# Hermitian vector bundles and Hilbert $C^*$-modules

Let $$X$$ be a compact Hausdorff space and $$C(X)$$ its algebra of continuous complex valued functions. The Gelfand-Naimark theorem tells us that we have a duality between commutative $$C^*$$-algebras and compact Hausdorff spaces: Send $$X$$ to its $$C^*$$-algebra of continuous functions.

The Swan-Serre theorem tells that there is a duality between finitely generated projective $$C(X)$$-modules and complex continuous vector bundles $$V$$ over $$X$$: Send a vector bundle $$V$$ to its space of sections $$\Gamma$$.

If we go further and put an Hermitian metric $$g$$ on $$V$$ then the space of sections $$\Gamma$$ is a Hilbert $$C(X)$$-module. In the other direction things are less clear to me. If we have a Hilbert module structure on $$\Gamma(V)$$, does it necessarily come from an Hermitian metric and if so does this give a duality between Hilbert $$C(X)$$-modules and Hermitian vector bundles over $$C(X)$$?

In addition to Nik Weaver's references, let me just sketch the proof which is in fact not very difficult:

1. A construction of Kaplansky (Rings of operators, Thm 26) shows that if $$\mathcal{A}$$ is a $$*$$-algebra with the property that for every $$A \in M_n(\mathcal{A})$$ the matrix $$1 + A^*A$$ is invertible, then every idempotent in $$M_n(\mathcal{A})$$ is equivalent to a projection. A $$C^*$$-algebra and in particular $$C(M)$$ clearly has this feature by spectral calculus for every $$n \ge 1$$.

Suppose now that $$\mathcal{A}$$ is such a $$^*$$-algebra satisfying this condition for all $$n$$ (†).

1. On every finitely generated projective module $$\mathcal{E}$$ over $$\mathcal{A}$$, written without restriction as $$\mathcal{E} = P \mathcal{A}^n$$ with $$P = P^2 = P^* \in M_n(\mathcal{A})$$, the restriction $$\langle . , .\rangle$$ of the canonical Hermitian algebra-value inner product of the free module $$\mathcal{A}^n$$ is still positive (in the strongest possible sense) and non-degenerate (in the strongest possible sense that the musical homomorphism is in fact an anti-isomorphism to the dual module). This gives immediately the existence of positive algebra-valued inner products on finitely generated projective modules over such algebras.

2. Now suppose the algebra satisfies an additional feature: suppose $$H \in M_n(\mathcal{A})$$ is an invertible positive element. Then $$H$$ has a positive square root $$H = \sqrt{H}^2$$ with the property that $$\sqrt{H}$$ commutes with all matrices which commute with $$H$$ (‡). Again, a $$C^*$$-algebra clearly satisfies this for all $$n$$ by spectral calculus.

Now suppose that on the fgpm $$\mathcal{E} = P \mathcal{A}^n$$ you have another positive algebra-valued inner product, denoted by $$h$$. On the complement $$P^\bot = (1 - P)\mathcal{A}^n$$ we use the restriction of the canonical algebra-valued inner product to obtain a new inner product, still denoted by $$h$$, on the direct sum. This is still positive and has all required (very strong) non-degeneracy properties.

1. By matrix calculus with free modules one obtains then a matrix $$H \in M_n(\mathcal{A})$$ with $$$$h(x, y) = \langle x, Hy\rangle$$$$ for all $$x, y \in \mathcal{A}^n$$. Since by assumption $$H = \sqrt{H}^2$$ the square root $$U = \sqrt{H}$$ provides an isometry between the two inner products on the free module. Since by assumption the square root commutes with everything commuting with $$H$$ and since by construction $$H$$ commutes with $$P$$ (check this!) $$U$$ restrict to an isometry between $$h$$ and $$\langle ., .\rangle$$ on the original $$\mathcal{E}$$.

In conclusion: for $$^*$$-algebras (like e.g. $$C^*$$-algebras) satisfying the above two properties (†) and (‡), every finitely generated projective module carries a unique-up-to-isometry positive algebra-valued inner product.

As a side remark: many other types of $$^*$$-algebras satisfy these properties as well like e.g. the smooth functions on a manifold etc. So the same proof also implies (via Serre-Swan in the smooth context) that on smooth vector bundles you always have a unique-up-to-isometry positive fiber metric.

Now, to answer you question: the uniqueness gives you the desired result that every algebra-valued inner product comes from a positive fiber metric. Indeed, a positive fiber metric meets the above properties and the isometry $$U$$ maps fiber metrics to fiber metrics as it is algebra-linear.

• I put little signposts on the two conditions you require for the result, they didn't stand out among the various nesting of itemization. Please check I got them right. Commented Oct 10, 2021 at 2:54
• @DavidRoberts Thanks! looks fine with me ;) Commented Oct 10, 2021 at 9:04

Yes, every finitely generated Hilbert module comes from a hermitian complex vector bundle in this way. In fact more is true: arbitrary Hilbert modules over $$C(X)$$ correspond to continuous (in an appropriate sense) bundles of Hilbert spaces over $$X$$. Unfortunately, the only place I've seen this proven is in a dissertation (Takehashi, Fields of Hilbert Modules, 1971). The finitely generated case might be in Rieffel's paper "Morita equivalence for operator algebras", but I don't have access to it right now. Hopefully someone else can give a better reference.

• Rieffel’s paper gives a proof of the other direction in the vector bundle case: that the global sections of a vector bundle over $X$ yield a finitely generated projective Hilbert module over $C(X)$. Commented Oct 9, 2021 at 14:21