Geometric interpretation for partial trace? This MO question asks for a geometric interpretation of the trace of a linear transformation. I'm wondering about a geometric interpretation of partial trace.
Given a linear transformation $f: X\otimes U\to Y\otimes U,$ the partial trace is a linear transformaton $\text{Tr}^U_{X,Y}\;\;(f):X\to Y$ satisfying certain properties. Basically, if you think of $f$ as a matrix with $|X|\times|Y|$-many $|U|\times|U|$ blocks, then $\text{Tr}(f)$ is given by taking the trace of each block.
So how can I visualize this operation? How can I tell a story about it, especially without resorting to a choice of bases?  
 A: We observe that if $Z$ is a finite-dimensional inner product space, and $f:Z\rightarrow Z$, then $$\text{Tr}(f)=n\cdot\int_{z\in S}\langle f(x),x\rangle dm(x)$$
whenever $m$ is the normalized probability measure on the unit ball $S$ of $Z$ and $n$ is the dimension of $Z$.
This observation was given by Yemon Choi as an answer to the corresponding question for a geometric interpretation of the trace. This formula can be obtained from the formula for the trace using bases simply by integrating over all bases. This observation extends to the following coordinate free formula for the partial trace.

Proposition: Suppose that $X,Y,U$ are finite-dimensional inner
product spaces. If $f:X\otimes U\rightarrow Y\otimes U$ is a linear
mapping, then
$$\langle\text{Tr}_{X,Y}^{U}(f)x,y\rangle=n\cdot\int_{z\in S}\langle
 f(x\otimes z),y\otimes z\rangle dm(z)$$ whenever $x,y\in X$, $S$ is
the unit ball in $U$, $n=\dim(U)$, and $m$ is the normalized probability measure on
$S$.

Proof: For this proof, we shall define the partial trace as the unique linear operator $\text{Tr}_{X,Y}^{U}:L(X\otimes U,Y\otimes U)\rightarrow L(X,Y)$ where
$\text{Tr}_{X,Y}^{U}(R\otimes T)=R\cdot\text{Tr}(T)$ whenever $T\in L(U,U),R\in L(X,Y)$.
It suffices to prove our equation in the case when $f=R\otimes T$ for some linear operators $R:X\rightarrow Y,T:U\rightarrow U$ since the general case follows from linearity. In this special case, we have
$$\langle \text{Tr}_{X,Y}^{U}(f)x,y\rangle=\langle\text{Tr}_{X,Y}^{U}(R\otimes T)(x),y\rangle$$
$$=\langle R(x)\text{Tr}(T),y\rangle=\langle R(x),y\rangle\text{Tr}(T)
=\langle R(x),y\rangle\cdot n\cdot\int_{z\in S}\langle T(z),z\rangle dm(z)$$
$$=n\int_{z\in S}\langle \langle R(x),y\rangle\langle T(z),z\rangle dm(z)
=n\int_{z\in S}\langle \langle (R\otimes T)(x\otimes z),y\otimes z\rangle dm(z)$$
$$=n\int_{z\in S}\langle \langle f(x\otimes z),y\otimes z\rangle dm(z)$$
Q.E.D.
This result can also be obtained by integrating the right hand side in the following formula/definition for the partial trace for a finite dimensional inner product space that holds whenever $(e_{1},\dots,e_{n})$ is a basis of $U$:
$$\langle\text{Tr}_{X,Y}^{U}(f)(x),y\rangle=\sum_{k=1}^{n}\langle f(x\otimes e_{i}),y\otimes e_{i}\rangle.$$
Observe that the above formula does have some coordinates, but it only has the coordinates for the space $U$, so it is coordinate free for $X,Y$ but not for $U$.
Now, we can even get an explicit basis-free formula for $\text{Tr}_{X,Y}^{U}(f)(x)$ at the expense of having to compute a double integral. For the following formula, $n_{Z}$ is the dimension of a vector space $Z$, $S_{Z}$ is the unit ball in $Z$, and $m_{Z}$ is the normalized probability measure on $S_{Z}$.
Observe that if $f_{1},\dots,f_{m}$ is a basis for $Y$, then $y=\sum_{k=1}^{m}\langle y,e_{k}\rangle e_{k}$. Therefore,
$y=n_{Y}\int_{S_{Y}}\langle y,z\rangle\cdot z dm_{Y}(z)$ for each $y\in Y$. Therefore,
$$\text{Tr}_{X,Y}^{U}(f)(x)=n_{Y}\int_{S_{Y}}\langle\text{Tr}_{X,Y}^{U}(f)(x),y\rangle\cdot y\cdot dm_{Y}(y)$$
$$=n_{Y}\int_{S_{Y}}n_{U}\int_{S_{U}}\langle f(x\otimes z),y\otimes z\rangle dm_{U}(z)ydm_{Y}(y)$$
$$=n_{Y}n_{U}\int_{S_{Y}}\int_{S_{U}}\langle f(x\otimes z),y\otimes z\rangle\cdot y dm_{U}(z)dm_{Y}(y).$$
Of course, this formula can also be obtained from the definition of the partial trace involving coordinates where one averages over all choices of bases.
