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If you consider pairs of unitaries instead, and the group, that you get out of a similar construction, the answer is negative. That is an analogous question but in a slightly different context, since the . The new question is only interesting if one studies it for all dimensions at once.

More precisely

To be more precise, in http://arxiv.org/abs/1003.4093, it was shown that there are exotic families of continuous maps $$\phi_n \colon U(n) \times U(n) \to U(n),$$ such that:

1) $\phi_{n+m}(U \oplus V,U' \oplus V') = \phi_n(U,V) \oplus \phi_n(U',V')$,

2) $\phi_{nm}(U \otimes V,U' \otimes V') = \phi_n(U,V) \otimes \phi_n(U',V')$, and

3) $\phi_n(AUA^{-1},AVA^{-1}) = A\phi_n(U,V)A^{-1}$.

Here, exotic means that $\phi_n(U,V)$ is not given by evaluating the pair of unitaries at a word $w \in {\mathbb F}_2$, where ${\mathbb F}_2$ denotes the free group on two generators.

1

If you consider pairs of unitaries instead, and the group, that you get out of a similar construction, the answer is negative. That is analogous but a slightly different context, since the question is only interesting if one studies it for all dimensions at once.

More precisely, in http://arxiv.org/abs/1003.4093, it was shown that there are exotic families of continuous maps $$\phi_n \colon U(n) \times U(n) \to U(n),$$ such that:

1) $\phi_{n+m}(U \oplus V,U' \oplus V') = \phi_n(U,V) \oplus \phi_n(U',V')$,

2) $\phi_{nm}(U \otimes V,U' \otimes V') = \phi_n(U,V) \otimes \phi_n(U',V')$, and

3) $\phi_n(AUA^{-1},AVA^{-1}) = A\phi_n(U,V)A^{-1}$.

Here, exotic means that $\phi_n(U,V)$ is not given by evaluating the pair of unitaries at a word $w \in {\mathbb F}_2$, where ${\mathbb F}_2$ denotes the free group on two generators.