Moving one family of commuting self-adjoint operators to another without losing commutativity on the way 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"?
 A: 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.
A: Here is a "scratch of a proof". It might be completely wrong since I though about it in 1am.


*

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

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

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

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

*We do the same for $B_i$.

*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
