# Graded $*$-homomorphisms where are homotopic to $0$

This is an Exercise 3.15, pg 42.

1. 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.
2. 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?

• Scaling is not a good idea here, the maps $H(\cdot,t)$ are no homomorphisms for $t\in (0,1)$. – MaoWao Mar 4 at 10:15
• Ah, you mean, they don't respect multiplication? That is true. I will edit: I also notice my original grading seems to be wrong. – CL. Mar 4 at 10:46

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.

• Hi s.harp, in retrospect I don't really understand your second argument of, I don't really understand what is meant by characters, is there a reference for this? – CL. Apr 15 at 11:25
• A character is a (non-zero) $*$-morphism into $\mathbb C$, alternatively a multiplicative probability state. The book by Murphy on C* algebras contains the definition. They are useful since for an abelian $C^*$ algebra $A$ you have $A\cong C_0(\sigma(A))$ where $\sigma(A)$ is the set of characters with the weak* topology. The details of this tell you that for $A=C_0(X)$ the only characters are the point evaluations $\psi_z: C_0(X)\to\Bbb C, f\mapsto f(z)$. – s.harp Apr 15 at 11:32
• Ok, thanks, I will read your proof again! I wonder if you may have a look at my most recent question on Cayley Transform. – CL. Apr 15 at 11:39