Let $\mathcal H$ be a separable Hilbert space, with inner product $\langle\cdot,\cdot\rangle$, and with orthonormal basis $(e_i)_{i\in\mathbb N}$. Consider a continuous linear embedding $A\colon\mathcal H\to\mathcal H$. Then one can apply the Gram-Schmidt process to the (linearly independent) vectors $Ae_i$. That is, we define recursively $$ f_i=\frac{Ae_i-\sum_{j<i}\langle Ae_i,f_j\rangle f_j}{\|Ae_i-\sum_{j<i}\langle Ae_i,f_j\rangle\|}. $$ Define a new operator $GS(A)$ by $Ae_i=f_i$. Since the $f_i$ are an orthonormal family of vectors, $GS(A)$ will be an isometric embedding $\mathcal H\to\mathcal H$.

Let $\mathcal E$ be the space of all continuous linear embeddings $\mathcal H\to\mathcal H$, and $\mathcal I$ the space of all isometric embeddings $\mathcal H\to\mathcal H$. Then the above construction defines a map $GS\colon\mathcal E\to\mathcal I$.

Is this map $GS$ continuous with respect to the operator norm topologies on both domain and target? If the answer is no, what is the largest subspace of $\mathcal E$ such that the restricted map $GS$ is continuous? (obviously $GS|_{\mathcal I}$ is the identity) Also if the answer is no: Is there a continuous alternative, i.e. a continuous retraction $\mathcal E\to\mathcal I$?

If it were, then the in particular the $f_i$ would depend continuously on $A$ uniformly in $i$, and I do not even see if this is true.

  • $\begingroup$ If $A=0$, then how is $GS(A)$ defined? Is $GS(0)$ the identity operator? $\endgroup$
    – Skeeve
    Feb 28, 2019 at 8:08
  • $\begingroup$ Does "Linear embedding" mean injective bounded linear operator with closed range"? $\endgroup$ Feb 28, 2019 at 8:15
  • $\begingroup$ @Pietro: I did not want to include "closed range" in the definition of embedding. However, it follows from Nik's answer that I better should have: Put $Ae_i=e_{2i}$. Then the linear homotopy joining $A$ and the identity runs through injective bounded linear operators with non-closed range. If there were a continuous retraction as in my question, then $A$ and the identity would be homotopic in $\mathcal I$. However, $ind(A)=\infty$ whereas $ind(id)=0$. $\endgroup$ Feb 28, 2019 at 10:26
  • $\begingroup$ @Skeeve: $0$ is not an embedding. $\endgroup$ Feb 28, 2019 at 10:26

1 Answer 1


The answer to the main question is no. Working on $l^2$, let $A$ be the operator $A: e_n \mapsto \frac{1}{n}e_n$ and for each $i$ let $A_i$ be $A$ followed by the unitary $U_i$ that switches $e_i$ and $e_{i+1}$ and fixes the other standard basis vectors. Then $A_i \to A$ in norm but $(U_i)$ does not converge in norm.

To the second question, there is no "largest" subspace on which the map is continuous, however for each $N$ its restriction to the set of operators for which $A^{-1}: {\rm ran}(A) \to H$ is bounded, with norm at most $N$, is continuous.

To the third question, I'm pretty sure there is a continuous retraction, but this is infinite dimensional topology and I wouldn't know where to find this result. You can treat each Fredholm index separately since the isometries with index $n$, for $n=0,1,\ldots,\infty$, are the connected components of the set of all isometries.

  • $\begingroup$ Thank you for your answer. Do you have a reference for the statement about the connected components of the set of isometries? $\endgroup$ Feb 28, 2019 at 10:27
  • $\begingroup$ Oh, any book on K-theory should have it. I like the book by Wegge-Olsen but there are many others. $\endgroup$
    – Nik Weaver
    Feb 28, 2019 at 12:02
  • 1
    $\begingroup$ You may be right, but once you know it for unitaries it follows easily for isometries. Given two isometries $V_1$, $V_2$ whose ranges have codimension $n$, there is a unitary $U$ such that $V_2 = UV_1$. A path from $U$ to $I$ in the unitaries then yields a path from $V_1$ to $V_2$ in the isometries with codimension $n$. $\endgroup$
    – Nik Weaver
    Feb 28, 2019 at 14:02
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    $\begingroup$ Look, if ${\rm codim}({\rm ran}(V_1)) > {\rm codim}({\rm ran}(V_2))$ then there must be a unit vector in ${\rm ran}(V_2)$ that is orthogonal to ${\rm ran}(V_1)$. So $\|V_1-V_2\| \geq \sqrt{2}$. $\endgroup$
    – Nik Weaver
    Feb 28, 2019 at 15:50
  • 2
    $\begingroup$ Yes, since the linear homotopy $\gamma$ joining the isometry $A\colon e_i\mapsto e_{2i}$ to the identity is injective (and bounded) on every stage. But since $A$ has index $-\infty$ and $id$ has index $0$, they cannot be connected in $\mathcal I$. Now if $\phi\colon\mathcal E\to\mathcal I$ were a retraction then $\phi\circ\gamma$ would connect $A$ and $id$ in $\mathcal I$. $\endgroup$ Feb 28, 2019 at 18:21

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