Erratum to the paper “Distinguished Varieties, Acta Math. Vol. 194”

Jim Agler ^{∗}  John E. McCarthy ^{†}

July 7, 2009

Lemma 4.6 is incorrect. The theorem it supports, Theorem 4.1, is correct as stated, and can be proved with a slight modification of the argument in the paper.

The error in the lemma is that there can be points of the distinguished variety

V = {(z, w) ∈ 𝔻^{2} : det[Ψ(z) − wI] = 0}

where the dimension of the null space of Ψ(z) − wI is discontinuous, and at these points one may not be able to choose û^{1} continuously.

If these singularities are disjoint from the set of nodes {λ_{1}, …, λ_{N}}, then the proof of Theorem 4.1 is not materially affected. However, to include the case that a node be a singular point, we must modify the argument, and replace Lemma 4.6 with a correct version, Lemma 4.16 below.

Lemma 4.16 Every admissible kernel on a set {λ_{1}, …, λ_{N}} can be extended to an admissible kernel k on a distinguished variety V that contains the points λ_{1}, …, λ_{N}. V can be represented as

V = {(z, w) ∈ 𝔻^{2} : det[Ψ(z) − wI] = 0

∗Partially supported by National Science Foundation Grant

†Partially supported by National Science Foundation Grant DMS 0501079

for some matrix-valued inner function Ψ. Moreover, the extension can be chosen in such a way that

k(z, w) = sz̄ ⊗ û^{1}(z, w),

where each vector û^{1}(z, w) is in the null-space of det[Ψ(z)−wI], and so that, at each node λ_{j}, there are q_{j} sequences

(α_{m}, β_{p,m})_{m=1}^{∞} 1 ≤ p ≤ q_{j}

that converge to λ_{j} and such that the vector û^{1}(λ_{j}) = u_{j}^{1} is the limit of vectors in the linear span of

{û^{1}(α_{m}, β_{p,m}) : 1 ≤ p ≤ q_{j}}.

Proof of Lemma 4.16: Everything in the proof of Lemma 4.6 is correct except for the assertion that (û^{1}, û^{2}) can be chosen continuously. We wish to show that these vectors can be chosen so that the conclusion of Lemma 4.16 holds.

Fix some node, λ_{1} say. By Gaussian elimination, after permutation of the coordinates, there are analytic functions f_{ij}, 1 ≤ i ≤ j ≤ d_{1}, on a neighborhood of λ_{1} such that

Ker (Ψ(z) − wI) = Ker ( \begin{matrix}f_{11}(z, w) & f_{12}(z, w) & … & f_{1d_{1}}(z, w) \\ 0 & f_{22}(z, w) & … & f_{2d_{1}}(z, w) \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & f_{d_{1}d_{1}}(z, w) \end{matrix} ) ,

and, moreover, the diagonal functions f_{11} through f_{rr} do not vanish at λ_{1}, and the functions f_{jj} do vanish for r < j ≤ d_{1}.

Note that the zero sets of f_{jj} are (unions of) sheets of V near λ_{1}. Choose any sheet of the variety near λ_{1}. With the exception of a possible jump at λ_{1}, the dimension of Ker(Ψ(z) − wI) will be locally constant, t say. After a permutation of the last d_{1} − r coordinates if necessary, we can assume the sheet corresponds to the vanishing of f_{r+1,r+1} through f_{r+t,r+t} (we are not assuming that they are coprime). Now choose û^{1} on this sheet so that its (r + 1)^{st} through (r + t)^{th} coordinates agree with those of u_{1}^{1}, its (r + t + 1)^{th} through d_{1}^{th} coordinates are zero, and it lies in

Ker ( \begin{matrix}f_{11}(z, w) & f_{12}(z, w) & … & f_{1d_{1}}(z, w) \\ 0 & f_{22}(z, w) & … & f_{2d_{1}}(z, w) \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & f_{d_{1}d_{1}}(z, w) \end{matrix} ) .

Repeat this for each sheet, and the sum of the chosen vectors will converge to u_{1}^{1}. □

To prove Theorem 4.1 given Lemma 4.16, we will change the second and last paragraphs of Step 2.

We claim that at every node λ_{j}, the formula (4.10) uniquely defines the solution to the Pick problem on some sheet through λ_{j}. The union of all the irreducible components containing these particular sheets will give a distinguished variety containing every node on which the solution is unique.

Indeed, if the denominator in equation (4.10) vanished on every sheet of V that meets λ_{j}, then any linear combination of the q vectors

(w_{j}K̂((α_{m}, β_{p,m}), λ_{j}))_{j=1}^{N}

would be orthogonal to γ, and by taking a limit we obtain (4.11) again. The remainder of the proof proceeds as before, showing that (4.11) would contradict minimality of the problem. □