# Use of Uhlmann representation in proving the strong subadditivity of the von Neumann entropy

I am trying to prove strong subadditivity of the von Neumann-entropy, using joint convexity of the quantum relative entropy. The procedure is given in https://en.wikipedia.org/wiki/Strong_subadditivity_of_quantum_entropy. However, I do not have access to the cited paper by Uhlmann on "Endlich Dimensionale Dichtematrizen II (1973)".

In the wikipedia article, $$\text{tr}_B(\rho_{AB} ) = N^{-1} \sum_{j=1}^N (1_{A}\otimes U_j) \rho_{AB}(1_{A}\otimes U_j^\dagger)$$ for some unitaries $$\{U_j\}$$, is given as "the Uhlmann representation" of the partial trace.

I am not familiar with this representation and I would argue that the partial trace cannot be expressed via unitaries, since it maps into a smaller space.

How do I have to understand this decomposition to be able to continue with the proof?

• Note: I tried approaching the problem via the Kraus decomposition $$\text{tr}_B: \rho_{AB}\mapsto\sum_{j=1}^N (1_A \otimes \langle j|_B)\ \rho_{AB}\ (1_A\otimes|j\rangle_B).$$ Since the next step relies on unitary invariance of the quantum relative entropy, this does not seem to work.

Looking at the Wikipedia article, I would assume what is meant is $$\text{tr}_B(\rho_{AB} )\otimes 1\!\!1 = N^{-1} \sum_{j=1}^N (1_{A}\otimes U_j) \rho_{AB}(1_{A}\otimes U_j^\dagger)\ ,$$ in particular since it is suggested that one can take a Haar integral instead (which definitely gives the above formula). The l.h.s. can the in some sense be considered as "equivalent" to $$\mathrm{tr}_B(\rho_{AB})$$.