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Spectral convolution eigenspaces are finite-dimensional and invariant

Statement

Assume the Axiom of Choice. Let $G$ be a compact Lie group and $k \in C (G)$ .

The Hilbert adjoint of $T_{k}$ is $T_{k^{*}}$ with $k^{*} (x) = \overline{k (x^{- 1})}$ , and $T_{k}$ commutes with every left translation $L_{g}$ .
If $k^{*} = k$ , then $T_{k}$ is compact self-adjoint, its nonzero eigenspaces are finite-dimensional and left-invariant, and the closed span of the eigenspaces is the closure of the range of $T_{k}$ .
For arbitrary $k$ the operator $T_{k}^{*} T_{k}$ is compact, positive and self-adjoint; its nonzero eigenspaces are finite-dimensional and left-invariant, and their closed span is $\overline{ran (T_{k}^{*} T_{k})} = (\ker (T_{k}^{*} T_{k}))^{⊥}$ .

Facts & Assumptions

Given: Assume the Axiom of Choice, a compact Lie group $G$ with normalized Haar measure, and $k \in C (G)$ .

[A1]

The Axiom of Choice is The Axiom of Choice; it enters through the compactness and spectral theory cited.

[L1]

$T_{k}$ is a compact operator on $L^{2} (G)$ with kernel $K (x, y) = k (x^{- 1} y)$ ; the left and right regular representations are unitary and satisfy $L_{g} R_{h} = R_{h} L_{g}$ (Continuous convolution operators are Hilbert–Schmidt, Left and right regular representations on L2(G)).

[L2]

Hilbert adjoints satisfy $(T S)^{*} = S^{*} T^{*}$ , $T^{* *} = T$ and $∥ T^{*} T ∥ = ∥ T ∥^{2}$ ; for a compact self-adjoint operator the set of nonzero eigenvalues consists of real numbers with finite-dimensional eigenspaces and the closed span of the eigenspaces is the orthogonal complement of the kernel, equal to the closure of the range (Hilbert-adjoint identities, Spectral theorem for compact self adjoint operators).

[L3]

Integrals are invariant under left translation (Haar integration is translation and conjugation invariant), and Fubini applies to integrable functions on the finite product measure space (Fubini's theorem for L^1 functions on a sigma-finite product).

[L4]

The complex $L^{2}$ pairing is linear in its first variable and satisfies Cauchy–Schwarz (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz).

Proof

technique · direct

1.1

Since Haar measure has mass one, [L4] applied to $∣ f ∣, 1$ gives $∥ f ∥_{1} \leq ∥ f ∥_{2}$ , and similarly for $h$ . Thus $k (x^{- 1} y) f (y) \overline{h (x)}$ is absolutely integrable, with integral of its modulus at most $∥ k ∥_{\infty} ∥ f ∥_{1} ∥ h ∥_{1}$ . Fubini gives $⟨ T_{k} f, h ⟩ = \int_{G} f (y) \overline{\int_{G} \overline{k (x^{- 1} y)} h (x) d x} d y = ⟨ f, T_{k^{*}} h ⟩$ , because $k^{*} (y^{- 1} x) = \overline{k (x^{- 1} y)}$ . The continuous function $k^{*}$ defines a bounded operator by [L1], so uniqueness of the adjoint gives $T_{k}^{*} = T_{k^{*}}$ . No substitution reversing the two kernel arguments is made.

A1L1L2L3L4

1.2

For $g \in G$ , $(L_{g} T_{k} f) (x) = \int_{G} k (x^{- 1} g y) f (y) d y$ . Substituting $u = g y$ , using left invariance, gives $\int_{G} k (x^{- 1} u) f (g^{- 1} u) d u = (T_{k} L_{g} f) (x)$ . Hence $L_{g} T_{k} = T_{k} L_{g}$ .

L1L3

2.1

If $k = k^{*}$ , step 1.1 makes $T_{k}$ self-adjoint. Compactness [L1] and the spectral theorem [L2] give finite-dimensional nonzero eigenspaces whose closed span is $(\ker T_{k})^{⊥} = \overline{ran T_{k}}$ . If $T_{k} v = λ v$ , step 1.2 gives $T_{k} L_{g} v = λ L_{g} v$ ; applying this also to $g^{- 1}$ proves invariance of the eigenspace. The standing AC assumption supplies the countable choice required by the spectral theorem.

A1L1L2step 1.1step 1.2

3.1

Put $S = T_{k}^{*} T_{k}$ . The adjoint identities give $S^{*} = T_{k}^{*} T_{k} = S$ , and $⟨ S f, f ⟩ = ∥ T_{k} f ∥_{2}^{2} \geq 0$ . It is compact: the image under $T_{k}$ of the unit ball has compact closure, and its image under the bounded, hence continuous, operator $T_{k}^{*}$ is compact and contains $S$ of that ball. By steps 1.1–1.2, both factors commute with every $L_{g}$ , so $S$ does too. Apply the compact self-adjoint spectral theorem directly to $S$ : its nonzero eigenspaces are finite-dimensional, and their closed span is $(\ker S)^{⊥} = \overline{ran S}$ . Commutation and the inverse translation prove their left invariance exactly as for $T_{k}$ . If $k = 0$ , both operators vanish and the nonzero-eigenspace family is empty with closed span ${0}$ ; no finite-dimensionality assertion is made about a zero eigenspace.

A1L1L2L4step 1.1step 1.2∎

Depends on

Used by

Peter–Weyl theorem Theorem

Dependency tree · two levels

73 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)
Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)