This is an Exercise 3.15, pg 42.
- Show that the map between $\Bbb Z/2 \Bbb Z$- graded $C^*$-algebras $$C_0(\Bbb R) \rightarrow M_2(\Bbb C)$$ $$ \varphi: f \mapsto \begin{pmatrix} f(0) & 0 \\ 0 & f(0) \end{pmatrix} $$ is homotopic (through graded $*$-homomoprhisms) to the $0$ homomoprhisms.
- Whereas the $*$-homomorphism $$ \psi:f \mapsto \begin{pmatrix} f(0) & 0 \\ 0 & 0 \end{pmatrix} $$ is not null homotopic.
By definition in page 41, the even of $M_2(\Bbb C)$ consists of diagonal elemnts, the odd are those of off diagonals.
My questions would be: 1. How does one construction this homotopy? 2. How does one show the non-homotopy for second part?
Here is the construction for your first question.
Check that $$f\mapsto\begin{pmatrix}\frac{f(x)+f(-x)}2& \frac{f(x)-f(-x)}2\\ \frac{f(x)-f(-x)}2 &\frac{f(x)+f(-x)}2\end{pmatrix}$$ is a graded $*$ morphism, call it $\phi_x$. Now check that for each $f$ the map $$[0,1]\to M_2(\Bbb C), \qquad t\mapsto \begin{cases}\phi_{1-1/t}(f) & t>0\\ 0 & t=0\end{cases}$$ is continuous (the only difficulty is at $t=0$, but note that $f(1/t)\to 0$ as $t\to0$). This implies that this defines a homotopy on $\mathrm{Hom}_*^{\Bbb Z_2}(C_0(\Bbb R),M_2(\Bbb C))$ where this space has been given the strong operator topology. I assume that this is the topology you want, since it is impossible to have null-homotopic morphisms in the norm topology.
For the second part note that the images of $*$-morphism of $C_0(\Bbb R)$ in $M_2(\Bbb C)$ must be simultaneously diagonalisable. The projection onto the first and second eigenvalue (after choosing such a diagonalisation) will then be characters of $C_0(\Bbb R)$, these are always of the form $f\mapsto f(x)$.
Now if you want your mapping to respect the gradient, there are essentially three different possible scenarios. The first is that both characters are $f\mapsto f(0)$, second is that one is $f\mapsto f(0)$ and the other is the zero character and the third is that one character is $f\mapsto f(x)$ and the other is $f\mapsto f(-x)$. I did not think of a proof for this, but I don't believe it is difficult.
So any homotopy must be of the form $$t\times f\mapsto U(t)\begin{pmatrix} f(x_1(t)) &0 \\ 0 & f(x_2(t))\end{pmatrix} U(t)^*,$$ where $U(t)$ is unitary and $x_1(t), x_2(t)$ obey the above condition and $x_1(0)=0, x_2(0)=\infty$. But continuity will forbid you from changing this arrangement. Thus you cannot deform to zero.