H}$$, for every $$u\in V$$?

The answer is no.

Let be construct such a $$(V,\sigma)$$. Let $$X$$ be the set of functions $$\mathbf{N}\to\mathbf{R}$$. Let $$Y\subset X$$ be a subset such that

• $$Y$$ is linearly independent
• $$Y$$ contains all Dirac functions $$n\mapsto\delta_{m,n}$$
• for every $$g\in X$$, there exists $$f\in Y$$ such that $$\limsup(f/(|g|+1))=\infty$$.

Note that the last condition forces $$Y$$ to be uncountable. The existence of $$Y$$ is checked by considering a maximal linearly independent subset containing all Dirac functions.

Now consider the space $$V$$ with basis indexed by $$\mathbf{N}\sqcup Y$$, namely $$(e_n)_{n\in\mathbf{N}}$$ and $$(z_f)_{f\in Y}$$, and symmetric bilinear form $$\sigma(e_n,e_m)=\delta_{m,n},\quad \sigma(e_n,z_f)=f(n),\quad\sigma(z_f,z_g)=0.$$

I claim that $$\sigma$$ is non-degenerate; I'll use the first two axioms. Indeed, let $$v=\sum_na_ne_n+\sum_fb_fz_f$$ belong to the kernel of $$\sigma$$. Then $$0=\sigma(v,z_f)=\sum_na_nf(n)=0$$ for all $$f\in Y$$. Choosing $$f=\delta_n$$, we deduce $$a_n=0$$, for all $$n$$. So $$v=\sum_fb_fz_f$$. Then $$0=\sigma(v,e_n)=\sum_{f\in Y}b_ff(n)$$ for all $$n$$, which means that $$\sum_{f\in Y}b_ff=0$$. Since $$Y$$ is linearly independent, we deduce $$b_f=0$$ for all $$f$$, and hence $$v=0$$.

Now suppose the existence of $$H$$, $$S:V\to H$$ and $$T$$ as required. Write $$E_n=S(e_n)$$ and $$Z_f=S(z_f)$$, $$C_f=\|TZ_f\|$$, $$g(n)=\|E_n\|$$. Then for all $$f\in Y$$ and $$n$$, we have $$f(n)=\sigma(z_f,e_n)=\langle TZ_f,E_n\rangle$$. So $$|f(n)|\le C_fg(n)$$ for all $$n$$. Hence $$\limsup (f/(g+1))<\infty$$. Since this holds for each $$f\in Y$$, we contradict the third axiom, which ends the proof.

However, if $$V$$ has countable dimension, the answer is yes. Indeed, one can construct, by induction, an orthogonal basis and conclude in the same way as in finite dimension.

For the induction, one first observes that every finite-dimensional subspace $$W$$ is contained in a non-degenerate one. This allows to write $$V$$ as an increasing union of non-degenerate subspaces, and thereby construct the desired basis.

• PS at the moment I don't know if there's always a Hilbert representation when $\sigma>0$ (i.e., $\sigma(v,v)>0$ for all $v\neq 0$).
– YCor
Sep 22 '18 at 8:48
• PPS At the end of the post I used the fact that if $(V,\sigma)$ has an orthogonal basis, then it has a Hilbert representation. The converse is false. Indeed, consider an infinite dimensional separable Hilbert space itself $H$, with orthonormal Hilbert basis $(e_n)$. If $H$ had an (algebraic) orthogonal basis $(v_i)_{i\in I}$, then $I$ has to be uncountable. Write $J_n=\{i:\langle e_n,v_i\rangle\neq 0\}$. Then $J_n$ is a finite subset of $I$, so $J=\bigcup_nJ_n$ is countable. Pick $i\notin J$: then $\langle v_i,e_n\rangle=0$ for all $n$, which forces $v_i=0$, a contradiction.
– YCor
Sep 22 '18 at 8:54
• If $\sigma > 0$ then $(V,\sigma)$ is an inner product space and we can simply take $H$ to be its completion, with $S$ the inclusion map and $T=I$. Sep 22 '18 at 14:06
• @NateEldredge oh thanks, so simple :)
– YCor
Sep 22 '18 at 14:17
• So, well, a necessary condition for the existence of a representation can be written as: for each sequence $(v_n)$ in $V$, there exists a positive real sequence $(t_n)$ such that for every $v\in V$, one has $|\sigma(v_n,v)|=O(t_n)$. I don't know if it's sufficient.
– YCor
Sep 22 '18 at 14:26