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A Hilbert space $H$ is a real or complex vector space endowed with an inner product such that $H$ is a complete metric space when endowed with the norm induced by this inner product.

8 votes
0 answers
193 views

Hilbert spaces over the semi-field $\mathbb R_+$

Let $\mathbb R_+$ be the semi-field of non-negative real numbers. Definition (preliminary): A Hilbert space over $\mathbb R_+$ is a pair $(H,P)$, where $H$ is a complex Hilbert space, and $P\subset H …
André Henriques's user avatar
64 votes
6 answers
10k views

Are dagger categories truly evil?

Recall that a dagger category is a category equipped with an involution $*:Hom(x,y)\to Hom(y,x)$ that satisfies $f^{**}=f$ and $f^* g^*=(gf)^*$. A prominent example of a dagger category is the categor …
André Henriques's user avatar
18 votes
2 answers
1k views

compact-open topology on $B(H)$

In topology, it is common to use the compact-open topology on the set of continuous maps between two given topological spaces. Let now $H$ be a Hilbert space and $B(H)$ the set of continuous linear m …
André Henriques's user avatar
22 votes
2 answers
3k views

Does it make sense to talk about smooth bundles of Hilbert spaces?

Is there a notion of "smooth bundle of Hilbert spaces" (the base is a smooth finite dimensional manifold, and the fibers are Hilbert spaces) such that: 1• A smooth bundle of Hilbert spaces over a poi …
André Henriques's user avatar
9 votes
2 answers
470 views

why is this a sufficient condition for a domain to be a core of an unbounded operator?

Let $\alpha:\mathbb R\to U(H)$ be a strongly continuous action of the reals on some Hilbert space, and let $A=-i\frac d{dt}\alpha(t)|_{t=0}$ be its infinitesimal generator, so that $\alpha(t)=e^{itA}$ …
André Henriques's user avatar
18 votes
2 answers
1k views

Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform

The Hilbert transform on the real Hilbert space $L^2(\mathbb R)$ is the singular integral operator $$ \mathcal H(f)(x) := \frac{1}{\pi} \int_{-\infty}^\infty \frac{1}{x-y} f(y) dy. $$ It satisfies $\ …
André Henriques's user avatar
11 votes
0 answers
526 views

Contraction semigroup on Hilbert space

I'd like to know whether a certain unbounded operator on a Hilbert space is the generator of a strongly continuous contraction semigroup. (Such operators are known as maximally dissipative operators.) …
André Henriques's user avatar
10 votes
3 answers
1k views

ordered exponential of unbounded operators

Let $H$ be a Hilbert space, and let $A_t$ be a family of unbounded positive (self-adjoint) operators on $H$ parametrized by $\mathbb t\in R_{\ge 0}$. Consider the ordinary differential equation $$ \qq …
André Henriques's user avatar
18 votes
0 answers
368 views

Can Rep(G) tell us whether G is discrete?

Given a locally compact group $G$, let $$\mathrm{Rep}(G)$$ be its category of unitary representations. The objects of that category are strongly continuous unitary representations of $G$ on Hilbert sp …
André Henriques's user avatar