# I have an embedding $\iota$ between two Hilbert spaces and want to know if $\iota\iota^\ast$ is something simple like an orthogonal projection

I'm reading A Concise Course on Stochastic Partial Differential Equations. In Proposition 2.5.2 the authors define the notion of a cylindrical $Q$-Wiener process $W$. It turns out that $W$ is just a $Q_1$-Wiener process on a "larger" Hilbert space. I've got the feeling that $Q_1$ is actually a really "simple" operator (like the identity or some kind of projection), especially in the popular case where $Q$ is the identity, but I got problems to figure that out.

Let me introduce the necessary objects: Let$^1$

• $U$ and $V$ be real Hilbert spaces
• $Q\in\mathfrak L(U)$ be nonnegative and symmetric
• $U_0:=Q^{1/2}U$ be equipped with $$\langle u,v\rangle_{U_0}:=\langle Q^{-1/2}u,Q^{-1/2}v\rangle_U\;\;\;\text{for }u,v\in U_0$$
• $\iota\in\operatorname{HS}(U_0,V)$ be an embedding
• $C:=\iota\iota^\ast$
• $V_0:=C^{1/2}V$ be equipped with $$\langle u,v\rangle_{V_0}:=\langle C^{-1/2}u,C^{-1/2}v\rangle_V\;\;\;\text{for }u,v\in V_0$$

In Proposition 2.5.2 the authors show that $V_0=\iota U_0$ and that $\iota$ is an isometry between $U_0$ and $V_0$.

If we treat $\iota$ as being an element of $\mathfrak L(U_0,V_0)$, it's easy to see that $\iota$ is unitary and hence $$C=\text{id}_{V_0}\;.$$ However, in the context of the book, I need to treat $\iota$ as an element of $\mathfrak L(U_0,V)$ (in Proposition 2.5.2 the authors show that $\iota$ is nonnegative and symmetric with finite trace). I think that even in that case $C$ is something really "simple", but I don't know what I need to do to reveal that.

$^1$ Let $\mathfrak L(A,B)$ and $\operatorname{HS}(A,B)$ denote the space of bounded, linear operators and Hilbert-Schmidt operators from $A$ to $B$, respectively, and $\mathfrak L(A):=\mathfrak L(A,A)$.

• I think you mean "the authors show that $\iota\iota^*$ [i.e., $C$] is nonnegative and symmetric with finite trace". If all you know about $\iota$ is that it is Hilbert-Schmidt and has no kernel, there is nothing else you can say about $\iota\iota^*$. – Nik Weaver Jun 24 '16 at 0:21