Perturbing a partial isometry to part of a unitary

In Narutaka Ozawa's solution over at Mathoverflow, the following result is implicitly used:

Let $$M\subseteq B (H)$$ be a von Neumann algebra, and let $$\xi\in H$$. If $$u\in M$$ is a partial isometry, then for $$\epsilon>0$$ we can find a partial isometry $$v\in M$$ with $$\|(u-v)\xi\| <\epsilon$$ and with $$1-v^*v \sim 1-vv^*$$.

Here $$\sim$$ is Murray-von Neumann equivalence. It's easy to see that if $$1-v^*v \sim 1-vv^*$$ then $$v$$ can be extended to a unitary in $$M$$. Why is this claimed result true?

If $$M$$ is finite, then as indicated in this answer we always have that if $$p\sim q$$ then $$1-p\sim 1-q$$ (see also, for example, Exercise 6.9.6 in Kadison and Ringrose, Volume 2). By definition $$u^*u\sim uu^*$$ so the claim follows with $$v=u$$. Indeed, this would also hold if $$u^*u$$ (equivalently, $$uu^*$$) were finite, even if $$M$$ were not.

Ozawa's wording suggests to me that in general, we might seek a projection $$p\in M$$ with $$p\leq u^*u$$ and set $$v=up$$. Then $$v^*v = pu^*up = p$$, and we want $$\|u(1-p)\xi\|<\epsilon$$, which seems plausible to achieve, perhaps? I have no clue how to get $$1-v^*v \sim 1-vv^*$$? Maybe instead a type-decomposition argument could work, but again I don't see how to get started.

As stated, this is not true. Let $$M=B(H)$$, $$\xi=e_1$$, $$u=S^*$$ the adjoint of the unilateral shift. Then, for any unitary $$v$$, $$\|(u-v)\xi\|=\|v\xi\|=1.$$

On the other hand, if I'm not wrong what Ozawa is claiming is that if $$\|u\xi-\eta\|<\delta$$, with $$\|\xi\|=\|\eta\|=1$$ and $$u\in M$$, then there exists $$v\in M$$, unitary, with $$\|u\xi-v\xi\|<\varepsilon$$. Below is what I could come up with, which no doubt is much less fancier than how Ozawa thought about it.

We may assume that $$M$$ has no finite-dimensional summand, as that case is easily dealt with. When $$M$$ has no finite-dimensional summand, the unitary group is sequentially wot dense in the unit ball (see Conway and Szücs, Indiana Univ. Math. J. 22 (1972/73), 763–768. Article and zbMath). So there exists a sequence $$\{v_n\}\subset M$$ with $$v_n$$ unitary and $$v_n\to u$$ wot. Then, noting that $$\|u\xi\|>\|\eta\|-\delta=1-\delta$$ \begin{align} \limsup_n\|u\xi-v_n\xi\|^2&=\limsup_n\|u\xi\|^2+\|\xi\|^2-2\operatorname{Re}\langle u\xi,v_n\xi\rangle\\[0.3cm] &\leq \limsup \delta+2-2\operatorname{Re}\langle u\xi,v_n\xi\rangle\\[0.3cm] &=\delta+2(1-\|u\xi\|)<\delta+2\delta=3\delta. \end{align} Choosing $$n$$ big enough, we can get a unitary $$v=v_n$$ such that $$\|u\xi-v\xi\|<\sqrt{4\delta}=2\sqrt\delta.$$ Some can take $$\varepsilon=2\sqrt\delta$$.

