# Partition of the identity in a simple $C^*$-Algebra

The following question is from $$C^*$$-Algebras by Examples written by Kenneth R. Davidson, Problem V.9. Given a unital simple $$C^*$$-Algebra $$\mathfrak{A}$$ and a positive element $$A$$, how to show there exists finitely many $$\{X_i\}_{i \leq n}$$ such that $$\sum_{i \leq n} X_i^* A X_i = I$$? Beside the original question, is it possible that all these $$X_i$$ are unitary?

The given hint is $$X A Y + Y^* A X^* \leq X A X^* + Y^* A Y$$ iff $$(X - Y^*) A (X^* - Y) \geq 0$$ which is true in this case. I have no idea where to come up with a $$\{X_i\}_{i \leq n}$$ such that $$\sum_{i \leq n} X_i^* A X_i$$ and how to show it is the identity. One of my attempt is to define $$\mathcal{A} = \{X Z Y\,\vert\,Z \in \overline{A \mathfrak{A} A}, X, Y \in \mathfrak{A}\}$$. This is an non-zero ideal and hence equal to the whole $$\mathfrak{A}$$.

I wish I can define $$\mathcal{A}_n = \{ \sum_{i \leq n}X A X^*\,\vert\,X \in \mathfrak{A}\}$$ and show that for some $$n$$ it would be a hereditary subalgebra. Then $$0 \leq Y \leq X A X^*, \exists B \in \mathfrak{A}\,\implies\,\sqrt{Y} = B (X A X^*)^{\frac{1}{4}}\,\implies\,Y = B (X A X^*)^{\frac{1}{4}} B^* \in \mathcal{A}_n$$. If this can be done, then according to the provided inequality $$2 \vert\,X A Y\,\vert \leq \sum_{i \leq n} X_i^* A X_i\,\implies X A Y \in \mathcal{A}_n$$ and I can replace $$A$$ by other elements in $$\overline{A \mathfrak{A} A}$$. However, it seems to me that $$\mathcal{A}_n$$ can hardly be an algebra... If I am on the right track where can I find such an algebra that contains finite sums of $$X A X^*$$?

Let $$\mathfrak I=\operatorname{span}\{XAY:X,Y\in\mathfrak A\}$$. Then $$\mathfrak I$$ is a nonzero ideal in $$\mathfrak A$$, hence is dense in $$\mathfrak A$$ as $$\mathfrak A$$ is simple. But then there is some $$\tilde X_k,Y_k\in\mathfrak A$$ such that $$\|I-\sum_{k=1}^n\tilde X_kAY_k\|<1$$, so $$Z=\sum_{k=1}^n \tilde X_kAY_k$$ is invertible in $$A$$, and thus $$I=Z^{-1}Z=\sum_{k=1}^n(Z^{-1}\tilde X_k)AY_k$$. Now write $$X_k=\frac12Z^{-1}\tilde X_k$$, so that $$I=2\sum_{k=1}^nX_kAY_k$$. Thus $$I=\frac{I+I^*}{2}=\sum_{k=1}^nX_kAY_k+Y_k^*AX_k^*\leq\sum_{k=1}^nX_kAX_k^*+Y_k^*AY_k.$$ Now let $$B=\sum_{k=1}^nX_kAX_k^*+Y_k^*AY_k$$, and put $$W_k=B^{-1/2}X_k$$ for $$k=1,\ldots,n$$ and $$W_k=B^{-1/2}Y_k^*$$ for $$k=n+1,\ldots,2n$$. Then we have $$I=B^{-1/2}BB^{-1/2}=\sum_{k=1}^{2n}W_kAW_k^*.$$
As far as the second question, this fails to work for the case $$\mathfrak A=\mathbb C$$, as pointed out in the comments.
• Nice answer! Regarding if its possible to do with unitaries: Look at $A=5\,\Bbb 1$ as an example where its not possible. Commented Jul 14, 2020 at 21:02
• @Aweygan Very brilliant answer! Could you explain how you conclude $B$ is invertible? I tried to show $\|\,I - B\,\| \leq 2 \|\,\frac{I}{2} - \sum_{k = 1}^n X_k A Y_k\,\| < 1$ which will be equivalent to $\|\,I - \sum_{k = 1}^n Z^{-1} \tilde X_k A Y_k\,\| < 1$. But I need to adjust the norm of $\sum_{k = 1}^n \tilde X_k A Y_k$ before calling it $Z$. Commented Jul 14, 2020 at 21:31
• @s.harp Thank you for your answers. If $A$ is just a big enough constant then my guess will be obviously wrong. Could that be true when $A$ is not a constant? If the identity can be partitioned in such way, maybe among those many invertible elements above we can build an arc to connect some of them to some unitaries. Commented Jul 14, 2020 at 21:41
• $B$ is invertible because $B\geq I$. This means $B-I\geq 0$, so $\sigma (B-I)\subset[0,\infty)$, so $\sigma(B)\subset[1,\infty)$, so in particular $0$ is not in the spectrum of $B$ Commented Jul 14, 2020 at 21:41