# sub-algebra of inductive limit(II)

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?

• The inverse image of a subalgebra under a homomorphism is always a subalgebra. – Ruy Feb 22 at 20:47
• Yes, I am asking if we do not assume $\beta_{i}$ are surjective, does all subalgebra in $A$ lies in some image of $\beta_{i}$ – Ken.Wong Feb 22 at 22:12
• 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 $\beta_j$. – Ruy Feb 23 at 0:14
• 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. – Alex Kruckman Feb 23 at 2:53
• Maybe the confusion is this: the notation $f^{-1}(X)$ makes sense even when $X$ is not contained in the image of $f$. – 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.

Then

• 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 $$\beta _j$$.

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