# Inner product spaces without symmetry/hermitian axiom

Consider a vector space $X$ over $\mathbb R$ and a bilinear form $\langle \cdot, \cdot \rangle : X \times X \rightarrow \mathbb R$.

We assume furthermore that for any $x \in X$ there exists $y \in X$ such that $\langle x, y \rangle \neq 0$. In other words, the inner product is non-degenerate.

However, we do not assume any symmetry (or hermitianness if the field would have been $\mathbb C$) as an axiom.

Question: How close to Hilbert space theory can we develop a theory for such spaces? It should be possible to turn $X$ into a topological vector space with such a product given, and a Riesz representation theorem should be possible.

• By "turn into a topological vector space" you mean "find a vector space topology such that $\langle\cdot,\cdot\rangle$ becomes continuous" ? – Christian Remling Apr 27 '15 at 0:03
• Do you want to assume some sort of positivity condition? Even restricting to symmetric bilinear forms, if you don't require the form to be definite then it seems unlikely to me that there is any natural associated vector space topology generalizing the norm in the definite case, essentially because there can be too many automorphisms that preserve the bilinear form (consider an infinite direct sum of hyperbolic planes, for instance). – Eric Wofsey Apr 27 '15 at 3:24

Elaborating on my comment, here is a negative answer to the question as asked. I claim that there is no way to assign a vector space topology $\tau_X$ to every real vector space $X$ with a nondegenerate bilinear form which simultaneously has the following three properties:
1. For any $X$, the bilinear form is continuous as a map $X\times X\to\mathbb{R}$ with respect to $\tau_X$.
2. If the bilinear form on $X$ is symmetric and either positive definite or negative definite, then $\tau_X$ is the usual norm topology.
3. If $T:X\to Y$ is a linear injection preserving the bilinear form, then $T$ is continuous with respect to $\tau_X$ and $\tau_Y$.
In fact, this holds even if you restrict to symmetric bilinear forms; all that matters is that they are not required to be definite. To show this, let $X$ be a direct sum of countably many copies of $\mathbb{R}^2$ with bilinear form $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Let $\{e_n, f_n\}$ be the standard basis of the $n$th summand of $X$. If $Y$ is the subspace spanned by the vectors $e_n+f_n$, then the bilinear form is positive definite when restricted to $Y$. In the norm topology on $Y$, the sequence $((e_n+f_n)/n)$ converges to $0$, and so by (2) and (3) this sequence must also converge to $0$ with respect to $\tau_X$. Similarly, the vectors $e_n-f_n$ span a negative definite subspace, so $((e_n-f_n)/n)$ must converge to $0$. It follows that $(e_n/n)$ and $(f_n/n)$ also converge to $0$. Now for any sequence $a_n$ of nonzero scalars, the automorphism $T$ of $X$ that sends $e_n$ to $a_ne_n$ and $f_n$ to $f_n/a_n$ preserves the bilinear form. By (3), it now follows that $(a_n e_n)$ and $(a_n f_n)$ converge to $0$ for any scalars $a_n$. In particular, this implies that $(e_n+f_n)$ must converge to $0$. But this contradicts (1), since $\langle e_n+f_n,e_n+f_n\rangle=2$ for all $n$.