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I have a questions from representation theory. ($G$ is finite group)

Fulton and Harris in "Representation Theory. A first Course" write that: the character defines a map $$\chi : R(G) \to \mathbb{C}_{class}(G)$$ from $R(G)$ to the ring of complex-valued functions on G; this map is in fact a ring homomorphism. The statement that a representation is determined by its character then says that $\chi$ is injective.

Why this map is not isomorphism?

Are the $R(G)$ and $\mathbb{C}_{class}(G)$ a vector spaces?

(and then authors write) Because characters ${\chi_V}$ of irreducible representations of $G$ form an orthonormal basis for $\mathbb{C}_{class}(G)$ then $\chi$ induces an isomorphism: $$\chi_{\mathbb{C}} : R(G)\otimes \mathbb{C} \to \mathbb{C}_{class}(G)$$ Why we must have tensor product with $\mathbb{C}$ to obtain a isomorphism? (i'm not familiar with tensor products)

Representation ring $R(G)$ - as a group we take $R(G)$ to be the free abelian group generated by all (isomorphism classes of) representations of $G$, and mod out by the subgroup generated by elements of the form $V + W- (V \oplus W)$. Equivalently, given the statement of complete reducibility, we can just take all integral linear combinations $\sum a_i V_i$ of the irreducible representations $V$ of $G$; The ring structure is then given simply by tensor product, defined on the generators of $R(G)$ and extended by linearity.

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    $\begingroup$ $R(G)$ is not the ring of complex-valued functions on $G$. It is, as you write, a free abelian group, not a $\mathbb C$-vector space. So the image of $\chi$ misses all the class functions which are not $\mathbb Z$-linear combinations of the irreducible characters. $\endgroup$ – darij grinberg Nov 20 '14 at 20:54
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    $\begingroup$ $\mathbb C_{\operatorname{class}} (G)$ is a $\mathbb C$-vector space. The tensor product with $\mathbb C$ is needed to get a $\mathbb C$-vector space out of the $\mathbb Z$-module $R(G)$. You can understand $R(G) \otimes \mathbb C$ as the free $\mathbb C$-vector space (rather than free $\mathbb Z$-module) generated by all isoclasses of representations of $G$ modulo the $\mathbb C$-vector subspace spanned by elements of the form $V + W - (V \oplus W)$. It has $\mathbb C$-linear combinations of the irreps, not just $\mathbb Z$-linear combinations. $\endgroup$ – darij grinberg Nov 20 '14 at 20:56
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As darij writes in the comments, $R(G)$ is free abelian generated by irreps, so its image in the space of class-functions consist of integer-linear combinations of irreducible characters. One must tensor against the complex numbers $\Bbb C$ so that we have formal $\Bbb C$-linear combinations of irreps, which will then be sent off to $\Bbb C$-linear combinations of irreducible characters, which indeed spans all class functions.

Take $G$ to be the trivial group, for example. Then the original map is simply the inclusion $\Bbb Z\hookrightarrow\Bbb C$; clearly this is not an isomorphism!

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