# Is the homomorphism $\mathbb{Q}G\to \prod M_{\chi_i(1)}(\mathbb{Q})$ given by $x \mapsto (\rho_i(x))_i$ an isomorphism?

If we have the group algebra $$\mathbb{Q}G$$ and $${\chi_1,...,\chi_n}$$ the irreducible characters of $$G$$ afforded by the representation $$\rho_1,...,\rho_n$$, is it true that the map:

$$\mathbb{Q}G\to \prod M_{\chi_i(1)}(\mathbb{Q})$$ sending $$x\in \mathbb{Q}G$$ to $$(\rho_i(x))_i$$

is an isomorphism?

We know that $$\mathbb{Q}G$$ can be decomposed into simple algebras, and every simple algebra produces an irreducible character. Can I mix these facts to give a positive answer to the question?

No. For a very simple example, let $$G$$ be cyclic of order 3. Then $$G$$ has only two irreducible representations over $$\mathbb{Q}$$: the trivial representation $$\rho_1$$, and the quotient $$\rho_2$$ of the regular representation by the trivial representation (which is an irreducible representation of degree $$2$$). Your map $$\mathbb{Q}G\to M_1(\mathbb{Q})\times M_2(\mathbb{Q})$$ then cannot be surjective, since $$\mathbb{Q}G$$ is $$3$$-dimensional and $$M_1(\mathbb{Q})\times M_2(\mathbb{Q})$$ is $$5$$-dimensional. What's going on here is that the subring generated by the image of $$\rho_2$$ is not the full matrix ring $$M_2(\mathbb{Q})$$; instead it's a $$2$$-dimensional subring which is isomorphic to the field $$\mathbb{Q}(\zeta)$$ where $$\zeta$$ is a primitive cube root of $$1$$.

In general, the image of $$\rho_i$$ is a matrix ring over the endomorphism ring $$D_i$$ of $$\rho_i$$. This endomorphism ring $$D_i$$ is a division algebra since $$\rho_i$$ is irreducible. If $$D_i$$ is just $$\mathbb{Q}$$, then the image of $$\rho_i$$ will be all of $$M_{\chi_i(1)}(\mathbb{Q})$$, but if $$D_i$$ is larger than $$\mathbb{Q}$$ then the image of $$\rho_i$$ is a proper subring of $$M_{\chi_i(1)}(\mathbb{Q})$$ (namely, the subring of elements that commute with every element of $$D_i$$).

Your statement would be correct if you were working over $$\mathbb{C}$$ instead of $$\mathbb{Q}$$. Over $$\mathbb{C}$$, there are no finite-dimensional division algebras besides $$\mathbb{C}$$ itself, so the image of every irreducible representation is the full matrix ring $$M_{\chi_i(1)}(\mathbb{C})$$. In general, the Artin-Wedderburn theorem says a semisimple ring is isomorphic to the product of the images of its irreducible representations, which are matrix rings over the endomorphism rings of those representations (and those endomorphism rings are division algebras).

• Thanks for your time, you help me a lot :D – Alopiso Mar 24 at 16:08

This is not true. The group algebra of the cyclic group of order three is commutative, and decomposes as $$\mathbb{Q}C_3\cong \mathbb{Q}\oplus\mathbb{Q}(\omega)$$, where $$\omega$$ is a primitive third root of unity.

If you replace $$\mathbb{Q}$$ with $$\mathbb{C}$$, then the statement is true (in the latter case, $$\mathbb{C}C_3\cong \mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}$$).

• Why is true for $\mathbb{Q}S_3 \cong \mathbb{Q}\oplus \mathbb{Q}\oplus M_2(\mathbb{Q})$? – Alopiso Mar 24 at 15:49
• $\mathbb{Q}C_3\cong \mathbb{Q}[x]/(x^3-1)$ – David Hill Mar 24 at 16:11