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Are there examples of (distinct) Hilbert spaces $H_1$=$(H,\langle\cdot,\cdot\rangle_1)$, $H_2 $=$(H,\langle\cdot,\cdot\rangle_2)$ and a linear operator $V: H_1\to H_2$ such that $V^n: H_1\to H_2$ is an isometry $\mathbf{??}$ for every $n=1,2,...$.

Note that:

1) $H_1$ and $H_2$ have the same underlying linear space structure, and we look for $V$: $||V^n(x)||_2=||x||_1$ for all $x\in H_1$ with $||x||_k=\sqrt{\langle x,x\rangle_k}$, $k=1,2$)

2) if $V: K\to K$ is an isometry on a Hilbert space $K$, so is $V^n$ (i.e. $||V^n(x)||=||x||$, for all $x\in K$), so the question is if we can find such isometries between $\underline{\rm distint}$ Hilbert space structures.

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    $\begingroup$ How do you define $V^n$ if $H_1$ and $H_2$ are entirely distinct? $\endgroup$ Jan 25, 2014 at 17:27
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    $\begingroup$ @MichaelRenardy: the vector spaces are the same, only the scalar products are asked to be different. Paulo should probably have written $V:H \to H$ to make that clearer. $\endgroup$ Jan 25, 2014 at 17:45

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The answer depends on whether you ask $V$ to be invertible (you do not seem to do, but the usual definition of an isometry includes it).

(Edit: paragraph corrected thanks to Nick Weaver's comment) If $V$ is not asked to be invertible, then the answer is yes: take $H=\ell^2(\mathbb{N})$, $\langle\cdot,\cdot\rangle_1$ be the usual $\ell^2$ product, $V$ be the right shift and $\langle\cdot,\cdot\rangle_2$ be equal to $\langle\cdot,\cdot\rangle_1$ except that it is rescaled on the first coordinate (e.g. $\langle u,v\rangle_2=2u_1v_1+\sum_{i>1} u_iv_i$).

If $V$ is asked to be invertible, then the answer is no: simply look at the action of $V$ on scalar product $$V\cdot\langle\cdot,\cdot\rangle = \langle V^{-1}\cdot,V^{-1}\cdot\rangle.$$ Then you ask for two distinct points (in the space of scalar products) such that one is fixed and the other is mapped to the first one. This is clearly impossible for an invertible map.

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    $\begingroup$ The left shift is a co-isometry, not an isometry. Interchange $H_1$ and $H_2$ and use the right shift. $\endgroup$
    – Nik Weaver
    Jan 25, 2014 at 19:00
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    $\begingroup$ (Also your comment when $V$ is invertible is a little weird ... why not just say that if $V$ is a bijection and preserves the inner product then $H_1$ and $H_2$ cannot be distinct?) $\endgroup$
    – Nik Weaver
    Jan 25, 2014 at 19:01
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    $\begingroup$ Final comment --- in the context of Hilbert space operators, the usual definition of an isometry does not include invertibility. Isometry is $V^*V = I$, co-isometry is $VV^* = I$, unitary is $V^*V = VV^* = I$. $\endgroup$
    – Nik Weaver
    Jan 25, 2014 at 19:03
  • $\begingroup$ @NikWeaver: thanks for the correction (now edited) and the terminology, which I didn't know. For the weirdness of the invertible part, I do not quite get what you write: there are a priori two inner products. One can certainly state this part in many ways. $\endgroup$ Jan 25, 2014 at 22:36
  • $\begingroup$ No problem. I withdraw the "weirdness" criticism, there is nothing wrong with your comment. $\endgroup$
    – Nik Weaver
    Jan 25, 2014 at 23:57

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