This question is similar to my question here, but not the same question.

Let $(A_{i},\alpha_{i})$ be directed system of C* algebras and *-homomorphisms. Let the $\beta_{i}:A_{i}\rightarrow A$ are the canonical *-homomorphisms. Consider some subalgebra $B$ belongs to direct limit $A$. Does it always exists a $j\in I$ such that $B\subseteq \beta_{j}(A_{j})$?

My guess is that is true. Since $A$ is quotient algebra of disjoint union of algebras, we can write every elements $x\in A$ in the form $x=(x_{1},x_{2},...)+N$, with $x_{i}\in A_{i}$, $N$ is the ideal form by equivalence relation. If $B$ is a sub-algebra, then every component of $B$ form a sub-algebra. Is my thought correct?

  • $\begingroup$ The inverse image of a subalgebra under a homomorphism is always a subalgebra. $\endgroup$ – Ruy Feb 22 at 20:47
  • $\begingroup$ Yes, I am asking if we do not assume $\beta_{i}$ are surjective, does all subalgebra in $A$ lies in some image of $\beta_{i}$ $\endgroup$ – Ken.Wong Feb 22 at 22:12
  • 1
    $\begingroup$ Well, this is quite different from asking if $\beta^{-1}_j(B)$ is a subalgebra. So I guess you should have asked whether $B\subseteq \beta_j(A_j)$, for some $j$. If this is what you want to know then the answer is no: just take $B$ to be the algebra generated by some element not in the union of the ranges of the $\beta_j$. $\endgroup$ – Ruy Feb 23 at 0:14
  • 1
    $\begingroup$ The question you have written is actually weaker than your preveious question. Of course there is a $j$ such that the preimage of $B$ under $\beta_j$ is a subalgebra - this is true for every $j$! If you meant to ask a different question, you should edit to clarify. $\endgroup$ – Alex Kruckman Feb 23 at 2:53
  • 1
    $\begingroup$ Maybe the confusion is this: the notation $f^{-1}(X)$ makes sense even when $X$ is not contained in the image of $f$. $\endgroup$ – Alex Kruckman Feb 23 at 2:55

The inductive limit of C$^*$-algebras is often not the union of the ranges of the $\beta _j$.

For instance, consider the C$^*$-algebra $K$, formed by all compact operators on $\ell ^2$. Also, for each $n$, consider the subset $K_n\subseteq K$, formed by all operators whose matrix $(a_{i, j})_{i, j}$, relative to the canonical basis of $\ell ^2$, have nonzero entries only in the top left $n\times n$ block.


  • each $K_n$ is a closed $^*$-subalgebra of $K$ (isomorphic to $M_n(\mathbb C)$),

  • $K_n\subseteq K_{n+1}$, and

  • the union $\bigcup_nK_n$ is dense in $K$.

With this much information you are able to deduce that $K$ is the inductive limit of the $K_n$, with the connecting maps being the inclusions $K_n\hookrightarrow K_{n+1}$.

Observing that every operator in $\bigcup_nK_n$ has finite rank, we see that $\bigcup_nK_n$ is not equal to $K$. It is only a dense $^*$-subalgebra (and hence can't be the inductive limit in the category of C$^*$-algebras since we want the inductive limit to be a C$^*$-algebra).

Any infinite rank compact operator will therefore generate a subalgebra that is not contained in the union of the ranges of the $\beta _j$.

  • $\begingroup$ Thank you for the great answer. I would like to ask if we require $B$ to be dense also, will the answer changes? $\endgroup$ – Ken.Wong Feb 23 at 13:04
  • 1
    $\begingroup$ No, the answer will still be negative. In fact the worst case is when $B$ is the whole inductive limit! $\endgroup$ – Ruy Feb 23 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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