• Great! Thanks not only for the counter-example, but especially for working out what I should have asked, and then answering that instead. Jan 15 at 9:55
• I guess it's also worth stressing that this argument works for any contraction $u\in M$ while "my" application only needed it with $u$ a partial isometry; so this is very nice! Jan 15 at 10:10
• Indeed. Many many years ago, the exercises in Kadison-Ringrose's chapter 5 about showing that in $B(H)$ the projections are wot-dense in the positive unit ball, and the unitaries in the full unit ball were eye-opening for me. Jan 15 at 12:38

Here is a (sketch) proof of how to use Comparison Theory. We start with a partial isometry $$u\in M$$ and a vector $$\xi\not=0$$, and for $$\epsilon>0$$ we seek a partial isometry $$v$$ with $$\|(u-v)\xi\| > (1-\epsilon) \|u\xi\|$$. In fact, we will construct $$v$$ with $$v^*v \leq u^*u$$. First a couple of reductions. If $$\xi'=u^*u\xi$$ so $$\|\xi'\| = \|u\xi\|$$, and we can find $$v$$ with $$\|(u-v)\xi'\| > (1-\epsilon)\|\xi'\|$$ then as $$v\xi = v(v^*v)\xi = v(v^*v)(u^*u)\xi = v(u^*u)\xi = v\xi'$$, and $$u\xi=u\xi'$$, we have $$\|(u-v)\xi\| = \|(u-v)\xi'\| > (1-\epsilon)\|\xi'\| = (1-\epsilon)\|u\xi\|$$ as required. So wlog $$u^*u\xi=\xi$$ and $$\xi$$ is a unit vector. The problem also respects direct sums, so we may decompose $$M$$ as a direct sum and solve the problem in each part.

If $$u^*u$$ is finite, then as already observed in the OP, we're done. Otherwise $$u^*u$$ is infinite, so there is a central projection $$z$$ with $$zu^*u$$ finite and $$(1-z)u^*u$$ properly infinite. We have solved the problem in the summand $$zM$$, so we can focus on $$(1-z)M$$. So wlog assume that $$e = u^*u$$ is properly infinite.

By the "halving lemma" there is $$g\leq e$$ with $$g\sim e$$ and $$e-g\sim e$$. Notice that we can then further decompose (say) $$e-g$$, and so by induction, for any $$n$$ find orthogonal $$g_1,\cdots,g_n$$ with $$\sum_i g_i=e$$ and $$g_i\sim e$$ for each $$i$$. Notice also that if $$f_1,f_2$$ are orthogonal with $$f_i\sim e$$ for $$i=1,2$$ then $$f_1\sim e \sim g$$ and $$f_2\sim e \sim e-g$$ so $$f_1+f_2\sim g+(e-g)=e$$. An induction shows that if $$f_1,\cdots,f_n$$ are orthogonal with $$f_i\sim e$$ for each $$i$$ then $$\sum_i f_i\sim e$$. In particular, for any $$i$$, we find that $$\sum_{i\not=j} g_j \sim e$$.

Set $$f=uu^*$$. For any $$i$$, set $$\hat g_i = e - \sum_{j\not=i} g_j$$, and observe that $$1-\hat g_i = (1-e) + (e-\hat g_i) = (1-e) + g_i \sim (1-e)+e = 1.$$ Using the partial isometry $$u$$ which by definition gives $$e\sim f$$, we can transport the decomposition $$e = \sum_j g_j$$ to $$f = \sum_j g_j'$$ say. Set $$\hat g_i' = f - \sum_{j\not=i} g'_j \sim \hat g_i$$. So the same argument gives $$1-\hat g_i' \sim 1$$ and so $$1-\hat g_i \sim 1-\hat g_i'$$.

As $$1 = \|\xi\|^2 = \|u^*u\xi\|^2 = \sum_j \|g_j\xi\|^2$$ we see that for some $$i$$, we must have $$\|g_i\xi\|^2 \leq 1/n$$, and so $$\|\hat g_i\xi\|^2 \geq 1-1/n$$. Taking $$n$$ large enough and setting $$v = u(\hat g_i^*\hat g_i)$$ gives the required conclusion.

• So this is not as general as Martin Argerami's answer, but we do prove a little more, namely that a "cut down" of $u$ may be extended to a unitary (rather than finding some unitary which does the job). Jan 16 at 11:42