# Is $\mathbb{C}[G]$-module a vector space?

Let $$G$$ be a finite group and the $$\mathbb{C}[G]=\{\sum_{g\in G}c_gg\}$$ be the group ring and $$V$$ be a $$\mathbb{C}[G]$$ module. My question is whether $$V$$ is always a vector space( have a basis)? If not, then why we can always view $$V$$ as a linear representation of $$G$$ and how we calculate the character the representation without a basis?

$$\mathbb{C}$$ sits inside $$\mathbb{C}[G]$$, so any $$\mathbb{C}[G]$$-module is also a $$\mathbb{C}$$-module, and $$\mathbb{C}$$-modules are the same thing as complex vector spaces.

• Ah I see, I would call that a free $\mathbb{C}[G]$-module. In that case the answer is basically never. Even the trivial representation doesn't have a $\mathbb{C}[G]$-basis (unless $G$ is trivial).
– Nate
Mar 30, 2019 at 16:25
• Thanks, I got it! A $\mathbb{C}$-v.s. is enough since a representation only needs to be a $\mathbb{C}$-v.s. not a $\mathbb{C}[G]$ one! Mar 30, 2019 at 16:27

This first part is just Nate's answer in a bit more detail. A question about whether something is a vector space is incomplete if you do not include over what field it is meant to be a vector space. The group algebra $$\mathbb{C}[G]$$ is a $$\mathbb{C}$$-algebra, meaning it is equipped with a ring monomorphism $$\mathbb{C}\hookrightarrow \mathbb{C}[G]$$ (thus it is a $$\mathbb{C}$$-vector-space itself), so there is a natural sense in which any $$\mathbb{C}[G]$$-module is a $$\mathbb{C}$$-module as well: restrict the action to $$\mathbb{C}$$.

There are some unnatural yet reasonable ways in which a $$\mathbb{C}[G]$$-module can be made into a vector space. If $$G$$ is a finite group, then by the Artin-Wedderburn theorem the group algebra decomposes into a direct sum of matrix rings over $$\mathbb{C}$$. Corresponding to the $$1$$-dimensional irreducible representations of $$G$$, some of these matrix rings are of $$1\times 1$$ matrices. Hence $$\mathbb{C}[G]$$ has $$\mathbb{C}$$ summands for each $$1$$-dimensional irreducible representation of $$G$$. Given such a summand, one can restrict a $$\mathbb{C}[G]$$-module to an $$F$$-module where $$F\approx\mathbb{C}$$. The natural $$\mathbb{C}$$-module structure does not come from these if $$G\neq 1$$.

Regarding one of your comments, suppose $$V$$ is a $$\mathbb{C}[G]$$-module with $$G$$ finite. There is a minimal generating set $$\{v_\alpha\}_{\alpha\in I}$$ of vectors $$v_\alpha\in V$$ since $$\mathbb{C}[G]$$ is a semisimple algebra. The vectors give a direct sum decomposition $$V\cong \bigoplus_{\alpha\in I}\mathbb{C}[G]v_\alpha$$, where each $$\mathbb{C}[G]v_\alpha$$ is an irreducible $$G$$-representation. This is not a basis as a module (unless $$G=1$$) since $$\mathbb{C}[G]\to V$$ defined by $$g\mapsto gv_\alpha$$ has a nontrivial kernel, a maximal left ideal in fact.

As far as calculating characters without a $$\mathbb{C}$$-basis goes (though one wonders how you managed to define the representation at all without one!), perhaps the best bet would be to identify the irreducible subrepresentations and then sum up characters from a character table.