# The inverse limit of C$^*$-algebras and whether it commutes with taking the minimal tensor product

Suppose we are given a C$^*$-algebra $A$ and a family of C$^*$-ideals $\mathfrak{I}$ that is upwards directed when ordered by reverse inclusion (i.e. for any $I_1,I_2\in\mathfrak{I}$ there exists a $J\in\mathfrak{I}$ such that $J\subseteq I_1\cap I_2$). Moreover the intersection $I_0=\bigcap_{I\in\mathfrak{I}} I$ is nonempty. To simplify the notation let's write $A_0$ for $A/I_0$ and $A_I$ for $A/I$. Furthermore we will write $p_I:A\twoheadrightarrow A_I$ and $\tilde{p}_I:A_0\twoheadrightarrow A_I$ for the natural quotient maps. Consider the inverse limit $\lim\limits_{\leftarrow} A_I$ of the $A_I$. This inverse limit can be defined by a universal property (see for instance the diagram here: http://mathworld.wolfram.com/InverseLimit.html, which we consider for C$^*$-algebras).

Concretely the inverse limit is given by \begin{equation} \lim\limits_{\leftarrow} A_I = \{ (a_I)_I \in \prod_{I\in\mathfrak{I}} A_I \mid \sup\|a_I\|<\infty, p_{IJ}(a_J)=a_I \text{ if } J\subseteq I\} \end{equation} where $p_{IJ}$ is the natural quotient map $A_J\twoheadrightarrow A_I$ when $J\subseteq I$. My questions are the following:

(1) Is it true that $A_0\cong \lim\limits_{\leftarrow} A_I$? It is easy to see that the map \begin{equation} P:A_0\rightarrow \lim\limits_{\leftarrow} A_I: a\mapsto (\tilde{p}_I(a))_I \end{equation} is an injection. It seems logical that $A_0\cong \lim\limits_{\leftarrow} A_I$, but I have not been able to show that the above map is surjective as well.

(2) Given any other C$^*$-algebra $B$, is it true that $B\otimes_{min}\lim\limits_{\leftarrow} A_I \cong \lim\limits_{\leftarrow} (B\otimes_{min} A_I)$? Maybe this is even true in general, but if not, it maybe is in this specific situation. Together with (1) this would imply that $B\otimes_{min} A_0 \cong \lim\limits_{\leftarrow} (B\otimes_{min} A_I)$. Again we have a canonical map \begin{equation} \Psi:B\otimes_{min} A_0\rightarrow \lim\limits_{\leftarrow} (B\otimes_{min} A_I): x\mapsto ((id\otimes\tilde{p}_I)(x))_I \end{equation} but I have not been able to prove that it is an isomorphism (assumed it really should be, which I am not sure of either).

## 1 Answer

Concerning 1, yes. This is an example where abstract nonsense is the most useful: An inverse limit is a categorical limit in the category of $C^*$-algebras, and those are unique up to unique isomorphism. If you check that $A_0$ fulfills the universal property, that means automatically that $A_0 \cong \lim{A_I}$.

A proof goes like that: Assume there's a $C^*$-algebra $C$ and maps $f_I : C \rightarrow A_I$ such that $f_J = p_{IJ} \circ f_I$. You can define $f : C \rightarrow A_0$ by $f(x) := y$ for any $y$ such that $\tilde{p}_I(y) = f_I(x)$. This is well-defined (since $\tilde{p}_I$ is surjective such a $y$ must exist and the directedness of the family of ideals guaranties that the choice of $y$ doesn't matter). It is also the only possible option under the conditions $\tilde{p}_I \circ f = f_I$, hence unique.

Concerning 2, this holds by the adjoint functor theorem iff the tensor product functor $B \otimes_{min} -$ has a left adjoint, i.e., there's a functor such that $\begin{equation} \text{hom}(C, B \otimes_{min} A) \cong \text{hom}(F(C), A). \end{equation}$ I do not have much experience with tensorproducts in $C^*$-algebras, but maybe you are aware of such a property. It is usually easier to find the left adjoint $F$ (which is unique up to natural isomorphism, if it exists) than to try to prove the limit preservation property directly.

• Concerning 1: I don't see why the existence of such a $y$ is assured. We need that $\tilde{p}_I(y) = f_I(x)$ for all $I$ and the surjectivity of the individual $\tilde{p}_I$'s seems not enough to conclude that. On the other hand, if we fix an $I$, we can find $y$, but then it seems not necessary that $\tilde{p}_J(y) = f_J(x)$ for other $J$. Concerning 2: I'm quite new to the subject, so I'm not aware of such a result, but I'll have a look at it. – This Is Me Jul 28 '15 at 10:37
• We have $f_J(x) = p_{IJ} \circ f_I (x) = p_{IJ} \circ \tilde{p}_I (y) = \tilde{p}_J (y)$, since $p_{IJ} \circ \tilde{p}_I = \tilde{p}_J$ for $J \subset I$. – Georg Lehner Jul 28 '15 at 10:42
• But $p_{IJ}$ is a map from $A_J$ to $A_I$ if $J\subset I$, so in this case $p_{IJ}\circ\tilde{p}_J = \tilde{p}_I$ and we can only conclude that $p_{IJ}(\tilde{p}_J(y))=p_{IJ}(f_J(x))$. – This Is Me Jul 28 '15 at 12:23
• The condition on the $f_I$ was that $f_J = p_{IJ} \circ f_I$. I should have been more clear, my apologies: The intuition is that for any candidate for a limit, $A_0$ is the best possible one, that means more precisely, for any other candidate $C$ there is a map $f : C \rightarrow A_0$ "commuting with everything in sight". Now, what does candidate for a limit mean? It's an object $C$ together with maps $f_I$ that "commute with every thing in sight". That's the condition I wrote here. – Georg Lehner Jul 28 '15 at 14:29