Lemma 1

Let $\mathcal{A}$ be a Banach algebra with unit $\mathbf{1}$. Suppose $x$ and $y$ are idempotent and

\[\lVert x - y \rVert < \frac{1}{\lVert 2x - \mathbf{1} \rVert}.\]

Then $x$ and $y$ are conjugate, i.e., $x = aya^{-1}$ for some $a \in \mathcal{A}$.

In other words, conjugacy classes partition the set of idempotent elements into open subsets.

Proof

Displayed below is our candidate $a$. $a = \mathbf{1} - x - y + 2xy.$ Observe

\[xa = x - x^{2} - xy + 2x^{2}y = xy\]

and

\[ay = y - xy - y^{2} + 2xy^{2} = xy.\]

If $a$ is invertible, then $x = aya^{-1}.$ It therefore remains only to prove the invertibility of $a$, and we do so by invoking Proposition 1 of the previous post.

I compute:

\[\mathbf{1} - a = x + y - 2xy = (x - y)(2x - \mathbf{1}).\]

This readily implies

\[\lVert a - \mathbf{1} \rVert \leq \lVert x - y \rVert \lVert 2x - \mathbf{1} \rVert < 1.\]

Now apply Proposition 1 as planned. $\tag*{$\blacksquare$}$

Note that $\lVert 2x - \mathbf{1} \rVert$ is nonzero for idempotent $x$, since $x = (2x - \mathbf{1})x$ implies $1 \leq \lVert 2x - \mathbf{1} \rVert.$

I now specialize to the following case. Let $X$ be a compact Hausdorff space. Let $M(n,C(X))$ be the Banach algebra of $n \times n$ matrices with entries in the ring of continuous $\mathbf{C}$-valued functions on $X$ equipped with the supremum norm.

Lemma 2

Elements $E$ of $M(n,C(X))$ determine continuous functions $E\colon X \to M(n,\mathbf{C})$.

Proof

Fix $A \in M(n,\mathbf{C})$ and $\epsilon > 0$. Consider the open ball $U_{\epsilon}(A)$. I claim $E^{-1}(U_{\epsilon}(A))$ is open in $X$. To this end, fix $p \in E^{-1}(U_{\epsilon}(A))$.

\[\lVert E(p) - A \lVert < \epsilon.\]

Define

\[\delta(p) = \frac{\epsilon - \lVert E(p) - A \lVert}{n^{2}}.\]

I claim that

\[V(p) = \bigcap_{1 \leq i,j \leq n} E_{ij}^{-1}(U_{\delta(p)}(E_{ij}(p)))\]

is an open neighborhood of $p$ contained within $E^{-1}(U_{\epsilon}(A))$.

Certainly $p$ lies in $V(p)$. Suppose $q \in X$ too lies in $V(p)$. For any vector $v \in \mathbf{C}^{n}$ of unit length,

\[\left| (E(q) - E(p))v \right| \leq n \max_{1 \leq i \leq n} \left| \sum_{j=1}^{n} (E_{ij}(q) - E_{ij}(p))v_{j} \right|\] \[\leq n \max_{1 \leq i \leq n} \sum_{j=1}^{n} \left| E_{ij}(q) - E_{ij}(p) \right| |v_{j}|\] \[< n \max_{1 \leq i \leq n} \sum_{j=1}^{n} \delta(p) |v_{j}|\] \[\leq \epsilon - \lVert E(p) - A \rVert.\]

This computation reveals

\[\lVert E(q) - A \rVert \leq \lVert E(q) - E(p) \rVert + \lVert E(p) - A \rVert < \epsilon.\]

That is, $q \in E^{-1}(U_{\epsilon}(A))$. $\tag*{$\blacksquare$}$

Proposition 1

Let $E\colon X \to \operatorname{M}(n,\mathbf{C})$ be an idempotent element of $\operatorname{M}(n,C(X))$. Then $\operatorname{Ran} E$ is a subbundle of $\Theta^{n}(X)$.

Proof

Fix $x \in X$. Define

\[V = \left\{ F \in \operatorname{M}(n,\mathbf{C}) : \lVert E(x) - F \rVert < \frac{1}{\lVert 2E(x) - I_{n} \rVert} \right\}.\]

Define $U = E^{-1}(V)$. By Lemma 2, $U$ is an open subset of $X$. We prove $U$ is a trivial neighborhood of $x$.

Define $S\colon U \to \operatorname{GL}(n,\mathbf{C})$ by the formula below.

\[S(y) = I_{n} - E(x) - E(y) + E(x)E(y).\]

By Lemma 1, $E(y) = S(y)^{-1}E(x)S(y).$ Hence, the function $S(y)\colon (\operatorname{Ran} E)_{y} \to (\operatorname{Ran} E)_{x}$ is an isomorphism. $\tag*{$\blacksquare$}$

The obvious generalization to endomorphisms of vector bundles does not hold. Consider for instance the trivial bundle over the unit interval together with the endomorphism

\[(t,\lambda) \mapsto (t,t\lambda),\]

where $(t,\lambda) \in [0,1] \times \mathbf{R}$. The stalk at $0$ is zero-dimensional.