Suspension of Cuntz algebra is traceless I saw a conclusion in a reference book: the suspension of the Cuntz algebra $C_0((0,1))\otimes \mathcal O_2$ has no tracial states.
My thought: there are many tracial states on $C_0((0,1))$. We take a tracial state $\tau$ on $C_0((0,1))$, then we can define a tracial state $\tilde{\tau}$ on $C_0((0,1))\otimes \mathcal O_2$ as follows:
$$\tilde{\tau} (x\otimes y)=\tau(x),\forall x\in C_0((0,1)),y\in \mathcal O_2.$$
Can anyone point out the mistake, thanks!

Your $\tilde\tau$ is not well-defined. You have
$$
\tilde\tau(x\otimes 1)=\tau(x)=\tilde\tau(x\otimes (-1))=-\tilde\tau(x\otimes 1)=-\tau(x), 
$$
a contradiction unless $\tau=0$. 
If you had a trace $\varphi$ on $A\otimes \mathcal O_2$, it induces a trace $\psi$ on $\mathcal O_2$ by 
$$
\psi(x)=\varphi(1\otimes x). 
$$
So $\psi=0$, and one can show that this implies that $\varphi=0$. 
