# The proof of Stinespring dilation

(Stinespring dilation) Let $A$ be a unital C*-algebra and $\phi: A \rightarrow B(H)$ be a completely positive map. Then, there exist a Hilbert space $H_{1}$, and a *-representation $\pi: A \rightarrow B(H_{1})$ and operator $V:H \rightarrow H_{1}$ such that $\phi(.)=V^{\ast}\pi(.)V$. In particular, $||\phi||=||V||^{2}=||\phi(1)||$.

In the first step of the proof of this theorem, defining a sesquilinear form $< , >$ on $A \odot H$ (this is the algebraic tensor product) by $<\Sigma_{j}~b_{j}\otimes \eta_{j},~\Sigma_{i}~a_{i}\otimes \xi_{i}>=\Sigma_{i, j} <\phi(a_{i}^{\ast}b_{j})\eta_{j},~\xi_{i}>_{H}$.

I can not understand the form of the element in $A \odot H$ (that is $\Sigma_{i}~a_{i}\otimes \xi_{i}$). why not $a_{i}\otimes \xi_{i}$? I mean how to explain the sum of $a_{i}\otimes \xi_{i}$.

If you only consider elementary tensors, then what would their sum be? Remember that you want $A\odot H$ to be a vector space.
• But how to compute the sum in $A \odot H$? I mean, if two elements $\Sigma_{i=1}^{n} b_{i} \otimes \eta_{i}$ and $\Sigma_{j=1}^{k} a_{j} \otimes \xi_{j}$, then what is their sum? – Yan kai Mar 10 '14 at 15:57
• A sum of elementary tensors is a sum of elementary tensors. If you add two sums of elementary tensors, you get a sum of elementary tensors: $\sum_ib_i\otimes\eta_i+\sum_ja_j\otimes\xi_j$ is just that, a sum of elementary tensors. – Martin Argerami Mar 10 '14 at 16:05
• I still have some doubt. If the sum of elemetary tensors is just $\Sigma_{i} b_{i} \otimes \eta_{i}+ \Sigma_{j} a_{j} \otimes \xi_{j}$, how to verify the $<. , .>$ defined above is sesquilinear ? – Yan kai Mar 11 '14 at 15:00