Left Ideals in Finite Dimensional $\mathrm{C}^\ast$-Algebras

Question

Let $A$ be a finite dimensional $\mathrm{C}^\ast$-algebra.

Let $I$ be a left ideal in $A$. I believe there is a projection $p\in A$ (an element such that $p=p^2=p^*$) such that:

$$I=Ap.\qquad(1)$$

Is there an easy-to-write down proof, or easy reference for this?

I have:

As a (norm-)closed left ideal of a finite dimensional $\mathrm{C}^\ast$-algebra, $I$ is a weakly closed left ideal of a von Neumann algebra, and thus $I$ must be of the form $F(G)p$ for a projection $p$.

This is obvious overkill (or am I actually mistaken about (1)?)

How do you define projection $p \in A$? – J. De Ro Mar 23 '20 at 17:35 — J. De Ro, Mar 23 '20 at 17:35
An element such that $p=p^2=p^*$. – JP McCarthy Mar 23 '20 at 17:41 — JP McCarthy, Mar 23 '20 at 17:41

Martin Argerami · Accepted Answer · 2020-03-24T20:47:05.653

1

Here some steps to get $p$.

$I$ contains a nonzero projection. Indeed, if $I\ne\{0\}$, take $a\in I$ nonzero. Then $a^*a\in I$. Then $f(a^*a)\in I$ for all $f\in\mathbb C[x]$. Because $A$ is finite-dimensional, any spectral projection of $A$ can be obtained as $f(a^*a)$.
$I$ contains a nonzero minimal projection. Indeed, given a projection $q\in I$, let $q_0\leq q$ be a minimal projection in $A$ (which exists because $A$ is finite-dimensional). Then $q_0=q_0 q\in I$.
There exist minimal projections $q_1,\ldots,q_m\in I$ such that $I=A(q_1+\cdots+q_m)$. Indeed, let $q_1\in I$ be a nonzero minimal projection. Consider the left ideal $I(1-q_1)$. Note that $I(1-q_1)\subset I$: if $b\in I$ then $b-bq_1\in I$. The steps above apply to $I(1-q_1)$, if nonzero, so there exists a minimal projection $q_2\in I(1-q_1)\subset I$. Then $q_2(1-q_1)=q_2$, which gives $q_2q_1=0$. We can repeat this process with $q_1+q_2$, etc., until at some point $I(1-q_1-\cdots-q_m)=\{0\}$.
Take $p=q_1+\cdots+q_m\in I$. We have $I=Ip$. Then $Ip\subset Ap\subset I=Ip$.

edited Mar 24 '20 at 20:47

answered Mar 24 '20 at 15:54

Martin Argerami

205,756

Are there well-known counterexamples for infinite dimensional algebras? – user760 Jun 21 '23 at 11:14
Yes, $K(H)$ inside $B(H)$. – Martin Argerami Jun 21 '23 at 17:50
Thanks. What about counterexamples of only one-sided (but not two-sided) ideals? Should I open a new question for this? – user760 Jun 22 '23 at 04:12
1

Take an invertible operator $T$ and let $J=K(H)T$. – Martin Argerami Jun 22 '23 at 15:48

score 1 · Answer 2 · answered Mar 24 '20 at 16:09

Since $A$ is finite-dimensional, it is of the form $A = \oplus_{i=1}^r M_{k_i}$. Below its shown (after referring to other answers) that every left ideal of $M_n$ is of the form $M_np$ for some projection $p$. The general result follows from the direct sum decomposition of $A$: one has $I = \oplus I_{i=1}^r$, and $I_i = M_{k_i}p_i$, so that $I = Ap$, where $p = \oplus_{i=1}^r p_i$.

One can see an answer like (Any left ideal of $M_n(\mathbb{F})$ is principal) or (What are the left and right ideals of matrix ring? How about the two sided ideals?) to figure out what left ideals of $M_n$ looks like. They are just left annihilators of column spaces.

Let $I$ be a left ideal in $M_n$. So $I = \{T \in M_n \mid TB = 0\}$ for some matrix $B$. Let $p$ be the projection onto the range of $B$, and let $J = M_n(1-p)$. Then $T \in J$ implies that $T = S(1-p)$, so that $S(1-p)B = 0$ as $(1-p)$ is orthogonal to the range of $B$; this gives that $T \in I$. Moreover if $T \in I$, we have that $TB = 0$. Notice that $T = T(1-p) + Tp$, and that $Tp = 0$ since $p$ is the projection onto the range of $B$, so that $T = T(1-p) \in J$.

This gives that $I = J = M_nq$ for the projection $q = 1-p$.

Left Ideals in Finite Dimensional $\mathrm{C}^\ast$-Algebras

2 Answers2