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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?

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

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  • $\begingroup$ 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? $\endgroup$ – CL. Apr 15 at 11:25
  • $\begingroup$ 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 $\psi_z: C_0(X)\to\Bbb C, f\mapsto f(z)$. $\endgroup$ – s.harp Apr 15 at 11:32
  • $\begingroup$ Ok, thanks, I will read your proof again! I wonder if you may have a look at my most recent question on Cayley Transform. $\endgroup$ – CL. Apr 15 at 11:39

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