Direct limit with constant homomorphisms?

Question

I have the following direct system of finite-dimensional C*-algebras:

$$\mathbb{C} \to \mathbb{C} \oplus \mathbb{C} \to \mathbb{C} \oplus \mathbb{C} \oplus \mathbb{C} \to \mathbb{C} \oplus \mathbb{C} \oplus \mathbb{C} \to \mathbb{C} \oplus \mathbb{C} \oplus \mathbb{C} \to \cdots$$

where the first two maps are $\begin{pmatrix}1 \\ 1\end{pmatrix}$ and $\begin{pmatrix}1&0 \\ 1&1 \\ 0&1\end{pmatrix}$, and all the other maps are constantly $\begin{pmatrix}1&1&0 \\ 1&0&1 \\ 0&1&1\end{pmatrix}$.

First of all, am I right in thinking that the first two maps have no influence on the direct limit? And secondly, is there an easy way to describe the resulting algebra?

Answer 1

You are right: When taking a direct limit like $$A_1\overset{T_1}{\longrightarrow}A_2\overset{T_2}{\longrightarrow}\cdots,$$ the first elements of your direct sequence do not matter. In fact, you can take any subsequence $(n_k)$ (and compose the maps appropriatelly to get a directed sequence $A_{n_1}\to A_{n_2}\to\cdots$) and you still get the same direct limit. You can show this using the universal property.

As for what is the direct limit in your case, we have to make a guess somehow. Essentially, direct limits work in the following way: When the $T_1$ are injective, we may identify $A_i$ with its image $T_i(A_i)$ inside $A_{i+1}$, so we have a sequence $A_1\subseteq A_2\subseteq\cdots$. "Taking a limit" in this case would mean taking the union of all $A_i$, which is an algebra (and put an appropriate norm and close to get a C*-algebra).

Now suppose we have the same algebra over and over again: $$A\overset{T_1}{\longrightarrow}A\overset{T_2}{\longrightarrow}\cdots,$$\ and each $T_i$ is an isomorphism. Then identifying $A$ with its image under $T_i$ yields $A=A=A=\cdots$, so the union is again $A$.

Now to make things formal: Suppose $A$ is a C*-algebra, $A_i=A$ and and $T_i:A_i\to A_{i+1}$ are isomorphisms as above (it's useful to leave the indices). Let's show that $\lim(A_i,T_i)=A$. For this we need to find morphisms $\psi_i:A_i\to A$ which satisfy the universal property. Set $\psi_1=\operatorname{id}$ (this is the identification of $A_1$ with $A_1$), and for $i\geq 2$, set $\psi_i=T_1^{-1}\cdots T_{i-1}^{-1}$ (this is how we identify $A_1$ inside $A_i$). Then I'll leave it to you to show that these $\psi_i$ satisfy the universal property.

In particular, the direct limit in your example is $\mathbb{C}^3$.

Answer 2

It's true that any finite number of maps at the beginning don't affect the direct limit, so you may ignore the first two.

Then the direct limit is the stabilization of the image of your $3\times 3$-matrix. Explicitly, if you call $A$ this matrix, then the image of $A^n$ stabilizes for $n$ big enough, and the limit is this image.

Now just looking at its discriminant shows that $A$ is invertible, so the limit is just $\mathbb{C}\oplus \mathbb{C}\oplus \mathbb{C}$ (with whatever algebra structure it's supposed to have).