embedding of $\prod_{n\in\mathbb{N}}M_{n}(\mathbb{C})$ in a type $II_{1}$ factor Suppose $M$ is a type $II_{1}$ factor with trace $\tau$. Let $\lbrace p_{n}\rbrace_{n\in\mathbb{N}}$ be an increasing sequence of projections such that $\tau(p_{n})\rightarrow 1$. Now, let's consider $\lbrace q_{n}\rbrace_{n\geq 2}$ such that for $n\geq 2$ we define $q_{n}=p_{n}-p_{n-1}$ and finally let's consider $Q_{j}=q_{j}Mq_{j}$, for $j\geq 2$. This is the setting and I am reading a paper where now is said that is a "well-known fact" that there is a unital embedding $\prod_{n\in\mathbb{N}}M_{n}(\mathbb{C})\hookrightarrow Q_{j}\subset M$.
Can someone explain me this well known fact ? Thank you
 A: You can do this in any II$_1$-factor. Note that $Q_j$ is a II$_1$-factor. 
In any II$_1$-factor $M$ you can always get a sequence of pairwise orthogonal projections that add to the identity (those could be the $q_j$ in your setup). So now you want to embed $M_n(\mathbb C)\hookrightarrow q_jMq_j$. Since you are in a II$_1$-factor, you can divide $q_j$ as a sum of $n$ pairwise equivalent projections $r_1,\ldots,r_n$. Let $e_{1r}$ be a partial isometry with $e_{1k}e_{1k}^*=r_1$, $e_{1k}^*e_{1k}=r_k$. Now define $e_{kj}=e_{1k}^*e_{1j}$. It is easy to check that this elements behave like matrix units, and so you get a unital copy of $M_n(\mathbb C)$ in $q_jMq_j$. 
Edit: To be more explicit. Fix a II$_1$ factor $M$. Let $\{q_n\}$ be a sequence of pairwise orthogonal projections with $\sum q_n=I$. If we apply the above to the II$_1$-factor $q_nMq_n$ we have a $*$-monomorphism $\pi_n:M_n(\mathbb C)\to q_nMq_n$ given by 
$$
\pi_n(a)=\sum_{k,j=1}^na_{kj}\,e^*_{1k}e_{1j}.
$$
Now define $\pi:\prod_nM_n(\mathbb C)\to M$ by
$$
\pi(\{a_n\})=\sum_n\pi_n(a_n).
$$
This last sum converges in the two norm and it defines a $*$-monomorphism, because $\pi_n(a_n)=q_n\pi_n(a_n)q_n$ and $\sum q_n=I$. 
