Let us consider the space $L^2:=L^2(\mathbb{R}^n,\mathbb{C})$ and the associated scalar product $S(f,g):=\int f \overline g$. In distribution theory, we have a situation where we have to deal with two different identifications (which makes things a little bit tricky) :

  • if we identify $L^2$ with its antidual, an element $f\in L^2$ is an antilinear form $g\mapsto S(f,g)$

  • if we identify $L^2$ as a distribution space, an element $f\in L^2$ is a linear form $g\mapsto \int f g $.

I am aware of the following facts :

  • we can anti-identify (through a anti-isomorphism) $L^2$ to its dual (as an Hilbert space can always be), but this just shifts the problem, since the first identification becomes the map $f\mapsto (g\mapsto \int \overline f g)$.

  • We can define distributions as antilinear maps (and not linear), and define the identification of a function as a distribution by : $f\mapsto \int f \overline g$. This way, the two identifications coincide, and there is no need to keep track of the two different identifications made.

These observations made me think that maybe the discrepancy can be solved in another way, where instead of redefining distributions, we redefine (or more precisely refine) the abstract definition of Hilbert spaces.

More precisely, let us say that a (complex) Hilbert space with real structure is a triple $(H,b,C)$ where $C$ is a antilinear involution on $H$, and $b$ is a bilinear map such that $\langle \cdot,\cdot\rangle = b(\cdot,C(\cdot))$ is a scalar product on $H$ giving $H$ the structure of a Hilbert space. With this definition it seems that a Hilbert space with real structure is naturally isomorphic (through a bijective linear isometry, namely $f\mapsto b(f,\cdot)$) to its dual (and not its antidual), which is indeed the natural situation we have in the case of $L^2$ (where $C:f\mapsto \overline f$ and $b(f,g):=\int f g$).

My question is then : why do we study (complex) Hilbert spaces in general and not (complex) Hilbert spaces with real structure (as previously defined) ? Are there Hilbert spaces that appear naturally without an obvious real structure ?

Or again, formulated differently : if we formulated the whole theory of Hilbert space with the definition given before (with a real structure), would we gain anything in applications to abstract it further to the case of Hilbert space without real structure ?

  • $\begingroup$ The question seems to only involve (finite or infinite dimensional) complex Hermitian spaces and real inner product spaces, independently from their completeness with respect to the corresponding norms. That is, it seems purely a matter of linear algebra rather than functional analysis. $\endgroup$
    – Qfwfq
    Jan 7, 2017 at 22:00
  • 1
    $\begingroup$ @Qfwfq I agree with the fact that one could generalise the question to the general setting of inner product spaces, but I'm mainly interested in the case where we do deal with Hilbert spaces. In other words, if there are indeed natural noncomplete inner product spaces without any real structure, this would not directly answer the question. $\endgroup$
    – LCO
    Jan 7, 2017 at 22:09
  • 3
    $\begingroup$ My instinctive thought is: the issue is not that one might have Hilbert spaces without real structures, but that one might wish to consider several different real structures on the same Hilbert space (just as: every Hilbert space has an o.n. basis, but we don't make a choice of basis part of the definition) $\endgroup$
    – Yemon Choi
    Jan 7, 2017 at 23:05
  • 1
    $\begingroup$ For instance, the obvious real structure on $L^2({\bf R})$ gets taken (I think) to a non-obvious real strucure on $L^2({\bf R})$ by the Fourier transform. $\endgroup$
    – Yemon Choi
    Jan 7, 2017 at 23:07
  • $\begingroup$ Further to @NikWeaver's examples/remarks below: in general there are Hilbert spaces of analytic functions (e.g. Hardy space, Dirichlet space) and it seems to me that the only "obvious" real structures come from taking some "usual" choice of orthonormal basis, then transferring the real structure on $\ell^2$. $\endgroup$
    – Yemon Choi
    Jan 8, 2017 at 2:52

1 Answer 1


Good question. It does seem like most naturally occurring complex Hilbert spaces come equipped with a natural real structure. Incidentally, I pointed out in this paper that the same is true of Hilbert modules over C*-algebras, and in this case you can use the natural involution to turn a left action into a right action and vice versa. So many naturally occurring Hilbert modules are automatically Hilbert bimodules.

The first example that comes to mind of a Hilbert space without an obviously natural real structure is the Bargmann-Segal space of entire analytic functions on $\mathbb{C}^n$ whose integral against $e^{-|z|^2}$ is finite. I guess there is a natural isomorphism of this space with $L^2(\mathbb{R}^n)$, and you can then import the real structure of the latter, so maybe it's not a completely convincing example.

Edit: better example, the Bergman space of holomorphic functions on a domain in $\mathbb{C}$ which are square integrable against Lebesgue measure. I don't see any obvious real structure for a general domain.

  • $\begingroup$ I think it's still a good example of a naturally-occuring complex Hilbert space whose "natural real form", if any, is not obvious. $\endgroup$ Jan 7, 2017 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.