Associativity of polar decomposition By polar decomposition, every continuous linear function $f \colon H \to K$ between Hilbert spaces can be written uniquely as $f = \widehat{f} \circ |f|$ for a positive operator $|f| \colon H \to H$ and a partial isometry $\widehat{f} \colon H \to K$ with $\ker(\widehat{f})=\ker(|f|)$. The binary operation $(f,g) \mapsto |g \circ f|$ on the set of positive operators is not associative. Is the binary operation $(f,g) \mapsto \widehat{g \circ f}$ on the set of partial isometries associative? 
 A: I have not fully checked this idea, but here goes.  I prefer the notation $P(T)$ for the partial isometry occurring in the polar decomposition of $T$.  I also got lost with three Hilbert spaces in the mix, so this answer is only for the case where $A,B,C$ all operate on the same fixed Hilbert space $H$.
In this notation, I believe the question is whether or not
$$
P(P(AB)C) = P(A P(BC))
$$
holds for all partial isometries $A,B,C$ on $H$.
I think it is easy to see that if $U$ is unitary, then for all $X$ we have $P(UX) = U P(X)$, and that if $T$ is a partial isometry, then $P(T) = T$.  From this, it seems to follow that if $A$ is assumed unitary, and $B$ and $C$ are partial isometries, we have
$P(P(AB)C) = P(AP(B)C) = P(ABC) = A P(BC)$ and $P(A P(BC)) = A P(P(BC)) = A P(BC)$ so the desired result holds.
If $A$ is not unitary, it still has a unitary dilation.  This means there is another Hilbert space $K$ and operators $X: K \to H$ and $Y: K \to K$ with the property that the operator $A'$ on $H \oplus K$ given by the block operator matrix $A' = \begin{pmatrix} A & X \cr 0 & Y \end{pmatrix}$ is unitary.  So consider the operators $B' = \begin{pmatrix} B & 0 \cr 0 & 0 \end{pmatrix}$ and $C' = \begin{pmatrix} C & 0 \cr 0 & 0 \end{pmatrix}$ on $H \oplus K$.  The operators $B'$ and $C'$ are partial isometries on $H \oplus K$, so by the work above,
$$
P(P(A'B')C') = P(A' P(B'C')).
$$
Now calculate: $A'B' = \begin{pmatrix} AB & 0 \cr  0 & 0 \end{pmatrix}$ and so a moment's thought ought to show that $P(A'B') = \begin{pmatrix} P(AB) & 0 \cr 0 & 0 \end{pmatrix}$, making the left hand side of the above equal to
$$
P(\begin{pmatrix} P(AB) C & 0 \cr 0 & 0 \end{pmatrix}) = \begin{pmatrix} P(P(AB) C) & 0 \cr 0 & 0 \end{pmatrix}.
$$
The right hand side is almost the same, but with $P(A P(BC))$ in the upper left corner, and unless I made a ridiculous error, you have what you want.  (There is a sightly more complicated unitary dilation theorem for operators between different spaces, so if there were no mistakes in this approach, maybe the same idea works in the general case too.)
A: Ah, (quite) some fiddling with Mathematica gave a counterexample.
In the notation of anon's answer, take 
$$
  A' = \begin{pmatrix} 1 & 1 & 1 & 1 \\\\ 1 & 1 & 1 & 0 \end{pmatrix}, \qquad
  B = \begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}, \qquad
  C = \begin{pmatrix} 0 & 1/\sqrt{2} \\\\ 1 & 0 \\\\ 0 &
      1/\sqrt{2} \end{pmatrix}.
$$
Then $A=P(A')$, $B$ and $C$ are partial isometries, but $P(P(CB)A) \neq P(CP(BA))$.
Notice that $C$ is a partial isometry that is not an isometry, i.e. has nontrivial kernel.
