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Assume we have a homomorphism $\phi: C(S^{1},M_{n}(\mathbb{C}))\rightarrow C(S^{1},M_{m}(\mathbb{C}))$ where $n$ divides $m$. Under what conditions does $\phi$ send constant functions to constant functions?

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You should explain your notations (and the notion of homomorphism). –  Sergei Akbarov May 23 '13 at 15:01
    
Well, $S^{1}$ is the unit circle. $C(S^{1},M_{n}(\mathbb(C))$ is the algebra of continuous functions from $S^{1}$ to the matrix algebra $M_{n}(\mathbb(C)$. $\phi$ is a $\ast$-homomorphism. –  David May 23 '13 at 15:14
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up vote 0 down vote accepted

This is perhaps just a partial answer; but I feel that Will's answer is incomplete, and this is a more interesting question that he is suggesting.

Given a *-homomorphism $\phi$, composing with point-evaluation at any point $x \in S^1$ gives us a representation $$\phi_x: C(S^1, M_n) \to M_m, $$ which is therefore of the form $$ \phi_x(f) = \alpha_x(\text{diag}(f(y(x,1)), \dots, f(y(x,k)))), $$ where $\alpha_x$ is an automorphism of $M_m$ and $y(x,1),\dots,y(x,k) \in S^1$ (and $k=m/n$).

Now, $\phi$ sends constant functions to constant functions if and only if $\alpha_x \circ \alpha_y^{-1}$ acts as the identity on the $M_m \otimes 1_k \subset M_n$.

I'm not sure if there is a more explicit characterization. Some remarks:

  • Certainly, $\phi$ preserving constant functions doesn't imply that $\alpha_x = \alpha_y$.

  • $\alpha_x$ isn't uniquely determined (when there are repeats among $y(x,1),\dots,y(x,k)$, and related to this, needn't be continuous.

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Does the diagonal form of the representation come from the fact that any representation is unitary equivalent to a direct sum of irreducible representations? I don't know how you got the automorphism $\alpha_{x}$. –  David May 26 '13 at 22:28
    
Yes, this is how the diagonal form arises. The automorphism $\alpha_x$ is given by conjugating by the unitary. –  Aaron Tikuisis May 30 '13 at 12:19
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Basically never. Take a typical homomorphism, then conjugate it by a very non-constant function in $C(S^1, GL_m(\mathbb C))$. Conjugation is always a homomorphism, and satisfies every niceness condition I can think of. Any constant matrix that become a constant non-scalar matrix after the first homomorphism will become a non-constant matrix after conjugation.

Even constant scalar matrices can become constant non-scalar matrices and then non-constant matrices. For instance, you can send $M$ to

$\left(\begin{array}{cc} M & 0 \\ 0 & 0 \end{array}\right)$ or $\left(\begin{array}{cc} M & 0 \\ 0 & \overline{M} \end{array}\right)$

But this can be avoided by requiring the homomorphism to be unital and complex-linear.

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Thanks. Yes, I realized that it must depend on the type of unitary conjugation. –  David May 23 '13 at 17:42
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