# First isomorphism theorem for maps between Hilbert modules?

Let

• $X$ be a compact Hausdorff topological space,
• $H,K$ be Hilbert modules over the $C^*$-algebra $C(X)$,
• $T:H\rightarrow K$ be a bounded $C(X)$-linear map such that ran($T$) is a Hilbert module over $C(X)$.

Then is there an analogue of the First Isomorphism Theorem saying that ran($T$) is isomorphic as a Hilbert module to the Hilbert module $H/\ker T$ over $C(X)$?

• Why should the range of $T$ be closed? And if you take the closure of $T(H)$ as ran$(T)$, the induced map is clearly not onto. Feb 28, 2017 at 14:38
• Yes, sorry. I meant with possibly extra assumptions. I will add this and repost. Feb 28, 2017 at 14:42
• By Hilbert module you mean representations on Hilbert spaces or actual Hilbert modules (with a scalar product taking values in $C(X)$ ?) because quotienting Hilbert modules is not possible in general and "isomorphisms" is too vague (do you mean isometry ? or continuous isomorphism of $C(X)$-modules ? do you want the isomorphism to be induced by $T$ ?). What you are asking is clearly false (even for Hilbert spaces) for isometry induced by $T$ as there is bounded isomorphism that are not isometry. Feb 28, 2017 at 15:00
• then $H/ker T$ can fail to be a Hilbert module, but you will have an isomorphisms at the level of Banach $C(X)$-modules. Feb 28, 2017 at 15:05
• I see. Thanks. Is there a reference for these things? I am not a functional analyst, and so would appreciate any help! Feb 28, 2017 at 15:06

What happens when $$X$$ is a single point? Then we obtain Hilbert spaces $$H,K$$ and a bounded linear map $$T:H\rightarrow K$$ which has closed range. Can we turn $$H/\ker T$$ into a Hilbert space? To do this, we need an inner-product on $$H/ker T$$; the naive definition would be $$(x+\ker T|y+\ker T) = (x|y)$$ but of course this is not remotely well-defined. In an attempt to make it well-defined, we probably need to have a "distinguished" way to represent an equivalence class $$x+\ker T$$. Using the "projection theorem" any $$x\in H$$ can be written as $$x_0 + x_1$$ where $$x_0\in\ker T$$, $$x_1\in (\ker T)^\perp$$; of course $$(x_0|x_1)=0$$. Then represent $$x+\ker$$ by $$x_1$$, and define $$(x+\ker T|y+\ker T) = (x_1|y_1).$$ This then works. Indeed, all we have done is to write $$H$$ as the orthogonal direct sum $$\ker T \oplus (\ker T)^\perp$$ and then to identify $$H/\ker T$$ with $$(\ker T)^\perp$$. We hence convert the consideration of quotients to the consideration of subspaces, and subspaces of a Hilbert space are themselves Hilbert spaces.
With $$X$$ general, it is usual to consider only adjointable maps $$T:H\rightarrow K$$, that is, as assume the existence of $$T^*:K\rightarrow H$$. This is not automatic: let $$H=C(X)$$ and let $$Y\subseteq X$$ be closed non-empty with dense complement and let $$K=\{f\in C(X) : f(y)=0 \ (y\in Y) \}$$. If $$T:K\rightarrow H$$ is the inclusion, then you can check that if $$T^*$$ existed then $$T^*(1)=1\not\in K$$ a contradiction.
The problem we'll run into is that for a Hilbert C$$^*$$-module we do not have orthogonal decompositions. For example, with the example above, $$K^\perp=\{0\}$$ in $$H$$ but of course $$K\not=H$$.
A theorem of Miščenko says that if $$T:H\rightarrow K$$ is adjointable with closed range, then $$\ker T$$ is complemented: we have infact that $$\ker T \oplus \operatorname{im}(T^*) = H$$. Thus we can identify $$H/\ker T$$ with $$\operatorname{im}(T^*)$$ and proceed as in the Hilbert space case.
Here I have been following Lance's loverly little book "Hilbert $$C^*$$-modules: A toolkit for operator algebraists".