the standard even grading on $M_2(A)$ and $A\otimes \mathbb{K}$ I have a question about a passage in Blackadar's book about K-Theory. Let $A$ be a (ungraded) $C^*$-algebra. There is a grading on $M_2(A)$ with $M_2(A)^{(0)}$ the diagonal matrices and $M_2(A)^{(1)}$ the matrices with zero diagonal. This is an even grading with grading operator $diag(1,-1)\in M(M_2(A))$, where $M(M_2(A))$ denotes the multiplier algebra of $M_2(A)$. We call this the standard even grading. By identifying $A\otimes \mathbb{K}$ with $M_2(A)$, we obtain the standard even grading on $A\otimes \mathbb{K}$. My questions are: 1. How to understand the operator $diag(1,-1)$, i.e. what is meant here with 1? Is this the identity operator $M_2(A)\to M_2(A)$? But I'm not sure if it fits. 2. How does the standard even grading on $A\otimes \mathbb{K}$ look like ?

Here $$g=\text{diag}\,(1,-1)=\begin{bmatrix}1&0\\\0&-1\end{bmatrix}.$$ You can easily check that $$ M_2(A)^{(0)}=\\{a:\ gag=a\\},\ \ \ \ M_2(A)^{(1)}=\\{a:\ gag=-a\\}. $$

The standard even grading on $A\otimes\mathbb K$ is obtained by doing the above on $M_2(A\otimes\mathbb K)$ (diagonal matrices and matrices with diagonal zero). How it looks like depends on how you write the isomorphism $A\otimes\mathbb K\simeq M_2(A\otimes\mathbb K)$.

If we write $\mathbb K$ in terms of the canonical matrix units $\\{E_{kj}\\}$ and realize the isomorphism by decomposing $H$ as $(\sum E_{2k,2k})H\oplus (\sum E_{2k+1,2k+1})H$, then $$ A\otimes\mathbb K^{(0)}=\\{\sum_{k,j}a_{kj}\otimes E_{kj}:\ a_{kj}\in A,\ \ a_{kj}=0\ \text{ if }k+j\text{ is odd }\\}, $$ $$ A\otimes\mathbb K^{(1)}=\\{\sum_{k,j}a_{kj}\otimes E_{kj}:\ a_{kj}\in A,\ \ a_{kj}=0\ \text{ if }k+j\text{ is even }\\}, $$