18.727, Topics in Algebraic Geometry (rigid analytic geometry) Kiran S. Kedlaya, fall 2004 A little p-adic functional analysis (part 2 of 2)

Thanks to Abhinav Kumar for providing the corrections incorporated into this version (of 16 Sep 04).

Orthonormal bases

An orthogonal basis of a Banach space V is a subset {e_{i}}_{i∈I} of V with the property that each m ∈ M has a unique representation as a convergent sum ∑_{i∈I} c_{i}e_{i}, and if one always has ∥m∥ = max_{i∈I}{∥c_{i}e_{i}∥}. (Note that “convergent” only makes sense if I is at most countable; but note also that there is no distinction between “convergent” and “absolutely convergent” in the nonarchimedean setting!) The basis is orthonormal if ∥e_{i}∥ = 1 for each i. Lemma 1 from last time says that the norm on a Banach space of countable type can be approximated by equivalent norms which admit orthogonal bases; see below for an example.

Banach algebras

A Banach algebra (over K) is a Banach space A over K which is also a commutative Kalgebra, and which satsifies the following additional restrictions.

(a) ∥1∥ = 1.

(b) for x, y ∈ A, ∥xy∥ ≤ ∥x∥ · ∥y∥.

A Banach module over A is a Banach space M equipped with an A-module structure, such that ∥am∥ ≤ ∥a∥ · ∥m∥ for a ∈ A and m ∈ M.

Here’s an example where life turns out to be easier than in the real/complex case [FvdP, Lemma 1.2.3]. I’ll start next time with an example of this.

Lemma 1. Let A be a Banach algebra over K which is noetherian as a ring. Let M be a Banach module over A which is module-finite over A. Then any A-submodule of M is closed.

Proof. Let N be a submodule of M and let Ñ be the closure of N. Since A is noetherian, Ñ is module-finite over A; choose generators e_{1}, …, e_{n} of Ñ over A. Consider the A-module homomorphism A^{n} → Ñ defined by (a_{1}, …, a_{n}) ↦ ∑ a_{i}e_{i}, where A^{n} is equipped with the norm ∥(a_{1}, …, a_{n})∥ = max_{i}{∥a_{i}∥}. By the open mapping theorem, there exists c ∈ (0, 1) such that each x ∈ Ñ can be written as ∑ a_{i}e_{i} with c max_{i}{∥a_{i}∥} ≤ ∥x∥. Choose f_{1}, …, f_{n} ∈ N with ∥e_{i} − f_{i}∥ ≤ c^{2}; we show that f_{1}, …, f_{n} also generate Ñ, which implies that N = Ñ.

Given x ∈ Ñ, define the sequence x_{0}, x_{1}, … as follows. Set x_{0} = x; given x_{j}, write x_{j} = ∑_{i} a_{j,i}e_{i} with c max_{i}{∥a_{j,i}∥} ≤ ∥x_{j}∥, and put

x_{j+1} = ∑ a_{j,i}(e_{i} − f_{i}),

so that ∥x_{j+1}∥ ≤ c∥x_{j}∥. This means x_{j} → 0 and so for each i, the series ∑_{j} a_{j,i} converges to a limit a_{i} satisfying x = ∑ a_{i}f_{i}. Thus N = Ñ, as desired.

Tensor products

The tensor product of two Banach spaces V and W is not complete, so instead we will typically work with the completed tensor product V ˆ⊗W, which as the name suggests is just the completion of the ordinary tensor product as K-vector spaces.

The completed tensor product is also a Banach space, but this takes a bit of work to check. One gets a seminorm on V ⊗ W from the formula

∥x∥ = inf{max_{i}{∥v_{i}∥ · ∥w_{i}∥}},

the infimum taken over all presentations x = ∑_{i=1}^{m} v_{i}⊗w_{i}, and this extends to the completion. To check that it’s a norm, one needs to check that if {x_{j}}_{j=1}^{∞} is a sequence of elements of V ⊗ W, with x_{j} = ∑_{j=1}^{m_{j}} v_{ij} ⊗ w_{ij}, and max_{i}{∥v_{ij}∥ · ∥w_{ij}∥} → 0 as j → ∞, then x_{j} → 0 in V ⊗ W. We may check this after replacing V and W by the span of the v_{ij} and w_{ij}, respectively, dropping us into the countable type case. Then writing everything in terms of an orthogonal basis of an equivalent norm (using Lemma 1 from last time) yields the claim.

The completed tensor product is exact in the following sense. We say a sequence

0 → M_{1} →^{f} M_{2} →^{g} M_{3} → 0

of maps between Banach spaces is exact if it is exact in the usual sense, and also f and g are isometric. (For f, this means the norm on M_{1} is the restriction from M_{2}; for g, the norm on M_{3} is the quotient norm from M_{2}.) Then for any Banach space N, the sequence

0 → M_{1} ˆ⊗N →^{f} M_{2} ˆ⊗N →^{g} M_{3} ˆ⊗N → 0

is exact.

Warning: if A is a Banach algebra, the product seminorm on a tensor product M ⊗_{A} N of Banach modules M and N over A may not be a norm! But see exercises for an important case where this is okay.

Exercises

1. (Leftover from last time) Prove that ∪_{n=1}^{∞}k((t^{1/n})) is algebraically closed for k = ℂ, but not for any field k of positive characteristic. (Hint: look at P (x) = x^{p} − x − t^{−1}.)

2. Suppose K is a complete discretely valued (ultrametric) field. Let M be a Banach space over K such that ∥m∥ ∈ |K| for each m ∈ M. Put 𝔬_{M} = {m ∈ M : ∥m∥ ≤ 1} and \overline{M} = 𝔬_{M} ⊗_{𝔬_{K}} k. Prove that a subset of M forms an orthonormal basis if and only if its image in \overline{M} is a basis of \overline{M} as a k-vector space. (Hint: see [FvdP, Lemma 1.2.2].)

3. Let A be a Banach algebra which is noetherian as a ring, and let M and N be Banach modules over A which are module-finite over A. Show that the product seminorm

∥x∥ = inf{max_{i}{∥m_{i}∥ · ∥n_{i}∥ : x = ∑ m_{i} ⊗ n_{i}}

is a norm on M ⊗ N, and that M ⊗ N is complete for this norm. (Hint: choose finite presentations of M and N and reduce to the case of free modules.)