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This is actually not a question of mine, so I'll be short on motivation and say nothing beyond that if this were true, a few fancy harmonic analysis techniques that a colleague of mine used in proving his recent results could be replaced by the mean value theorem.

Suppose that $A_1,\dots,A_n$ and $B_1,\dots,B_n$ are two commuting families of self-adjoint operators in a Hilbert space $H$ (that is all $A$'s commute, all $B$'s commute, but $A$'s may not commute with $B$'s). Assume that $\|A_k-B_k\|\le 1$ for all $k$. Is it true that there exists a one-parameter family $C_k(t)$ of self-adjoint commuting (for each fixed $t$) operators such that $C_k(0)=A_k$, $C_k(1)=B_k$ and $\int_0^1\left\|\frac d{dt}C_k(t)\right\|dt\le M(n)$ where $M(n)$ is a constant depending on $n$ only? In other words, is the set of commuting $n$-tuples of self-adjoint operators a "chord-arc set"?

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@fedja: is your question clear for n=1? if A1 and B1 are orthogonal projections of different ranks, why do we have C1? maybe you mean $||A_k-B_k||<\epsilon<1$? –  Kate Juschenko Jan 6 '11 at 22:11
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Is this known when $H$ is finite dimensional? –  Andrey Rekalo Jan 6 '11 at 22:46
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@Andrey: I assume that in finite dimension you just take the orthonormal frame of eigenvectors of all $A$s and connect them by a geodesic (in the orthogonal/unitary group) to the orthogonal frame of eigenvectors of the $B$s. I did not do the computation, I confess, but it seems reasonably clear that the conditions are satisfied. This seems less obvious in the infinite dimensional case. –  Igor Rivin Jan 6 '11 at 23:46
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@Igor: But then the length of the path can be as large as the diameter $d_m$ of $SU_m$, $m$ the dimension of $H$. If the answer to the question is yes, we must have $d_m\le M(n)$, and therefore $\sup_md(m)<\infty$. Is this true ? –  Denis Serre Jan 7 '11 at 7:28
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@Martin: $C_k(t)$ should commute for each t. It seems that if norms of $A_k$ and $B_k$ are bounded then the statement is true, just taking homotopy of $A_k$ to $0$ and then $0$ to $B_k$, i.e $C_k(t)=(1-2t) A_k$ if $t\in [0,1/2]$ and $C_k(t)=(2t-1)B_k$ if $t\in[1/2,1]$ – Kate Juschenko 2 mins ago –  Kate Juschenko Jan 8 '11 at 17:06
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2 Answers

Here is a "scratch of a proof". It might be completely wrong since I though about it in 1am.

  1. We can attach to the family $A_i$ protectors $P_\lambda$ where $\lambda \in \mathbb R^n$.

  2. For $J=(j_1,...,j_n)\in \mathbb Z^n$ Let $V_J=Im P_J \cap \bigcap Ker P_{j_1,..., j_i-1,...j_n}$.

  3. We have an $A_i$ invariant decomposition $V=\bigoplus V_J$ and $||(A_i-j_i)|_{V_J}|| \leq 1$.

  4. we can assume that $(A_i)|_{V_J}=j_i$ (by connecting it by strait line).

  5. We do the same for $B_i$.

  6. Now the problem should be similar to the f.d. case. This step I did not think through, but I hope it will be OK.

Good luck

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Could you elaborate 6? Already having a proof in finite dimension with constant $M(n)$ not depending on the dimension does not seem clear to me. –  Mikael de la Salle Jan 11 '13 at 12:23
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I am not sure it is relevant exactly, or if you still have interest in this problem, but perhaps you might find this comment useful or interesting.

$\mathrm{Hom}(\mathbb{Z}^r,G)$ is generally NOT path-connected. For instance, if $r\geq 3$ and $G=\mathsf{SO}(n)$, over $\mathbb{R}$ or $\mathbb{C}$, for $n\geq 4$ then it is disconnected. Thus, in this setting, there are commuting collections $\lbrace A_i\rbrace$ and $\lbrace B_j\rbrace$ that are NOT connected by a path through commuting elements.

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I certainly still have interest in this problem (as well as in the other ones I posted). However I'm not sure how what you say helps because in my setting you can always transform commuting families by moving common orthogonal bases of eigenvectors and adjusting eigenvalues on the go. If nothing better, you can shrink everything to $0$ and then expand from there in the new direction. –  fedja May 25 '13 at 1:20
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